Example 1 Defrosting ice from an automobile windshield To defrost ice accumulated on the outer surface of an automobile windshield, warm air is blown over the inner surface of the windshield. Consider an automobile windshield (kw = 1.4 W/m.K). With an overall height of 0.5m and thickness of 5mm. The outside air (1 atm) ambient temperature is -20°C and the average air flow velocity over the outer windshield surface is 80 km/h, while the ambient temperature inside the automobile is 25°C. Determine the value of the convection heat transfer coefficient, for the warm air blowing over the inner surface of the windshield, necessary to cause the accumulated ice to begin melting. Assume the windshield surface can be treated as aa flat plate surface. Assumptions: the outside air pressure is 1 atm and the critical Reynolds number is Recr = 5 x105

.KW/m75.510.5m

.K0.02288W/m1131

L

k1131h

11310.738787110875.8037.0Pr871Re037.0k

LhNu

is windshield theof surface outer the on tcoefficien

transfer heataverage the number,Nusselt the for relation proper theUsing

flow. turbulent and laminar combined a is flow the 10Re105 Since

10875.8s/m10252.1

0.5m80/3.60m/sVL Re

table) (from 0.7387 Pr and /sm101.252v .K,0.02288W/m k are

C102

C0C20Tof etemperatur film the at airof properties The

2

0

3

18.053

18.0

L0

o

7

L

5

5

25L

25-

f

e.temperatur and rate

flow air warm theadjusting by done is This well. as varied be should ice the melt to

necessaryair warm theof tcoefficien transfer heatconvection the Therefore, speed.

automobile the and conditions weather with vary automobile the outside tcoefficien

transfer heatconvective the and etemperatur ambient the situation practical In

.KW/m6.481.4W/m.K

m005.0

C020

C250

.KW/m75.51

1

k

t

TT

TT

h

1h

be; must shield wind theof surface inner the overblowing air warm the

for tcoefficien transfer heatconvection the Then .C0 least at be should )(T

windshield theof etemperatur surface outer the melting, begin to ice the For

h1kt

TT

h1

TT

as; written be can

thickness windshield the through transfer heatthe balance energy the From

2

1

2

1

wo,so,

i,o,s

i

os,

iw

i,o,s

o

o,so,

Example 2

Cooling of Plastic Sheets by Forced Air The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 0.1 cm thick at a velocity of 9 m/min. The temperature of the plastic sheet is 95°C when it is exposed to the surrounding air, and a 0.6-m-long section of the plastic sheet is subjected to air flow at 25°C at a velocity of 3 m/s on both sides along its surfaces normal to the direction of motion of the sheet. Determine (a) the rate of heat transfer from the plastic sheet to air

by forced convection and radiation (b) the temperature of the plastic sheet at the end of the

cooling section. Take the density, specific heat, and emissivity of the plastic sheet to be ρ = 1200 kg/m3, Cp = 1.7 kJ/kg · °C, and ε = 0.9.

is sheet plastic the

across flow air theof end the at numberReynolds the m, 1.2 L thatNoting sheet. plastic

theof drop etemperatur the for account to ryif necessa nscalculatio the repeat will We

started. get to C95 at isothermal be to sheet plastic the assume we Therefore, drop.

thatof magnitude the know notdo we point this at but section,cooling long -m-0.6 the

through flows it as somewhat drop to sheet plastic theof etemperatur the expect We

/sm 101.896

0.7202 Pr

C · W/m 0.02808 k

table) in (given are pressure atm 1 and

C60 25)/2 (95 )/2T (T Tof etemperatur film the at airof properties The

air. room theof etemperatur the at are surfacesg surroundin the and atm 1

is pressure catmospheri local The .10 5 Re is numberReynolds Critical

25-

sf

5

cr

1380WW687612WQQQ

is radiation

and convection combined by sheet plastic theof cooling of rate the Therefore,

W687K298368m44.1K.W/m10.6759.0TTAQ

612WC2595m44.1C.W/m07.6TTAhQ

m44.1sides2m2.1m6.0A

C.W/m07.63.2591.2m

C.0.02808W/mNu

L

kh

Then,

3.2590.720210899.1664.0PrRe664.0k

hLNu

be to plate flat a for relations flow

laminar the from determined is numberNusselt the and sheet, entire the over

flow laminar havewe Thus, number.Reynolds critical the than less is which

10899.1s/m10896.1

1.2mm/s 3VL Re

radconvtotal

442428-4

surr

4

ssrad

22

ssconv

2

s

2

3

15.053

15.0

L

5

25L

Example 3 Heat Loss from a Steam Pipe in Windy Air A long 10-cm-diameter steam pipe whose external surface temperature is 110°C passes through some open area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit of its length when the air is at 1 atm pressure and 10°C and the wind is blowing across the pipe at a velocity of 8 m/s.

