veltech vel multimedia vel hightech - · web viewfind the heat loss per square meter and...

UNIT – ICONDUCTION

PART – A

1. Define Heat Transfer.

Heat transfer can be defined as the transmission of energy from one region to another region due to temperature difference.

2. What are the modes of Heat Transfer?

Conduction Convection Radiation

3. Define Conduction.

Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low temperature within a medium (solid, liquid or gases) or between different medium in direct physical contact.

In condition energy exchange takes place by the kinematic motion or direct impact of molecules. Pure conduction is found only in solids.

4. Define Convection.

Convection is a process of heat transfer that will occur between a solid surface and a fluid medium when they are at different temperatures.

Convection is possible only in the presence of fluid medium.

5. Define Radiation.

The heat transfer from one body to another without any transmitting medium is known as radiation. It is an electromagnetic wave phenomenon.

6. State Fourier’s Law of conduction. (Apr. ’97, M.U. Oct. ’98 M.U)

The rate of heat conduction is proportional to the area measured – normal to the direction of heat flow and to the temperature gradient in that direction.

Q - A

where A – are in m2

1

- Temperature gradient in K/mK – Thermal conductivity W/mK.

7. Define Thermal Conductivity. [Apr. ’97 M.U., Oct. ’99 M.U]

Thermal conductivity is defined as the ability of a substance to conduct heat.

8. Write down the equation for conduction of heat through a slab or plane wall.

Heat transfer Where

T = T1 – T2

- Thermal resistance of slab L = Thickness of slabK = Thermal conductivity of slabA = Area

9. Write down the equation for conduction of heat through a hollow cylinder.

Heat transfer Where

T = T1 – T2

thermal resistance of slab

L – Length of cylinderK – Thermal conductivityr2 – Outer radius r1 – inner radius

10. Write down equation for conduction of heat through hollow sphere.

Heat transfer Where

T = T1 – T2

- Thermal resistance of hollow sphere.

11. State Newton’s law of cooling or convection law.

Heat transfer by convection is given by Newton’s law of cooling

2

Q = hA (Ts - T)Where A – Area exposed to heat transfer in m2

h - heat transfer coefficient in W/m2KTs – Temperature of the surface in KT - Temperature of the fluid in K.

12. Write down the equation for heat transfer through a composite plane wall. [April ’97 M.U.]

Heat transfer Where

T = Ta– Tb

L – Thickness of slabha – heat transfer coefficient at inner diameterhb – heat transfer coefficient at outer side.

13. Write down the equation for heat transfer through composite pipes or cylinder. [April ’98 M.U]

3

Heat transfer Where

T = Ta– Tb

14. Write down one dimensional, steady state conduction equation without internal heat generation.

15. Write down steady state, two dimensional conduction equation without heat generation.

16. Write down the general equation for one dimensional steady state heat transfer in slab or plane wall without heat generation.

17. Define overall heat transfer co-efficient. [April ’97 M.U.]

The overall heat transfer by combined modes is usually expressed in terms of an overall conductance or overall heat transfer co-efficient ‘U’.

Heat transfer Q = UA T.

18. Write down the general equation for one dimensional steady state heat transfer in slab with heat generation. [Oct. ’99 M.U.]

19. What is critical radius of insulation (or) critical thickness. [Nov. ’96 M.U., Oct. ’97 M.U.]

4

Critical radius = rc

Critical thickness = rc – r1

Addition of insulating material on a surface does not reduce the amount of heat transfer rate always. In fact under certain circumstances it actually increases the heat loss up to certain thickness of insulation. The radius of insulation for which the heat transfer is maximum is called critical radius of insulation, and the corresponding thickness is called critical thickness.

20. Define fins (or) Extended surfaces.

It is possible to increase the heat transfer rate by increasing the surface of heat transfer. The surfaces used for increasing heat transfer are called extended surfaces or sometimes known as fins.

21. State the applications of fins.

The main application of fins are

1. Cooling of electronic components 2. Cooling of motor cycle engines.3. Cooling of transformers4. Cooling of small capacity compressors

22. Define Fin efficiency. [Nov. ’96 M.U., Oct ’97 M.U.]

The efficiency of a fin is defined as the ratio of actual heat transfer by the fin to the maximum possible heat transferred by the fin.

23. Define Fin effectiveness. [Nov. ’96 M.U. Apr. 2001 M.U.]

Fin effectiveness is the ratio of heat transfer with fin to that without fin

5

Fin effectiveness =

24. What is meant by steady state heat conduction?

If the temperature of a body does not vary with time, it is said to be in a steady state and that type of conduction is known as steady state heat conduction.

25. What is meant by Transient heat conduction or unsteady state conduction?

If the temperature of a body varies with time, it is said to be in a transient state and that type of conduction is known as transient heat conduction or unsteady state conduction.

26. What is Periodic heat flow?

In periodic heat flow, the temperature varies on a regular basis.

Example:1. Cylinder of an IC engine.2. Surface of earth during a period of 24 hours.

27. What is non periodic heat flow?

In non periodic heat flow, the temperature at any point within the system varies non linearly with time.

Examples :

1. Heating of an ingot in a furnace.2. Cooling of bars.

28. What is meant by Newtonian heating or cooling process?

The process in which the internal resistance is assumed as negligible in comparison with its surface resistance is known as Newtonian heating or cooling process.

29. What is meant by Lumped heat analysis? [Oct. ’98 M.U.]

In a Newtonian heating or cooling process the temperature throughout the solid is considered to be uniform at a given time. Such an analysis is called Lumped heat capacity analysis.

30. What is meant by Semi-infinite solids? [Oct.’99 M.U.]

6

In a semi infinite solid, at any instant of time, there is always a point where the effect of heating or cooling at one of its boundaries is not felt at all. At this point the temperature remains unchanged. In semi infinite solids, the biot number value is .

31. What is meant by infinite solid?

A solid which extends itself infinitely in all directions of space is known as infinite solid.

In semi infinite solids, the biot number value is in between 0.1 and 100.0.1< Bi < 100.

32. Define Biot number.

It is defined as the ratio of internal conductive resistance to the surface convective resistance.

Bi =

Bi = .

33. What is the significance of Biot number? [Nov. ’96 M.U. Apr. 2002 M.U.]

Biot number is used to find Lumped heat analysis, semi infinite solids and infinite solids

If Bi < 0.1 L Lumped heat analysis Bi = Semi infinite solids

0.1< Bi < 100 Infinite solids.34. Explain the significance of Fourier number.

It is defined as the ratio of characteristic body dimension to temperature wave penetration depth in time.

Fourier Number =

It signifies the degree of penetration of heating or cooling effect of a solid.

35. What are the factors affecting the thermal conductivity? [April ’97 M.U.]

1. Moisture 2. Density of material 3. Pressure4. Temperature

7

5. Structure of material

36. Explain the significance of thermal diffusivity. [Oct. ’98 M.U.]

The physical significance of thermal diffusivity is that it tells us how fast heat is propagated or it diffuses through a material during changes of temperature with time.

37. What are Heisler charts? [Oct. ’99 M.U.]

In Heisler chart, the solutions for temperature distributions and heat flows in plane walls, long cylinders and spheres with finite internal and surface resistance are presented. Heisler charts are nothing but a analytical solutions in the form of graphs.

PART – B

1. A wall of 0.6m thickness having thermal conductivity of 1.2 w/Mk. The wall is to be insulated with a material having an average thermal conductivity of 0.3 W/mK. Inner and outer surface temperatures are 1000 C and 10C. Heat transfer rate is 1400 W/m2 calculate the thickness of insulation.

Given Data

Thickness of wall L1 = 0.6 mThermal conductivity of wall K1 = 1.2 W/mK.Thermal conductivity of insulation K2 = 0.3 W/mK. Inner surface Temperature T1 = 1000C + 273 = 1273 KOuter surface Temperature T3 = 10C + 273 = 283 KHeat transfer per unit area Q/A = 1400 W/m2.

Solution:

8

Let the thickness of insulation be L2

We know [From equation (13)] (or) [HMT Data book page No. 34]

Where

T = Ta– Tb (or) T1 – T3

Heat transfer coefficient ha, hb and thickness L3 are not given. So neglect that terms.

2. The wall of a cold room is composed of three layer. The outer layer is brick 30cm thick. The middle layer is cork 20 cm thick, the inside layer is cement 15 cm thick. The temperatures of the outside air is 25C and on the inside air is -20C. The film co-efficient for outside air and brick is 55.4 W/m2K. Film co-efficient for inside air and cement is 17 W/m2K. Find heat flow rate.

Take K for brick = 2.5 W/mKK for cork = 0.05 W/mKK for cement = 0.28 W/mK

Given Data

Thickness of brick L3 = 30 cm = 0.3 mThickness of cork L2 = 20 cm = 0.2 mThickness of cement L1 = 15 cm = 0.15 mInside air temperature Ta = -20C + 273 = 253 KOutside air temperature Tb = 25C + 273 = 298 KFilm co-efficient for inner side ha = 17 W/m2K

9

Film co-efficient for outside hb = 55.4 W/m2KKbrick = K3 = 2.5 W/mKKcork = K2 = 0.05 W/mK.Kcement = K1 = 0.08 W/mK.

Solution:

Heat flow through composite wall is given by

[From equation (13)] (or) [HMT Data book page No. 34]

Where

T = Ta– Tb

The negative sign indicates that the heat flows from the outside into the cold room.

3. A wall is constructed of several layers. The first layer consists of masonry brick 20 cm. thick of thermal conductivity 0.66 W/mK, the second layer consists of 3 cm thick mortar of thermal conductivity 0.6 W/mK, the third layer consists of 8 cm thick lime stone of thermal conductivity 0.58 W/mK and the outer layer consists of 1.2 cm thick plaster of thermal conductivity 0.6 W/mK. The heat transfer coefficient on the interior and exterior of the wall are 5.6 W/m2K and 11 W/m2K respectively. Interior room temperature is 22C and outside air temperature is -5C.

10

Calculate

a) Overall heat transfer coefficientb) Overall thermal resistancec) The rate of heat transferd) The temperature at the junction between the mortar and the limestone.

Given Data

Thickness of masonry L1 = 20cm = 0.20 mThermal conductivity K1 = 0.66 W/mKThickness of mortar L2 = 3cm = 0.03 mThermal conductivity of mortar K2 = 0.6 W/mKThickness of limestone L3 = 8 cm = 0.08 mThermal conductivity K3 = 0.58 W/mKThickness of Plaster L4 = 1.2 cm = 0.012 mThermal conductivity K4 = 0.6 W/mKInterior heat transfer coefficient ha = 5.6 W/m2KExterior heat transfer co-efficient hb = 11 W/m2KInterior room temperature Ta = 22C + 273 = 295 KOutside air temperature Tb = -5C + 273 = 268 K.

Solution:

Heat flow through composite wall is given by

[From equation (13)] (or) [HMT Data book page No. 34]Where

T = Ta– Tb

11

We knowHeat transfer Q = UA (Ta – Tb) [From equation (14)]

Where U – overall heat transfer co-efficient

We know Overall Thermal resistance (R)

For unit Area

Interface temperature between mortar and the limestone T3

Interface temperatures relation

12

Temperature between Mortar and limestone (T3 is 276.5 K)

4. A steam to liquid heat exchanger area of 25.2 m2 is constructed with 0.5cm nickel and 0.1 cm plating of copper on the steam sides. The resistivity of a water-scale deposit on the steam side is 0.0015 K/W. The steam and liquid surface conductance are 5400 W/m2K ad 560 W/m2K respectively. The heated steam is at 110C and heated liquid is at 70C.

13

Calculate

1. Overall steam to liquid heat transfer co-efficient2. Temperature drop across the scale deposit

Take K(Copper) = 350 W/mK and K (Nickel) = 55 W/mK.

Given

Area A = 25.2 m2

Thickness of Nickel L1 = 0.5 cm = 0.5 10-2 mThickness of Copper L2 = 0.1 cm = 0.1 10-2 mResistivity of scale R3 = 0.0015 K/WLiquid surface conductance ha = 560 W/m2KSteam surface conductance hb = 5400 W/m2KSteam temperature Tb = 110C + 273 = 383 K Liquid temperature Ta = 70C + 273 = 343 K K2 (Copper) = 350 W/mK K1 (Nickel) = 55 W/mK

Solution:

Heat transfer through composite wall is given by [From equation (13)] (or) [HMT Data book page No. 34]

WhereT = Ta– Tb = 343 – 383 = -40 K

R3 value is given, R3 = 0.0015 K/W

14

[-ve sign indicates that the heat flows from, outside to inside]we know Heat transfer Q = UA (Ta – Tb) [From equation No. (14)]

Temperature drop (T3 – T4) across the scale is given by

5. A surface wall is made up of 3 layers one of fire brick, one of insulating brick and one of red brick. The inner and outer surface temperatures are 900C and 30C respectively. The respective co-efficient of thermal conductivity of the layers are 1.2, 0.14 and 0.9 W/mK and the thickness of 20cm, 8 cm and 11 cm. Assuming close bonding of the layers at the interfaces. Find the heat loss per square meter and interface temperatures.

[M.U. Oct. – 97]

Given

Inner temperature T1 = 900C + 273 = 1173 KOuter temperature T4 = 30C + 273 = 303 KThermal conductivity of fire brick K1 = 1.2 W/mKThermal conductivity of insulating brick K2 = 0.14 W/mKThermal conductivity of red brick K3 = 0.9 W/mKThickness of fire brick L1 = 20 cm = 0.2 mThickness of insulating brick L2 = 8 cm = 0.08 mThickness of red brick L3 = 11 cm = 0.11 m

15

Solution :

(i) Heat loss per square metre (Q/A)

Heat transfer [From equation (13)] (or) [HMT Data book page No. 34]

WhereT = Ta– Tb = T1 – T4

[Convective heat transfer co-efficient ha, hb are not given. So neglect that terms]

(ii) Interface temperatures (T2 and T3)

We know that, interface temperatures relation

16

where

6. A furnace wall made up of 7.5 cm of fire plate and 0.65 cm of mild steel plate. Inside surface exposed to hot gas at 650C and outside air temperature 27C. The convective heat transfer co-efficient for inner side is 60 W/m2K. The convective heat transfer co-efficient for outer side is 8W/m2K. Calculate the heat lost per square meter area of the furnace wall and also find outside surface temperature. [M.U. April – 98]

Given Data

Thickness of fire plate L1 = 7.5 cm = 0.075 mThickness of mild steel L2 = 0.65 cm = 0.0065 mInside hot gas temperature Ta = 650C + 273 = 923 KOutside air temperature Tb = 27C + 273 = 300KConvective heat transfer co-efficient for Inner side ha = 60W/m2KConvective heat transfer co-efficient for Outer side hb = 8 W/m2K.

Solution:(i) Heat lost per square meter area (Q/A)

17

Thermal conductivity for fire plate K1 = 1035 10-3 W/mK

[From HMT data book page No.11]

Thermal conductivity for mild steel plate

K2 = 53.6W/mK [From HMT data book page No.1]Heat flow Where T = Ta– Tb

[The term L3 is not given so neglect that term]

(ii) Outside surface temperature T3

We know that, Interface temperatures relation

18

7. A mild steel tank of wall thickness 10mm contains water at 90 C. Calculate the rate of heat loss per m2 of tank surface area when the atmospheric temperature is 15C. The thermal conductivity of mild steel is 50 W/mK and the heat transfer co-efficient for inside and outside the tank is 2800 and 11 W/m2K respectively. Calculate also the temperature of the outside surface of the tank. [M.U. April 2000]

Given Data

Thickness of wall L1 = 10mm = 0.01 mInside temperature of water Ta = 90 + 273 = 363 KAtmospheric temperature Tb = 15C + 273 = 288 KHeat transfer co-efficient for inside ha = 2800 W/m2KHeat transfer co-efficient for outside hb = 11 W/m2KThermal conductivity of mild steel K = 50 W/mK

To find

19

(i) Rate of heat loss per m2 of tank surface area (Q/A)(ii) Tank outside surface temperature (T2)

Solution:

Heat loss

Where T = Ta– Tb

[L3, L2 not given so neglect L2 and L3 terms]

We know

20

8. A composite slab is made of three layers 15 cm, 10 cm and 12 cm thickness respectively. The first layer is made of material with K = 1.45 W/mK, for 60% of the area and the rest of material with K = 2.5 W/mK. The second layer is made of material with K = 12.5 W/mK for 50% of area and rest of material with K = 18.5 W/mK. The third layer is made of single material of K = 0.76 W/mK. The composite slab is exposed on one side to warn at 26C and cold air at -20C. The inside heat transfer co-efficient is 15 W/m2K. The outside heat transfer co-efficient is 20 W/m2K determine heat flow rate and interface temperatures. [M.U. Nov. - 96]

Given Data

L1 = 15 cm = 0.15 mL2 = 10 cm = 0.1 mL3 = 12 cm = 0.12 mK1a = 1.45 W/mK, A1a = .60K1b = 2.5 W/mK A1b = .40K2a = 12.5 W/mK A2a = .50K2b = 18.5 W/mK A2b = .50K3 = 0.76 W/mKTa = 26C + 273 = 299 KTb = -20C + 273 = 253 Kha = 15 W/m2Khb = 20 W/m2K

Solution :

Heat flow Where

T = Ta– Tb

21

fig

Substitute R1a and R1b value in (1)

22

Similarly,

(ii) Interface temperatures (T1, T2, T3 and T4)

We know

23

9. An external wall of a house is made up of 10 cm common brick (K = 0.7 W/mK) followed by a 4 cm layer of zibsum plaster (K = 0.48 W/mK). What thickness of loosely packed insulation (K = 0.065 W/mK) should be added to reduce the heat loss through the wall by 80%. [M.U. Oct. 99 & Oct. 2001]

Given Data

Thickness of brick L1 = 10 cm = 0.1 mThermal conductivity of brick K1 = 0.7 W/mKThickness of zibsum L2 = 4 cm = 0.04 mThermal conductivity of zibsum K2 = 0.48 W/mKThermal conductivity of insulation K3 = 0.065 W/mK

To find

Thickness of insulation to reduce the heat loss through the wall by 80% (L3)

24

Solution:Heat flow rate

[From HMT data book Page No.34]Where

Heat loss is reduced by 80% due to insulation, so heat transfer is 20 W.

10. A furnace wall consists of steel plate of 20 mm thick, thermal conductivity 16.2 W/mK lined on inside with silica bricks 150 mm thick with conductivity 2.2 W/mK and on the outside with magnesia brick 200 mm thick, of conductivity 5.1 W/mK. The inside and outside surfaces of the wall are maintained at 650C and 150C respectively. Calculate the heat loss from the wall per unit area. If the heat loss is reduced to 2850W/m2 by providing an air gap between steel and silica bricks, find the necessary width of air gap if the thermal conductivity of air may be taken as 0.030 W/mK. [Madurai Kamaraj University April ‘97]

Given Data

25

Steel plate thickness L1 = 20 mm = 0.02 mThermal conductivity of steel K1 = 16.2 W/mKThickness of the silica L2 = 150 mm = 0.150 mThermal conductivity of silica K2 = 2.2 W/mKThickness of the magnesia L3 = 200 mm = 0.2 mThermal conductivity of magnesia K3 = 5.1 W/mKInner surface temperature T1 = 650C + 273 = 923Outer surface temperature T4 = 150C + 273 = 423 KHeat loss reduced due to air gap is 2850 W/m2

Thermal conductivity of the air gap Kair = 0.030 W/mK

Solution :

Heat transfer through composite wall is given by [without considering air gap]

Where T = T1– T4

Neglecting unknown terms (ha and hb)

Heat loss is reduced to 2850 W/m2 due to air gap. So the new thermal resistance is

Thermal resistance of air gap

26

Rair = Rnew – R = 0.1754 – 0.1086

We know

Thickness of the air gap = 1.98 10-3 m

11. A thick walled tube of stainless steel [K = 77.85 kJ/hr mC] 25 mm ID and 50 mm OD is covered with a 25 mm layer of asbestos [K = 0.88 kJ/hr mC]. If the inside wall temperature of the pipe is maintained at 550 C and the outside of the insulator at 45C. Calculate the heat loss per meter length of the pipe. [Madras University April 1995.EEE]Given Data Inner diameter of steel d1 = 25 mmInner radius r1 = 12.5 mm 0.0125 mOuter diameter D2 = 50 mmOuter radius r2 = 25 mm 0.025 mRadius r3 = r2 + 25 mm = 50 mm 0.05 mThermal conductivity of stainless steel

K1 = 77.85 kJ/hr mC = kJ/sec mC = 0.0216 kJ/sec mC 0.0216 kW/mC

Similarly,

Thermal conductivity of asbestos K2 = 0.88 kJ/hr mC

Solution :

Heat flow through composite cylinder is given by [From equation No.(19) or HMT

data book Page No.35]Where T = Ta– Tb

27

Convective heat transfer co-efficient are not given so neglect ha and hb terms.

12. A steel tube (K = 43.26 W/mK) of 5.08 cm inner diameter and 7.62 cm outer diameter is covered with 2.5 cm layer of insulation (K = 0.208 W/mK) the inside surface of the tube receivers heat from a hot gas at the temperature of 316C with heat transfer co-efficient of 28 W/m2K. While the outer surface exposed to the ambient air at 30C with heat transfer co-efficient of 17 W/m2K. Calculate heat loss for 3 m length of the tube.

[Madras University Oct – 1998]Given

Steel tube thermal conductivity K1 = 43.26 W/mKInner diameter of steel d1 = 5.08 cm = 0.0508 mInner radius r1 = 0.0254 mOuter diameter of steel d2 = 7.62 cm = 0.0762 mOuter radius r2 = 0.0381 mRadius r3 = r2 + thickness of insulation Radius r3 = 0.0381 + 0.025 m

r3 = 0.0631 m

Thermal conductivity of insulation K2 = 0.208 W/mKHot gas temperature Ta = 316C + 273 = 589 KAmbient air temperature Tb = 30C + 273 = 303 KHeat transfer co-efficient at inner side ha = 28 W/m2K

28

Heat transfer co-efficient at outer side hb = 17 W/m2KLength L = 3 m

Solution :

Heat flow [From equation No.(19) or HMT data book Page No.35]Where T = Ta– Tb

[The terms K3 and r4 are not given, so neglect that terms]

Heat loss Q = 1129.42 W.

13. A hollow sphere (K = 65 W/mK) of 120 mm inner diameter and 350 mm outer diameter is covered 10 mm layer of insulation (K = 10 W/mK). The inside and outside temperatures are 500C and 50C respectively. Calculate the rate of heat flow through this sphere.

Given

Thermal conductivity of sphere K1 = 65 W/mKInner diameter of sphere d1 = 120 mmRadius r1 = 60 mm = 0.060 mOuter diameter of sphere d2 = 350 mmRadius r2 = 175 mm = 0.175 m

29

Radius r3 = r2 + thickness of insulation R3 = 0.175 + 0.010

Thermal conductivity of insulation K2 = 10 W/mKInside temperature Ta = 500C + 273 = 773 KOutside temperature Tb = 50C + 273 = 323 K

Solution:

Heat loss through hollow sphere is given by [From equation No.(19) or HMT

data book Page No.34 & 35]Where T = Ta– Tb

ha, hb not given so neglect that terms.

30

Heat transfer = Q = 28361 W

Radius r3 = r2 + thickness of insulation = 0.0455 + 90 10-3 m

r3 = 0.1355 m

Radius r4 = r3 + thickness of insulation = 0.1355 + 40 10-3 m r4 = 0.1755 m

Thermal conductivity of pipe K1 = 47 W/mKThermal conductivity of insulation (I) K2 = 0.5 W/mKThermal conductivity of insulation (II) K3 = 0.25 W/mKOutside temperature T4 = 20C + 273 = 293 K

Solution :

Heat flow through composite cylinder is given by [From equation No.(19) or HMT

data book Page No.35]Where T = Ta– Tb (or) T1 –T4

31

Heat transfer Q/L = 448.8 W/m.

14. A hollow sphere has inside surface temperature of 300C and then outside surface temperature of 30C. If K = 18 W/mK. Calculate (i) heat lost by conduction for inside diameter of 5 cm and outside diameter of 15 cm (ii) heat lost by conduction, if equation for a plain wall area is equal to sphere area. [Madras University April ‘97]

Given Data :

T1 = 300C + 273 = 573 KT2 = 30C + 273 = 303 KK1 = 18 W/mKd1 = 5 cm = 0.05 mr1 = 0.025 md2 = 15 cm = 0.15 mr2 = 0.075 m

Solution:(i) Heat lost (Q)

Heat flow [From HMT data book Page No.34 & 35]

Where T = Ta– Tb (or) T1 – T2

32

[The terms ha, hb not given so neglect that terms].

(ii) Heat loss (If the area is equal to the plain wall area) Q1

L = r2 – r1

= 0.075 – 0.025

Derive an expression of Critical Radius of Insulation For A Cylinder.

33

Consider a cylinder having thermal conductivity K. Let r1 and r0 inner and outer radii of insulation.

Heat transfer [From equation No.(3)]

Considering h be the outside heat transfer co-efficient.

To find the critical radius of insulation, differentiate Q with respect to r0 and equate it to zero.

15. A wire of 6 mm diameter with 2 mm thick insulation (K = 0.11 W/mK). If the convective heat transfer co-efficient between the insulating surface and air is 25 W/m2L, find the critical thickness of insulation. And also find the percentage of change in the heat transfer rate if the critical radius is used.

