03.conduction part1

8/13/2019 03.Conduction Part1

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One-Dimensional Steady-State Conduction

Conduction problems may involve multiple directions andtime-dependent conditions

Inherently complex Difficult to determine temperature

distributions

One-dimensionalsteady-statemodels can representaccurately numerous engineering systems

In this chapter we will Learn how to obtain temperature profiles for common geometries with

and without heat generation.

Introduce the concept of thermal resistance and thermal circuits


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Chapter 2 : Introduction to Conduction

2

For cartesian coordinates

(2.17)


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Chapter 3 : One-dimensional, Steady state conduction

(without thermal generation)

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3.1 Methodology of a conduction analysis1. Specify appropriate form of the heat equation

2. Solve for the temperature distribution

3. Apply Fouriers law to determine the heat flux

Simplest case:

- One-dimensional, steady state conduction with no thermal energygeneration

Common geometries:

i. The plane wall: described in rectangular (x)

coordinate. Area perpendicular to direction of heattransfer is constant (independent of x).

ii. Cylindrical wall : radial conduction through tubewall

iii. Spherical wall : radial conduction through shell wall


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3.2 The plane wall temperature distribution

assuming steady-state conditions and nointernal heat generation (i.e. q = 0), thenthe 1-D heat conduction equation reducesto:

For constant kandA, second orderdifferential equation:

.

This mean:Heat flux (qx) is independent of xHeat rate (qx) is independent of xBoundary conditions: T(0)= Ts,1

T(L)= Ts,2

Using Eq. (2.2) in Chapter 2, by


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1-D heat conduction equation for steady-stateconditions and no internal heat generation (i.e. q= 0), is

.

for constant kandA

Integrate twice to get T(x)

For boundary conditions: T(0)= Ts,1andT(L)= Ts,2

at x = 0, T(x) = Ts,1 and C2= Ts,1at x = L, T(x) = Ts,2 and Ts,2= C1L + C2 = C1L + Ts,1

this gives, C1= (Ts,2 Ts,1)/2

and

Using value of C1and C2, the function of T(x)is

*From here, apply Fouriers

law to get heat transfer, qx


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Heat rate for plane wall (simplestcase):

Heat flux for plane wall (simplestcase):


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Example: Temp distribution problem

Consider a large plane wall of thickness L = 0.2 m, thermal conductivity k = 1.2W/mK, and surface area, A = 15m2. The two sides of the wall are maintained atconstant temperatures of T1= 120C and T2= 50C. Determine,

a) The temperature distribution equation within the wall

b) Value of temperature at thickness of 0.1mc) The rate of heat conduction through the wall under steady conditions


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Thermal Resistance

Based on the previous solution, the conduction heat transfer rate can be

calculated:

kAL

TTTT

L

kA

dx

dTkAQ

ssssx

/

2,1,2,1,

Recall electric circuit theory - Ohms law for electrical resistance:

Similarly for heat convection, Newtons law of cooling applies:

Resistance

eDifferencPotentialcurrentElectric

hATTTThAQ SSx

/1)()(

And for radiation heat transfer:

Ah

TT

TTAhQ r

surs

sursrrad /1

)(

)(

(3.2a)

(3.2b)

(3.2c)

.

.

.


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Thermal Resistance

Compare with equations 3.2a-3.2cThe temperature difference is the potential or driving force

for the heat flow and the combinations of thermalconductivity, convection coefficient, thickness and area ofmaterial act as a resistanceto this flow:

We can use this electrical analogy to represent heat transfer problems

using the concept of a thermal circuit(equivalent to an electrical circuit).

AhR

hAR

kA

LR

rradtconvtcondt

1,

1, ,,,

R

TQ overall

Resistance

ForceDrivingOverall.


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3.2.1 Thermal resistances & Thermal circuits

- Interestingly, there exists an analogy between the diffusion of heat and electricalcharge. For example if an electrical resistance is associated with the conduction ofelectricity, a thermal resistance may be associated with the conduction of heat.

- Defining thermal resistance for conduction in a plane wall:

- For convection :

- For previous simplest case, thermal circuit for plane wallwith adjoining fluids:


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3.2.1 Thermal resistances & Thermal circuits

- In case of radiation :

where,

Surface temperatureSurrounding temperature

(3.13)

(1.9)


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Example: (Problem 3.2a)

The rear window of an automobile is defogged by passing warm air over its innersurface. If the warm air is at T

,i= 40C and the corresponding convectioncoefficient is hi= 30 W/m

2K, what are the inner and outer surface temperatures of4-mm thick window glass, if the outside ambient air temperature is T

,o= -10Cand the associated convection coefficient is ho= 65 W/m

2K.


