Combined Convection and Conduction in Walls

Assumptions: Steady state, one-dimensional heat transfer through a plane wall that is exposed to convection on both sides, no heat losses between wall and fluids, T1>T2>T3>T4. Isothermal surfaces, constant K and A.

Rate ofheat convection

into the wall

Rate ofheat conductionthrough the wall

Rate ofheat convectionfrom the wall

= =

21 TTAhq i A

A xTTAk

32 43 TTAho

In many practical situations, the temperatures at surfaces (B.C.s) are not known. However, T for fluids present on both sides are known.

hi and ho are inside and outside convective transfer coefficients, respectively. They are empirical parameters that we will estimate later.

,21 AhqTTi

Ah

qTTo

43,32 AkxqTT

A

A

RT

RRRTT

AhAkx

Ah

TTq overall

oA

A

i

321

4141

11)(

Like we did before, add (T1-T2)+(T2-T3)+(T3-T4) to get (T1-T4) andrearrange:

R1 and R3 are convective resistances and R2 is conductive resistance.

Overall heat transfer can be described using an overall heat transfercoefficient (U) as:

)( 41 TTUATUAq overall By comparison,

202 .. 11

1ftFhr

BtuorKm

W

hkx

h

U

oA

A

i

Combined Convection and Conduction in Cylinders

Assumptions: Steady state, one-dimensional heat transfer (radial), cylinder is exposed to convection on both sides, no heat losses between wall and fluids, T1>T2>T3>T4. Isothermal surfaces, constant K and A.

Rate ofheat convection

into the wall

Rate ofheat conductionthrough the wall

Rate ofheat convectionfrom the wall

= =

21 TTAhq i 43 TTAho

AlmA

io

AkrrTT )( 32

Like we did before, add (T1-T2)+(T2-T3)+(T3-T4) to get (T1-T4) andrearrange:

RT

RRRTT

AhAkrr

Ah

TTq overall

oAlmA

io

i

321

4141

11)(

R1 and R3 are convective resistances and R2 is conductive resistance.

Overall heat transfer can be described using an overall heat transfercoefficient (U) as:

)( 41 TTUATUAq overall We can use inside or outsideareas to calculate U

202 .. )(1

1ftFhr

BtuorKm

W

hAA

AkArr

h

U

oo

i

AlmA

iio

i

i

Based on Ai:

Based on Ao: 202 .. 1)(

1ftFhr

BtuorKm

W

hAkArr

AhAU

oAlmA

oio

ii

oo

RAU

ii

1

RAU

oo

1

Example 4.3-3: Heat Loss by Convection and Conduction and Overall USaturated steam at 267 oF is flowing inside a 0.75 in steel pipe having an ID of 0.824 in and an OD of 1.05 in. The pipe is insulated with 1.5 in of insulation on the outside. hi=1000 Btu/oF.ft2.hr and ho=2 Btu/oF.ft2.hr. The mean thermal conductivity of steel is 26 btu/hr.ft.oF and for insulation is 0.037 btu/hr.ft.oF.(a) Calculate the heat loss for 1 ft of pipe using

resistances if the surrounding air is at 80 oF.(b) Repeat, using overall Ui.

4321 RRRRR

iilmsteelsteelilmioo AhAK

rrAk

rrAh

R 11 1223

2

3

23

lnAAAAAilm

1

2

12

lnAAAAA lmsteel

RAU

ii

1

)( oiii TTAUq Insulation

Stainless steel pipe

Di=0.0254 mm

t=1.5”

A B

1 ft

80 oF

SaturatedsteamT1=2670F Di=0.824” and Do=1.05”hi=1000 Btu/oF.ft2.hr

h0=2 Btu/oF.ft2.hr

k=0.037

k=26

Combined Conduction and Convection in a Sphere

Steady state, one-dimensional heat transfer through a spherical shell that is exposed to convection on both sides

22

221

12

12

1

2,1,

41

441

hrkrrrr

hr

RRRR convconductionconvtotal

totalRTT

q 2,1,

q

Combined Series-Parallel ArrangementThe total rate of heat transfer through the composite system

31 21 2 3

1 1 2 2 3 3 3

1 ; ; ; convLL LR R R R

k A k A k A hA

1 212 3 3

1 2total conv conv

R RR R R R R RR R

Where

andtotalR

TTq 1

Conduction with Internal Heat Generation in a Wall

In certain systems, heat is generated in the conductingmedium (body with reactions, nuclear fuel rods, sanitationfields, and food processing heaters).

Assumptions: Steady state, one-dimensional heat transfer (axial), heat is generated in the center of the wall uniformly, insulated walls (no convection), constant wall T, A and k.