Given Data

d1= 6 mm

34

r1 = 3 mm = 0.003 mr2 = r1 + 2 = 3 + 2 = 5 mm = 0.005 mK = 0.11 W/mKhb = 25 W/m2K

Solution :

1. Critical radius [From equation No.(21)]

Critical thickness = rc – r1

2. Heat transfer through an insulated wire is given by

Heat flow through an insulated wire when critical radius is used is given by

35

Percentage of increase in heat flow by using

Critical radius =

Internal Heat Generation – Formulae used For plane wall :1. Surface temperature

2. Maximum temperature

whereT - Fluid temperature, Kq - Heat generation, W/m3

L – Thickness, mh - Heat transfer co-efficient, W/m2KK – Thermal conductivity, W/mK.

For Cylinder 1. Heat generation

2. Maximum temperature

36

3. Surface temperature Where

V – Volume - r2 L r – radius – m

For sphere

1. Temperature at the centre

16. A current of 200 A is passed through a stainless steel wire (K = 19 W/mK) 3 mm in diameter. The resistivity of the steel may be taken as 70 cm and the length of the wire is submerged in a liquid at 110C with heat transfer co-efficient h = 4 kW/m2C. Calculate the centre temperature of the wire.

[Madras University April 2000]Given

Current A = 200 AThermal conductivity K = 19 W/mKDiameter d = 3 mm = 3 10-3 mResistivity = 70 - cm Liquid temperature Tw = 110C + 273 = 383 KHeat transfer co-efficient h = 4 kW/m2C = 4 10-3 W/m2CSolution:

The maximum temperature in the wire occurs at the centre.

We know that Q = I2R = (200)2 (0.099)

Heat generated

37

Substituting q value in Equation (A)

17. A sphere of 100 mm diameter, having thermal conductivity of 0.18 W/mK. The outer surface temperature is 8C and 250 W/m2 of energy is released due to heat source. Calculate 1. Heat generated 2. Temperature at the centre of the sphere.

Given Diameter of sphere d = 100 mm

r = 50 mm = 0.050 mThermal conductivity K = 0.18 W/mKSurface temperature Tw = 8C + 273 = 281 KEnergy released Q = 250 W/m2

Solution Heat generated

Temperature at the centre of the sphere

38

18. One end of the long solid rod of 50 mm diameter is inserted into a furnace with the other end is projecting the atmosphere at 25C. Once the steady state is reached, the temperature of the rod is measured at two points 20 cm apart are found to be 150C and 100C. The convective heat transfer co-efficient between the rod and the surrounding air is 30 W/m2K. Calculate the thermal conductivity of the rod material. Given Data :

Atmospheric Temperature T = 25C + 273 = 298 KDistance x = 20 cm = 0.20 mBase temperature Tb = 150C + 273 = 423 KIntermediate temperature T = 100C + 273 = 373 KHeat transfer co-efficient h = 30 W/m2K.

Solution :

Since the rod is long, it is treated as long fin. So, temperature distribution

[From HMT data book (CPK)

Page No.41]

39

We know that,

h – heat transfer co-efficient = 30 W/m2KP – Perimeter = d = 0.050

19. An aluminium alloy fin of 7 mm thick and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mK respectively. Determine

1. Temperature at the end of the fin.2. Temperature at the middle of the fin.3. Total heat dissipated by the fin.

Given

Thickness t = 7mm = 0.007 m

40

Length L= 50 mm = 0.050 mBase temperature Tb = 120C + 273 = 393 KAmbient temperature T = 22 + 273 = 295 KHeat transfer co-efficient h = 140 W/m2KThermal conductivity K = 55 W/mK.

Solution :

Length of the fin is 50 mm. So, this is short fin type problem. Assume end is insulated.

We know Temperature distribution [Short fin, end insulated]

[From HMT data book Page No.41]

(i) Temperature at the end of the fin, Put x = L

A – Area = Length thickness = 0.050 0.007

41

(ii) Temperature of the middle of the fin,

Put x = L/2 in Equation (A)

Temperature at the middle of the fin

(iii) Total heat dissipated


20. Ten thin brass fins (K = 100 W/mK), 0.75 mm thick are placed axially on a 1 m long and 60 mm diameter engine cylinder which is surrounded by 35C. The fins are extended 1.5 cm from the cylinder surface and the heat transfer

42

co-efficient between cylinder and atmospheric air is 15 W/m2K. Calculate the rate of heat transfer and the temperature at the end of fins when the cylinder surface is at 160C. [MU April 2000]

Given

Number of fins = 10Thermal conductivity K = 100 W/mKThickness of the fin t = 0.75 mm = 0.75 10-3 mLength of engine cylinder = 1mDiameter of the cylinder d = 60 mm = 0.060 mAtmosphere temperature T = 35C + 273 = 300 KLength of the fin L = 1.5 cm = 1.5 10-2 mHeat transfer co-efficient h = 15 W/m2KCylinder surface temperaturei.e. Base temperature Tb = 160C + 273 = 433 K

Solution

Assuming that the fin end is insulated and length of the fin is 1.5 cm. So this is short fin end insulated type problem.

We knowHeat transferred Q = (hPKA)1/2 (Tb - T) tan h (mL)….(A)

[From HMT data book Page No.41]Where

P – Perimeter = 2 Length of the cylinder = 2 1

A = Area = length of the cylinder thickness = 1 0.75 10-3 m

43

Heat transferred per fin = 58.1 WThe heat transfer for 10 fins = 58.1 10

Heat transfer from unfinned surface due to convection is

So, Total heat transfer Q = Q1 + Q2

Q = 581 + 375.8

We know that,

Temperature distribution [short fin, end insulated]


Temperature at the end of fin, so put x = L

44

21. Aluminium fins 1.5 cm wide and 10 mm thick are placed on a 2.5 cm diameter tube to dissipate the heat. The tube surface temperature is 170C ambient temperature is 20C. Calculate the heat loss per fin. Take h = 130 W/m2 C and K = 200 W/m2 C for aluminium. [Madras University Oct. -99, Oct. 2001]Given

Wide of the fin b = 1.5 cm = 1.5 10-2 mThickness t = 10 mm = 10 10-3 mDiameter of the tube d = 2.5 cm = 2.5 10-2 mSurface temperature Tb = 170C + 273 = 443 KAmbient temperature T = 20C + 273 = 293 KHeat transfer co-efficient h = 130 W/m2 CThermal conductivity K = 200 W/mC

Solution

Assume fin end is insulated, so this is short fin end insulated type problem.Heat transfer [short fin, end insulated]

Q = (hPKA)1/2 (Tb - T) tan h (mL) ……..(1) [From HMT data book Page No.41]Where

A – Area = Breadth thickness

45

22. A straight rectangular fin has a length of 35 mm, thickness of 1.4 mm. The thermal conductivity is 55W/mC. The fin is exposed to a convection environment at 20C and h = 500 W/m2C. Calculate the heat loss for a base temperature of 150C. [Madras University April 2002]GivenLength L = 35 mm = 0.035 mThickness t = 1.4 mm = 0.0014 mThermal conductivity K = 55 W/mCFluid temperature T = 20C + 273 = 293 KBase temperature Tb = 150C + 273 = 423 KHeat transfer co-efficient h = 500 W/m2K.

Solution

Length of the fin is 35 mm, so this is short fin type problem. Assume end is insulated.

Heat transferred [Short fin, end insulated]Q = (hPKA)1/2 (Tb - T) tan h (mL) …….(1)

[From HMT data book Page No.41] Where

P – Perimeter = 2 Length (Approximately) = 2 0.035

A – Area = Length thickness

46

= 0.035 0.0014

Substituting h, p, K, A, Tb, T, m, L values in equation (1)

23. A heating unit made in the form of a cylinder is 6 cm diameter and 1.2 m long. It is provided with 20 longitudinal fins 3 mm thick which protrude 50 mm from the surface of the cylinder. The temperature at the base of the fin is 80C. The ambient temperatures is 25C. The film heat transfer co-efficient from the cylinder and fins to the surrounding air is 10 W/m2K. Calculate the rate of heat transfer from the finned wall to the surrounding. Take K = 90 W/mK.[Manonmanium Sundaranar University Nov. – 96]

Given Diameter of the cylinder d = 6 cm = 0.06 mLength of the cylinder = 1.2 mNumber of fins = 20Thickness of fin (t) = 3 mm = 0.003 mLength of fin L = 50 mm = 0.050 mBase temperature Tb = 80C + 273 = 353 KAmbient temperature T = 25C + 273 = 298 KFilm heat transfer co-efficient h = 10 W/m2KThermal conductivity K = 90 W/mK.

47

Solution

Length of the fins is 50 mm. Assume end is insulated. So this is short fin end insulated type problem.

We know Heat transferred [short fin, end insulated]

Q = (hPKA)1/2 (Tb - T) tan h (mL) ……..(1) [From HMT data book Page No. 41]

Where P – Perimeter = 2 Length of the cylinder

= 2 1.2

A – Area = Length of the cylinder thickness of fin

= 1.2 0.003

Heat transferred per fin = 62.16 WNumber of fins = 20So, Total heat transferred Q1 = 62.16 20

Q1 = 1243.28 W

48

Heat transfer from unfinned surface due to convection is Q2 = h A T

So, Total heat transfer Q = Q1 + Q2

Q = 1243.28 + 122.75

Total heat transfer

24. An aluminium cube 6 cm on a side is originally at a temperature of 500C. It is suddenly immersed in a liquid at 10C for which h is 120 W/m2K. Estimate the time required for the cube to reach a temperature of 250 C. For aluminium = 2700 kg/m3, C = 900 J/kg K, K = 204 W/mK. [Oct. 2002 M.U.]Given Thickness of cube L = 6 cm = 0.06 mInitial temperature T0 = 500C + 273 = 773 KFinal temperature T = 10C + 273 = 283 KIntermediate temperature T = 250C + 273 = 523 KHeat transfer co-efficient h = 120 w/m2KDensity = 2700 kg/m3

Specific heat C = 900 J/Kg kThermal conductivity K = 204 W/mK

Solution

For Cube,

Characteristic length

We know

Biot number

49

Bi =

Biot number value is less than 0.1. So this is lumped heat analysis type problem

For lumped parameter system,

[From HMT data book Page No.48]We know,

Characteristics length

Time required for the cube to reach 250C is 144.86 s.

25. A copper plate 2 mm thick is heated up to 400 C and quenched into water at 30C. Find the time required for the plate to reach the temperature of 50C. Heat transfer co-efficient is 100 W/m2K. Density of copper is 8800 kg/m3. Specific heat of copper = 0.36 kJ/kg K.Plate dimensions = 30 30 cm. [Oct. 97 M.U. April ’97 Bharathiyar University]

Given

Thickness of plate L = 2 mm = 0.002 mInitial temperature T0 = 400C + 273 = 673 KFinal temperature T = 30C + 273 = 303 KIntermediate temperature T = 50C + 273 = 323 KHeat transfer co-efficient h = 100 W/m2KDensity = 8800 kg/m3

Specific heat C= 360 J/kg kPlate dimensions = 30 30 cm

To find

50

Time required for the plate to reach 50C. [From HMT data book Page No.2]Solution:

Thermal conductivity of the copper K = 386 W/mKFor slab,

Characteristic length

We know,

Biot number

Bi =

Biot number value is less than 0.1. So this is lumped heat analysis type problem.


……….(1)


Characteristics length Lc =

Time required for the plate to reach 50C is 92.43 s.

26. A 12 cm diameter long bar initially at a uniform temperature of 40 C is placed in a medium at 650C with a convective co-efficient of 22 W/m2K. Determine the time required for the center to reach 255C. For the material of the bar, K = 20 W/mK, Density = 580 kg/m3, specific heat = 1050 J/kg K.

[Oct. ’98 M.U.]

51

Given :

Diameter of bar, D = 12 cm = 0.12 mRadius of bar, R = 6 cm = 0.06 mInitial temperature T0 = 40C + 273 = 313 KFinal temperature T = 650C + 273 = 923 KIntermediate temperature T = 255C + 273 = 528 KHeat transfer co-efficient h = 22 W/m2KThermal conductivity K = 20 W/mKDensity = 580 kg/m3

Specific heat C = 1050 J/kg k

Solution

For cylinder,

Characteristic Length

We know,

Biot number

Bi = 0.033 < 0.1



……….(1)



52

Time required for the cube to reach 255C is 360.8 s.

27. A steel ball (specific heat = 0.46 kJ/kgK. and thermal conductivity = 35 W/mK) having 5 cm diameter and initially at a uniform temperature of 450C is suddenly placed in a control environment in which the temperature is maintained at 100C. Calculate the time required for the balls to attained a temperature of 150C. Take h = 10W/m2K.[M.U. April-2000, 2001, 2002, Bharathiyar Uni. April 98] Bharathiyar Uni. April 98]

Given Specific heat C = 0.46 kJ/kg K = 460 J/kg KThermal conductivity K = 35 W/mKDiameter of the sphere D = 5 cm = 0.05 mRadius of the sphere R = 0.025 mInitial temperature T0 = 450C + 273 = 723 KFinal temperature T = 100C + 273 = 373 KIntermediate temperature T = 150C + 273 = 423 KHeat transfer co-efficient h = 10 W/m2K

To find

Time required for the ball to reach 150C[From HMT data book Page No.1]

Solution

Density of steel is 7833 kg/m3

For sphere,


We know,

53

Biot number

Bi = 2.38 10-3 < 0.1



……….(1)



Time required for the ball to reach 150C is 5840.54 s.

28. An aluminium sphere mass 5.5 kg and initially at a temperature of 290o

is suddenly immersed in a fluid at 15C with heat transfer co-efficient 58 W/m3K. Estimate the time required to cool the aluminium to 95C. For aluminium take = 2700 kg/m3, C = 900 J/kg K, K = 205 W/mK. [M.U. Oct. -99, Bharathiyar Uni. Nov. 96] GivenMass, m = 5.5 kgInitial temperature T0 = 290C + 273 = 563 KFinal temperature T = 15C + 273 = 288 KIntermediate temperature T = 95C + 273 =368 KHeat transfer co-efficient h = 58 W/m2KThermal conductivity K = 205 W/mKDensity = 2700 kg/m3

Specific heat C = 900 J/kg K.

54

Solution

We know,

Density =

We know,

Volume of sphere

For sphere,


We know,

Biot number

Bi = 7.41 10-3 < 0.1



……….(1)


55


Time required to cool the aluminium to 95C is 1355.6 s.

29. Alloy steel ball of 2 mm diameter heated to 800C is quenched in a bath at 100C. The material properties of the ball are K = 205 kJ/m hr K, = 7860 kg/m3, C = 0.45 kJ/kg K, h = 150 KJ/ hr m2 K. Determine (i) Temperature of ball after 10 second and (ii) Time for ball to cool to 400 C. [April 99 M.U.]Given

Diameter of the ball D = 12 mm = 0.012 mRadius of the ball R = 0.006mInitial temperature T0 = 800C + 273 = 1073 KFinal temperature T = 100C + 273 = 373 KThermal conductivity K = 205 kJ/m hr K

Density = 7860 kg/m3

Specific heat C = 0.45 kJ/kg K = 450 J/kg K

Heat transfer co-efficient h = 150 kJ/hr m2 K

Solution

Case (i) Temperature of ball after 10 sec.

For sphere,


56

We know,

Biot number

Bi = 1.46 10-3 < 0.1



……….(1)



Case (ii) Time for ball to cool to 400C

T = 400C + 273 = 673 K

57

30. A large wall 2 cm thick has uniform temperature 30C initially and the wall temperature is suddenly raised and maintained at 400C. Find

1. The temperature at a depth of 0.8 cm from the surface of the wall after 10 s.

2. Instantaneous heat flow rate through that surface per m2 per hour. Take = 0.008 m2/hr, K = 6 W/mC. [April 97 M.U.]

Given Thickness L = 2 cm = 0.02 mInitial temperature Ti = 30C + 273 = 303 KSurface temperature T0 = 400C + 273 = 673 KThermal diffusivity = 0.008 m2/h

= 2.22 10-6 m2/sThermal conductivity K = 6 W/mC.

Case (i)Depth 0.8 cm = 0.8 10-2 m

= 0.008 m Time t = 10 s

Case (ii)

Time t = 1 h = 3600 sSolution

In this problem heat transfer co-efficient h is not given. So take it as . i.e. h .

We know that,

Biot number Bi =

h =

Bi value is . So this is semi infinite solid type problem.

Case (i)

58

For semi infinite solid.

[From HMT data book Page No. 50]

Where,

Put x = 0.008 m, t = 10 s, = 2.22 10-6 m2/s.

X = 0.848, corresponding erf (X) is 0.7706

[Refer HMT data book Page No.52]

Case (ii)

Instantaneous heat flow

[From HMT data book Page No.50]t = 3600 s (Given)

59

Intermediate temperature Tx = 387.85 K

Heat flux qx = 13982.37 W/m2.

31. A large cast iron at 750C is taken out from a furnace and its one of its surface is suddenly lowered and maintained at 45C. Calculate the following:

1. The time required to reach the temperature 350C at a depth of 45 mm from the surface.

2. Instantaneous heat flow rate at a depth of 45 mm and on surface after 30 minutes.

3. Total heat energy after 2 hr for ingot, Take = 0.06 m2/hr, K = 48.5 W/mK.

Given

Initial temperature Ti = 750C + 273 = 1023 KSurface temperature T0 = 45C + 273 = 318 KIntermediate temperature Tx = 350C + 273 = 623 KDepth x = 45 mm = 0.045 mThermal diffusivity = 0.06 m2/hr = 1.66 10-5 m2/sThermal conductivity K = 48.5 W/mK.

Solution

In this problem heat transfer co-efficient h is not given. So take it as , i.e. h .

We know that,

Biot number Bi =

h =

Bi value is . So this is semi infinite solid type problem.

1. For semi infinite solid.

60

[From HMT data book Page No. 50]

0.432 = erf (X)

erf (X) = 0.432

erf (X) = 0.432, corresponding X is 0.41

We know

Time required to reach 350C is 181.42 s.

2. Instantaneous heat flow

[From HMT data book Page No.50]t = 30 minutes (Given)t = 1800 s

61

[Negative sign shows that heat lost from the ingot].

3. Total heat energy

[Time is given, 2 hr = 7200 s]

[Negative sign shows that heat lost from the ingot]

32. A large steel plate 5 cm thick is initially at a uniform temperature of 400C. It is suddenly exposed on both sides to a surrounding at 60 C with convective heat transfer co-efficient of 285 W/m2K. Calculate the centre line temperature and the temperature inside the plate 1.25 cm from themed plane after 3 minutes.

Take K for steel = 42.5 W/mK, for steel = 0.043 m2/hr.[Nov. ’96 M.U.]

Given

Thickness L = 5 cm = 0.05 mInitial temperature Ti = 400C + 273 = 673 KFinal temperature T = 60C + 273 = 333 KDistance x = 1.25 mm = 0.0125 mTime t = 3 minutes = 180 sHeat transfer co-efficient h = 285 W/m2KThermal diffusivity = 0.043 m2/hr = 1.19 10-5 m2/s.Thermal conductivity K = 42.5 W/mK.

Solution

For Plate :Characteristic Length

We know,

62

Biot number

0.1 < Bi < 100, So this is infinite solid type problem. Infinite Solids

Case (i)

[To calculate centre line temperature (or) Mid plane temperature for infinite plate, refer HMT data book Page No.59 Heisler chart].

X axis value is 3.42, curve value is 0.167, corresponding Y axis value is 0.64

63

Case (ii)

Temperature (Tx) at a distance of 0.0125 m from mid plane

[Refer HMT data book Page No.60, Heisler chart]

X axis value is 0.167, curve value is 0.5, corresponding Y axis value is 0.97.

64

Temperature inside the plate 1.25 cm from the mid plane is 544 K. 33. A 10 cm diameter apple approximately spherical in shape is taken from a 20C environment and placed in a refrigerator where temperature is 5C and average heat transfer coefficient is 6 W/m2K. Calculate the temperature at the centre of the apple after a period of 1 hour. The physical properties of apple are density = 998 kg/m3. Specific heat = 4180 J/kg K, Thermal conductivity = 0.6 W/mK. [April ’98 M.U.]

Given:

Diameter of sphere D = 10 cm = 0.10 mRadius of sphere R = 5 cm = 0.05 mInitial temperature Ti = 20C + 273 = 293 KFinal temperature T = 5C + 273 = 278 KTime t = 1 hour = 3600 sDensity = 998 kg/m3

Heat transfer co-efficient h = 6 W/m2KSpecific heat C = 4180 J/kg KThermal conductivity K = 0.6 W/mK

Thermal diffusivity =

Solution

For Sphere,


We know,

Biot number

65

0.1 < Bi < 100, So this is infinite solid type problem.

Infinite Solids

[To calculate centre line temperature for sphere, refer HMT data book Page No.63].

X axis value is 0.20, curve value is 0.5, corresponding Y axis value is 0.86.

Center line temperature T0 = 290.9 K.

34. A long steel cylinder 12 cm diameter and initially at 20C is placed into furnace at 820C with h = 140 W/m2K. Calculate the time required for the axis temperature to reach 800C. Also calculate the corresponding temperature at a radius of 5.4 cm at that time. Physical properties of steel are K = 21 W/mK, = 6.11 10-6 m2/s. [Oct. ’99 M.U.]

66

Given:Diameter of cylinder D = 12 cm = 0.12 mRadius of sphere R = 6 cm = 0.06 mInitial temperature Ti = 20C + 273 = 293 KFinal temperature T = 820C + 273 = 1093 KHeat transfer co-efficient h = 140 W/m2K

Intermediate radius r = 5.4 cm = 0.054 mThermal diffusivity = 6.11 10-6 m2/s.Thermal conductivity K = 21W/mK

To find

1. Time (t) required for the axis temperature to reach 800C.2. Corresponding temperature (Tt) at a radius of 5.4 cm.

Solution

For Cylinder,


We know,

Biot number Bi =

0.1 < Bi <100, So this is infinite solid type problem.

Infinite Solids

Case (i)

To = 800C + 273 = 1073 KTime (t) ?

[Refer HMT data book Page No.61. Heisler chart]

67

Curve value is 0.4, Y axis 0.025, corresponding X axis value is 5.

Case (ii)

Intermediate radius r – 5.4 cm = 0.054 m

[Refer HMT data book Page No.62]

68

Curve value is 0.9, X axis value is 0.4, corresponding Y axis value is 0.84.

1. Time required for the axis temperature to reach 800C is 2945.9 s.2. Temperature (Tr) at a radius of 5.4 cm is 1076.2 K

UNIT – IICONVECTION

PART – A

1. What is dimensional analysis?

Dimensional analysis is a mathematical method which makes use of the study of the dimensions for solving several engineering problems. This method can be applied to all types of fluid resistances, heat flow problems in fluid mechanics and thermodynamics.

2. State Buckingham theorem.

Buckingham theorem states as Follows: “If there are n variables in a dimensionally homogeneous equation and if these contain m fundamental dimensions, then the variables are arranged into (n – m) dimensionless terms. These dimensionless terms are called terms.

3. What are all the advantages of dimensional analysis?

69

1. It expresses the functional relationship between the variables in dimensional terms.

2. It enables getting up a theoretical solution in a simplified dimensionless form. 3. The results of one series of tests can be applied to a large number of other sim-

ilar problems with the help of dimensional analysis.

4. What are all the limitations of dimensional analysis?

1. The complete information is not provided by dimensional analysis. It only indi-cates that there is some relationship between the parameters.

2. No information is given about the internal mechanism of physical phenome-non.

3. Dimensional analysis does not give any clue regarding the selection of vari-ables.

5. Define Reynolds number (Re).

It is defined as the ratio of inertia force to viscous force.

6. Define prandtl number (Pr).

It is the ratio of the momentum diffusivity of the thermal diffusivity.

7. Define Nusselt number (Nu).

It is defined as the ratio of the heat flow by convection process under an unit temperature gradient to the heat flow rate by conduction under an unit temperature gradient through a stationary thickness (L) of metre.

Nusselt number (Nu) =

8. Define Grash of number (Gr).

It is defined as the ratio of product of inertia force and buoyancy force to the square of viscous force.

9. Define Stanton number (St).

It is the ratio of nusselt number to the product of Reynolds number and prandtl number.

70

10. What is meant by Newtonion and non – Newtonion fluids?

The fluids which obey the Newton’s Law of viscosity are called Newtonion fluids and those which do not obey are called non – newtonion fluids.

11. What is meant by laminar flow and turbulent flow?

Laminar flow: Laminar flow is sometimes called stream line flow. In this type of flow, the fluid moves in layers and each fluid particle follows a smooth continuous path. The fluid particles in each layer remain in an orderly sequence without mixing with each other.

Turbulent flow: In addition to the laminar type of flow, a distinct irregular flow is frequency observed in nature. This type of flow is called turbulent flow. The path of any individual particle is zig – zag and irregular. Fig. shows the instantaneous velocity in laminar and turbulent flow.

12. What is hydrodynamic boundary layer?

In hydrodynamic boundary layer, velocity of the fluid is less than 99% of free stream velocity.

13. What is thermal boundary layer?

In thermal boundary layer, temperature of the fluid is less than 99% of free stream velocity.

14. Define convection.

Convection is a process of heat transfer that will occur between a solid surface and a fluid medium when they are at different temperatures.

71

15. State Newton’s law of convection.

Heat transfer from the moving fluid to solid surface is given by the equation Q = h A (Tw – T)

This equation is referred to as Newton’s law of cooling. Where

h – Local heat transfer coefficient in W/m2K.A – Surface area in m2

Tw – Surface (or) Wall temperature in KT - Temperature of fluid in K.

16. What is meant by free or natural convection?

If the fluid motion is produced due to change in density resulting from temperature gradients, the mode of heat transfer is said to be free or natural convection.