4

25L

25-

f

10219.4s/m10896.1

0.1m8m/sVL Re

0.7202 Pr

/sm101.896v

C.0.02808W/m k

table) (from are pressure atm 1 and

C602

C10C110Tof etemperatur film the at airof properties The

m. in pipe theof length the by above value

theg multiplyin by obtained be can pipe entire the from loss of heat rate The

W0931C10110.314m0C.4.8W/m3TTAhQ

m314.0m1m1.0DLpLA

becomes

length itsof unitper pipe the from transfer of heat rate the Then

C.W/m8.341240.1m

C.0.02808W/mNu

D

kh

Then,

124282000

10219.41

7202.04.01

7202.010219.462.03.0

282000

Re1

Pr4.01

PrRe62.03.0

k

hDNu

from determined is numberNusselt The

22

ssconv

2

s

2

5

4

8

54

4

1

3

2

3

12

14

5

4

8

5

4

1

3

2

3

1

2

1

Example 4 Cooling of a Steel Ball by Forced Air A 25-cm-diameter stainless steel ball ( 8055 kg/m3, Cp 480 J/kg·°C) is removed from the oven at a uniform temperature of 300°C. The ball is then subjected to the flow of air at 1 atm pressure and 25°C with a velocity of 3 m/s. The surface temperature of the ball eventually drops to 200°C. Determine the average convection heat transfer coefficient during this cooling process and estimate how long the process will take.

Assumptions • Radiation effects are negligible. • The outer surface temperature of the ball is uniform at all times. • The surface temperature of the ball during cooling is changing. Therefore, the

convection heat transfer coefficient between the ball and the air will also change. To avoid this complexity, we take the surface temperature of the ball to be constant at the average temperature of (300 + 200)/2 = 250°C in the evaluation of the heat transfer coefficient and use the value obtained for the entire cooling process.

1351076.2

10849.10.729610802.406.010802.44.02

PrRe06.0Re4.02k

hDNu

is numberNusselt The

10802.4s/m10562.1

0.25m3m/sVL Re

0.7296 Pr

/sm101.562

skg/m101.849v

C.0.02551W/m k

are atm 1 and C25of etemperatur

stream-free the at airof properties The kg/m·s. 10 2.76

is etemperatur surface average the at airof viscosity dynamic The

41

5

54.0324214

41

s

4.03221

4

25L

25-

5-

5

C250 @s

J3,163,000 C200) - C)(300 J/kg·kg)(480 (65.9 )T -(TmC Q

kg 65.9 m) (0.25 )kg/m (8055 D6

1 V m

:C200 to C300 from cools it as ball theof energy the in change

the simply is which ball, the from dtransferre heattotal the determine we Next

W 610 C25) - )(250m C)(0.1963 · W/m (13.8 )T -(T hAQ

m 0.1963 m) (0.25 D A

is, That e.temperatur surface average the

usingbycooling of law sNewton’ from transfer of heat rate average the

determine we C,200 to C300 from ball theof cooling of time the estimate to order In

C.W/m8.131350.25m

C.0.02551W/mNu

D

kh

becomes tcoefficien transfer heatconvection average the Then

12ptotal

333

22

ave s,save

22

2s

2

min 26 h1 s 5185 J/s610

J3,163,000

Q

Qt

ave

The time of cooling could also be determined more accurately using the transient temperature charts or relations introduced previously. But the simplifying assumptions we made above can be justified if all we need is a ballpark value. It will be naive to expect the time of cooling to be exactly 1 h 26 min, but, using our engineering judgment, it is realistic to expect the time of cooling to be somewhere between one and two hours.

In this calculation, we assumed that the entire ball is at 200°C, which is not necessarily true. The inner region of the ball will probably be at a higher temperature than its surface. With this assumption, the time of cooling is determined to be

Example 5 Heating of water in a tube by steam Water enters a 2.5 cm internal diameter thin cooper tube of a heat exchanger at 15 °C at a rate of 0.3kg/s and is heated by steam condensing outside at 120 °C. If the average heat transfer coefficient is 800W/m2. °C, determine the length of the tube required in order to heat the water to 115 °C.