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Example (problem 3.5):

The walls of a refrigerator are typically constructed by sandwiching a layer ofinsulation between sheet metal panels. Consider a wall made from fibreglassinsulation of thermal conductivity, ki= 0.046 W/mK and thickness Li= 50 mmand steel panels, each of thermal conductivity kp= 60 W/mK and thickness Lp= 3mm. If the wall separates refrigerated air at T

,o = 25C, what is the heat gain perunit surface area ?

Coefficients associated with natural convection at the inner and outer surfaces canbe approximated as hi= ho= 5 W/m

2K


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3.2.2 The composite wall (with negligible contact resistance)


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Composite wall with negligible contactresistance:

where,

Overall heat transfer coefficient:

* A modified form of Newtons Law of coolingto encompass multiple resistances to heattransfer

The composite wall (series type)


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Composite Walls

What is the heat transfer rate for this sys tem?

AlternativelyUAq

TRR

TUAQ

ttot

x

1

where U is the overall heat transfer coefficientand T the overall

temperature difference.

)]/1()/()/()/()/1[(

11

41 hkLkLkLhARU

CCBBAAtot

.


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The composite wall (parallel type)


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The composite wall (parallel type)

Electric analogy of thermalcircuits

- To solve a parallel resistancenetwork like that shown opposite,we can reduce the network to and

equivalent resistance

For electrical circuits:

For thermal circuits:


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Example: parallel resistances

*IR (infrared) photos show that the heattransfer through the built-up walls is morecomplex than predicted by a simple parallel-resistance.


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Example: (3.15)

Consider a composite wall that includes an 8-mm thick hardwood siding, 40 mmby 130 mm hardwood studs on 0.65 m centers with glass fibre insulation (paperfaced, 28 kg/m3) and a 12 mm layer of gypsum wall board.

What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 mwide (having 10 studs, each 2.5 m high)


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Example of resistance network with both radiativeand convectiveboundary (Example 3.1)

C t t R i t


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Contact Resistance


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3.3 Contact resistance

It is important to recognise that, in composite systems, thetemperature drop across the interface between material maybe appreciable (present analysis is neglected).

This attributed is due to thermal contact resistance Rt,c

*values depend on:materials A and B, surfacefinishes, interstitialconditions and contactpressure

Composite Walls with contact resistances


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Composite Wallswith contact resistances


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3.3 Radial systems: cylindrical wall

General heat equation for cylinder (from Chap. 2)

For 1-D steady state, with no heat generation

Integrate twice to get temperaturedistribution, T(r). For example, forconstant temperature boundary:

From T(r), heat flux for cylinder


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The thermal resistance for radial conduction

In case of cylinder with composite wall (negligible contact resistance)


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Critical radius for insulation

Adding more insulation to a walldecrease heat transfer

The thicker the insulation, the lower the heat transfer through the wall

However, adding insulation to a cylindrical pipe or a spherical shell is adifferent matter.

Additional insulation increase the conduction resistance of the insulationlayer but decrease the convection resistance of the surface because of the

increase in the outer surface area for convection

Hence, knowledge of critical radius of insulationis required


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Critical radius for insulation: see example 3.5 in Textbookfor details

If ri< rcr, Rtotdecreases and the heatrate thereforeincreases withinsulation

If ri> rcr, Rtotincreases

and therefore heat ratedecreases withinsulation

Insulation prop.

Outside conv. coeff.


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Example 3.39: cylinder

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has aninner diameter of 36 mm and a wall thickness of 2 mm. The pharmaceutical andambient air are at temperatures of 6C and 23C, respectively, while thecorresponding inner and outer convection coefficients are 400 W/m2K and 6W/m2K, respectively.

i) What is the heat gain per unit tube length (W/m) ?

ii) What is the heat gain per unit length if a 10-mm thick layer of calcium silicateinsulation (kins= 0.050 W/mK) is applied to the tube. Discuss the result withthe knowledge of rcrit.

(12.6 W/m, 7.7 W/m)


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3.4 Radial systems: spherical wall

General heat equation for sphere (from Chap. 2)

For 1-D steady state, with no heat generation

Integrate twice to get temperature distribution forconstant k, T(r)

From T(r), heat flux for sphere


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The thermal resistance for radial conduction in

sphere

In case of sphere with composite shell (negligible contact resistance)

The total thermal resistance due to conduction and convection in sphere


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Summary


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Example 3.54:

A storage tank consists of a cylindrical section that has a length and inner diameterof L=2m and Di=1m, respectively, and two hemispherical end sections. The tank isconstructed from 20 mm thick glass (Pyrex) and is exposed to ambient air forwhich the temperature is 300K and the convection coefficient is 10 W/m2K. Thetank is used to store heated oil, which maintains the inner surface at a temperatureof 400K. Determine the electrical power that must be supplied to a heatersubmerged in the oil if the prescribed conditions are to be maintained. Radiationeffects may be neglected, and the Pyrex may be assumed to have a thermalconductivity of 1.4 W/mK.

03.conduction part1

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