17. What is forced convection?

If the fluid motion is artificially created by means of an external force like a blower or fan, that type of heat transfer is known as forced convection.

18. According to Newton’s law of cooling the amount of heat transfer from a solid surface of area A at temperature Tw to a fluid at a temperature T is given by _____________.

Ans : Q = h A (Tw – T)

19. What is the form of equation used to calculate heat transfer for flow through cylindrical pipes?

Nu = 0.023 (Re)0.8 (Pr)n

n = 0.4 for heating of fluidsn = 0.3 for cooling of fluids

20. What are the dimensionless parameters used in forced convection?

1. Reynolds number (Re)2. Nusdselt number (Nu)3. Prandtl number (Pr)

21. Define boundary layer thickness.

The thickness of the boundary layer has been defined as the distance from the

72

surface at which the local velocity or temperature reaches 99% of the external velocity or temperature.

PART – B

1. Air at 20C, at a pressure of 1 bar is flowing over a flat plate at a velocity of 3 m/s. if the plate maintained at 60C, calculate the heat transfer per unit width of the plate. Assuming the length of the plate along the flow of air is 2m.

Given : Fluid temperature T = 20C, Pressure p = 1 bar, Velocity U = 3 m/s,

Plate surface temperature Tw = 60C, Width W = 1 m, Length L = 2m.

Solution : We know,

Film temperature

Properties of air at 40C:

Density = 1.129 Kg/m3

Thermal conductivity K = Kinematic viscosity v = Prandtl number Pr = 0.699

We know,

Reynolds number Re =

Reynolds number value is less than 5 105, so this is laminar flow.

For flat plate, Laminar flow, Local Nusselt Number Nux = 0.332 (Re)0.5 (Pr)0.333

73

Local heat transfer coefficient hx = 2.327 W/m2KWe know,

Average heat transfer coefficient h = 2 hx

h = 4.65 W/m2K

Heat transfer Q = h A (Tw - T)

2. Air at 20C at atmospheric pressure flows over a flat plate at a velocity of 3 m/s. if the plate is 1 m wide and 80C, calculate the following at x = 300 mm.

1. Hydrodynamic boundary layer thickness, 2. Thermal boundary layer thickness, 3. Local friction coefficient,4. Average friction coefficient, 5. Local heat transfer coefficient6. Average heat transfer coefficient,7. Heat transfer.

Given: Fluid temperature T = 20C Velocity U = 3 m/s Wide W = 1 m

Surface temperature Tw = 80CDistance x = 300 mm = 0.3 m

Solution: We know

Film temperature

74

We know,


Since Re < 5 105, flow is laminar For Flat plate, laminar flow,

1. Hydrodynamic boundary layer thickness:

2. Thermal boundary layer thickness:

3. Local Friction coefficient:

4. Average friction coefficient:

75

5. Local heat transfer coefficient (hx):

Local Nusselt NumberNux = 0.332 (Re)0.5 (Pr)0.333

We know

Local Nusselt Number

6. Average heat transfer coefficient (h):

7. Heat transfer:

We know that,

3. Air at 30C flows over a flat plate at a velocity of 2 m/s. The plate is 2 m long and 1.5 m wide. Calculate the following:

1. Boundary layer thickness at the trailing edge of the plate,2. Total drag force,

76

3. Total mass flow rate through the boundary layer between x = 40 cm and x = 85 cm.

Given: Fluid temperature T = 30C Velocity U = 2 m/s Length L = 2 m Wide W W = 1.5 m

To find:1. Boundary layer thickness2. Total drag force.3. Total mass flow rate through the boundary layer between x = 40 cm and

x = 85 cm.

Solution: Properties of air at 30C

We know,

Reynolds number

For flat plate, laminar flow, [from HMT data book, Page No.99]

Hydrodynamic boundary layer thickness

Thermal boundary layer thickness,

We know,Average friction coefficient,

77

We know

Total mass flow rate between x = 40 cm and x= 85 cm.

Hydrodynamic boundary layer thickness

78

4. Air at 30C, Flows over a flat plate at a velocity of 4 m/s. The plate measures 50 30 cm and is maintained at a uniform temperature of 90C. Compare the heat loss from the plate when the air flows (a) Parallel to 50 cm,(b) Parallel to 30 cmAlso calculate the percentage of heat loss.

Given: Fluid temperature T = 30C Velocity U = 4 m/s Plate dimensions = 50 cm 30 cm

Surface temperature Tw = 90C

Solution: Film temperature

Properties of air at 60C,

Case (i) : When the flow is parallel to 50 cm.

79

Reynolds number

Local nusselt number NUx = 0.332(Re)0.5(Pr)0.333

NUx =0.332

Local nusselt number NUx =95.35We know

We know

Average heat transfer coefficient

Case (ii) : When the flow is parallel to 30 cm side.


For flat plate, laminar flow, Local Nusselt Number

80

We know that, NUx =

Average heat transfer coefficient h = 2hx

Case (iii):

5. Air at 40C is flows over a flat plate of 0.9 m at a velocity of 3 m/s. Calculate the following:

1. Overall drag coefficient2. Average shear stress, 3. Compare the average shear stress with local shear stress (shear stress

at the trailing edge)

Given : Fluid temperature T = 40C Length L = 0.9 m Velocity U = 3 m/s.

Solution:

81

Properties of air at 40C:

We know,

Reynolds number

For plate, laminar flow,

Drag coefficient (or) Average skin friction coefficient

We know

Average friction coefficient

We know,

Local skin friction coefficient

we know

Local skin friction coefficient

82

6. Air at 290C flows over a flat plate at a velocity of 6 m/s. The plate is 1m long and 0.5 m wide. The pressure of the air is 6 kN/2. If the plate is maintained at a temperature of 70C, estimate the rate of heat removed form the plate.

Given : Fluid temperature T = 290C Velocity U = 6 m/s. Length L = 1 m

Wide W = 0.5 m Pressure of air P = 6 kN/m2

Plate surface temperature Tw = 70CTo find: Heat removed from the plate

Solution:

We know

Film temperature

Properties of air at 180C (At atmospheric pressure)

Note: Pressure other than atmospheric pressure is given, so kinematic viscosity will vary with pressure. Pr, K, Cp are same for all pressures.

Kinematic viscosity

83

We know,

Reynolds number

For plate, laminar flow, Local nusselt number

We know

NUx =

We know

Average heat transfer coefficient h = 2hx

84

Heat transfer from both side of the plate = 2 254.1

= 508.2 W.

7. Air at 40C flows over a flat plate, 0.8 m long at a velocity of 50 m/s. The plate surface is maintained at 300C. Determine the heat transferred from the entire plate length to air taking into consideration both laminar and turbulent portion of the boundary layer. Also calculate the percentage error if the boundary layer is assumed to be turbulent nature from the very leading edge of the plate.

Given : Fluid temperature T = 40C Length L = 0.8 m Velocity U = 50 m/s

Plate surface temperature Tw = 300C

To find :

1. Heat transferred for: i. Entire plate is considered as combination of both laminar and turbulent flow.ii. Entire plate is considered as turbulent flow.

2. Percentage error.

Solution: We know

Film temperature T

We know

85

Case (i): Laminar – turbulent combined. [It means, flow is laminar upto Reynolds number value is 5 105, after that flow is turbulent]

Average nusselt number = Nu = (Pr)0.333 (Re)0.8 – 871 Nu = (0.6815)0.333 [0.037 (1.26 106)0.8 – 871 Average nusselt number Nu = 1705.3

Case (ii) : Entire plate is turbulent flow:

Local nusselt number} Nux = 0.0296 (Re)0.8 (Pr)0.333

NUx = 0.0296 (1.26 106)0.8 (0.6815)0.333

NUx = 1977.57We know

Local heat transfer coefficient hx = 91.46 W/m2K

Average heat transfer coefficient (for turbulent flow)

h = 1.24 hx

= 1.24 91.46

Average heat transfer coefficient} h = 113.41 W/m2K

We know Heat transfer Q2 = h A (Tw + T)

86

= h L W (Tw + T) = 113.41 0.8 1 (300 – 40)Q2 = 23589.2 W

8. Air at 20C flows over a flat plate at 60C with a free stream velocity of 6 m/s. Determine the value of the average convective heat transfer coefficient upto a length of 1 m in the flow direction.

Given : Fluid temperature T = 20C

Plate temperature Tw = 60C Velocity U = 6 m/s Length L = 1 m

To find : Average heat transfer coefficient

Solution : We know

We know


For flat plate, laminar flow

87

Local nusselt number} Nux = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (3.53 105)0.5 (0.699)0.333

NUx = 175.27

We know,

Local nusselt number}

9. Air at 25C at the atmospheric pressure is flowing over a flat plate at 3 m/s. If the plate is 1 m wide and the temperature Tw = 75C. Calculate the following at a location of 1m from leading edge.

i. Hydrodynamic boundary layer thickness, ii. Local friction coefficient,

iii. Thermal boundary layer thickness,iv. Local heat transfer coefficient

Given : Fluid temperature T = 25C Velocity U = 3 m/s Wide W = 1 m

Plate surface temperature Tw = 75CDistance = 1 m

To find:

1. Hydrodynamic boundary layer thickness.2. Local friction coefficient 3. Thermal boundary layer thickness4. Local heat transfer coefficient

Solution: We know

88

We know,

Reynolds number Re=

For flat plate, laminar flow,

1. Hydrodynamic boundary layer thickness,

2. Local friction coefficient

3. Thermal boundary layer thickness,

4. Local heat transfer coefficient (hx):

We know

Local nusselt number} NUx = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (1.67 105)0.5 (0.698)0.333

NUx = 120.415

89

10. Atmospheric air at 300 K with a velocity of 2.5 m/s flows over a flat plate of length L = 2m and width W = 1m maintained at uniform temperature of 400 K. Calculate the local heat transfer coefficient at 1 m length and the average heat transfer coefficient from L = 0 to L = 2m. Also find the heat transfer,

Given : Fluid temperature T = 300 KVelocity U = 2.5 m/sTotal Length L = 2 mWidth W = 1 m

Surface temperature Tw = 400 K

To find:

1. Local heat transfer coefficient at L = 1 m2. Average heat transfer coefficient at L = 2 m3. Heat transfer Q

Solution:

Case (i): Local heat transfer coefficient at L = 1m

90


Local Nusselt number} NUx = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (118539.5)0.5 (0.692)0.333

NUx = 101.18We know,

Local nusselt number}

101.18 =

hx = 3.0832 W/m2K

Local heat transfer coefficient} hx = 3.08 W/m2K

Case (ii): Average heat transfer coefficient at L = 2m



NUx = 0.332 (Re)0.5 (Pr)0.333

= 0.332 (237079.18)0.5 (0.692)0.333

NUx = 143

Local heat transfer coefficient} hx = 2.17 W/m2KWe know that,

Average heat transfer coefficient} h = 2 hx

h = 2 2.17

91

h = 4.35 W/m2K

Average heat transfer coefficient} h = 4.35 W/m2K

Case (iii) : Heat transfer Q = h A (Tw - T)= 4.35 2 1 (400 – 300)

Q = 870 W.

11. For a particular engine, the underside of the crank case can be idealized as a flat plat measuring 80 cm 20 cm. The engine runs at 80 km/hr and the crank case is cooled by air flowing past it at the same speed. Calculate the loss of heat from the crank case surface of temperature 75C to the ambient air temperature 25C. Assume the boundary layer becomes turbulent from the loading edge itself.

Given : Area A = 80 cm 20 cm= 1600 cm2 = 0.16m2

Velocity U = 80 Km/hr

Flow is turbulent from the leading edge, i.e,. flow is fully turbulent.

To find:1. Heat loss

We know

92

For flat plate, turbulent flow, [Fully turbulent from leading edge – given]

Local Nusselt number} NUx = 0.0296 (Re)0.8 (Pt)0.333

= 0.0296 [9 105]0.8 (0.698)0.33

NUx = 1524.6

We know that,

Local heat transfer coefficient} hx = 53.85 W/m2KFor turbulent flow, flat plate

Average heat transfer coefficient} h = 1.24 hx

h = 1.24 53.85h = 66.78 W/m2KWe know, Heat loss Q = h A (Tw - T)= 66.78 0.16 (75 – 25)Q = 534.2 W

Formula used for Flow over cylinders and spheres

1. Film temperature

Where T - Fluid temperature C Tw – Plate surface temperature C

2. Reynolds number

Where U – Velocity, m/s D - Diameter, m - Kinematic viscosity, m2/s

3. Nusselt number NU = C (Re)m (Pr)0.333

93

4. Nusselt number NU =

5. Heat transfer Q = h A (Tw - T)Where A =

For sphere:

Nusselt number NU = 0.37 (Re)0.6

Heat transfer Q = h A (Tw - T)Where A 4r2

12. Air at 15C, 30 km/h flows over a cylinder of 400 mm diameter and 1500 mm height with surface temperature of 45C. Calculate the heat loss.

Given : Fluid temperature T = 15C Velocity U = 30 Km/h

Diameter D = 400 mm = 0.4 mLength L = 1500 mm = 1.5 mPlate surface temperature Tw = 45CTo find: Heat loss.Solution: We know

Film temperature

Properties of air at 30C : [From HMT data book, Page No.22]Density = 1.165 Kg/m3

Kinematic viscosity v = 16 10-6 m2/sPrandtl Number Pr = 0.701

Thermal conductivity K = 26.75 10-3 W/mK

We know

Reynolds Number Re =

94

We know

Nusselt Number Nu = C (Re)m (Pr)0.333

[From HMT data book, Page No.105]

ReD value is 2.08 105, so C value is 0.0266 and m value is 0.805.

[From HMT data book, Page No.105] NU = 0.0266 (2.08 105)0.805 (0.701)0.333

We know that,

Nusselt Number NU =

13. Air at 30C, 0.2 m/s flows across a 120W electric bulb at 130C. Find heat transfer and power lost due to convection if bulb diameter is 70 mm.

Given : Fluid temperature T = 30C Velocity U = 0.2 m/s Heat energy Q1 = 120 W

Surface temperature Tw = 130CDiameter D = 70 mm = 0.070 m

To find:

1. Heat Transfer2. Power lost due to convection

Solution:1. Film temperature

95

We know

Nusselt Number Nu = 0.37 (Re)0.6

= 0.37 (663.82)0.6

Nu = 18.25We know

Nusselt number

Heat transfer coefficient h = 7.94 W/m2KWe know

Heat transfer Q2 = h A (Tw - T)

2. % of heat lost =

96

14. Air at 40C flows over a tube with a velocity of 30 m/s. The tube surface temperature is 120C. Calculate the heat transfer for the following cases.

1. Tube could be square with a side of 6 cm.2. Tube is circular cylinder of diameter 6 cm

Given : Fluid temperature T = 40C Velocity U = 30 m/s

Tube surface temperature Tw = 120CTo find: Heat transfer coefficient (h)

Solution: We know

Case (i): Tube is considered as square of side 6 cmi.e., L = 6cm = 0.06m


97

Case (ii)

Tube diameter D = 6cm = 0.06 m

so corresponding C and m values are 0.0266 and 0.805 respectively.

Formulae Used for Flow Over Bank of Tubes

98

1. Maximum velocity Umax =

Where Sn – Transverse pitch, m.

2. Reynolds Number Re =

3. Nusselt Number, NU = 1.13 (Pr)0.33 [C Ren][From HMT data book, Page No.114]

15. In a surface condenser, water flows through staggered tubes while the air is passed in cross flow over the tubes. The temperature and velocity of air are 30C and 8 m/s respectively. The longitudinal and transverse pitches are 22 mm and 20 mm respectively. The tube outside diameter is 18 mm and tube surface temperature is 90C. Calculate the heat transfer coefficient.

Given: Fluid temperature T = 30C Velocity U = 8 m/s

Longitudinal pitch, Sp = 22mm = 0.022 mTransverse pitch, Sn = 20mm = 0.020 mDiameter D = 18mm = 0.018 mTube surface temperature Tw = 90C

Solution:

We know

Maximum velocity Umax =

We know

99

corresponding C, n values are 0.518 and 0.556 respectively.

[From HMT data book, Page No.114]C = 0.518n = 0.556

We know,

Nusselt Number Nu = 1.13 (Pr) 0.333[C (Re)n][From HMT data book, Page No.114]

We know

Nusselt Number

Heat transfer coefficient h = 428.6 W/m2K.

Formulae used for flow through Cylinders (Internal flow)

1. Bulk mean temperature

Tmi = Inlet temperature C,Where

Tmo = Outlet temperature C.

100

2. Reynolds Number

If Reynolds number value is less than 2300, flow is laminar. If Reynolds number values is greater than 2300, flow is turbulent.

3. Laminar Flow: Nusselt Number NU – 3.66


4. Turbulent Flow (General Equation)Nusselt Number Nu = 0.023 (Re)0.8 (Pr)n

n = 0.4 – Heating processn = 0.3 – Cooling process


This equation is valid for 0.6 < Pr < 160,Re < 10000

For turbulent flow,

This equation is valid for

5. Equivalent diameter for rectangular section,

Where A – Area, m2,P – Perimeter, mL – Length, m,W – Width, m.

6. Equivalent diameter for hollow cylinder

101

7. Heat transfer

Q = h A (Tw – Tm) where A = D L (or) Q = m Cp (Tmo – Tmi)

Where Tw – Tube wall temperature C, Tm – Mean temperature C. Tmi – Inlet temperature C Tmo – Outlet temperature C.

8. Mass flow ratem - A U Kg/s

Where - Density, Kg/m3

A – Area,

U – Velocity, m/s

16. When 0.6 Kg of water per minute is passed through a tube of 2 cm diameter, it is found to be heated from 20C to 60C. The heating is achieved by condensing steam on the surface of the tube and subsequently the surface temperature of the tube is maintained at 90C. Determine the length of the tube required for fully developed flow.Given : Mass m = 0.6 Kg/min =

= 0.01 Kg/s

To find: length of the tube (L)

Solution:Bulk mean temperature

102

Let us first determine the type of flow

For laminar flow,

Nusselt number NU = 3.66We know

103

17. Water at 50C enters 50 mm diameter and 4 m long tube with a velocity of 0.8 m/s. The tube wall is maintained at a constant temperature of 90C. Determine the heat transfer coefficient and the total amount of heat transferred if exist water temperature is 70C.

Given: Inner temperature of water Tmi = 50C

Diameter D = 50mm = 0.05 mLength L = 4 mVelocity U = 0.8 m/s

Total wall temperature Tw = 90CExit temperature of water Tmo = 70C

To find:

1. Heat transfer coefficient (h)2. Heat transfer (Q)

Solution:Bulk mean temperature

Let us first determine the type of flow:

104

ratio is greater than 60. Re value is greater than 10,000 and Pr value is in

between 0.6 and 160 so,Nusselt number NU = 0.023 (Re)0.8 (Pr)n

[Inlet temperature 50C, Exit temperature 70C Heating Process, So n = 0.4]

Heat transfer coefficient h = 4039.3 W/m2KHeat transfer Q = h A (Tw – Tm)

18. What flows through 0.8 cm diameter, 3m long tube at an average temperature of 40C. The flow velocity is 0.65 m/s and tube wall temperature is 140C. Calculate the average heat transfer coefficient.

Given : Diameter of tube D = 0.8 cm = 0.008 m Length L = 3 m Average temperature Tm = 40C Velocity U = 0.65 m/s Tube wall temperature Tw = 140C

To find: Heat transfer coefficient (h)

105

ratio is in between 10 and 400, Re < 10000, so Nusselt Number Nu = 0.036 (Re)0.8

(Pr)0.33

We know

Nusselt number NU=

19. Air at 15C, 35 m/s, flows through a hollow cylinder of 4 cm inner diameter and 6 cm outer diameter and leaves at 45C. Tube wall is maintained at 60C. Calculate the heat transfer coefficient between the air and the inner tube.

Given: Inner temperature of air Tmi = 15C Velocity U = 35 m/s Inner diameter Di = 4 cm = 0.04m Outer diameter Do = 6 cm = 0.06m Exit temperature of air Tmo = 45C Tube wall temperature Tw = 60C


Solution: We know

106

Hydraulic of Equivalent diameter

Since Re > 2300, flow is turbulent

For turbulent flow, general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

This is heating process so, n = 0.4

107

20. Air at 30C, 6 m/s flows over a rectangular section of size 300 800 mm. Calculate the heat leakage per meter length per unit temperature difference.

Given : Air temperature Tm = 30C

Velocity U = 6 m/s Area A = 300 800 mm2

A = 0.24 m2

To find: 1. Heat leakage per metre length per unit temperature difference.

Solution:

Equivalent diameter for 300 800 mm2 cross section is given by

We know

Since Re > 2300, flow is turbulent.

108

For turbulent flow general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

Assuming the pipe wall temperature to be higher than a temperature. So heating process n = 0.4

We know

Heat transfer coefficient h = 18.09 W/m2KHeat leakage per unit per length per unit temperature difference Q = h P =Q = 39.79 W

21. Air at 333K, 1.5 bar pressure, flow through 12 cm diameter tube. The surface temperature of the tube is maintained at 400K and mass flow rate is 75 kg/hr. Calculate the heat transfer rate for 1.5 m length of the tube.

Given : Air temperature Tm = 333 K = 60C Diameter D = 12 cm = 0.12 m

Surface temperature Tw = 400 K = 127C

Mass flow rate m = 75 kg/hr =

m = 0.020 Kg/sLength L = 1.5 m

To find:1. Heat transfer rate (Q)

Solution:

Since the pressure is not much above atmospheric, physical properties of air may be taken at atmospheric condition

109

We know

Since Re > 2300, so flow is turbulent For turbulent flow, general equation is (Re>10000)

Nu = 32.9

22. 250 Kg/hr of air are cooled from 100C to 30C by flowing through a 3.5 cm inner diameter pipe coil bent in to a helix of 0.6 m diameter. Calculate the value of air side heat transfer coefficient if the properties of air at 65C are

K = 0.0298 W/mK

110

= 0.003 Kg/hr – mPr = 0.7 = 1.044 Kg/m3

Given : Mass flow rate in = 205 kg/hr

Inlet temperature of air Tmi = 100COutlet temperature of air Tmo = 30CDiameter D = 3.5 cm = 0.035 m

Mean temperature


Solution:Reynolds Number Re =

Kinematic viscosity

Since Re > 2300, flow is turbulent For turbulent flow, general equation is (Re > 10000)

111

We know that,

Heat transfer coefficient h = 2266.2 W/m2K

23. In a long annulus (3.125 cm ID and 5 cm OD) the air is heated by maintaining the temperature of the outer surface of inner tube at 50 C. The air enters at 16C and leaves at 32C. Its flow rate is 30 m/s. Estimate the heat transfer coefficient between air and the inner tube.

Given : Inner diameter Di = 3.125 cm = 0.03125 m Outer diameter Do = 5 cm = 0.05 m

Tube wall temperature Tw = 50CInner temperature of air Tmi = 16COuter temperature of air tmo = 32CFlow rate U = 30 m/s


Solution:Mean temperature Tm =

We know,

Hydraulic or equivalent diameter

112

= 0.05 – 0.03125Dh = 0.01875 m

Re = 35.3 10-6

Since Re > 2300, flow is turbulent

For turbulent flow, general equation is (Re > 10000)Nu = 0.023 (Re)0.8 (Pr)n

This is heating process. So n = 0.4

24. Engine oil flows through a 50 mm diameter tube at an average temperature of 147C. The flow velocity is 80 cm/s. Calculate the average heat transfer coefficient if the tube wall is maintained at a temperature of 200C and it is 2 m long.

Given : Diameter D = 50 mm = 0.050 mAverage temperature Tm = 147C

Velocity U = 80 cm/s = 0.80 m/sTube wall temperature Tw = 200C

Length L = 2m

To find: Average heat transfer coefficient (h)

Solution : Properties of engine oil at 147C

113

We know

Since Re < 2300 flow is turbulent

For turbulent flow, (Re < 10000)

25. A system for heating water from an inlet temperature of 20 C to an outlet temperature of 40C involves passing the water through a 2.5cm diameter steel pipe. The pipe surface temperature is maintained at 110C by condensing steam on its surface. For a water mass flow rate of 0.5 kg/min, find the length of the tube desired.

Given : Inlet temperature Tmi = 20C Outlet temperature Tmo = 40C Diameter D = 2.5 cm = 0.025 m

Piper surface temperature Tw = 110CMass flow rate m = 0.5 Kg/min = 8.33 10-3 Kg/s

To find: Length of the tube (L)

Solution: We know

114

We know

For laminar flow,

Nusselt number Nu = 3.66


115

Formulae used for free convection

1. Film temperature

where Tw – Surface temperature in C T - Fluid temperature in C

2. Coefficient of thermal expansion

3. Nusselt Number Nu =

Where h – Heat transfer coefficient W/m2K L – Length, m

K – Thermal conductivity, W/mK

4. Grashof number for vertical plate

5. If GrPr value is less than 109, flow is laminar. If GrPr value is greater than 109, flow is turbulent.

i.e., GrPr > 109, Laminar flow GrPr > 109, Turbulent flow 6. For laminar flow (Vertical plate):

116

Nusselt number Nu = 0.59 (GrPr)0.25

This expression is valid for,104 < Gr Pr < 109

7. For turbulent flow (Vertical plate):Nusselt Number Nu = 0.10 [Gr Pr]0.333

8. Heat transfer (vertical plate):Q = h A (Tw - T)

9. Grashof number for horizontal plate:

Where Lc – Characteristic length =

W – Width of the plate.