The conduction resistance of a cooper tube is negligible so that the inner surface temperature of the tube is equal to the condensation temperature of steam. The specific heat of water at the bulk mean temperature of (15+115)/2=65 °C is 4187J/kg. °C. The heat of condensation of steam at 120 °C is 2203kJ/kg

m61

m025.0

4.78m

D

ALDLA

becomes length tube required Then

m78.4C85.32C0.8kW/m

kW 125.6

Th

QAT hAQ

is area surface transfer heatThe

C85.321055ln

1055

TTln

TTT

C510C15-C120TTT

C5C115-C120TTT

etemperatur mean clogarithmi The

kW 125.6 C15) - C)(115kJ/kg· )(4.187skg/ (0.3 )T -(Tcm Q

2

ss

2

2

ln

aveslnsave

ie

ieln

isi

ese

iep

The bulk mean temperature of water during this heating process is 65°C and thus the arithmetic mean temperature difference is ΔTam=120-65=55°C. Using ΔTam instead of ΔTln would give L=36m which is grossly in error. This shows the importance of using the logarithmic mean temperature in calculation.

Example 6 Flow of Oil in a Pipeline through a Lake Consider the flow of oil at 20°C in a 30-cm-diameter pipeline at an average velocity of 2 m/s. A 200-m-long section of the pipeline passes through icy waters of a lake at 0°C. Measurements indicate that the surface temperature of the pipe is very nearly 0°C. Disregarding the thermal resistance of the pipe material, determine a) the temperature of the oil when the pipe b) leaves the lake c) the rate of heat transfer from the oil d) the pumping power required to overcome the

pressure losses and to maintain the flow of the oil in the pipe

Assumptions The thermal resistance of the pipe is negligible. The inner surfaces of the pipeline are smooth. The flow is hydrodynamically developed when the pipeline reaches the lake.

We do not know the exit temperature of the oil, and thus we cannot determine the bulk mean temperature, which is the temperature at which the properties of oil are to be evaluated. The mean temperature of the oil at the inlet is 20°C, and we expect this temperature to drop somewhat as a result of heat loss to the icy waters of the lake. We evaluate the properties of the oil at the inlet temperature, but we will repeat the calculations, if necessary, using properties at the evaluated bulk mean temperature. At 20°C we read

666

/sm10019

0.3m2m/sDV Re

C880J/kg1 c

10400 Pr

/sm10019

C0.145W/m. k

888kg/m

26-

hmL

p

26-

3

Note that this Nusselt number is considerably higher than the fully developed value of 3.66. Then,

C71.19C) · J/kg kg/s)(1880 (125.5

)m C)(188.5 · W/m (18.0exp C]20) - [(0 - C0

cmhA-)expT -(TTT

from oilof etemperatur exit the determine we Next

kg/s 125.5 s/m2m) (0.34

1 )kg/m (888 VA m

m 188.5 m) (200 m3.0DL Lp A

,Also

C.W/m0.183.370.3m

C0.145W/m.Nu

D

kh

22

psisse

23

mc

2

s

2

Thus, the mean temperature of oil drops by a mere 0.29°C as it crosses the lake. This makes the bulk mean oil temperature 19.86°C, which is practically identical to the inlet temperature of 20°C. Therefore, we do not need to reevaluate the properties.

422

lnsave

is

es

eiln

1074.6C85.19188.5mC18.0kW/m T hAQ

C85.19

020

71.910ln

71.9102

TT

TTln

TTT

The logarithmic mean temperature difference and the rate of heat loss from the oil are;

Therefore, the oil will lose heat at a rate of 67.4 kW as it flows through the pipe in the icy waters of the lake. Note that ΔTln is identical to the arithmetic mean temperature in this case, since Δ Ti ≈Δ Te. The laminar flow of oil is hydrodynamically developed. Therefore, the friction factor can be determined from

0961.0666

64

Re

64f

kW 16.1kg/m 888

)N/m 10 kg/s)(1.14 (125.5PmW

m/N1014.12

m/s) )(2kg/m (888

m 0.3

m 2000.0961

2

V

D

LfP

3

25

pump

25

232

m

Then the pressure drop in the pipe and the required pumping power become

We will need a 16.1-kW pump just to overcome the friction in the pipe as the oil flows in the 200-m-long pipe through the lake. If the pipe is long, a more optimal design to reduce pump work would be to increase the pipe diameter D, since pump work is inversely proportional to D5. Optimal pipe sizing requires an economic life-cycle cost analysis

Example 7 Heating of Water by Resistance Heaters in a Tube Water is to be heated from 15°C to 65°C as it flows through a 3-cm-internaldiameter 5-m-long tube. The tube is equipped with an electric resistance heater that provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 10 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.