Rate ofheat in

Rate ofheat out

Rate ofheat accumulation- =

Rate ofheat generation+

0 xAqqq xxxxx Divide by x and take the limit as x approaches zero to get:

0 Aqdxdq

However, ,dxdTkAq 02

2

kq

dxTd

kq

dxTd

2

2

Assume rate of heat generation to be constant, We have learned how to solve this ODE (integrate twice). We need two B.C.s

Integrate once to get:1Cx

kq

dxdT

Integrate again to get: 212

2CxCx

kqT

B.C.sx=0, T=T0

20 CT ow TLCLkqT 1

2

2

Solve for C1 to get:

LkLqTT

Cw 2

2

0

1

oow Tx

kLq

LTTx

kqT

222

Substitute C1 and C2 in the T profile to get:

x=L, T=Twx=-L, T=Tw

This is dependent on the 2nd

B.C.

Total heat loss from the two faces of the wall at steady state is equal to the heat generated Tq

qLAqT 2

Generation with Heat Conduction in Walls

1) Considerable temperature difference between the inner and the outer surfaces of the wall (significant temperature gradient in the xdirection).

2) The wall surface is nearly isothermal.3) The wall has constant area A normal to heat flow. 4) Ignore convection, friction energies and work against P. 5) Steady state6) One dimensional7) Constant Properties

][

zTv

yTv

xTv

tTc zyxv

qzT

yTTk

]

x[ 2

2

2

2

2

2

)

x(

TPT-

zv

yvv zyx

P

ConvectiveTransient

Conductive FrictionWork done againstpressure

qTk ]

x[ 2

2

GenerationkqT

]

x[ 2

2

Conduction with Internal Heat Generation in a Cylinder

Assumptions: Steady state, one-dimensional heat transfer (radial), heat is generated in the center of the cylinder uniformly, insulated walls (no convection), constant wall T and k.

Rate ofheat in

Rate ofheat out

Rate ofheat accumulation- =

Rate ofheat generation+

0 VqAqAq rrrrr ,0222 qrrLrLqrLq rrrrr

rAVrLA ,2

0][lim 0

qrrrqrq rrrrr

r

0)( rq

drqrd

Divide by 2Lr and take the limit as r approaches zero to get:

r=0 T=Tw

Integrate once to get: ,2 1

2

Crqrq ,

21

rCrqq

rq

drqrd

)(

However, we know that at r=0, q is finite. This requires C1 to be 0.

,2rqq

drdTkq However:

Assume rate of heat generation to be constant, K constant Our second B.C. is at r=R, T=Tw

wTrRkqT )(

422

Substitute C2 in the T profile to get: When r=o, T =To

drdTkrq

2

Separations of variables, integrate as well ,2

kdTdrrq

2

2

4CkTrq

Apply B.C. ,4 2

2

CkTRqw

wTRkqT 2

0 4

2

2

4CkTRq

w

Heat Conduction Through a Hollow Cylinder

Consider the long cylindrical layer, r2-r1=tAssumptions:

The two surfaces of the cylindrical layer are maintained at constant temperatures T1 and T2, T1>T2

Constant thermal conductivity, One-dimensional heat conduction (radial).

Examples: Pipes, insulated pipes, and blood vessels.q

tT

kq

zTT

rrTr

rrgen

111

2

2

2

2

2

,2

2 rkq

drdTr

kq

drdTr

drd

r

1

,2 r

rkq

drdT

rr

kqT ln

42

rkq

drdTr

drd

0

Conduction with Internal Heat Generation in a Sphere

Assumptions: Steady state, one-dimensional heat transfer (radial), heat is generated in the center of the sphere uniformly, insulated walls (no convection), constant wall T and k.

Rate ofheat in

Rate ofheat out

Rate ofheat accumulation- =

Rate ofheat generation+

0 VqAqAq rrrrr ,0444 222 qrrqrqr rrrrr

rAVrA ,4 2

0][lim 222

0

qrr

qrqr rrrrrr

0)( 22

rqdrqrd

Divide by 4r and take the limit as r approaches zero to get:

Integrate once to get: ,3 1

32 Crqqr

,3 2

1

rCrqq

22 )( rq

drqrd

However, we know that at r=0, q is finite. This requires C1 to be 0.