10. For horizontal plate, upper surface heated,

Nusselt number Nu = 0.54 [Gr Pr]0.25

This expression is valid for

11. For horizontal plate, lower surface heated

Nusselt Number Nu = 0.27 [Gr Pr]0.25

This expression is valid for 105 < Gr Pr < 1011

12. Heat transfer (Horizontal plate)

Q = (hu + hj) A (Tw - T)Where hu – Upper surface heated, heat transfer coefficient W/m2 KHi – Lower surface heated, heat transfer coefficient, W/m2K

13. For horizontal cylinder

Nusselt number Nu = C [Gr Pr]m

14. For horizontal cylinder,

Heat transfer Q = h A (Tw - T)

117

Where A - DL

15.For sphere,

Nusselt number Nu = 2 + 0.43 [Gr Pr]0.25

Heat transfer Q = h A (Tw - T)Where A - 4r2

16. Boundary layer thickness

26. A vertical plate of 0.75 m height is at 170 C and is exposed to air at a temperature of 105C and one atmosphere calculate:1. Mean heat transfer coefficient,2. Rate of heat transfer per unit width of the plate

Given : Length L = 0.75 mWall temperature Tw = 170CFluid temperature T = 105C

To find:

1. Heat transfer coefficient (h)2. Heat transfer (Q) per unit width

Solution: Velocity (U) is not given. So this is natural convection type problem.

We know that

118

We know

Grahsof number Gr =

Since Gr Pr < 109, flow is laminar

Gr Pr value is in between 104 and 109 i.e., 104 < Gr Pr < 109

So, Nusselt Number

Nu = 0.59 (Gr Pr)0.25

We know

Nusselt number Nu =

Heat transfer coefficient h = 4.24 W/m2KWe know

27. A large vertical plate 4 m height is maintained at 606C and exposed to atmospheric air at 106C. Calculate the heat transfer is the plate is 10 m

119

wide.

Given : Vertical plate length (or) Height L = 4 mWall temperature Tw = 606CAir temperature T = 106CWide W = 10 m

To find: Heat transfer (Q)

Solution:

Gr = 1.61 1011

Gr Pr = 1.61 1011 0.676Gr Pr = 1.08 1011 Since Gr Pr > 109, flow is turbulent For turbulent flow, Nusselt number Nu = 0.10 [Gr Pr]0.333

We know that,

120

Nusselt number


Heat transfer Q = h A T

28. A thin 100 cm long and 10 cm wide horizontal plate is maintained at a uniform temperature of 150C in a large tank full of water at 75C. Estimate the rate of heat to be supplied to the plate to maintain constant plate temperature as heat is dissipated from either side of plate.

Given :Length of horizontal plate L = 100 cm = 1mWide W = 10 cm = 0.10 mPlate temperature Tw = 150CFluid temperature T = 75C

To find: Heat loss (Q) from either side of plate

Solution:

121

Lc = 0.05 m

Gr Pr = 5.29 109

Gr Pr value is in between 8 106 and 1011

i.e., 8 106 < Gr Pr < 1011

For horizontal plate, upper surface heated:

Nusselt number Nu = 0.15 (Gr Pr)0.333

We know that,

Upper surface heated, heat transfer coefficient hu = 3543.6 W/m2KFor horizontal plate, lower surface heated:

Nusselt number Nu = 0.27 [Gr Pr]0.25

122

Lower surface heated, heat transfer coefficient h1 = 994.6 W/m2K

Total heat transfer Q = (hu + h1) A T

= (hu + h1) W L (Tw - T)= (3543.6 + 994.6) 0.10 (150 – 75)Q = 34036.5 W

29. A hot plate 20 cm in height and 60 cm wide is exposed to the ambient air at 30C. Assuming the temperature of the plate is maintained at 110C. Find the beat loss from both surface of the plate. Assume horizontal plate.

Given :

Height (or) Length of the Plate L = 20 cm = 0.20 mWide W = 60 cm = 0.60 mFluid temperature T = 30CPlate surface temperature Tw = 110C

To find: Heat loss from both the surface of the plate (Q)

Solution:

We know

Coefficient of thermal expansion}

123

We know

Gr Pr value is in between 8 106 and 1011

i.e., 8 106 < Gr Pr < 1011

For horizontal plate, Upper surface heated,


Upper surface heated, heat transfer coefficient hu = 6.99 W/m2K

For horizontal plate, lower surface heated:


= 0.277 [1.06 108]0.25

Nu = 28.15We know that,

124

Lower surface heated, heat transfer coefficient h = 2.78 W/m2K

30. A vertical pipe 80 mm diameter and 2 m height is maintained at a consent temperature of 120C. The pipe is surrounded by still atmospheric air at 30C. Find heat loss by natural convection.

Given : Vertical pipe diameter D = 80 mm = 0.080 mHeight (or) Length L = 2 mSurface temperature Tw = 120 CAir temperature T = 30C

To find: Heat loss (Q)

Solution: We know

125

We know

For turbulent flow,

Nu = 0.10 [Gr Pr]0.333

= 0.10 [3.32 1010]0.333

Nu = 318.8

We know that,

UNIT – IIIPHASE CHANGE HEAT TRANSFER AND HEAT EXCHANGERS

PART – A

126

1. Define boiling.

The change of phase from liquid to vapour state is known as boiling.

2. What is meant by condensation?

The change of phase from vapour to liquid state is known as condensation.

3. Give the applications of boiling and condensation.

Boiling and condensation process finds wide applications as mentioned below.

1. Thermal and nuclear power plant.2. Refrigerating systems3. Process of heating and cooling4. Air conditioning systems

4. What is meant by pool boiling?

If heat is added to a liquid from a submerged solid surface, the boiling process referred to as pool boiling. In this case the liquid above the hot surface is essentially stagnant and its motion near the surface is due to free convection and mixing induced by bubble growth and detachment. 5. What are the modes of condensation?

There are two modes of condensation

1. Film wise condensation2. Drop wise condensation

6. What is meant by Film wise condensation? [April 2000 MU Oct. 2000 MU]

The liquid condensate wets the solid surface, spreads out and forms a continuous film over the entire surface is known as film wise condensation.

7. What is meant by Drop wise condensation? [April 2000 MU Oct 2000 MU]

In drop wise condensation the vapour condenses into small liquid droplets of various sizes which fall down the surface in a random fashion.

8. Give the merits of drop wise condensation? [April 1999 MU]

In drop wise condensation, a large portion of the area of the plate is directly exposed to vapour. The heat transfer rate in drop wise condensation is 10 times higher than in film condensation.

9. Draw different regions of boiling and what is nucleate boiling?

127

[April 1999 MU April 2002 MU]

Nucleate boiling exists in regions II and III. The nucleate boiling begins at region. As the excess temperature is further increased, bubbles are formed more rapidly and rapid evaporation takes place. This is indicated in region III. Nucleate boiling exists up to T = 50C.

I - Free convectionII - Bubbles condense in super heated liquid III - Bubbles raise to surfaceIV - Unstable filmV - Stable filmVI - Radiation coming into play

10. What is heat exchanger?

A heat exchanger is defined as an equipment which transfers the heat from a hot fluid to a cold fluid.

11. What are the types of heat exchangers?

The types of heat exchangers are as follows1. Direct contact heat exchangers2. Indirect contact heat exchangers3. Surface heat exchangers4. Parallel flow heat exchangers5. Counter flow heat exchangers6. Cross flow heat exchangers7. Shell and tube heat exchangers8. Compact heat exchangers.

128

12. What is meant by Direct heat exchanger (or) open heat exchanger?

In direct contact heat exchanger, the heat exchange takes place by direct mixing of hot and cold fluids.

13. What is meant by Indirect contact heat exchanger?

In this type of heat exchangers, the transfer of heat between two fluids could be carried out by transmission through a wall which separates the two fluids.

14. What is meant by Regenerators?

In this type of heat exchangers, hot and cold fluids flow alternately through the same space.Examples: IC engines, gas turbines.

15. What is meant by Recupcradors (or) surface heat exchangers?

This is the most common type of heat exchangers in which the hot and cold fluid do not come into direct contact with each other but are separated by a tube wall or a surface.

16. What is meant by parallel flow heat exchanger?

In this type of heat exchanger, hot and cold fluids move in the same direction.

17. What is meant by counter flow heat exchanger?

In this type of heat exchanger hot and cold fluids move in parallel but opposite directions.

18. What is meant by cross flow heat exchanger?

In this type of heat exchanger, hot and cold fluids move at right angles to each other.

19. What is meant by shell and tube heat exchanger?

In this type of heat exchanger, one of the fluids move through a bundle of tubes enclosed by a shell. The other fluid is forced through the shell and it moves over the outside surface of the tubes.

20. What is meant by compact heat exchangers? [Nov 1996 MU]

There are many special purpose heat exchangers called compact heat exchangers. They are generally employed when convective heat transfer coefficient associated with one of the fluids is much smaller than that associated with the other fluid.

129

21. What is meant by LMTD?

We know that the temperature difference between the hot and cold fluids in the heat exchanger varies from point in addition various modes of heat transfer are involved. Therefore based on concept of appropriate mean temperature difference, also called logarithmic mean temperature difference, also called logarithmic mean temperature difference, the total heat transfer rate in the heat exchanger is expressed as Q = U A (T)mWhere U – Overall heat transfer coefficient W/m2KA – Area m2

(T)m – Logarithmic mean temperature difference.

22. What is meant by Fouling factor?

We know the surfaces of a heat exchangers do not remain clean after it has been in use for some time. The surfaces become fouled with scaling or deposits. The effect of these deposits the value of overall heat transfer coefficient. This effect is taken care of by introducing an additional thermal resistance called the fouling resistance.

23. What is meant by effectiveness?

The heat exchanger effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer.

Effectiveness

=

PART – B

1. Water is to be boiled at atmospheric pressure in a polished copper pan by means of an electric heater. The diameters of the pan is 0.38 m and is kept at 115C. Calculate the following 1. Power required to boil the water 2. Rate of evaporation 3. Critical heat flux

Given : Diameter d = 0.38 mSurface temperature Tw = 115C

130

Solution:

We know saturation temperature of water is 100Ci.e. Properties of water at 100C

(From HMT data book Page No.13)

From steam table R.S. Khurmi Steam table Page No.4 At 100CEnthalpy of evaporation hfg = 2256.9 kj/kghfg = 2256.9 103 j/kg

Specific volume of vapour vg = 1.673 m3/kg

Density of vapour

131

So this process is nucleate pool boiling. 1. Power required to boil the water. For nucleate boiling

We know (From HMT data book Page No.142)Where = surface tension for liquid vapour interface.At 100C

From HMT data book Page No.147


Substitute and Pr values in Equation (1)

2. Rate of evaporation (m)

132

We know heat transferred Q = m hfg

3. Critical heat flux

We know for nucleate pool boiling. Critical heat flux


2. Water is boiled at the rate of 24 kg/h in a polished copper pan, 300 mm in diameter, at atmospheric pressure. Assuming nucleate boiling conditions calculate the temperature of the bottom surface of the pan.

Given :m = 24 kg / h

d = 300 mm = .3m

Solution:

We know saturation temperature of water is 100Ci.e. Tsat = 100C

Properties of water at 100CFrom HMT data book Page No.13

133

From steam table (R.S. Khumi Steam table Page No.4)At 100C

Enthalpy of evaporation hfg = 2256.9 kj/kg

Specific volume of vapourVg = 1.673 m3/kg

Density of vapour

For nucleate boiling

134

At 100C (From HMT data book Page No.147)

For water – copper – Csf = Surface fluid constant = 013


Substitute, and Pr values in Equation (1)

3. A nickel wire carrying electric current of 1.5 mm diameter and 50 cm long, is submerged in a water bath which is open to atmospheric pressure. Calculate the voltage at the burn out point, if at this point the wire carries a current of 200A.

Given :D = 1.5mm = 1.5 10-3 m; L = 50 cm = 0.50m; Current I = 200A

Solution


Properties of water at 100C

135


R.S. Khurmi Steam table Page No.4

= Surface tension for liquid – vapour interface At 100C

(From HMT data book Page No.147)For nucleate pool boiling critical heat flux (AT burn out)


Substitute hfg, values in Equation (1)

We know Heat transferred Q = V I

136

4. A heating element cladded with metal is 8 mm diameter and of emissivity is 0.92. The element is horizontally immersed in a water bath. The surface temperature of the metal is 260C under steady state boiling conditions. Calculate the power dissipation per unit length for the heater.

Given :Diameter D = 8mm = 8 10-3 m; Emissivity E = 0.92Surface temperature Tw = 260C.To find:

Power dissipation

Solution

We know that saturation temperature of water is 100Ci.e. Tsat = 100C

Excess temperature

So this is film pool boiling

Film temperature Tf =

Properties of water vapour at 180C


137

Properties of saturated water at 100C

In film pool boiling heat is transferred due to both convection and radiation.



Stefan boltzmann constant

138

Substitute (2) (3) in (1)

OrPower dissipation = 1753.34 W/m.

5. Water is boiling on a horizontal tube whose wall temperature is maintained ct 15C above the saturation temperature of water. Calculate the nucleate boiling heat transfer coefficient. Assume the water to be at a pressure of 20 atm. And also find the change in value of heat transfer coefficient when

1. The temperature difference is increased to 30C at a pressure of 10 atm.2. The pressure is raised to 20 atm at T = 15C

Given :

Wall temperature is maintained at 15C above the saturation temperature.

= p = 10 atm = 10 bar case (i)

case (ii)p = 20 atm = 20 bar; T - 15CSolution:

We know that for horizontal surface, heat transfer coefficient

h = 5.56 (T)3 From HMT data book Page No.128h = 5.56 (Tw – Tsat)3 = 5.56 (115 – 100)3

Heat transfer coefficient other than atmospheric pressure

hp = hp0.4 From HMT data book Page No.144 = 18765 100.4

139

Case (i)

P = 100 bar T = 30C From HMT data book Page No.144

Heat transfer coefficient

Heat transfer coefficient other than atmospheric pressure hp = hp0.4

Case (ii)P = 20 bar; T = 15C

Heat transfer coefficient h = 5.56 (T)3 = 5.56 (15)3

Heat transfer coefficient other than atmospheric pressure hp = hp0.4

= 18765 (20)0.4

Boiling correlations

It is obvious from the boiling curve that various physical mechanisms are involved in different regions and there will be correspondingly many types of correlations for the boiling process. Some of them are given below.

1. Nucleate pool Boiling (From HMT data book Page No.142)a. Heat flux

140

b. Critical heat Flux

c. Excess temperature

d. Heat transfer Q = mX hfg --------------32. Film pool boiling

(From HMT data book Page No.143)a. Heat transfer coefficient h = hconv + ¾ hrad ………………4

where Kv – Thermal conductivity of vapour, w/mkv – Density of vapour kg/m3

l – Density of liquid kg/m3

g – Acceleration due to gravity 9.81 m/s2

hfg – Enthalpy of evaporation j/kgCpv – Specific heat of vapour at constant pressurev – Dynamic viscosity of vapour Ns/m2

D – Diameter, mT – Excess temperature = Tw - Tsat

141

Stefan Boltzmann constant = 5.67 10-8 w/m2K4

E – EmissivityTw – Surface temperature, CTsat – Saturation temperature C

b. Excess temperature T = Tw – Tsat > 50C for film pool boiling

6. Correlation for film wise condensing process


a. Film thickness for laminar flow vertical surface

where

b. Local transfer coefficient hx for vertical surface, laminar flow

c. Average heat transfer coefficient h for vertical surface laminar flow

The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc adams

d. Average heat transfer coefficient for horizontal surface laminar flow

142

e. Average heat transfer coefficient for bank of tubes, laminar flow

f. For Laminar flow Re < 1800.g. For turbulent flow Re > 1800.h. Average heat transfer coefficient for vertical surface turbulent flow

Solved Problems of Laminar flow vertical surfaces.7. Dry saturated steam at a pressure of 3 bar, condenses on the surface of a vertical tube of height 1m. The tube surface temperature is kept at 110 C. Calculate the following 1. Thickness of the condensate film2. Local heat transfer coefficient at a distance of 0.25mGiven Assume Laminar flow

Pressure P = 3 bar surface temperature Tw = 110C

Distance x = 0.25 m

Solution:

Properties of steam at 3 bar From steam table R.S. Khurmi steam table Page No.10)

We know that

Properties of saturated water at 121.75C


143

For vertical surfaces,


Local heat transfer coefficient


8. A vertical tube of 65 mm outside diameter and 1.5 m long is exposed to steam at atmospheric pressure. The outer surface of the tube is maintained at a temperature of 60C by circulating cold water through the tube. Calculate the following.1. The rate of heat transfer to the coolant.2. The rate of condensation of steam.

Given :

Diameter D = 65 mm = 0.65m; Length L = 1.5mSurface temperature Tw = 60C

Solution:


Properties of steam at 100C

144

From R.S. Khurmi steam table Page No.4

Enthalpy of evaporation

We know

Properties of saturated water at 80CFrom HMT data book Page No.13

Assuming that the condensate film is laminar

For laminar flow, vertical surface heat transfer coefficient


The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc Adams

145

ii) The rate of condensation of steam m

We know

Heat transfer Q = m hfg

Let us check the assumption of laminar film condensation.We know

P = Perimeter = D = 0.0065 = 0.204 m

So our assumption laminar flow is correct.

9. A vertical flat plate in the form of fin is 500m in height and is exposed to steam at atmospheric pressure. If surface of the plate is maintained at 60C. calculate the following.

1. The film thickness at the trailing edge 2. Overall heat transfer coefficient 3. Heat transfer rate4. The condensate mass flow rate.

Assume laminar flow conditions and unit width of the plate.

Given :

Height ore length L = 500 mm = 5mSurface temperature Tw = 60C

Solution


146

(From R.S. Khurmi steam table Page No.4

hfg = 2256.9kj/kghfg = 2256.9 103 j/kg

We know

Properties of saturated water at 80C(From HMT data book Page No.13)

1. Film thickness x

We know for vertical plate Film thickness

WhereX = L = 0.5 m

2. Average heat transfer coefficient (h)

For vertical surface Laminar flow

The factor 0.943 may be replace by 1.13 for more accurate result as suggested by

147

Mc Adams

3. Heat transfer rate Q

We know

4. Condensate mass flow rate mWe know

10. Steam at 0.080 bar is arranged to condense over a 50 cm square vertical plate. The surface temperature is maintained at 20C. Calculate the following.

a. Film thickness at a distance of 25 cm from the top of the plate. b. Local heat transfer coefficient at a distance of 25 cm from the top of

the plate.c. Average heat transfer coefficient.d. Total heat transfere. Total steam condensation rate. f. What would be the heat transfer coefficient if the plate is inclined at

30C with horizontal plane.

Given :

Pressure P = 0.080 barArea A = 50 cm 50 cm = 50 050 = 0.25 m2 Surface temperature Tw = 20CDistance x = 25 cm = .25 m

148

Solution

Properties of steam at 0.080 bar(From R.S. Khurmi steam table Page no.7)

We know

Properties of saturated water at 30.76C = 30CFrom HMT data book Page No.13

a. Film thickness We know for vertical surfaces

b. Local heat transfer coefficient hx Assuming Laminar flow

149

c. Average heat transfer coefficient h(Assuming laminar flow)

The factor 0.943 may be replaced by 1.13 for more accurate result as suggested by Mc adams

Where L = 50 cm = .5 m

d. Heat transfer (Q)We know

Q = hA(Tsat – Tw)

e. Total steam condensation rate (m)We know

Heat transfer

150

f. If the plate is inclined at with horizontal

Let us check the assumption of laminar film condensation We know

So our assumption laminar flow is correct.

11. The outer surface of a cylindrical vertical drum having 25 cm diameter is exposed to saturated steam at 1.7 bar for condensation. The surface temperature of the drum is maintained at 85C. Calculate the following

1. Length of the drum2. Thickness of condensate layer to condense 65 kg/h steam.

Given:

Diameter D = 25 cm = 0.25m; Pressure p = 1.7 barSurface temperature Tw = 85C

Mass m = 65kg/h =

M = .0180 kg/s.

Solution

Properties of steam at 1.7 barFrom R.S. Khurmi steam table Page No.9

151

We know that

Film temperature


For vertical surfaces (Assuming Laminar flow)Average heat transfer coefficient


Heat transfer Q

152

We know that

Substitute h value

2. Film thickness

Let us check the assumption of laminar flow We know that

Where

P = Perimeter =

So our assumption laminar flow is correct. 12. A horizontal tube of outer diameter 2.2 cm is exposed to dry steam at 100C. The pipe surface is maintained at 62C by circulating water through it. Calculate the rate of formation of condensate per metre length of the pipe.Given:

153

Solution

Properties of steam at 100 CFrom R.S. Khurmi steam table Page No.4

At 100Chfg = 2256.9 kj/kg = 2256.9 103 j/kg

We know



For horizontal tubes heat transfer coefficient.


154

We knowHeat transfer Q

We know

Q = mhfg

13. A steam condenser consisting of a square array of 900 horizontal tubes each 6 mm in diameter. The tubes are exposed to saturated steam at a pressure of 0.18 bar and the tube surface temperature is maintained at 25C calculate. 1. Heat transfer coefficient2. The rate at which steam is condensed

Given :

Horizontal tubes = 900Diameter D = 6mm = 6 10-3mPressure P = 0.18 barSurface temperature Tw = 23C

Solution Properties of steam P = 0.18 bar

From R.S. Khurmi steam table Page No.8Tsat = 57.83Chfg = 2363.9 kj/kghfg = 2363.9 103 j/kg

We know

155



with 900 tubes a 30 30 tube of square array could be formed

i.e.

For horizontal bank of tubes, heat transfer coefficient


Heat transfer Q = hA (Tsat – Tw)

We know that

156

For complete array the rate of condensation is

14. A condenser is to designed to condense 600 kg/h of dry saturated steam at a pressure of 0.12 bar. A square array of 400 tubes, each of 8 mm diameter is to be used. The tube surface is maintained at 30C. Calculate the heat transfer coefficient and the length of each tube.

Given :

Pressure P – 0.12 barNo. of tubes = 400

Solution

Properties of steam at 0.12 barFrom R.S. Khurmi steam table Page No.7

We know

157


with 400 tubes a 20 20 tube of square array could be formed i.e.

For horizontal bank of tubes heat transfer coefficient.


We know

Heat transfer

We know

158

Problems on Parallel flow and Counter flow heat exchangersFrom HMT data book Page No.135

Formulae used 1. Heat transfer Q = UA (T)m

Where U – Overall heat transfer coefficient, W/m2KA – Area, m2

(T)m – Logarithmic Mean Temperature Difference. LMTD For parallel flow

In Counter flow

Where

T1 – Entry temperature of hot fluid CT2 – Exit temperature of hot fluid CT1 – Entry temperature of cold fluid CT2 – Exit temperature of cold fluid C

2. Heat lost by hot fluid = Heat gained by cold fluid Qh = Qc

Mh – Mass flow rate of hot fluid, kg/sMc – Mass flow rate of cold fluid kg/sCph – Specific heat of hot fluid J/kg KCpc – Specific heat of cold fluid J/kg L

3. Surface area of tube

159

A = D1 LWhere D1 Inner din

4. Q = m hfgWhere hfg – Enthalpy of evaporation j/kg K

5. Mass flow rate m = AC

15. In a counter flow double pipe heat exchanger, oil is cooled from 85C to 55C by water entering at 25C. The mass flow rate of oil is 9,800 kg/h and specific heat of oil is 2000 j/kg K. the mass flow rate of water is 8,000 kg/h and specific heat of water is 4180 j/kg K. Determine the heat exchanger area and heat transfer rate for an overall heat transfer coefficient of 280 W/m2 K.

Given :

Hot fluid – oil Cold fluid – water T1,T2 t1, t2

Entry temperature of oil T1 = 85CExit temperature of oil T2 = 55CEntry temperature of water t1 = 25CMass flow rate of oil (Hot fluid) mh = 9,800 kg/h

Specific heat of oil Cph = 2000 j/kg K

Mass flow of water (Cold fluid mc = 8,000 kg/h)

Specific heat of water Cpc – 4180 j/kg KOverall heat transfer coefficient U = 280 W/m2K

To find 1. Heat exchanger area (A)2. Heat transfer rate (Q)

Solution We know that

160

Heat lost by oil Hot fluid = Heat gained by water cold fluid Qh = Qc


For counter flow

Substitute (T)m U and Q values in Equation (1)

16. Water flows at the rate of 65 kg/min through a double pipe counter flow heat exchanger. Water is heated from 50C to 75C by an oil flowing through the tube. The specific heat of the oil is 1.780 kj/kg K. The oil enters at 115C and leaves at 70C. The overall heat transfer coefficient is 340 W/m2 K. Calculate the following

1. Heat exchanger area2. Rate of heat transfer

161

Given :

Hot fluid – oil Cold fluid – water T1, T2 t1, t2

Mass flow rate of water cold fluid mc = 65 kg/min

Entry temperature of water t1 = 50CExit temperature of water t2 = 75CSpecific heat of oil (Hot fluid) Cph = 1.780 kj/kg K

= 1.780 103 j/kg K

Entry temperature of oil T1 = 115CExit temperature of oil T2 = 70COverall heat transfer coefficient U = 340 W/m2/KTo find

1. Heat exchanger area (A)2. Heat transfer rate (Q)

Solution

We know

We know

Heat transfer Q = U A (T)m

From HMT data book Page No., 154 Where

(T)m – Logarithmic Mean Temperature Difference. LMTD

For Counter flow

162

Substitute (T)m Q and U values in Equation (1)

17. In a double pipe heat exchanger hot fluid with a specific heat of 2300 j/kg K enters at 380C and leaves at 300C. cold fluid enters at 25C and leaves at 210C. Calculate the heat exchanger area required for 1. Parallel flow 2. Counter flowTake overall heat transfer coefficient is 750 w/m2 K and mass flow rate of hot fluid is 1 kg/s.