The properties of water at the bulk mean temperature of Tb=(Ti + Te)/2 = (15+ 65)/2 = 40°C.

kg/s 0.1654 kg/min 9.921 /min)m )(0.01kg/m (992.1 Vm

becomes rate flow mass the

Then /min.m 0.01 L/min 10 V as given is waterof rate flow volume The

m 0.471 m) m)(5 (0.03 DL pL A

m 10 7.069m03.04

1D

4

1A

are areas surface transfer heatand sectional cross The

C 4179J/kgc

4.32 Pr

/sm 10 0.658

C0.631W/m. k

992.1kg/m

33

3

2

s

2422

c

p

26

3

2

2

c

s

s

msmss

iep

kW/m 73.46m 0.471

kW 34.6

A

Qq

from determined be can value

its and case, this in constant is flux heatsurface The location. that at fluid

theof etemperatur mean the is Tm and tcoefficien transfer heatthe is hwhere

h

qT T )T - h(Tq

from

determined be can location any at tube theof Ts etemperatur surface The

kW. 34.6 be must heatertheof rating

power the Therefore, heater.resistance the from come must energy thisof All

kW 34.6 kJ/s 34.6 C15) - C)(65kJ/kg 9kg/s)(4.17 (0.1654 )T - (TCmQ

of rate a at water the to

supplied be must heatC,65 to C15 from rate flow mass this at water the heatTo

69.5 (4.34) 60)0.023(10,7 Pr Re 0.023k

hD Nu

from number

Nusselt the determine and pipe entire the in flow turbulent developed fully

assume can we Therefore, pipe. theof length total the than shorter much is which

m 0.3 0.03 10 10D L L

roughly is length

entry the and turbulent is flow the Therefore, 10,000. than greater is which

10,760/sm 10 0.658

m) m/s)(0.03 (0.236DVRe

m/s 0.236 m/min 14.15 m 10 7.069

/minm 0.010

A

VV

: numberReynolds the and waterof

velocity mean the find to needfirst we t,coefficien transfer heatthe determine To

0.40.80.40.8

th

26-

m

24-

3

c

m

C115C · W/m 1462

W/m 73,460 C65

h

qT T

becomes exit the at pipe theof etemperatur surface the and

C W/m 1462 (69.5)m 0.03

C · W/m 0.631Nu

D

kh

Then;

2

2

s

ms

2

Discussion Note that the inner surface temperature of the pipe will be 50°C higher than the mean water temperature at the pipe exit. This temperature difference of 50°C between the water and the surface will remain constant throughout the fully developed flow region.

Example 8 Heat Loss from the Ducts of a Heating System Hot air at atmospheric pressure and 80°C enters an 8–m-long uninsulated square duct of cross section 0.2 m 0.2 m that passes through the attic of a house at a rate of 0.15 m3/s. The duct is observed to be nearly isothermal at 60°C. Determine the exit temperature of the air and the rate of heat loss from the duct to the attic space.

We do not know the exit temperature of the air in the duct, and thus we cannot determine the bulk mean temperature of air, which is the temperature at which the properties are to be determined. The temperature of air at the inlet is 80°C and we expect this temperature to drop somewhat as a result of heat loss through the duct whose surface is at 60°C. At 80°C and 1 atm we read;

C J/kg1008 c

0.7154 Pr /sm 10 2.097

C0.02953W/m k kg/m 0.99994

p

25-

3

The characteristic length (which is the hydraulic diameter), the mean velocity, and the Reynolds number in this case are

91.4 (0.7154) 65)0.023(35,7 Pr Re 0.023k

hD Nu

from numberNusselt

the determine and duct entire the in flow turbulent developed fully assume

can we Therefore, duct. theof length total the than shorter much is which

m 2 m 0.2 10 10D L L

roughly is length

entry the and turbulent is flow the Therefore, 10,000. than greater is which

35,765/sm 10 2.097

m) m/s)(0.2 (3.75DVRe

m/s 3.75 m) (0.2

/minm 0.15

A

VV

m 0.2 a4a

4a

p

4AD

0.30.80.30.8h

th

25-

m

2

3

c

m

2

ch

Then,

C3.71C) · J/kg kg/s)(1008 (0.151

)m C)(6.4 · W/m (13.5exp C]80) - [(60 - C06

cmhA-)expT -(TTT

from oilof etemperatur exit the determine we Next

kg/s 0.151 /s)m )(0.15kg/m (1.009 V m

m 6.4 m) m)(8 (0.2 4 4aLLp A

,Also

C.W/m5.1391.40.2m

C.0.02953W/mNu

D

kh

22

psisse

33

2

s

2

h

W 1313 - C))(-15.2m C)(6.4W/m (13.5T hAQ

C2.15

80 - 60

71.3 - 60ln

3.1708

T - T

T - Tln

TTT

become air the from

loss of heat rate the and difference etemperatur mean clogarithmi The

22

lns

e

is

es

eiln

Therefore, air will lose heat at a rate of 1313 W as it flows through the duct in the attic. Discussion The average fluid temperature is (80 + 71.3)/2 = 75.7°C, which is sufficiently close to 80°C at which we evaluated the properties of air. Therefore, it is not necessary to re-evaluate the properties at this temperature and to repeat the calculations.