,3rqq

drdTkq However:

Assume rate of heat generation to be constant, K constant Our second B.C. is at r=R, T=Tw

wTrRkqT )(

622

Substitute C2 in the T profile to get: When r=o, T =To

drdTkrq

3

Separations of variables, integrate as well ,3

kdTdrrq

2

2

6CkTrq

Apply B.C. ,6 2

2

CkTRqw

wTRkqT 2

0 6

2

2

6CkTRq

w

Heat Conduction with Generation In a Hollow Sphere

Consider a layer of sphere, r2-r1=tAssumptions:

The two surfaces of the spherical layer are maintained at constant temperatures T1 and T2, T1>T2

Constant thermal conductivity, One-dimensional heat conduction (radial).

q

tT

kqT

rT

rrTr

rrgen

1

sin1sin

sin11

2

2

2222

2

kq

rTr

rr

2

2

122 r

kq

drdTr

drd

32

3r

kq

drdTr

23 r

rkq

drdT

rr

kqT 2

6

0

Example 4.3-4 Heat Generation in a CylinderAn electric current of 200A is passed through a stainless-steel wire having a radius R of 0.001268 m. The wire length is 0.091 m and has a resistance R=0.126 ohms. The outer surface temperature Tw is held constant at 422.1 K. the average thermal conductivity k is 22.5 W/m.K. Calculate the center temperature.

wTRkqT 2

0 4

Solution We have shown before that:

We know:Tw= 422.1 K, K=22.5 W/m.K, R=0.0112368 mWe only need the heat generation rate

39

2222

22

2

10096.1

126.0200)()91.0()001268.0(

mWq

AWqmm

RIqLR

RIqV

e

e

Substitute to get T0=441.7 K

Critical Thickness of Insulation for a Cylinder Assumptions: Steady state, one-dimensional heat transfer (radial), ignore convection of fluid inside (r1), high k, outside convection is important, T1, T2 and k are constants. T1>T2>T0. Example: A metal pipe with steam inside it. We have convection and conduction due to insulation. As insulation is added, T2 will decrease.

),( 2 oo TTAhq LrA 22

1

2

12

12

21

ln,)(

AAAAA

kArrTTq lm

lm

LrA 22 2LrA 11 2

Assume no heat losses between insulator and outside (qConvection = qConduction),

substitute for A, A1 and A2, rearrange equations to get (T2-T0) and (T1-T2) and add them to get q.

2

12

01

1)/ln()(2

rhkrr

TTLq

o

In order for us to determine the critical thickness of insulation, or to investigate the effect of r2 over q, we need to evaluate dq/dr2.

2

2

12

222

01

2 ]1)/ln([

11)](2[

rhkrr

rhkrTTL

drdq

o

o

When dq/dr2=0, we can get critical thickness.

oo hkr

rhkr Critical 22

22

,11 If r2<r2C, adding more insulation will increase heat transfer and vice versa.For common ho and k, r2C is ~mm’s.

For small electric wires, insulation increase heat loss

For large pipes, insulation decrease heat loss

Adding layers in the winter may not always be a good idea

k of insulation materials really matters

Example 4.3-5 Insulating an electrical wire and critical radius An electric wire having a diameter of 1.5 mm and covered with a plastic insulation (k=0.4 W/m.K) with a 2.5 mm thickness is exposed to air at 300 K with ho = 20 W/m2.K. Assume that the wire surface temperature is constant at 400 K and is not affected by covering.(a)Calculate the value of critical radius.(b)Calculate the heat loss per m of wire length with no

insulation. (c)Repeat (b) for wire with insulation.

Solution

mmhkro

Critical 202 (a)

(b) Without insulation, heat loss is due to convection. Ignore conduction for the wire thickness

WTTAhq airwireo 42.9)(

(c) With insulation, heat loss is due to conduction and convection

2

12 1)/ln()(2

rhkrr

TTLq

o

airwire

Wq 98.32

201025.31

4.0)75.0/25.3ln((

)300400)(1(2

3

Since r2 is smaller than r2C, adding insulation will increase heat loss.

Contact Resistance at the Interface In reality, surfaces have some roughness. When two surfaces are pressed against each other,

the peaks form good material contact but the valleys form voids filled with stagnant air. No perfect contact.

As a result, an interface contains numerous air gaps of varying sizes are developed that act as insulation because of the low k of air.

An interface offers some resistance to heat transfer, which is termed the thermal contact resistance, Rc.

This means that T at the interfaces are not constant.

In nuclear plants, when heat fluxes are present, significant T drop may occur at the interface.

R is a function of:RMS, Pressure holding surfaces,T at the interface, Type of trapped fluid

c

c

c RT

Ah

TTAhq

1

Heat transfer can take place in fluid via conduction, convection, and radiation as well as through conduction at solid joints.

No reliable theories or empirical correlations exist to predict Rc.

However,

hc is the contact heat transfer coefficientand ranges from 0.2 ×104 to 1× 104

W/m2.K between two metals. A is the contact area.

• If known, we can add Rc to other resistances in the system and study itseffect on heat flux.

• If we know the maximum gap between two surfaces (x), then the maximum Rc can be calculated.

• We assume that conduction dominates heat transfer in the gap,• If radiation or convection (not stagnant fluid) are present, they will

decrease Rc. x

khc