Given :

Specific heat of hot fluid Cph = 2300 j/kg KEntry temperature of hot fluid T1 = 380CExit temperature of hot fluid T2 = 300CEntry temperature of cold fluid t1 = 25CExit temperature of cold fluid t2 = 210COverall heat transfer coefficient U = 750 W/m2KMass flow rate of fluid mh = 1 kg/s

Solution

Case (i)

For parallel flow

163


We know that

Heat transfer Area for parallel flow A = 1.27 m2

Case (ii)

For counter flow

We know that,Heat transfer Q = UA (T)m

18. In a counter flow single pass heat exchanger is used to cool the engine

164

oil from 150C with water available at 23c as the cooling medium. The specific heat of oil is 2125 J/kg K. The flow rate of cooling water through the inner tube of 0.4 m diameter is 2.2 Kg/s. the flow rate of oil through the outer tube of 0.75m diameter is 2.4 kg/s. If the value of the overall heat transfer coefficient is 240 W/m2 how long must the heat exchanger be to meet its cooling requirement?

Given :

Hot fluid oil Cold fluid waterT1, T2 (t1, t2)

Entry temperature of oil T1 = 150CExit temperature of oil T2 = 55Entry temperature of water t1 = 23CSpecific heat of oil hot fluid Cph = 2125 J/Kg KInner Diameter D1 = 0.4 mFlow rate of water cooling fluid mc = 2.2 kg/sOuter diameter D2 = 0.75 mFlow rate of oil Hot fluid mh = 2.4 kg/sOver all heat transfer coefficient U= 240 W/m2K

Solution

We knowHeat lost by oil Hot fluid = Heat gained by water (cold fluid)

[Specific heat of water Cpc = 4186 J/Kg K]

We know

Heat transfer Q = mc Cpc (t2 – t1) (or)MhCph (T1 – T2)

We know Heat transfer Q = UA (T)m …………(1)

165

[From HMT data book Page No.154]where

(T)m – Logarithmic Mean Temperature Difference (LMTD).

For Counter flow,


19. Saturated steam at 126C is condensing on the outer tube surface of a single pass heat exchanger. The heat exchanger heats 1050 kg/h of water from 20C to 95C. The overall heat transfer coefficient is 1800 W/m2K. Calculate the following.1. Area of heat exchanger2. Rate of condensation of steamTake hfg = 2185 kj/kg

Given :

Hot fluid – steam Cold fluid – water T1, T2 t1, t2

Saturated steam temperature T1 = T2 = 126CMass flow rate of water mc = 1050 kg/h

166

Entry temperature of water t1 = 20CExit temperature of water t2 = 95COver all transfer coefficient U = 1800 W/m2KEnthalpy of evaporation hfg = 2185 kg/jg

= 2185 103 j/kg

Solution

We knowHeat transfer

We know Heat transfer

Rate of condensation of steam mh = 0.0416 kg/sWe know that

Heat transfer Q = UA (T)m ………….1From HMT data book Page No.154

Where

167

20. An oil cooler of the form of tubular heat exchanger cools oil from a temperature of 90C to 35C by a large pool of stagnant water assumed constant temperature of 28C. The tube length is 32 m and diameter is 28 mm. The specific heat and specific gravity of the oil are 2.45 kj/kg K and 0.8 respectively. The velocity of the oil is 62 cm/s. Calculate the over all heat transfer coefficient.

Given :

Hot fluid – Oil Cold fluid – Water T1, T2 t1, t2

Entry temperature of oil T1 = 90CExit temperature of oil T2 = 35CEntry and Exit temperature of water t1=t2=28CTube length L = 32 mDiameter D = 28 mm = 0.028 mSpecific heat of oil Cph = 2.45 kj/kg/KCph = 2.45 103 j/kg KSpecific gravity of oil = 0.8Velocity of oil C = 62 cm/s = 0.62 m/s.

To find

We know

Mass flow rate of oil

168


We know Heat transfer Q = U A (T)m ………..1 (From HMT data book Page No.154)Where (T)m – Logarithmic Mean Temperature Difference LMTD.For parallel flow

Substitute (T)m Q value in equation 1

Result

U = 577.9 W/m2K

169

Problems on cross flow heat exchangers (or) shell and tube heat exchangers. From HMT data book Page No.154

Formulae used

1. Q = FU A (T)m (Counter Flow)Where

F – Correction factor – From data bookU – Overall heat transfer coefficient W/m2K(T)m – Logarithmic mean temperature difference

For counter flow

where T1 – Entry temperature of hot fluid CT2 – Exit temperature of hot fluid CT1 – Entry temperature of cold fluid CT2 – Exit temperature of cold fluid C

2. Heat lost by hot fluid = Heat gained by cold fluid

Qh = Qc

21. In a cross heat exchangers both fluids unmixed hot fluid with a specific heat of 2300 J/kg K enters at 380C and leaves at 300C cold fluids enters at 25C and leaves at 210C. Calculate the required surface of heat exchanger. Take overall heat transfer coefficient is 750 W/m2 K. Mass flow rate of hot fluid is 1 kg/s

Given :

Specific heat of hot fluid Cph = 23000 J/kg KEntry temperature of hot fluid T1 = 380CExit temperature of heat fluid T2 = 300CEntry temperature of cold fluid t1 = 25CExit temperature of cold fluid t2 = 210COverall heat transfer coefficient U = 750 W/m2KMass flow rate of hot fluid mh= 1 kg/s

170

To find

Heat exchanger area (A)

Solution

This is cross flow both fluids unmixed type heat exchanger. For cross flow heat exchanger.

Q = FU A (T)m (Counter flow)……….1From HMT data book Page No.154

Where F – correction factor

(T)m – Logarithmic Mean Temperature Difference for Counter Flow

For Counter flow


To find correction factor E refer HMT data book Page No.164.

Single pass cross flow heat exchanger – Both fluids unmixed.

From graph

171

Value is 0.432 corresponding Yaxis value is 0.97 i.e. F = 0.07

Substitute Q, F (T)m and U value in Equation (1)

22. In a refrigerating plant water is cooled from 20C to 7C by brine solution entering at -2C and leaving at 3C. The design heat load is 5500 W and the overall heat transfer coefficient is 800 W/m2 K. What area required when using a shell and tube heat exchanger with the water making one shell pass and the brine making two tube passes.

Given :

Hot fluid – Water Cold fluid – brine solution (T1, T2) (t1, t2)

Entry temperature of water T1 = 20CExit temperature of water T2 = 7CEntry temperature of brine solution t1 = -2CExit temperature of brine solution t2 = 3CHeat load Q = 5500 WOverall heat transfer coefficient U = 8000 W/m2 K

To find Area required A

Solution

Shell and tube heat exchanger – one shell pass and two tube passes For shell and tube heat exchanger or cross heat exchanger.

Q = F U A (T)m (Counter flow)(From HMT data book Page No.154)

172

Where F – Correction factor(T)m – Logarithmic mean temperature difference for counter flow For counter flow

To find correction factor F refer HMT data book Page No.161One shell pass and two tube passes

From graph

Xaxis value is 0.22 curve value is 2.6 corresponding Yaxis value is 0.94

Substitute (T)m Q, U and F value is Equation (1)

23. Saturated steam at 120C is condensing in shell and tube heat exchanger. The cooling water enters the tuber at 25C and leaves at 80C. Calculate the logarithmic meant temperature difference if the arrangement is (a) Counter Flow (b) Parallel Flow (c) Cross Flow

Given :

Hot fluid steam Cold fluid water (T1, T2) (t1, t2)

173

Saturated steam temperature T1 = T2 = 120CEntry temperature of water t1 = 25CExit temperature of water t2 = 80C

To find (T)m for parallel flow counter flow and cross flow

Solution

Case (i)For parallel flow [From HMT data book Page No.154]

Case (ii)

For Counter Flow

Case (iii)For cross flow (T)m = F (T)m for Counter flow

Where F = Correction factor

Refer HMT data book Page No.163

Correction factor for single pass cross flow heat exchanger one fluid mixed other

174

unmixed.

Xaxis value is 0.578 curve value is 0So corresponding Yaxis value is 1

From (1) (2) and (4) we came to know when one of the fluids in a heat exchanger changes phase, the logarithmic mean temperature difference and rate of heat transfer will remain same for parallel flow counter flow and cross flow.

Solved problems – NTU method

Note NTU method is used to determine the inlet or exit temperatures of heat exchanger. 24. A parallel flow heat exchanger is used to cool. 4.2 kg/min of hot liquid of specific heat 3.5 kj/kg K at 130C. A cooling water of specific heat 4.18 kj/kg K is used for cooling purpose at a temperature of 15C. The mass flow rate of cooling water is 17 kg/min calculate the following.1. Outlet temperature of liquid2. Outlet temperature of water3. Effectiveness of heat exchanger

Take care,

Overall heat transfer coefficient is 1100 W/m2 K.Heat exchanger area is 0.30 m2

Given :

Mass flow rate of hot liquid mh = 4.2 kg/minmh = 0.07 kg/s

specific heat of hot liquid Cph = 3.5 kj/kg K

175

Inlet temperature of hot liquid T1 = 130CSpecific heat of water Cph = 4.18 kj/kg K

Cph = 4.18 103 j/kg KInlet temperature of cooling water t1 = 15CMass flow rate of cooling water mc = 17 kg/min

Overall heat transfer coefficient U = 1100 w/m2KArea A = 030 m2

To find

1. Outlet temperature of liquid (T2)2. Outlet temperature of water (t2)3. Effectiveness of heat exchanger ()

Solution:

Capacity rate of hot liquid


To find effectiveness refer HMT data book Page No.165

176

Parallel Flow heat exchanger

From graph

We know

Maximum possible heat transfer

Actual heat transfer rate

We know that

Heat transfer

177

We know that Heat transfer

25. In a counter flow heat exchanger water at 20C flowing at the rate of 1200 kg/h it is heated by oil of specific heat 2100 J/kg K flowing at the rate of 520 kg/h at inlet temperature of 95C. Determine the following.1. Total heat transfer2. Outlet temperature of water3. Outlet temperature of oilTake

Overall heat transfer coefficient is 1000 W/m2 K. Heat exchanger area is 1m2

Given:

Cold fluid – Water Hot fluid – OilInlet temperature of water t1 = 20CMass flow rate of water mc = 1200 kg/h

mc = 0.33 kg/sSpecific heat of oil Cph = 2100 J/kg KMass flow rate of oil = mh = 520 kg/h

Inter temperature of oil T1 = 95COverall heat transfer coefficient U = 1000 W/m2K

Heat exchanger area A = 1m2

To find

178

1. Total heat transfer (Q)2. Outlet temperature of water (T2)3. Outlet temperature of oil (t2)

Solution

Capacity rate of oil

Capacity rate of water

Specific heat of water Cpc = 4186 J/kg K

From Equation (1) and (2)

Number of transfer units NTU =


To find effectiveness refer HMT data book Page No.166

(Counter Flow heat exchanger)

From graph

179

We know Maximum possible heat transfer

We know




180

26. In a cross flow both fluids unmixed heat exchanger, water at 6C flowing at the rate of 1.25 kg/s. It is used to cool 1.2 kg/s of air that is initially at a temperature of 50C. Calculate the following.1. Exit temperature of air2. Exit temperature of water Assume overall heat transfer coefficient is 130 W/m2K and area is 23 m2.

Given :Cold fluid – water Hot fluid – airInlet temperature of water t1 = 6CMss flow rate of water mc = 1.25 kg/sMass flow rate of air mh = 1.2 kg/sInitial temperature of air T1 = 50COverall heat transfer coefficient U = 130 W/m2 KSurface area A = 23 m2

To find 1. Exit temperature of air (T2)2. Exit temperature of water (t2)

Solution

We know Specific heat of water Cpc = 4186 J/kg KSpecific heat of air Cph = 1010 J/kg K (constant)We know Capacity rate of water

From Equation (1) and (2) we know

181

(To find effectiveness refer HMT data book Page No.169)

(cross flow both fluids unmixed)

From graph

Maximum heat transfer


182


we know Heat transfer

UNIT – IVRADIATION

PART – A 1. Define Radiation.

The heat transfer from one body to another without any transmitting medium is known as radiation. It is an electromagnetic wave phenomenon.

2. Define emissive power [E] [Oct. 97, M.U.] ; [Oct. 2000 M.U.]

The emissive power is defined as the total amount of radiation emitted by a body per unit time and unit area. It is expressed in W/m2.

3. Define monochromatic emissive power. [Eb]

The energy emitted by the surface at a given length per unit time per unit area in all directions is known as monochromatic emissive power.

4. What is meant by absorptivity?

Absorptivity is defined as the ratio between radiation absorbed and incident

183

radiation.

Absorptivity

5. What is meant by reflectivity?

Reflectivity is defined as the ratio of radiation reflected to the incident radiation.

Reflectivity

Absorptivity

6. What is meant by transmissivity?

Transmissivity is defined as the ratio of radiation transmitted to the incident radiation.

Transmissivity

7. What is black body? [April.97, April 99, M.U.]

Black body is an ideal surface having the following properties.

1. A black body absorbs all incident radiation, regardless of wave length and di-rection.

2. For a prescribed temperature and wave length, no surface can emit more en-ergy than black body.

8. State Planck’s distribution law. [Oct. 97, April 2000, M.U.]

The relationship between the monochromatic emissive power of a black body and wave length of a radiation at a particular temperature is given by the following expression, by Planck.

Where Eb = Monochromatic emissive power W/m2

= Wave length – mc1 = 0.374 10-15 W m2

c2 = 14.4 10-3 mK

184

9. State Wien’s displacement law.

The Wien’s law gives the relationship between temperature and wave length corresponding to the maximum spectral emissive power of the black body at that temperature.

Where c3 = 2.9 10-3 [Radiation constant]

10. State Stefan – Boltzmann law. [April 2002, M.U.]

The emissive power of a black body is proportional to the fourth power of absolute temperature.

11. Define Emissivity. [Oct. 2000, April 2002, M.U.]

It is defined as the ability of the surface of a body to radiate heat. It is also defined as the ratio of emissive power of any body to the emissive power of a black body of equal temperature.

Emissivity

12. What is meant by gray body? [April, 2000, 2002, M.U.]

If a body absorbs a definite percentage of incident radiation irrespective of their wave length, the body is known as gray body. The emissive power of a gray body is always less than that of the black body.

13. State Kirchoff’s law of radiation. [April 2001 M.U.]

This law states that the ratio of total emissive power to the absorbtivity is constant for all surfaces which are in thermal equilibrium with the surroundings. This can be written as

185

It also states that the emissivity of the body is always equal to its absorptivity when the body remains in thermal equilibrium with its surroundings. 1 = E1; 2 = E2 and so on.

14. Define intensity of radiation (Ib). [Nov. 96, Oct. 98, 99 M.U.]

It is defined as the rate of energy leaving a space in a given direction per unit solid angle per unit area of the emitting surface normal to the mean direction in space.

15. State Lambert’s cosine law.

It states that the total emissive power Eb from a radiating plane surface in any direction proportional to the cosine of the angle of emission

Eb cos

16. What is the purpose of radiation shield? [Apr. 99, Oct. 99 Apr.2001 M.U.]

Radiation shields constructed from low emissivity (high reflective) materials. It is used to reduce the net radiation transfer between two surfaces.

17. Define irradiation (G) [Nov. 96, M.U.]

It is defined as the total radiation incident upon a surface per unit time per unit area. It is expressed in W/m2.

18. What is radiosity (J) [April 2001 M.U.]

It is used to indicate the total radiation leaving a surface per unit time per unit area. It is expressed in W/m2.

19. What are the assumptions made to calculate radiation exchange between the surfaces?

1. All surfaces are considered to be either black or gray 2. Radiation and reflection process are assumed to be diffuse. 3. The absorptivity of a surface is taken equal to its emissivity and independent of

temperature of the source of the incident radiation.

20. What is meant by shape factor?[M.U. EEE, Oct. 97, Apr. 98, Oct 2001 M.U.]

186

The shape factor is defined as the fraction of the radiative energy that is diffused from on surface element and strikes the other surface directly with no intervening reflections. It is represented by Fij. Other names for radiation shape factor are view factor, angle factor and configuration factor.

PART – B

1. A black body at 3000 K emits radiation. Calculate the following:

i) Monochromatic emissive power at 7 m wave length.ii) Wave length at which emission is maximum.iii) Maximum emissive power.iv) Total emissive power,v) Calculate the total emissive of the furnace if it is assumed as a real

surface having emissivity equal to 0.85. [Madras University, April 96]

Given: Surface temperature T = 3000K

Solution:

1. Monochromatic Emissive Power :

From Planck’s distribution law, we know

[From HMT data book, Page No.71]Where

c1 = 0.374 10-15 W m2

c2 = 14.4 10-3 mK = 1 10-6 m [Given]

2. Maximum wave length (max)

From Wien’s law, we know

187

3. Maximum emissive power (Eb) max:

Maximum emissive power (Eb)max = 1.307 10-5 T5

= 1.307 10-5 (3000)5

(Eb)max = 3.17 1012 W/m2

4. Total emissive power (Eb):

From Stefan – Boltzmann law, we know that Eb = T4


Where = Stefan – Boltzmann constant= 5.67 10-8 W/m2K4

Eb = (5.67 10-8) (3000)4

Eb = 4.59 106 W/m2

5. Total emissive power of a real surface:

(Eb)real = T4

Where = Emissivity = 0.85

(Eb)real =

2. A black body of 1200 cm2 emits radiation at 1000 K. Calculate the following:

1. Total rate of energy emission2. Intensity of normal radiation3. Wave length of maximum monochromatic emissive power. 4. Intensity of radiation along a direction at 60 to the normal.

Solution:

From Stefan – Boltzmann law.

1. Energy emission Eb = T4


188

Eb =

Here Area = 1200 10-4 m2, Eb = 5.67103 1200 10-4

Eb = 6804 W

2. Intensity of normal radiation

3. From Wien’s law, we know that

3. Assuming sun to be black body emitting radiation at 6000 K at a mean distance of 12 1010 m from the earth. The diameter of the sun is 1.5 109

m and that of the earth is 13.2 106 m. Calculation the following.

1. Total energy emitted by the sun.2. The emission received per m2 just outside the earth’s atmosphere.3. The total energy received by the earth if no radiation is blocked by the earth’s

atmosphere. 4. The energy received by a 2 2 m solar collector whose normal is inclined at

45 to the sun. The energy loss through the atmosphere is 50% and the diffuse radiation is 20% of direct radiation.

Given: Surface temperature T = 6000 KDistance between earth and sun R = 12 1010 m Diameter on the sun D1 = 1.5 109 m

Diameter of the earth D2 = 13.2 106 m

Solution:

1. Energy emitted by sun Eb = T4

189

2. The emission received per m2 just outside the earth’s atmosphere:

The distance between earth and sun R = 12 1010 m

3. Energy received by the earth:

190

Energy received by the earth

4. The energy received by a 2 2 m solar collector;

Energy loss through the atmosphere is 50%. So energy reaching the earth.

Energy received by the earth

Diffuse radiation is 20%

Total radiation reaching the collection

Energy received by the collector

4. A large enclosure is maintained at a uniform temperature of 3000 K. Calculate the following:

1. Emissive power2. The wave length 1 below which 20 percent of the emission is concentrated and

the wave length 2 above which 20 percent of the emission is concentrated. 3. The maximum wave length.4. Spectral emissive power.5. The irradiation incident.

Given : Surface temperature T = 3000 K

1. Emissive power Eb = T4

191

2. The wave length 1 corresponds to the upper limit, containing 20% of emitted radiation.

, corresponding

1T = 2666 K[From HMT data book, Page No.72]

The wave length 2 corresponds to the lower limit, containing 20% of emitted radiation.

3. Maximum wave length (max):

max T = 2.9 10-3 mK

4. Spectral Emissive Power:

From Planck’s distribution law, we know

192


5. Irradiation:

The irradiation incident on a small object placed within the enclosure may be treated as equal to emission from a black body at the enclosure surface temperature. So, G = Eb = .

5. The sun emits maximum radiation at = 0.52. Assuming the sun to be a black body, calculate the surface temperature of the sun. Also calculate the monochromatic emissive power of the sun’s surface. [April 98, M.U.]

Given :

To find : 1. Surface temperature T. 2. Monochromatic emissive power Eb.

Solution:

1. From Wien’s law, we know

2. Monochromatic emissive power (Eb):

From Planck’s law, we know

193


6. A furnace wall emits radiation at 2000 K. Treating it as black body radiation, calculate

1. Monochromatic radiant flux density at 1m wave length.2. Wave length at which emission is maximum and the corresponding

emissive power. 3. Total emissive power [April 98, MU]

Given: Temperature T = 2000 K; = 1 m = 1 10-6

Solution:

1. Monochromatic emissive power (Eb):


2. Maximum Wave Length (max):

From Wien’s Law, we know that

194

max T = [From HMT data book, Page No.71]

Corresponding emissive power

3. Total emissive power (Eb):

From Stefan – Boltzmann law, we know

Eb = T4

Where - Stefan – Boltzmann constant

7. The temperature of a black surface 0.25 m2 of area is 650C. Calculate,

1. The total rate of energy emission2. The intensity of normal radiation.

The wavelength of maximum monochromatic emissive power. [Oct. 96 EEE, MU]

Given : A = 0.25 m2

T = 650 + 273 = 923 KTo find : 1. Eb ; 2. In ; 3. max

195

Solution:

1. We know Emissive power Eb = T4

Here Area = 0.25 m2

2. We know

3. From Wien’s law,

8. Calculate the heat exchange by radiation between the surfaces of two long cylinders having radii 120mm and 60mm respectively. The axis of the cylinder are parallel to each other. The inner cylinder is maintained at a temperature of 130C and emissivity of 0.6. Outer cylinder is maintained at a temperature of 30C and emissivity of 0.5.

Given : r1 = 60 mm = 0.060 m r2 = 120 mm

= 0.12 T1 = 130C + 273 = 403 1 = 0.6 T2 = 30C + 273 = 303 K 2 = 0.5

To find : Heat exchange (Q)

Solution: Heat exchange between two large concentric cylinder is given by

196

9. Two concentric spheres 30 cm and 40 cm in diameter with the space between them evacuated are used to store liquid air at - 130C in a room at 25C. The surfaces of the spheres are flushed with aluminium of emissivity = 0.05. Calculate the rate of evaporation of liquid air if the latent heat of vaporization of liquid air is 220 kJ/kg.

Given: Inner diameter D1 = 30 cm = 0.30 m

Inner radius r1 = 0.15 mOuter diameter D2 = 40 cm

= 0.40 mOuter radius r2 = 0.20 m

T1 = - 130C + 273 = 143 K

T2 = 25C + 273 = 298 K = 0.05

Latent heat of vapourisation = 220kJ /kg = 220 103 J / kg

197

figTo find: Rate of evaporation

Solution: This is heat exchange between large concentric sphere problem. Heat transfer

Where

10. A pipe of outside diameter 30 cm having emissivity 0.6 and at a temperature of 600 K runs centrally in a brick of 40 cm side square section having emissivity 0.8 and at a temperature of 300K. Calculate the following:

198

1. Heat exchange per metre length.2. Convective heat transfer coefficient when surrounding of duct is 280 K.

Given :

Pipe diameter D1 = 30 cm D1 = 0.30 m

Surface area A1 = D1L = 0.30 1

A1 = 0.942 m2

1 = 0.6 T1 = 600 K

Brick duct side = 40 cm = 0.40 mSurface area A2 = (0.4 1) 4

[length L = 1m; No. of sides = 4]

To find: 1. Heat exchange (Q) 2. Convective heat transfer coefficient (h) when

T = 280 KSolution:

Case 1: We know that

Heat exchange

where

Case (ii) :

Heat transfer by convection Q = hA (T - T)

199

Q12 = h A (T2 - T) Q12 = h 1 (300 – 280)

Equating (2) and (3),

3569.2 = 20h

11. Emissivities of two large parallel plates maintained at 800C and 300C are 0.5 respectively. Find net radiant hat exchange per square metre for these plates. Find the percentage reduction in heat transfer when a polished aluminium radiation shield of emissivity 0.06 is placed between them. Also find the temperature of the shield.

Given : T1 = 800C + 273 = 1073 K T2= 300C + 273

= 573 K 1 = 0.3 2 = 0.5

Shield emissivity 3 = 0.06

To find:

1. Net radiant heat exchange per square metre. (Q/A)2. Percentage reduction in heat loss due to radiation shield.3. Temperature of the shield (T3).

Solution: Heat exchange between two large parallel plates without radiation shield is given by

200

Heat exchange between plate 1 and radiation shield 3 is given by

Heat exchange between radiation shield 3 and plate 2 is given by

We know Q13 = Q32

201

Substituting T3 value in equation (A) (or) equation (B), Heat transfer with radiation shield

Heat transfer with radiation shield

Reduction in heat loss due to radiation shield

12. A pipe of diameter 30 cm, carrying steam runs in a large room and is exposed to air at a temperature of 25C. The surface temperature of the pipe is 300C. Calculate the loss of heat of surrounding per meter length of pipe due to thermal radiation. The emissivity of the pipe surface is 0.8.

What would be the loss of heat due to radiation of the pipe is enclosed in a 55 cm diameter brick of emissivity 0.91?

Given :

202

Case 1: Diameter of pipe D1 = 30 cm = 0.30 mSurface temperature T1 = 300C + 273

= 573 KAir temperature T2 = 25C + 273

= 298 KEmissivity of the pipe 1 = 0.8

Case 2: Outer diameter D2 = 55 cm = 0.55m Emissivity 2 = 0.91

To find: 1. Loss of heat per metre length (Q/L). 2. Reduction in heat loss.

Solution:

Case 1: Heat transfer

Heat loss per metre length = 4271.3 W/m

Case 2: When the 30 cm dia pipe is enclosed in a 55 cm diameter pipe, heat exchange between two large concentric cylinder is given by

203

Reduction in heat loss = 4271.3 – 4057.8= 21.3.4

13. Emissivities of two large parallel plates maintained at T1 K and T2 K are 0.6 and 0.6 respectively. Heat transfer is reduced 75 times when a polished aluminium radiation shields of emissivity 0.04 are placed in between them. Calculate the number of shields required.

Given: 1 = 0.6 2 = 0.6

204

Heat transfer reduced = 75 times Emissivity of radiation shield, s = 3 = 0.04

To find: Number of screens require.

Solution: Heat transfer with n shield is given by

205

14. Find the relative heat transfer between two large plane at temperature 1000 K and 500 K when they are

1. Black bodies2. Gray bodies with emissivities of each surface is 0.7.

Given: T1 = 1000 K T2 = 500 K

1 = 0.7 2 = 0.7

Solution :

Case 1: Heat exchange between two large parallel plate is given by

206

Case 2:

15. The inner sphere of liquid oxygen container is 40 cm diameter and outer sphere is 50 cm diameter. Both have emissivities 0.05. Determine the rate at which the liquid oxygen would evaporate at -183C when the outer sphere at 20C. Latent heat of oxygen is 210 kJ/kg. [April 99, M.U.]

Given : Inner diameter D1 = 40 cm = 0.40 m Inner radius r1 = 0.20 m Outer diameter D2 = 50 cm = 0.50 m Outer radius r2 = 0.25 m Emissivity 1 = 0.05

2 = 0.05 Inner temperature T1 = -183C + 273 = 90K Outer temperature T2 = 20C + 273

= 293 K Latent heat of oxygen = 210 kJ / kg

= 210 103 J/kgTo find : Rate of evaporation

Solution :

207

This is heat exchange between two large concentric spheres problem. Heat transfer

[From equation No.27]

[Negative sign indicates heat is transferred from outer surface to inner surface.]

Rate of evaporation =

16. Emissivities of two large parallel plates maintained at 800C are 0.3 and 0.5 respectively. Find the net radiant heat exchange per square metre of the plates. If a polished aluminium shield ( = 0.05) is placed between them.

208

Find the percentage of reduction in heat transfer. [Oct. 99, M.U.]

Given : T1 = 800C + 273 = 1073 K T2 = 300C + 273 = 573 K 1 = 0.3 2 = 0.5

Radiation shield emissivity 3 = 0.05

To find:

1. Net radiant heat exchange per square metre

2. Percentage of reduction in heat loss due to radiation shield.

Solution:

Case 1 : Heat transfer without radiation shield:

Heat exchange between two large parallel plats without radiation shield is given by

209

Case 2: Heat transfer with radiation shield:

Heat exchange between plate 1 and radiation shield 3 is given by

Heat exchange between radiation shield 3 and plate 2 is given by

We know Q13 = Q32

Substitute T3 value in equation (A) or (B).

210

Substituting T3 value in equation (A) (or) equation (B),

17. The amount of radiant energy falling on a 50 cm 50 cm horizontal thin metal plate insulated to the bottom is 3600kJ /m2 hr. If the emissivity of the plate surface is 0.8 and the ambient air temperature is 30 C, find the equilibrium temperature of the plate.

[April 97, M.U.]

Given : Area A = 50 cm 50 cm = 0.5 0.5 m

Emissivity = 0.8

Ambient air temperature T2 = 30C + 273

211

= 303 K

To find : Plate temperature T1

Solution : We know

Heat transfer

18. Calculate the shape factors for the configuration shown in fig.

1. A black body inside a black enclosure.

2. A tube with cross section of an equilateral triangle.

3. Hemispherical surface and a plane surface

Solution:

Case 1:[All radiation emitted from the black surface 2 is absorbed by the enclosing

surface 1.]

212

We know F1-1 + F1 – 2 = 1 ….(1)

By reciprocity theorem A1F12 = A2F21

Now considering radiation from surface 2,

By reciprocity theorem, we know

A1F12 = A2F21

213

Result: F1 – 1 = 0, F21 = F12 = 0.5 F22 = 0

F1 – 2 = 0.5, F2 – 3 = 0.5 F1 – 3 = 0.5

Case 3: We know F1 – 1 + F1 – 2 = 1

By reciprocity theorem, A1 F1 – 2 = A2 F2 – 1

[Since all radiation emitting from the black surface 2 are absorbed by the enclosing surface 1]

19. Two black square plates of size 2 by 2 m are placed parallel to each other at a distance of 0.5 m. One plate is maintained at a temperature of 1000C and the other at 500C. Find the heat exchange between the plates.

Given: Area A = 2 2 = 4 m2

T1 = 1000C + 273 = 1273 K

214

T2 = 500C + 273 = 773 KDistance = 0.5 m

To find : Heat transfer (Q)

Solution : We know

Heat transfer general equation is

where

[From equation No.(6)]

For black body

Where F12 – Shape factor for square plates

In order to find shape factor F12, refer HMT data book, Page No.76.

Curve 2 [Since given is square plates]

215

X axis value is 4, curve is 2. So corresponding Y axis value is 0.62.i.e.,

20. Two circular discs of diameter 0.3 m each are placed parallel to each other at a distance of 0.2 m. one is disc is maintained at a temperature of 750C and the other at 350C and their corresponding emissivities are 0.3 and 0.6. Calculate heat exchange between the discs.

Given : D1 = 0.3 m D2 = 0.3 m

T1 = 750C + 273 = 1023 KT2 = 350C + 273 = 623 K1 = 0.3

2 = 0.6

Distance between discs = 0.2 m.

To find : Heat exchange between discs (Q),

Solution:Heat transfer by radiation general equation is

216

Where F12 – Shape factor for disc


X axis value is 1.5, curve is 1. So, corresponding Y axis value is 0.28.

21. Two parallel rectangular surfaces 1 m 2m are opposite to each other at a distance of 4m. The surfaces are black and at 300C and 200C. Calculate the heat exchange by radiation between two surfaces.

217


Distance = 4 m T1 = 300C + 273

= 573 K T2 = 200C + 273 = 473 K

To find: Heat exchange (Q12)

Solution : We know, Heat transfer general equation is

Where F12 – Shape factor for parallel rectangles

In order to find shape factor refer HMT data book, Page No.77 and 78.

218

From graph, we know,

22. Two parallel plates of size 3 m 2 m are placed parallel to each other at a distance of 1 m. One plate is maintained at a temperature of 550 C and the other at 250C and the emissivities are 0.35 and 0.55 respectively. The plates are located in a large room whose walls are at 35C. If the plates located exchange heat with each other and with the room, calculate.1. Heat lost by the plates.2. Heat received by the room.

Given: Size of the plates = 3 m 2 mDistance between plates = 1 m

First plate temperature T1 = 550C + 273 = 823 KSecond plate temperature T2 = 250C + 273 = 523 KEmissivity of first plate 1 = 0.35Emissivity of second plate 2 = 0.55Room temperature T3 = 35C + 273 = 308 K

To find: 1. Heat lost by the plates 2. Heat received by the room.

219

Solution: In this problem, heat exchange take place between two plates and the room. So this is three surface problem and the corresponding radiation network is given below.

Area A1 = 3 2 = 6 m2

Since the room is large

From electrical network diagram.

Apply values in electrical network diagram.

220

To find shape factor F12 refer HMT data book, Page No.78.

X value is 3, Y value is 2, corresponding shape factor [From table]

F12 = 0.47

We know that, F11 + F12 + F13 = 1But, F11 = 0

Similarly, F21 + F22 + F23 = 1We know F22 = 0

From electrical network diagram,

221

From Stefan – Boltzmann law, we know

[From diagram]

The radiosities, J1 and J2 can be calculated by using Kirchoff’s law.

The sum of current entering the node J1 is zero.At Node J1:

At node j2

222

-+*

Solving equation (7) and (8),

Heat lost by plate (1) is given by

Heat lost by plate 2 is given by

223

Total heat lost by the plates Q = Q1 + Q2

= 49.36 103 – 3.59 103

Heat received by the room

From equation (9), (10), we came to know heat lost by the plates is equal to heat received by the room.

23. Two black square plates of size 1 by 1 m are placed parallel to each other at a distance of 0.4 m. One plate is maintained at a temperature of 900 C and the other at 400. Find the net heat exchange of energy due to radiation between the two plates. [Oct. 99, M.U.]


Distance = 0.4 m T1 = 900C + 273

= 1173 KT2 = 400C + 273 = 673 K

To find: Heat exchange (Q)

224

Solution: Heat transfer by radiation general equation is

Where F12 – shape factor for square plates.


Curve 2 [since given is square plate]

X axis value is 2.5, curve is 2, so corresponding Y axis value is 0.42.i.e., F12 = 0.42

24. Two circular discs of diameter 20 cm each are placed 2 m apart. Calculate the radiant heat exchange for these discs if there are maintained

225

at 800C and 300C respectively and the corresponding emissivities are 0.3 and 0.5. [Apr. 2000, M.U.]

Given : D1 = 20 cm = 0.2 mD2 = 0.2 mT1 = 800C + 273 = 1073 KT2 = 300C + 273 = 573 K1 = 0.32 = 0.5

To find: Heat exchange (Q)

Solution: Area =

A1 = 0.031 m2

A2 = 0.031 m2

Heat transfer by radiation generation equation is

226

Where F12 = Shape factor for disc.

In order to find shape factor, F12 refer HMT data book, Page No.76.

Curve 1 [since given is disc]

X axis value is 0.1, curve is 1, so corresponding Y axis value is 0.01.

F12 = 0.01

F12 = 0.01

(1)

Q12 = 20.7 Watts.

25. A long cylindrical heater 30 in diameter is maintained at 700C. It has surface emissivity of 0.8. The heater is located in a large room whose wall are 35C. Find the radiant heat transfer. Find the percentage of reduction in heat transfer if the heater is completely covered by radiation shield ( = 0.05) and diameter 40 mm. [April 99, M.U.]

Given : Diameter of cylinder D1=30mm=0.030 mm Temperature T1=700C + 273 = 973 K Emissivity 1 = 0.8 Room temperature T2 = 35C + 273 = 308 K

227

Radiation Shield :

Emissivity 3 = 0.05Diameter D3 = 40 mm = 0.040 m

Solution:

Case 1 : Heat transfer without shield:

Heat transfer by radiation general equation is

Since room is large F12 = Shape factor

Small body enclosed by large body F12 = 1

[Refer HMT data book, Page No.73]

228

Heat transfer without shield

Case 2: Heat transfer with shield:

Heat transfer between heater (1) and radiation shield (3) is given by

Shape factor for concentric long cylinder F13 = 1[Refer HMT data book, Page No. 73]

Heat exchange between radiation shield (3) and Room (2) is given by

Since room is large, A2 =

Shape factor for small body enclosed by large body

229

F32 = 1[Refer HMT data book, Page No.73]

Substitute T3 value in (3) or (4).Heat transfer with radiation shield

26. A gas is enclosed in a body at a temperature of 727 C. The mean beam length of the gas body is 3 m. The partial pressure of water vapour is 0.2 atm and the total pressure is 2 atm. Calculate the emissivity of water vapour.

Given : Temperature T = 727C + 273 = 1000KMean beam length Lm = 3mPartial pressure of water vapour

Total pressure P = 2 atm.

To find : Emissivity of water vapour Solution:

From HMT data book, Page No.92, we can find emissivity of H2o.

230

From graph,

Emissivity of H2o = 0.3

To find correction factor for H20:

From HMT data book, Page No.94, we can find correction factor for H2o

From graph, Correction factor for H2o = 1.36

So, Emissivity of H2o,

231

27. A gas mixture contains 20% CO2 and 10% H2o by volume. The total pressure is 2 atm. The temperature of the gas is 927C. The mean beam length is 0.3 m. Calculate the emissivity of the mixture.

Given : Partial pressure of CO2, = 20% = 0.20 atm Partial pressure of H2o, = 10% = 0.10 atm.

Total pressure P = 2 atm Temperature T = 927C + 273

= 1200 KMean beam length Lm = 0.3 m

To find: Emissivity of mixture (mix).

Solution : To find emissivity of CO2

From HMT data book, Page No.90, we can find emissivity of CO2.

From graph, Emissivity of CO2 = 0.09

To find correction factor for CO2

Total pressure, P = 2 atm

From HMT data book, Page No.91, we can find correction factor for CO2

From graph, correction factor for CO2 is 1.25

232

To find emissivity of :

From HMT data book, Page No.92, we can find emissivity of

From graph Emissivity of = 0.048

To find correction factor for :

From HMT data book, Page No.92 we can find emission of H20

From graph,Correction factor for = 1.39

233

Correction factor for mixture of CO2 and H2O:

From HMT data book, Page No.95, we can find correction factor for mixture of CO2

and

From graph, Total emissivity of gascous mixture

28. A furnace of 25 m2 area and 12 m2 volume is maintained at a temperature of 925C over is entire volume. The total pressure of the combustion gases is 3 atm, the partial pressure of water vapour is 0.1 atm and that of CO2 is 0.25 atm.Calculate the emissivity of the gaseous mixture.

Given : Area A = 25 m2

Volume V = 12 m3

Temperature T = 925 + 273= 1198 K

Total pressure P = 3 atm

234

Partial pressure of water vapour, Partial pressure of CO2

To find: Emissivity of mixture

Solution : We know Mean beam length for gaseous mixture.

To find emissivity of CO2

From HMT data book, Page No.90, we can find emissivity of CO2.

From graph, Emissivity of CO2 = 0.15

To find correction factor for CO2:

Total pressure P = 3 atm.

From HMT data book, Page No.91, we can find correction factor for CO2.

235

From graph, we find

To find emissivity of H2O:

From HMT data book, Page No.92, we can find emissivity of H2O.

From graph, Emissivity of H2O = 0.15

To find correction factor for H2O:

From HMT data book, Page No.94, we can find correction factor for H2O.

236

From graph, we find

Correction Factor for mixture of CO2 and H2O:

From HMT data book, Page No.95 we can find correction factor for mixture of CO2 and H2O.

From graph, we find = 0.045.

Total emissivity of the gaseous mixture is

237

UNIT – VMASS TRANSFER

PART – A

1. What is mass transfer?

The process of transfer of mass as a result of the species concentration difference in a mixture is known as mass transfer.

2. Give the examples of mass transfer.

Some examples of mass transfer.1. Humidification of air in cooling tower2. Evaporation of petrol in the carburetor of an IC engine. 3. The transfer of water vapour into dry air.

3. What are the modes of mass transfer?

There are basically two modes of mass transfer,

1. Diffusion mass transfer2. Convective mass transfer

4. What is molecular diffusion?

The transport of water on a microscopic level as a result of diffusion from a region of higher concentration to a region of lower concentration in a mixture of liquids or gases is known as molecular diffusion.

5. What is Eddy diffusion?

When one of the diffusion fluids is in turbulent motion, eddy diffusion takes place.

6. What is convective mass transfer?

Convective mass transfer is a process of mass transfer that will occur between surface and a fluid medium when they are at different concentration.

238

7. State Fick’s law of diffusion.

The diffusion rate is given by the Fick’s law, which states that molar flux of an element per unit area is directly proportional to concentration gradient.

8. What is free convective mass transfer?

If the fluid motion is produced due to change in density resulting from concentration gradients, the mode of mass transfer is said to be free or natural convective mass transfer.Example : Evaporation of alcohol.9. Define forced convective mass transfer.

If the fluid motion is artificially created by means of an external force like a blower or fan, that type of mass transfer is known as convective mass transfer. Example: The evaluation if water from an ocean when air blows over it.

10. Define Schmidt Number.

It is defined as the ratio of the molecular diffusivity of momentum to the molecular diffusivity of mass.

11. Define Scherwood Number.

It is defined as the ratio of concentration gradients at the boundary.

239

PART – B

1. Explain FICK’s Law of Diffusion.

Consider a system shown in figure.

A partition separates the two gases a and b. When the partition is removed, the two gases diffuses through one other until the equilibrium is established throughout the system.

The diffusion rate is given by the Fick’s law, which states that molar flux of an element per unit area is directly proportional to concentration gradient.

where,

2. Explain steady diffusion through a plane membrane.

Consider a plane membrane of thickness L, containing fluid ‘a’. The concentrations of the fluid at the opposite wall faces are Ca1 and Ca2 respectively.

240

a b

Apply boundary condition

At, x = LCa1 = C2

Ca2 = C1L + C2

Ca2 = C1L + Ca1

From Fick’s law we know

Molar flux,

241

L – ThicknessFor cylinders.

for sphere,

L = r2 – r1

3. Helium diffuses through a plane membrane of 2 mm thick. At the inner side the concentration of helium is 0.25 kg mole/m3. At the outer side the concentration of helium is 0.007 kg mole/m3. What is the diffusion flux of helium through the membrane. Assume diffusion coefficient of helium with respect to plastic is 1 10-9 m2/s.

242

Given Data:

Thickness, L = 2mm = 0.002 m

Concentration at inner side,

To find

Solution :

We know, for plane membrane

4. Gaseous hydrogen is stored in a rectangular container. The walls of the container are of steel having 25 mm thickness. At the inner surface of the container, the molar concentration of hydrogen in the steel is 1.2 kg mole/m3

while at the outer surface of the container the molar concentration is zero, calculate the molar diffusion flux for hydrogen through the steel. Take diffusion coefficient for hydrogen in steel is 0.24 10-12 m2/s.Given data:

Thickness, L = 25 mm = 0.025 m

243

Molar concentration at inner side, Ca1 = 1.2

Concentration at outer side, Ca2 = 0

Diffusion coefficient, Hydrogen

To find:

Solution:

We know, for plain membrane,

5. Hydrogen gases at 3 bar and 1 bar are separated by a plastic membrane having thickness 0.25 mm. the binary diffusion coefficient of hydrogen in the plastic is 9.1 10-3 m2/s. The solubility of hydrogen in the membrane is 2.1

10-3 An uniform temperature condition of 20 is assumed.

Calculate the following

1. Molar concentration of hydrogen on both sides

244

2. Molar flux of hydrogen3. Mass flux of hydrogen

Given Data:

Inside pressure P1 = 3 bar Outside pressure P2 = 1 barThickness, L = 0.25 mm = 0.25 10-3 mDiffusion coefficient Dab =

To find 1. Molar concentration on both sides Ca1 and Ca2

2. Molar flux3. Mass flux

Solution :

1. Molar concentration on inner side,Ca1 = Solubility inner pressureCa1 = 2.1 10-3 3

Ca1 = 6.3 10-3

Molar concentration on outer side

Ca2 = solubility Outer pressureCa2 = 2.1 10-3 1

Ca2 = 2.1 10-3

2. We know

3. Mass flux = Molar flux Molecular weight

245

6. Oxygen at 25C and pressure of 2 bar is flowing through a rubber pipe of inside diameter 25 mm and wall thickness 2.5 mm. The diffusivity of O2 through rubber is 0.21 10-9 m2/s and the solubility of O2 in rubber is 3.12 10-3 . Find the loss of O2 by diffusion per metre length of pipe.

Given data:Temperature, T = 25C FigInside pressure P1 = 2 barInner diameter d1 = 25 mmInner radius r1 = 12.5 mm = 0.0125 mOuter radius r2 = inner radius + Thickness

= 0.0125 + 0.0025r2 = 0.015 m

Molar concentration on outer side, Ca2 = Solubility Outer pressure Ca2 = 3.12 10-3 0Ca2 = 0[Assuming the partial pressure of O2 on the outer surface of the tube is zero]

We know,

246

7. Explain steady state Equimolar counter diffusion.

Consider two large chambers a and b connected by a passage as shown in figure.

Na and Nb are the steady state molar diffusion rates of components a and b respectively.

Equimolar diffusion is defined as each molecule of ‘a’ is replaced by each molecule of ‘b’ and vice versa. The total pressure P = Pa + Pb is uniform throughout the system.

P = Pa + Pb

Differentiating with respect to x

Since the total pressure of the system remains constant under steady state conditions.

Under steady state conditions, the total molar flux is zero.

247

From Fick’s law,

We know

similarly,

where,

Dab = Diffusion coefficient – m2/s

G – Universal gas constant -

A – Area – m2

Pa1 – Partial pressure of constituent at 1 in N/m2

Pa2 – Partial pressure of constituent at 2 in N/m2

T – Temperature - K

248

8. Solved Problems on Equimolar Counter Diffusion Ammonia and air in equimolar counter diffusion in a cylindrical tube of 2.5 mm diameter and 15m length. The total pressure is 1 atmosphere and the temperature is 25 C. One end of the tube is connected to a large reservoir of ammonia and the other end of the tube is open to atmosphere. If the mass diffusivity for the mixture is 0.28 10-4 m2/s. Calculate the following a) Mass rate of ammonia in kg/hb) Mass rate of air in kg/h

Given data: Diameter, d = 2.5 mm 2.5 10-3 mLength (x2 – x1) = 15 mTotal pressure, P = 1 atm = 1 barTemperature, T = 25C + 273 = 298 KDiffusion coefficient, Dab = 0.28 0-4 m2/s.

To find 1. Mass rate of ammonia in kg/h2. Mass rate of air in kg/h

Solution :

We know that, Total pressure P = Pa1 + Pa2

1 bar = Pa1+ 0 [ open to atmosphere. So, Pa2 = 0]

For equimolar counter diffusion

Where, G – Universal gas constant = 8314

249

We know,

Mass transfer rate of ammonia = 3.69 10-13 17Molecular weight of ammonia = 17,

[refer HMT data, Page No.187]

We know,

Mass transfer rate of air =

9. Co2 and air experience equimolar counter diffusion in a circular tube whose length and diameter are 1.2 m, d is 60 mm respectively. The system is at a total pressure of 1 atm and a temperature of 273 k. The ends of the tube are connected to large chambers. Partial pressure of CO2

at one end is 200

250

mm og Hg while at the other end is 90mm of Hg. Calculate the following.

1. Mass transfer rate of Co2 and2. Mass transfer rate of air

Given data:

Diameter, d = 60 mm = .060 mLength, (x2 – x1) = 1.2 mTotal pressure, P = 1 atm = 1 bar Temperature T = 273 KPartial pressure of Co2 at one end

Pa1 = 200 mm of Hg =

Partial pressure of CO2 at other end

Pa2 = 90 mm of Hg =

To find1. Mass transfer rate of CO2

2. Mass transfer rate of air

Solution

We know, for equimolar counter diffusion

251

Where, Dab – Diffusion coefficient – m2sThe diffusion coefficient – m2/s

we know,

Area, A =

We know,

[ Molecular weight of CO2 = 44, refer HMT data book, Page No.187]

We know,

Molar transfer rate of air, mb =

252

10. Solved University Problems on Equimolar counter Diffusion. 1. Two large tanks, maintained at the same temperature and pressure are connected by a circular 0.15 m diameter direct, which is 3 k in length. One tank contains a uniform mixture of 60 mole % ammonia and 40 mole % air and the other tank contains a uniform mixture of 20 mole % air and the other tank contains a uniform mixture of 20 mole % ammonia and 80 mole % air. The system is at 273 K and 1.013 105 pa. Determine the rate of ammonia transfer between the two tanks. Assuming a steady state mass transfer.

Given :

Diameter d = 0.15 mLength (x2 – x1) = 3 mPa1 = 60/100 = 0.6 bar = 0.6 105 N/m2

Pb1 = 40/100 = 0.4 bar = 0.4 105 N/m2

Pa2 = 20/100 = 0.2 bar = 0.2 105 N/m2

Pb2 = 80/100 = 0.8 bar = 0.8 105 N/m2

T = 273 KP = 1.013 105 N/mm2

‘a’ – Ammonia‘b’ – Air

To find Rate of ammonia transfer

Solution:

We know, for equimolar counter diffusion,

253

Where,

G – Universal gas constant = 8314

Area, A =

Dab = Diffusion coefficient of ammonia with air = 77.8 10-3 m2/h [From HMT data book, Page No.185]

11. Determine the diffusion rate of water from the bottom of a test tube of 25 mm diameter and 35 mm long into dry air at 25C. Take diffusion coefficient of water in air is 0.28 10-4 m2/s.

Given :

Diameter d = 25 mm = .025 mLength (x2 – x1) = 35 mm = .035 mTemperature, T = 25C + 273 = 298 KDiffusion coefficient, Dab = 0.28 10-4 m2/s.

To find Diffusion rate of water

Solution :

We know, for isothermal evaporation.

254

Pw2 – Partial pressure at the top of the test tube, that is zero

We know that,

255

12. Estimate the rate of diffusion of water vapour from a pool of water at the bottom of a well which is 6.2 m deep and 2.2 m diameter to dry ambient air over the top of the well. The entire system may be assumed at 30C and one atmospheric pressure. The diffusion coefficient is 0.24 10-4 m2/s.

Given :Diameter d = 2.2 mDeep (x2 – x1) = 6.2 mTemperature, T = 30C + 273 = 303 KTotal Pressure, P = 1 bar = 1 105 N/m2

Diffusion coefficient Dab = 1.24 10-4 m2

To findDiffusion rate of water

Solution We know, for isothermal evaporation,

256

G – Universal gas constant – 8314

Pw1 – Partial pressure at the bottom of the test tube corresponding to saturation temperature 30CAt 30CPw1 = 0.04241 bar

Pw2 = 0

We know,

13. An open pan 210 mm in diameter and 75 mm deep contains water at 25C and is exposed to dry atmospheric air. Calculate the diffusion coefficient of water in air. Take the rate of diffusion of water vapour is 8.52 10-4 kg/h.

Given :

Diameter d = 210 = .210 m

257

Deep (x2 – x1) = 75 mm = .075 mTemperature, T = 25C + 273 = 298KDiffusion rate (or) mass rate, = 8.52 10-4 kg/h = 8.52 10-4 kg/3600s = 2.36 10-7 kg/sMass rate of water vapour = 2.36 10-7 kg/s

To find Diffusion coefficient (Dab)

Solution Dry atmospheric air

We know that, molar rate of water vapour.

We know that,

258

Pw2 = 0

14. Solved University problems on Isothermal Evaporation of water into air.Estimate the diffusion rate of water from the bottom of a test tube 10mm in diameter and 15cm long into dry atmospheric air at 25C. Diffusion coefficient of water into air is 0.255 10-4 m2/s.

Given :

Diameter d = 10mm = .010 mLength (x2-x1) = 15 cm = .15mTemperature, T = 25 + 273 = 298 KDiffusion coefficient (Dab) = Dry atmospheric air

259

To find Diffusion rate of water

Solution:We know that for isothermal evaporation.

Pw1 – Partial pressure at the bottom of the test tube corresponding to saturation temperature 25CAt 25C,Pw1 = 0.03166 barPw1 = 0.03166 105

N/m2

Pw2 = Partial pressure at the top of the test tube that is zero.Pw2 = 0

We know that,

260

2. Estimate the diffusion rate of water vapour from the bottom of a test tube 1.5 cm diameter and 15 cm long into dry air at 25C. Take D = 0.256 cm2/s.

Given :

Diameter d = 1.5 cm = .015Length (x2-x1) = 15 cm = .15 mTemperature T = 25 + 273 = 298 K

Diffusion coefficient (Dab)

To find Diffusion rate of water vapour

Solution We know for isothermal evaporation.

P – Total pressure = 1 bar = 1 105 N/m2

Pw1 – Partial pressure at the bottom of the test tube corresponding to saturation

temperature 25C

261

At 25C Pw1 = 0.03166 bar

Pw2 – Partial pressure at the top of the test tube, that is zero

We know,

15. An open pan of 150 mm diameter and 75 mm deep contains water at 25 C and is exposed to atmospheric air at 25C and 50% R.H. Calculate the evaporation rate of water in grams per hour.

Given :

Diameter, d = 150mm = .150mDeep (x2 –x1) = 75 mm = .075mTemperature, T = 25 + 273 = 298 KRelative humidity = 50%

To find Evaporation rate of water in grams per hour

Solution:

Diffusion coefficient (Dab) [water + air] at 25C

Atmospheric air 50% RH (2)

262

We know that, for isothermal evaporation,

Molar flux,

At 25CPw1 = 0.03166 barPw1 = 0.03166 105 N/m2

Pw2 = Partial pressure at the top of the test pan corresponding to 25C and 50% relative humidity. At 25C

263

For laminar flow :

Sherwood Number (Sh) = 0.664 (Re)0.5 (Sc)0.333


where, Sc – Schmidt Number =

Dab – Diffusion coefficient

Sherwood Number, Sh =

Where, hm – Mass transfer coefficient – m/sFor Turbulent flow :

Shedwood Number (Sh) = [.037 (Re)0.8 – 871] Sc0.333

Solved Problems on Flat Plate.

16. Air at 10C with a velocity of 3 m/s flows over a flat plate. The plate is 0.3 m long. Calculate the mass transfer coefficient.

Given :

Fluid temperature, T = 10CVelocity, U = 3 m/sLength, x = 0.3 m

264

To find:Mass transfer coefficient (hm)

Solution:Properties of air at 10C [From HMT data book, Page No.22]

Kinematic viscosity. V = 14.16 10-6 m2/sWe know that,

For Laminar flow, flat plate,

Sherwood Number (Sh) = 0.664 (Re)0.5 (Sc)0.333 ….(1)

[From HMT data book, Page No.179]Where,

Sc – Schmidt Number =

Dab – Diffusion coefficient (water+Air) at 10C = 8C

Substitute Sc, Re values in equation (1)

265

17. Dry air at 30C and one atmospheric pressure flows over a flat plate of 600 mm long at a velocity of 55 m/s. Calculate the mass transfer coefficient at the end of the plate.

Given :

Fluid temperature, T = 30CVelocity U = 55 m/sLength X = 600 mm = 0.6 m

To find

Mass transfer coefficient (hm)

Solution:-

Properties of air at 30CKinematic viscosity, v = 16 10-6 m2/s.

For combined Laminar – Turbulent flow, flat plate.

266

Sherwood Number (Sh) = [.037 (Re)0.8 – 871] Sc0.333….(1)Where,

We know that,


18. The water in a 6m 15m out door swimming pool is maintained at a temperature of 28C. Assuming a wind speed of 2.5 m/s in the direction of the long side of the pool. Calculate the mass transfer coefficient.

Given :

Size = 6 m 15 mFluid temperature, T = 28CSpeed, U = 2.5 m/s

Wind speed in the direction of the long side of pool, so x = 15 m.

To find:Mass transfer coefficient (hm)

Solution :

267

Properties of air at 28C 30C[From HMT data book, Page No.22]

Kinematic viscosity, v = 16 10-6 m2/sWe know,

Since, re > 5 105, flow is turbulent.

[Flow is laminar upto Re = 5 105, after that flow is turbulent].

For combined Laminar – Turbulent flow, flat plate

Sherwood Number (Sh) = [.037(Re)0.8 – 871]Sc0.333….(1)

[From HMT data book, Page No.180]where,


We know that,

268


19. Air at 25C flows over a tray full of water with a velocity of 2.8 m/s. The tray measures 30 cm along the flow direction and 40 cm wide. The partial pressure of water present in the air is .007 bar. Calculate the evaporation rate of water if the temperature on the water surface is 15C. Take diffusion coefficient is 4.2 10-5 m2/s.

Given:

Fluid temperature T = 25CSpeed, U = 2.8 m/s

Flow direction is 30 cm side, so x = 30cmX = 0.30 m

Area, A = 30 cm 40 cm = 0.30 0.40m2

Partial pressure of water, Pw2 = 0.0007 bar

Water surface temperature, Tw = 15CDiffusion coefficient Dab = 4.2 10-5 m2/s.

To find:Evaporation rate of water (mw)

Solution:

We know that,

Film temperature,

Properties of air 20C[From HMT data book, Page No.22]

Kinematic viscosity, v = 15.06 106 m2/s.

We know that,

269

For flat plate, laminar flow:

Sherwood Number (Sh) = [0.664(Re)0.5 (Sc)0.333]….(1)[From HMT data book, Page No.179]

Where,

Substitute Sc. Re values in equation (1)

We know that,

Mass transfer coefficient based on pressure difference is given by,

270

Formula used for internal Flow (Cylinders or Pipes)Problems

For laminar flow:

For Turbulent flow:

Sherwood Number (Sh) = 0.023 (Re)0.44


Where, Sc = Schmidt Number = .

Solved Problems on Internal Flow (Pipes and Cylinders)

20. Air at 30C and atmospheric pressure flows in a 12 mm diameter tube of 1 metre length with a velocity of 2.5 m/s. The inside surface of the tube contains a deposit of naphthalene. Determine the average mass transfer coefficient. Take diffusion coefficient Dab = 0.62 10-5 m2/s.Given :Fluid temperature, T = 30CVelocity, U = 2.5 m/sDiameter, D = 12 mm = 0.012 mLength x = 1m

271

Diffusion coefficient, Dab = 0.62 10-5 m2/s

To find

Average mass transfer hm

Solution

Properties of air at 30C (From HMT data book, Page No.22)Kinematic viscosity v = 16 10-6 m2/sWe know that

Reynolds Number

Since Re, 2000 Flow is laminar

For laminar Internal Flow

Sherwood number Sh = 3.66We know that

Sherwood number Sh =

21. Air at 20C and atmospheric pressure containing small quantities of iodine flows with a velocity of 4 m/s inside a 4 cm inner diameter tube. Determine the mass transfer co efficient. Assume Dab = 0.75 10-5 m2/s.Given Fluid temperature T = 20CVelocity U = 4 m/sDiameter D = 4 cm = 0.04 mDiffusion coefficient Dab = 0.75 10-5 m2/s

To find

Properties of air at 20C (From HMT data book Page No.22)

272

Kinematic viscosity v = 15.06 10-6 m2/sWe know that

2000Since re > 2000 flow is turbulent.

For turbulent, Internal flow

Sherwood Number Sh = 0.023 (Re)083 (Sc)044 …….1Where

Schmidt Number Sc =


We know that

Sherwood number Sh =

A.U. APRIL / MAY 2004ME 340 – HEAT AND MASS TRANSFER

(Use of Steam table Mollier chart and HMT data book is

273

permitted)Max : Marks : 100 on : 3 hr

Part – A ( 10 2 = 20 marks)

1. What is Fourier’s law of heat conduction?2. What is critical thickness of insulation?3. State Stefan Boltzmann law and Planck’s law. How they are related?4. What is radiation shield?5. Explain clearly the difference between natural convection heat transfer and forced

convection heat transfer?6. Define boundary layer thickness. 7. What is pool boiling?8. What are the modes of heat transfer present in steam generator?9. Give two examples of convective mass transfer?Ans : 1. Evaporation of alcohol 2. Evaporation of water from an ocean when air blows over it.

10. What is Sherwood number?

PART – B ( 5 16 = 80 Marks)

11. In a counter flow double pipe heat exchanger, water is heated from 25C to 65C by an oil with a specific heat of 1.45 kj/kg K and mass flow rate of 0.9 kg/s. the oil is cooled from 230C to 160C. If the overall heat transfer coefficient is 420 W/m2

C, calculate the following.i. The rate of heat transfer

ii. The mass flow rate of water iii. The surface area of the heat exchanger

Given Hot fluid – Oil Cold fluid – water (T1, T2) (t1, t2)

Entry temperature of water t1 = 25CExit temperature of water t2 = 65C

274

Solution


Where (T)m – Logarithmic Mean Temperature Difference LMTD for counter flow

Substitute (T)m Q, and U values in equation 1

We know that for cold fluid

275

Result i. Heat transfer Q = 91.35 103 Wii. Mass flow rate of water mc = 0.545 Kg/siii. Surface area of the heat exchanger A = 1.455 m2

12. (a) Derive general heat conduction equation in Cartesian coordinates.

Solution: Consider a small rectangular element of sides dx dy and dz as shown in figure.

The energy balance of this rectangular element is obtained from first law of thermodynamics.

Net heat conducted into element from all the coordinate direction

Let qx be the heat flux in a x direction at x, face ABCD and qx+dx be the heat flux at x+dx face EFGH

The rate of heat flow into the element in x direction through the face ABCD is

276

Where kx – Thermal conductivity - W/mm

The rate of heat flow out of the element in x direction through the face at x+dx EFGH is

Subtracting (2) – (3)

Net heat conducted

277

We know that

Heat generated with in the element q dx dydz ……8

Heat stored in the element

Substituting (7) (8) (9) in equation (1)

Considering the material is isotropic so

kx = ky = kz = k = constant

Divided by k

278

Where Thermal diffusivity =

Equation (10) is known as general three dimensional heat conduction equation.

12. (ii) An exterior wall of a house may be approximated by a 0.1 m layer of common brick (k = 0.7 W/m C) followed by a 0.04 m a layer of gypsum plaster (k = 0.48 W/m C). What thickness of loosely packed rock wool insulation (k = 0.065 W/mC) should be added to reduce the heat loss or gain through the wall by 80%.12. b) The temperature distribution across a large concrete slab (k = 1.2 W/mC = 1.77 10-3 m2/h) 500 mm thick heated from one side as measured by thermocouples approximates to the relation t = 60-50x + 12x2+20x3-15x4

where ‘t’ is in C and x is in meters. Considering an area of 5 m2 compute

i. The heat entering and leaving the slabs in unit time. ii. The heat energy stored in unit time.iii. The rate of temperature change at both sides of the stabs.iv. The point where the rate of heating or cooling is maximum.

Given :

0.04916mm = 0.5m

t = 60 – 50x + 12x2 + 20x3 – 15x4

A = 5m2

Solution

Heat entering the slab

279

Heat leaving the slab

(ii) The heat energy stored in unit time

Q = Qin – Qout = 300 – 183

(iii) The rate of temperature change at both sides of the slab

Substitute x = 0

Substitute x = 0.5 in equation (1)

iv. The point where the rate of heating or cooling is maximum

280

Result :

Qin = 300 W; Qout = 183 WQ = 117 W

13. (a) Two parallel plates of size1 m 1 m are spaced 0.5 m apart are located in a very large room, the walls of which are maintained at a temperature of 27C. One plate is maintained at a temperature of 900C and the other at 400C. Their emissivities are 0.2 and 0.5 respectively. If the plates exchange heat between themselves and surroundings, find the net heat transfer to each plate and to the room. Consider only the plate surfaces facing each other. 13. (b) (i) Write short notes on Gaseous emission and absorption. (14) Calculate the net radiant heat exchange per m2 area for two large parallel plates at temperature of 427C ad 27C respectively

If a polished aluminium shield is placed between them find the percentage reduction in the heat transfer .

15. (a) (i) Air at atmospheric pressure and 200C flows over a plate with a velocity of 5 m/s. The plate is 15 mm wide and is maintained at a temperature of 120C. Calculate the thickness of hydrodynamic and thermal boundary layers and the local heat transfer coefficient at a distance of 0.5 m from the leading edge. Assume that the flow is on one side of the plate.

(ii) A flat plate 1 m wide and 1.5 m long is to be maintained at 90 C in air

281

with a free steam temperature of 10C. Determine the velocity with which air must flow over flat plate along 1.5 m side. So that the rate of energy dissipation from the plate is 3.75 KW. Take the following properties of air at 50C.

Given

Wide W = 1mLength L = 1.5 m

Plate surface temperature Tw = 90CFluid temperature T = 10CHeat transfer or Energy transfer Q = 3.75 KW

= 3.75 103 W

To find : velocity of air U

Solution : We know that Heat transfer Q = h A (Tw - T)

We know that

Local heat transfer coefficient

Local Nusselt Number Nux = 0.332 (Re)0.5 (Pr)0.333

282

14. (b) A hot plate 1.2 m wide, 0.35 m high and at 115C is exposed to the ambient still air at 25C. Calculate the following.

i. Maximum velocity at 180 mm from the leading edge of the plate. ii. The boundary layer thickness at 180 mm from the leading edge of the

plate. iii. Local heat transfer coefficient at 180 mm from the leading edge of the

plate. iv. Average heat transfer coefficient over the surface of the plate. v. Total mass flow through the boundary.

vi. Heat loss from the plate.vii. Rise in temperature of the air passing through the boundary. Use ap-

proximate solution.

Given

Wide W = 1.2 mHeight or length L = 0.35 mPlate surface temperature Tw = 115CFluid temperature T = 25CDistance x = 180 mm = 0.180 mm

Solution :

We know that

Film temperature Tf =

Properties of air at 70C From HMT data book Page No.22

283

We know that

(i) Maximum velocity at 180 mm from the leading edge, Umax

(ii) The boundary layer thickness at 180 mm from the leading edge of the plate

(iii) Local heat transfer coefficient at 180 mm from the leading edge of the

284

plate hx

We know that

(iv) Average heat transfer coefficient over the surface of the plate h

Grashof Number grL For entire plate

Average heat transfer coefficient h

v. Total mass flow through the boundary (m) We know that

285

vi. Heat loss from the plate, Q

vii. Rise in temperature of the air passing through the boundary.

Result

15. (a) The molecular weights of the two components A and B of a gas mixture are 24 and 28 respectively. The molecular weight of gas mixture is found to be 30. If the mass concentration of the mixture is 1.2 kg m3

determine the following.i. Density of component A and Bii. Molar fractions

286

iii. Mass fractionsiv. Total pressure if the temperature of the mixture is 290 K.

Given: Molecular weight of component A1 MA = 24 Molecular weight of component B, MB = 48 Molecular weight of gas mixture M = 30 Concentration =1.2 kg/m3

Temperature T = 290 K To findi. Density of component A and B A, Bii. Molar fractions XA and XB

iii. Mass fractions ma and MB

iv. Total pressure p

Solution: Mobile concentration of the mixture

.

We know that

B = 48CB

We know thatA = B =

solving equation (1) and (2)

CA = 0.03 kg mole/m3

CB = 0.01 kg mole/m3

We know that

(i) Density A = 24 CA

= 24 x 0.03

287

A = 0.72 kg/m3

B = 48 CB

= 48 x 0.01

(ii) Molar fractions

(iii) Mass fractions

iv. Total pressure at 290 K

We know that gas law Pv = m RT

Result

288

15. (b) I state Fick’s law of diffusion

(iii) Air at 1 atm and 25C containing small quantities of iodine, flow with a velocity of 6.2 m/s inside a 35 mm diameter tube. Calculate mass transfer coefficient for iodine. The thermophysical properties of air are

v = D =

A.U. NOVEMBER / DECEMBER 2004ME 340 – HEAT AND MASS TRANSFER

Use of steam table Mollier chart and HMT data book is permittedMax. Marks : 100 Duration : 3 hrs

PART – A (10 2 = 20 marks)

1. What do you understand by critical radius of insulation and give its expression?2. Define fin efficient and effectiveness.3. What do you understand by gray body and block body?4. State Wien’s displacement law and Kirchoff’s law.5. Mention the difference between free and forced convection.6. What is the importance of boundary layer?7. State the difference between filmwise and dropwise condensation. 8. Sketch the temperature variations in parallel flow and counter flow heat exchang-

ers.

289

9. Define molar concentration and mass fraction.

Molar concentration: The molar concentration CA is the number of moles of moles of species ‘A’ per unit volume of the mixture.

Where MA – Molecular weight of component A.Mass fraction : Mass fraction mA is defined as the ratio of the concentration of species A to the total mass density of the mixture.

10. State Fick’s law of diffusion and give its expression.PART – B (5 16 = 80 marks)

11. (i) Discuss the general arrangement of parallel flow counter flow and cross flow heat exchangers.

(ii) In a double pipe counter flow heat exchanger 10,000 kg/h of an oil having a

290

specific heat of 2095 J/kg K is cooled from 80C to 50C by 8000 kg/h of water entering at 25C. Determine the heat exchangers area for an overall heat transfer coefficient of 300 W/m2K. Take Cp for water as 4180 J/kgK.

12. (a) (i) Explain the different modes of heat transfer with appropriate expressions.

(ii) A composite wall consists of 10 cm thick layer of building brick, K = 0.7 W/mK. An insulating material of K = 0.08 to be added to reduce the heat transfer through the wall by 40%. Find its thickness.

12. (b) Circumferential aluminium fins of rectangular profile 1.5 cm wide and 1 mm thick are fitted to a 90 mm engine cylinder with a pitch of 10 mm. The height of the cylinder is 120 mm. The cylinder base temperature before and after fitting the fins are 200C and 150C respectively. Take ambient at 30C and it (average) = 100 W/m2K. Estimate the heat dissipated from the finned and the unfinned surface areas of cylinder body.

13. (a) (i) Define emissivity, absorptivity and reflectivity.

(ii) Describe the phenomenon of radiation from real surfaces.

The real surfaces do not be have like black bodies. The radiative properties of real surfaces vary with the direction of emission, wave length, surface temperature and surface roughness.

Emissivity: The spectral radiation emitted by a real surface differs from the Planck’s distribution given for a black body. The emission from the real surface is much less than that of black body.

Wavelength: For conductors, the emissivity decreases with increase of wavelength. For non-metals the wavelength dependence of emissivity is quite weak.

Surface temperature: The total emissivity of metals increases with temperature for wavelengths greater than about 5. For shorter wavelengths the emissivity of metals decreases with temperature.

Surface roughness: The effects of surface roughness on radiation characteristic of surfaces are very complex. There is no proper way of defining surface roughness.

A surface can be optically smooth for longer wavelengths and rough at lower wavelengths.

[OR]

13. (b) (i) What are radiation view factors and why are they used?

(ii) Determine the view factor (F14) for the figure shown below.

291

From Fig. A5 = A1 + A

14. (a) Air at 25C flow over 1 m 3m (3m long) horizontal plate maintained at 200C at 10 m/s. Calculate the average heat transfer coefficients for both laminar and turbulent regions. Take Re (Critical) = 3.5 105.

Given :

Fluid temperature T = 25CLength L = 3 mPlate temperature Tw = 200CVelocity U = 10 m/sRe (critical) = 3.5 105

Solution

292

Properties of air at 112.5C: (From HMT data book, Page No.22)

Here Re (critical) = 3.5 105 i.e., flow is laminar upto Reynolds number value is 3 105 after than flow is turbulent.

Case (i) : For laminar flow,

Local Nusselt number Nux = 0.332 (Re)0.5 (Pr) 0.333[From HMT data book, Page No.99]

Average heat transfer coefficient for turbulent flow

[OR]

293

14. (b) (i) Define Reynold’s, Nusselt and Prandtl numbers. (ii) A steam pipe 10 cm outside diameter runs horizontally in a room at 23 C. Take the outside surface temperature of pipe as 165C. Determine the heat loss per metre length of the pipe.

15. (a) (i) Discuss the various regimes of pool boiling heat transfer.

(ii) Dry saturated steam at a pressure of 2.45 bar condenses on the surface of a vertical tube of height 1m. The tube surface temperature is kept at 117C. Estimate the thickness of the condensate film and the local heat transfer coefficient at a distance of 0.2 m from the upper end of the tube. [OR]

15. (b) (i) A mixture of O2 and N2 with their partial pressures in the a ratio 0.21 to 0.79 is in a container at 25C. Calculate the molar concentration, the mass density, the mole fraction and the mass fraction of each species for a total pressure of 1 bar. What would be the average molecular weight of the mixture?

We know that,

We know that

294

The overall mass density is

The average molecular weight

ME 340 – HEAT AND MASS TRANSFER A.U. MAY / JUNE 2006

PART - A (10 X 2 = 20 Marks)

1. State Fourier’s law of heat conduction.2. Write down the three-dimensional heat conduction equation in rectangular coordinate system.3. What do you understand by a graybody and blackbody?4. State Wien’s displacement law and Kirchoff’s law.5. An electrically heated plate dissipates heat by convection at a rate of 8000 W/m2

into the ambient air at 250C. If the surface of the hot plate is at 125oC, calculate the heat transfer coefficient for convection between the plate and the air.

q = h(Tw – Tt)h = 80 w/m2oC

6. Define Reynolds number and Prandtl number.7. What do you understand by free and forced convection?8. State the difference between filmwise and dropwise condensation.9. State Fick’s law of diffusion and give its expression.

295

10. Describe the two mechanisms of mass transfer.

PART – B

11. (i) Discuss how the radiation from gases differ from that of solids.(ii) Two very large parallel plates with emissivities 0.5 exchange heat. Determine the percentage reduction in the heat transfer rate if a polished aluminium radiation shield of = 0.04 is placed in between the plates.

12. (a) (i) A furnace wall consists of three layers. The inner layer of 10 cm thickness is made of firebrick (k = 1.04 W/mK). The intermediate layer of 25 cm thickness is made of masonry brick (k = 0.69 W/mK) followed by a 5 cm thick concrete wall (k = 1.37 W/mK). When the furnace is in continuous operation the inner surface of the furnace is at 800oC while the outer concrete surface is at 50oC. Calculate the rate of heat loss per unit area of the wall, the temperature at the interface of the firebrick and masonry brick and the temperature at the interface of the masonry brick and concrete.(ii) An electrical wire of 10 m length and 1 mm diameter dissipates 200 W in

air at 25oC. The convection heat transfer coefficient between the wire surface and air is 15 W/m2K. Calculate the critical radius of insulation and also determine the temperature of the wire if it is insulated to the critical thickness of insulation.

(b) (i) An aluminium rod (k =204 W/mK) 2 cm in a diameter and 20 cm long pro-trudes from a wall which is maintained at 300oC. The end of the rod is insu-lated and the surface of the rod is exposed to air at 30oC. The heat transfer co-efficient between the rod’s surface and air is 10 W/m2K. Calculate the heat lost by the rod and the temperature of the rod at a distance of 10 cm from the wall.

(ii) A large iron plate of 10 cm thickness and originally at 800oC is suddenly ex-posed to an environment at 0oC where the convection coefficient is 50 W/m2K. Calculate the temperature at a depth of 4 cm from one of the faces 100 seconds after the plate is exposed to the environment. How much energy has been lost per unit area of the plate during this time?

13. (a) (i) Write down the momentum equation for a steady, two dimensional flow of an incompressible, constant property newtonian fluid in the rectangular coordinate system and mention the physical significance of each term.

(ii) A large vertical plate 5 m high is maintained at 1000C and exposed to air at 30oC. Calculate the convection heat transfer coefficient.

(b) (i) Sketch the boundary layer development of a flow over a flat plate and explain the significance of the boundary layer.

296

(ii) Atmospheric air at 275 K and a free stream velocity of 20 m/s flows over a flat plate 1.5 m long that is maintained at a uniform temperature of 325 K. Calculate the average heat transfer coefficient over the region where the boundary layer is laminar, the average heat transfer coefficient over the entire length of the plate and the total heat transfer rate from the plate to the air over the length 1.5 m and width 1 m. Assume transition occurs at Rec = 2 x 105.

14. (a) (i) It is desired to boil water at atmosphericpressure on a copper surface which is electrically heated. Estimate the heat flux from the surface to the water, if the surface is maintained at 110oC and also the peak heat flux.

(ii) A tube of 2 m length and 25 mm OD is to be used to condense saturated steam at 100oC while the tube surface is maintained at 92oC. Estimate the average heat transfer coefficient and the rate of condensation of steam if the tube is kept horizontal. The steam condenses on the outside of the tube.

Or

(b)(i) Give the classification of heat exchangers.

(ii) It is desired to use a double pipe counter flow heat exchanger to cool 3 kg/2 of oil (Cp = 2.1 kJ/kgK) from 120oC. Cooling water at 20oC enters the heat exchanger at a rate of 10 kg/s. The overall heat transfer coefficient of the heat exchanger is 600 W/m2K and the heat transfer area is 6 m2. Calculate the exit temperatures of oil and water.

15. (a) (i) Define mass concentration, molar concentration, mass fraction and mole faction.(ii) The diffusivity of CCl4 in air is determined by observing the steady state evaporation of CCl4 liquid level is 10 cm below the top level of the tube. The system is held at 25oC and 1 bar pressure. The saturation pressure of CCl4 at 25oC is 14.76 kPa. If it is observed that the rate of evaporation of CCl4 is 0.1 g/hour determine the diffusivity of CCl4 into air.

Or

(b) (i) Dry air at 20oC ( = 1.2 kg/m3, v = 15 x 10-6 m2/s, D = 4.2 x 10-5 m2/s) flows over a flat plate of length 50 cm which is covered with a thin layer of water at a velocity of 1 m/s. Estimate the local mass transfer coefficient at a distance of 10 cm from the leading edge and the average mass transfer coefficient.

(ii) Discuss the analogy between heat and mass transfer.

ME 340 – HEAT AND MASS TRANSFERPART – A – (10 X 2 = 20 MARKS)

297

1. Rate of heat flow by conduction in a given direction is proportional to the area normal to the direction of heat flow and to the gradient of temperature in that direction.

2.

3. Grey body : is defined such that the monochromatic emissivity of the body is independent of wavelength.

Black body : is an ideal body that absorbs all incident radiant energy and reflects or transmits none.4. Wien’s displacement law : max . T = C.

Kirchoff’s law : establishes a relationship between the emissive power of a surface to its absorptivity.

5. q = h(Tw – Tf). h = 80 W / m2 oC.

6. Reynold’s no. = .

Prandtl no. =

7. Free convection : due to density variations associated with temperature gradients within the fluid. Forced convection : Fluid motion produced due to superimposed velocity field like fan, pump, etc

8. Film wise condensation : Condensate wets the surface forming a continuous film which covers the entire surface.

Dropwise condensation : Vapour condenses into small liquid droplets.

9. JAx = -DAB Where J Ax – molar fluxdCA / dx – concentration gradientDAB – diffusivity coefficient.

10. Diffusion : Mass transfer by random molecular motion.

Convective : Mass transfer by concentration gradient of species.

PART B – (5 X 16 = 80 Marks) 11. (i) The radiation from gases differs from solids in the following ways:

298

(1) The radiation from solids is at all wavelengths, whereas gases radiate our specific wavelength ranges ranges or bands within the thermal spec-trum.

(2) The intensity of radiation as it passes through an absorbing gas de-creases with the length of passage through the gas volume. This is un-like solids wherein the absorption of radiation takes place within a small distance from the surface.

(6)

(ii)

With radiation shield placed in between the plates F1S = FS2 = 1.

The resistances are

The total resistance is given by

The rate of radiation heat transfer in the presence of shield is given by

Percentage reduction in heat transfer rate

12. (a) (i) Thermal resistance of fire brick layer

Thermal resistance of masonry brick layer

Thermal resistance of concrete layer

299

R = 0.495 K/W.

Temperature difference across firebrick layer isqR1 = 145.7oC

Temperature difference across masonry brick layer isqR2 = 548.9oC.

Temperature at the interface of firebrick layer and masonry brick layer = 800-145.7 = 654.3oC.Temperature at the interface of masonry brick and concrete = 654.3 – 548.9 = 105.4oC.

(ii) At critical radius of insulation the thermal resistance for convection is taken as R2 rc = k/h = 38.8 mm.

Thermal resistance for conduction due to insulation

R = R1 + R2 = 0.1274 K/W.T = q . R.Ti = 50.5o C.

(b) (i) P = d = 6.283 x 10-2 m.

tan h(ml) = 30 W.

Temperature distribution is given by

300

T = 265.6o C.

(ii) Bi = = 0.034.

From Heisler charts

and F0 = 0.814.

Then

= 662.4oC.

Or T = 662.4oC

Fo (Bi)2 = 9.4 x 10-4

Q/Qi = 0.02 at (Fo) (Bi)2 = 9.4 x 10-4 and Bi = 0.034.

= 285.56 MJ.

Q = 0.02 Qi = 5.71 MJ.

13. (a) (i)

Where F – body forcesLHS represent inertia forces1st term on RHS is body force2nd term on RHS is pressure force

Last term on RHS in viscous forces acting on the fluid element.

301

(ii)

Properties of air at 338 K areK=0.02905, Pr=0.6999, v=19.4418x 10-6 m2/s.

Gr =

Gr Pr = 4.6977 x 1011.

Nu = C(Gr Pr)m.C = 0.021 and m = 2/5.

h = 5.69 w/m2 K.

Or(b) (i) Boundary layer development.

Two regions of boundary layer are present

(1)a thin region (boundary layer) where the velocity and temperature gradi-ents are large.

(2)The region outside the boundary layer where the velocity and tempera-ture gradients are very nearly equal.

(ii) = 300 K.

Physical properties of atmospheric air at 300 K. k = 0.026 w/moC, Pr = 0.078, v = 16.8 x 10-6 m2/s = 1.98 x 10-5 kg/m.s.

= 41.0 w/m20 C.

302

= 49.1 w/m2 . oC.Q = wLhm (Tw - T).

= 3683 W.

14. (a) (i) For water at 100oC, hfg = 2257 kJ / kg

= 958.4 kg/m3, C = 4.211 kJ / kg K, = 277.5 x 10-6 Ns/m2.

Pr = 1.75

= 58.9 x 10-3 N/m and csf = 0.013.

Rohsenow correlation is

Neglecting v in comparison to and n = 1 for water, we get q’ = 137.91 kw/m2.

Zuber correlation for peak heat flux is given by

At 100oC, v = 0.5977 kg/m2 from steam tables

q” = 1.108 MW / m2.(ii)

At 96oC, kf = 0.68 W/m . K, f = 293.4 x 10-6 Ns/m2

= 961 kg/m3 , hfg = 2257 kJ / kg at 100oC.

v << we can approximate ( - v) as 2

303

= 13.187 kw/m2k.

Q = hA(Ts – Tw)

= 16.571 kW.Rate of condensation

= 7.34 x 10-3 kg / s.

Or

(b) (i) Parallel flow, counterflow, crossflow-mixed and unmixed, shell and tube and compact heat exchangers.

(ii) c Cc = 42 kJ / K, hCh = 6.3 kJ / K

h Ch is minimum, hence Cmin = 6.3 kJ / K, Cmax = 42 kJ / K.

NTU =

from the chart or from the formula = 42%.

h =

Th2 = 78oC.

Q = Cmin (Th1 – Tc2) = 264.6 kW

TC1 = 26.3oC.

15. (a) (i) Mass concentration =

Molar concentration =

Mass fraction =

Mole fraction =

(ii) Molar mass of CCl4 = 154 g/mol.

304

C.S.A. of tube = = 7.85 x 10-5 m2.

=2.298x10-3 mol/m2s.

We know,

Or

At the liquid level, pB1 = P - PCCl4 = 100 -14.76 = 85.24 kPa(y = y1)

At the top of tube, PB2 = 100 kPa(y = y2)

DAB = 3.565 x 10-5 m2 / s.Or

(b) (i) Rex = 0.1

Sc = = 0.357.

Since Rex < 5 x 105, flow is in laminar region.

Shx = = 0.332 Rex0.8 Sc0.33

hx = 0.114 m/s.

m = 2hx = l.

Shx = l = = 0.332 Rel0.8 Sc0.33

hl = 0.082 m/s. m = 2x0.082 = 0.164 m/s.

(ii) Analogy between heat and mass transfer.ME 340 – HEAT AND MASS TRANSFER

(Use of HMT Data Hand Book and Steam Tables Permitted)

305

PART A (10 X 2 = 20 Marks)

1. Define fin efficiency and fin effectiveness.2. List down the three types of boundary conditions.3. What do you understand by black body and grey body?4. Define radiosity and emissive power.5. Indicate the concept of boundary layer.6. Define Nusselt number and Stanton number.7. What do you understand by filmwise and dropwise condensation?8. How are heat exchangers classified based on flow arrangement?9. State Fick’s law of diffusion and mention its importance.10. Define mass concentration and molal concentration.

PART - B – (5 X 16 = 80 Marks)

11. (i) A composite wall consists of 10 cm thick layer of brick, k = 0.7 W/mK and 3 cm thick plaster, k = 0.5 W/mK. An insulating material of k = 0.08 W/mK is to be added to reduce the heat transfer through the wall by 40%. Find its thickness.

(ii) An aluminium plate (k = 160 W/moC, = 2790 kg/m3, cp = 0.88 kJ/kgoC) of thickness L = 3 cm and at a uniform temperature of 225o C is suddenly immersed at time t = 0 in a well stirred fluid maintained at a constant temperature T = 25o C. Take h = 320 W/m2 oC. Determine the time required for the center of the plate to reach 50o C.

12. (a) (i)Explain absorptivity, reflectivity and transmissivity.

(ii) Two large parallel planes at 800 K and 600 K have emissivities of 0.5 and 0.8 respectively. A radiation shield having an emissivity of 0.1 on one side and an emissivity of 0.05 on the other side is placed between the plates. Calculate the heat transfer rate by radiation per square meter with and without the radiation shield. Comment on the results.

Or(b) (i) Determine the view factors F1-2 and F2-1 for the figure shown below.

306

(ii) Discuss the radiation characteristics of carbon dioxide and water vapour.

13. (a) (i) Consider laminar hydrodynamically fully developed coquette flow (that is flow between parallel plates) fluid being viscous. The upper plate at temperature T2 moves with a velocity U while the lower plate at T1 less than T2 is stationary. The distance between the plates is w. Write the appropriate governing flow and energy equations for the above and hence obtain expressions for the velocity and temperature profiles across the flow.

(ii) Air at 20oC is flowing along a heated plat at 134oC at a velocity of 3 m/s. The plate is 2 m long and 1.5 m wide. Calculate the thickness of the hydrodynamic boundary layer and the skin friction coefficient at 40 cm from the leading edge of the plate. The kinematic viscosity of air at 20oC is 15.06X10-6 m2/s.

Or

(b) (i) Distinguish between free and forced convection giving examples.

(ii) A steam pipe 10 cm OD runs horizontally in a room at 23oC. Take outside temperature of pipe as 165oC.

Determine the heat loss per unit length of the pipe. Pipe surface temperature reduces to 80oC with 1.5 cm insulation. What is the reduction in heat loss?

14. (a) (i) Discuss the various regimes of pool boiling heat transfer.

(ii) An aluminium pan of 15 cm diameter is used to boil water and the water depth at the time of boiling is 2.5 cm. The pan is placed on an electric stove and the heating element raises the temperature of the pan to 110o C. Calculate the power input for boiling and the rate of evaporation. Take Csf = 0.0132.

Or

(b) (i) Describe the principle of parallel flow and counter flow heat exchangers showing the axial temperature distribution.

307

(ii) A parallel flow heat exchanger has hot and cold water stream running through it, the flow rates are 10 and 25 kg/min respectively. Inlet temperatures are 75oC and 25oC on hot and cold sides. The exit temperature on the hot side should not exceed 50oC. Assume hi = ho = 600 W/m2K. Calculate the area of heat exchanger using - NTU approach.

15. (a) Atmospheric air at 40oC flows over a wet bulb thermometer and it shows 25oC. Calculate the concentration of water vapour in the free stream and also its relative humidity. Take D (air-water) = 0.256 X 10-4 m2 / s. If temperatures of dry and wet bulb are 30o C and 25o C respectively. What would be the corresponding values?

Or

(b) (i) A mixture of O2 and N2 with their partial pressures in the ratio 0.21 to 0.79 is in a container at 25oC. Calculate the molar concentration, the mass density, the mole fraction and the mass fraction of each species fro a total pressure of 1 bar. What would be the average molecular weight of the mixture?

(ii) Discuss the analogy between heat and mass transfer.

******

B.E./B.TECH. DEGREE EXAMINATION, APRIL/MAY 2008.Sixth Semester

Mechanical EngineeringME 340 – HEAT AND MASS TRANSFER

PART –A1. Calculate the critical radius of insulation for asbestos (k=0.172 W/mK) surrounding a pipe and exposed to room air at 300 K with h=2.8 W/m2C.2. Whether closely packed thin fins or loosely packed thick fins are preferred justify.3. What is thermal contact resistance?4. State Wien’s Displacement Law.5. Define thermal boundary layer thickness.6. Sketch temperature distribution in condensers and evaporators.7. In what circumstances, gas radiation takes a major role?8. A wire of 0.6 mm diameter and 150 mm length is submerged horizontally in water at 3 bar. The wire carries a heat flux of 390 W/m2. If the surface of wire is

308

maintained at 200C. calculate the boiling heat transfer coefficient.9. Define Fick’s law of diffusion.10. Calculate the diffusion coefficient for NH3 in air in 27C temperature and atmospheric pressure. NH3 (Gas A): Molecular weight = 17, Molecular volume = 26.43 cm3 /gm mole, Air (Gas B): Molecular weight = 29, Molecular volume = 30.6 cm3/gm mole.

PART – B11. (a). A pipe consist of 100 mm internal diameter and 8 mm thickness carries steam at 170 C. The convective heat transfer coefficient on the inner surface of pipe is 75 W/m2C. This pipe is insulated by two layers of insulation, the first layer of insulation is 46 mm in thickness have thermal conductivity of 0.14 W/mC. The second layer of insulation is also 46 mm in thickness having thermal conductivity of 0.46 W/m2C Ambient air temperature = 33C. the convective heat transfer coefficient from the outer surface of pipe =12 W/m2C. Thermal conductivity of steam pipe = 46 W/mC. Calculate the heat loss per unit length of pipe and determine the interface temperatures. Suggest the materials used for insulation.

Or(b) a long rod is exposed to air at 298 C. It is heated at one end. At steady state conditions, the temperature at two points along the rod separated by 120 mm are found to the 130C and 110C respectively. The diameter of the rod is 25mm and its thermal conductivity is 116 W/m0C. Calculate the heat transfer coefficient at the surface of the rod and also the heat transfer rate.

12. (a) Liquid Helium at 4.2 K is stored in a dewar flask of inner diameter = 0.48 m and outer diameter = 0.5 m. the dewar flask can be treated as a spherical vessel. The outer surface of the inner vessel and the inner surface of the outer vessel are well polished and the emissivity of these surfaces is 0.05. The space between the two vessels is thoroughly evacuated. The inner surface of the dewar flask ia 4.5 K while the outer surface is at 300 K. Estimate the rate of heat transfer between the surfaces.

Or(b) A thin aluminium sheet with an emissivity of 0.1 on both sides is placed between two very large parallel plates that are maintained at uniform temperatures T1= 800 mK and T2 =500 K and have emissivities 1 =0.2 and 2 =0.7 respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface are of the plates and compare the result to that without shield.13. (a) Steam condenses at atmospheric pressure on the external surface of the tubes of a steam condenser. The tubes are 12 in number and each is 30 mm in diameter and 10 m long. The inlet and outlet temperatures of cooling water flowing inside the tubes are 25C and 60C respectively. If the flow rate is 1.1 kg/s, calculate.

(i) The rate of condensation of steam(ii) The number of transfer units

309

(iii) The effectiveness of the condenser.Or

(b) A chemical having specific heat of 3.3 kJ/kgK flowing at the rate 20000 kg/h enters a parallel flow heat exchanger at 120C. The heat transfer area is 10 m2 and the overall heat transfer coefficient is 1050 W/m2K. find (i) The effectiveness of the heat exchanger (ii) The outlet temperature of water and chemical.

14.(a) A tube of 2 m length and 25 mm outer diameter is to be used to condense saturated steam at 100C while the tube surface is maintained at 92C. Estimate the average heat transfer coefficient and the rate of condensation of steam if the tube is kept horizontal . the steam condenses on the outside of the tube.

Or

(b) Air at 200 kPa and 200C is heated as it flows through a tube with a diameter of 25 mm at a velocity of 10 m/sec. The wall temperature is maintained constant and is 20C above the air temperature all along the length of tube. Calculate:

(i) The rate of heat transfer per unit length of the tube.(ii) Increase in the bulk temperature of air over a 3 m length of the tube.

15. (a) a steel sphere of radius 60 mm which is initially at a uniform temperature of 352C is suddenly exposed to an environment at 25C; with convection heat transfer coefficient 500 W/m2K. Calculate the temperature at a radius 36 mm and the heat transferred 100 seconds after the sphere is exposed to the environment.

Or(b) the tyre tube of vehicle has a surface area 0.62 m2 and wall thickness 12 mm.

the tube has air filled in it at a pressure 2.4 105 N/m2. The air pressure drops to 23 105 N/m2 in 10 days. The volume of air in the tube is 0.034 m3. Calculate the diffusion coefficient of air in rubber at the temperature of 315K, Gas constant value = 287. solubility of air in rubber tube = 0.0753 of air/m3 of rubber tube at one atmosphere.

*******

B.E./B.TECH. DEGREE EXAMINATION, APRIL/MAY 2008.Sixth Semester

Mechanical EngineeringME 340 – HEAT AND MASS TRANSFER

PART – A

1. A temperature difference of 500 C is applied across a fire-clay brick. 10 cm thick

310

having a thermal conductivity of 1.0 W/m.K. Find the heat transfer rate per unit area.2. Write the general 3-D heat conduction equation in cylindrical coordinates.3. Biot number is the ratio between ___________ and _____________4. Define bulk temperature.5. A vertical flat plate is maintained at a temperature lower than the surrounding fluid. Draw the velocity and temperature profiles assuming natural convection.6. What is burnout point ? why is it called so?7. What is a compact heat exchanger? Give examples.8.What is thermal radiation and what is its wavelength band?9. What are radiation shields?10. Explain the physical meaning of Schmidt number.

PART – B

11.(a) A composite wall is formed of a 2.5 cm copper plate (k=355 W/m.K), a 3.2 mm layer of asbestos (k=0.110 W/m.K) and a 5 cm layer of fiber plate (k=0.0049 W/m.K). the wall is subjected to an overall temperature difference of 560C (560 C on the Cu plate side and 0C on the fiber plate side). Estimate the heat flux through this composite wall and the interface temperature between asbestos and fiber plate.

Or(b) When a thermocouple is moved from one medium to another medium at a different temperature, sufficient time must be given to the thermocouple to come to thermal equilibrium with the new conditions before a reading is taken. Consider a 0.1-cm-diameter copper thermocouple wire originally at 150C. find the temperature response (i.e. an approximate plot of temperature Vs time for intervals of 0,40 and 120 seconds) when this wire is suddenly immersed in(i) Water at 40C (h+80 W/m2.K)(ii) air at 40C (h=40 W/m2.K)

Assume unit length of wire

12. (a) Air at 400 K and 1 atm pressure flows at speed of 1.5 m/s over a flat plate of 2m long. The plate is maintained at a uniform temperature of 300K. if the plate has width of 0.5 m, estimate the heat transfer coefficient and the rate of heat transfer from the air stream to the plate. Also estimate the drag force acting on the plate.

Or(b) Cylindrical can of 150 mm length and 65 mm diameter are to be cooled from an initial temperature of 20C by placing them in a cooler containing air at a temperature of 1C and a pressure of 1 bar. Determine the cooling rate when the

311

cans the kept in

(i) Horizontal position(ii) Vertical position

13. (a) Water is to be boiled at atmospheric pressure in a mechanically polished stainless steel pan placed on top of a heating unit. The inner surface of the bottom of the pan is maintained at 108C. the diameter of the bottom of the pan is 30 cm. Assuming Csf = 0.0130 . calculate

(i) The rate of heat transfer to the water , and(ii) the rate of evaporation of water

or(b) Define effectiveness of a heat exchanger. Derive an expression for the effectiveness of a double pipe parallel flow heat exchanger. State the assumptions made.

14. (a) (i) Discuss briefly the variation of black body emissive power with wavelength for different temperatures.

(ii) The spectral emissivity function of an opaque surface at 800 K is approximated as

Calculate the average emissivity of the surface and its emissive power

Or(b) Explain briefly the following:

(i) Specular and diffuse reflection(ii) Reflectivity and transmissivity(iii) Reciprocity rule and summation rule

15 (a) Discuss briefly the following:(i) Fick’s law of diffusion(ii) Equimolar counter diffusion(iii) Evaporation process in the atmosphere

Or(b) (i) What are the assumptions made in the 1-D transient mass diffusion problem?

(ii) An open pan, 20 cm diameter and 8 cm deep contains water at25Cand is exposed to dry atmospheric air. Estimate the diffusion coefficient of water in air, if the rate of diffusion of water is 8.54 10-4 kg/h.

*******

KeyMechanical Engineering

312

ME 1351 – HEAT AND MASS TRANSFERPART –A

1.

2.

3. 9 conv. In a fluid and 9 cond. in a solid.4. Average temp. of the fluid at any cross –section of flow

Point at which heat flux is maximum. Once we cross this point, large temp. difference is required to get the same heat flux and most material may burn out at this temp.

7. Area density 700 m2/m3. Car radiators; Gas turbine heat exchangers.8. Radiation due to temp.of bodies, =0.1 to 100 m.9. Highly reflective thin plates kept between 2 surfaces to reduce radiation.

10.

PART – B

11. (A) Resistance Newtork:

313

Figure

=532.5 /m2

for fiber plate :Q=

Or(b) for both water and air, Bi <<1.0 so, LSA

is applicable.

per unit length, A= D = 3.14 10-3 m2,

(i) air: Bi f0 = 0.0117 t T(C)150 109 67 T(sec) 0 40 120 (ii)water: Bif0 = 0.0936 t T(C) 150 42.6 40 T(sec) 0 40 120

12.

314

(b) Properties at 10C:

Rayleigh No:

(i) Horizontal position: L=D, Ra = 6.12 105.

(ii) Vertical position:L= 0.15 m. Ra = 7.52 106

Cooling rate Q =hA T =13. (a) (i) Properties at 100 C.

T= 8 C, so, Nucleate boiling.

=72.045 k/m2

Q = A 9 = 5093 .(ii)

Or

(b) Effectiveness,

Assumptions.

Derivation:

Where

14. (a) (i) Graph of Eb vs for different temps:Wien’s displacement law.

315

Solar radiation curve showing visible light region.(ii) Gives (T) = 0.521. Emissive Power, E(T) = T4 = 12,1002.

Or(b) Definitions.

15. (a) (i) Fick’s Law:

is, mass flux = const. of proportiondity concentration gradient.(ii) consider 2 reservoirs connectedC=CA +CB.NA =NB (or) NA + NB =0Explanation discussion

(b) (ii) .

At 25C , PA1 = 0.03165 bar, PA2 =0 (diff. into dry air)P-PA1 = 0.9815 barDAB = 0.259 10-4 m2/s.

***************

B.E./ B.Tech. DEGREE EXAMINATION, APRIL/MAY 2010Fourth Semester

Mechanical EngineeringME2251 – HEAT AND MASS TRANSFER(Common to Automobile Engineering)

PART – A

1. Write the Poisson’s equation for heat conduction.2. What is lumped heat capacity analysis?3. Define thermal boundary layer thickness.4. What do you understand by free and forced convection?5. What is effectiveness of a heat exchange?6. Give the expression for NTU.

316

7. Find the temperature of the sun assuming as a black body, if the intensity of radiation is maximum at the wave length of 0.5 µ.

8. State Kirchhoffs law.9. Define molar concentration.10. What is mass average velocity?

PART – B

11. (a) Derive the general heat conduction equation in cylindrical coordinates..Or

(b) Derive the general heat conduction equation for a hollow cylinder.

12. (a) Air at 20C at 3m/s flows over a thin plate of 2m long and 1m wide. Estimate the boundary layer thickness at the trailing edge, total drying force, mass flow of air between x=30cm and x= 80 cm. Take v = 15 × 104

and = 1.17 kg/m3

Or(b) Calculate the convective that transfer from a radiator 0.5m wide and 1m

high at 84C in a room at 20C. Treat the radiator as a vertical plate.

13. (a) Dry steam at a 2.45 bar condenses on a vertical tube of height of 1m at 117C. Estimate the thickness of the condensate film and the local heat transfer coefficient at a distance 0.2 m from the upper end of the plate.

Or(b) Derive the LMTD for a parallel flow heat exchanger stating that

assumptions.

14. (a) Derive the radiation exchange between(i) Large parallel gray surfaces and (ii) Small gray bodies.

Or(b) Two large parallel plates of 1m × 1m spaced 0.5 apart in a very large

room whose walls are at 27C. The plates are at 900C and 400C with emissivities 0.2 and 0.5 respectively. Find the net heat transfer to each plate and to the room.

15. (a) The temperature recorded by a thermometer whose bulb covered by a wet wick in dry air at atmospheric pressure is 22C. Estimate the true air temperature.

Or(b) Dry air at 27C and 1 bar flows over a wet plate of 50 cm at 50 m/s.

Calculate the mass transfer coefficient of water vapour in air at the end of the plate.

317

****************

318

veltech vel multimedia vel hightech - · web viewfind the heat loss per square meter and...

Documents