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Part 5

MULTIMODE HEAT TRANSFER

In the preceding chapters only a single mode of heat transfer was considered. The isolation of

each mode helps us to identify the mode of heat transfer and to develop the analytical and computa-

tional skills required to estimate the heat transfer by each mode. A few cases involving both conduc-tion and convection were considered in the chapters on conduction, convection, and heat

exchangers, but the resulting equations were usually linear. When radiative heat transfer occurs in

conjunction with conductive and (or) convective heat transfer, the resulting equations are nonlinear,

usually requiring numerical techniques for their solutions. Most practical problems involve multi-

mode heat transfer. We now consider such cases with emphasis on radiative heat transfer. The algo-rithms for the numerical solutions depend on the nature of the equations. We illustrate some of them

through examples. In this chapter we

Consider a few cases with heat transfer by more than one mode with emphasis on radiative

heat transfer. Illustrate a few algorithms to solve the resulting equations.

Give a brief introduction to solar collectors, which involve heat transfer by all three modes.

5.1 INTRODUCTION

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Multimode

1.2

In the previous chapters, solutions with only one mode of heat transfer was emphasized, as an

understanding of heat transfer by each mode is necessary before solving problems with more than

one mode of heat transfer. However, most practical problems involve more than one mode. In a few

problems involving both conduction and convection considered in the earlier chapters, the heat flux

is related to the temperature differences and the resulting equations were linear: the solutions were

straightforward unless the properties are strongly temperature dependent. If radiative heat transfer is

one of the modes, the heat flux is usually related to the difference in the fourth powers of the abso-lute temperatures leading to nonlinear equations.

In the majority of cases involving multimode heat transfer, the resulting equations depend on

the modeling. Therefore, no universal algorithm can be suggested for the solutions to such problems.

In many cases numerical methods are needed to solve the resulting equations. The solutions may

also need iterative methods. We now present solutions to several problems.___________________________________________________________________________

Example 5.1.1

Conduction, Convection, and Radiation

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Advanced Heat Transfer

1.3

A 20-cm long, 1-cm diameter cylindrical carbon

resistance heating element is placed inside a large duct

(Figure 5.1.1). Atmospheric air at 0 oC, 20 m/s flows

inside the duct. The outer surface of the electrically

heated element is maintained at 800 oC. The duct walls are

at 20 oC. Determine

(a) The heat transfer rate from the element.(b) The maximum temperature in the element.

Given

Material: CarbonL = 20 cm d= 1 cm

Fluid: Air V= 20 m/s

Find

(c) q

(d)

Assumptions

(1) Steady state.

(2) Air does not participate in radiation.

(3) Negligible heat transfer from the heating element end-surfaces.

Figure 5.1.1 Air in cross

Ts 800oC=

T 0oC= Tw 20

oC=

Tmax

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(4) Constant thermal conductivity of carbon heating element.

(5) Negligible blockage effect, i.e., negligible effect of the duct surfaces on the air flow pattern.

Solution

(a) The heat transfer rate from the element is by both by convection to the surrounding air and by

radiation to the duct surface,

Convective heat transfer rate: Atmospheric air properties at an approximate film temperature of

400 oC, (Table A7) are

= 0.5244 kg/m3 cp = 1069 J/kg K = 3.26110-5 Ns/m2

k= 0.05015 W/m K Pr = 0.6948

[1:1.1]

Employing Equation (4.2.19),

Radiative heat transfer rate: To compute the radiative heat transfer rate we consider the heating

element surface as one surface, completely enclosed by the much larger duct surface. The radiative

q qc

qR

+=

Red Vd---------- 3217= =

Nud 0.3

0.62Red1 2 Pr1 3

1 0.4 Pr( )2 3+[ ]1 4----------------------------------------------------+ 27.61= = hNudk

d------------ 138.4 W/m2 oC= =

qc hdL Ts T( ) 695.9 W= =

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1.5

heat transfer rate is given by Equation (1.5.2). From Table A10, the emissivity of carbon filament is

0.53. Therefore (with Tin K),

(b) To find the maximum temperature in the element, we consider the resistance heating as uniform

internal energy generation.

Assumptions(6) Constant thermal conductivity

(7) Negligible heat transfer from the end-surfaces and, therefore, one-dimensional temperature

distribution

Solution

With a total internal energy generation of 949.9 W, internal energy generation per unit volume is

Employing cylindrical coordinates (x = r), constant thermal conductivity, and simplifying Equation

(2.2.1) to the steady state, we obtain

qR dL Ts4 Tw

4( ) 249 W= = q qc qR+ 944.9 W= =

q '''q

d2L 4------------------- 6.015

710 W/m3= =

rd

dkr

rd

dT q '''r=

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Boundary conditions:

Integrating the equations twice and applying the boundary conditions,

From Table A3, k(carbon filament) = 8.49 W/m K

___________________________________________________________________________

Example 5.1.2 Conduction, Convection, and Radiation

rd

dT

r R=

finite or Tis finite= TR

Ts 800oC= =

Tq '''r2

4k----------- Ts

q '''R2

4k-------------+ + Ts

q '''R2

4k------------- 1

r2

R2------

+= =

Tmax Tr 0= Ts q '''R2

4k-------------+= =

Tmax

8006.015

710 0.0052

4 8.49-----------------------------------------------+ 844.3 oC= =

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1.7

Saturated steam at 200 oC flows in a thin walled, 5-cm

diameter horizontal tube insulated with 5-cm thick

rockwool and covered with a thin painted metal cladding

( = 0.8). The surrounding air and surfaces are at 25 oC(Figure 5.1.2). Determine the heat loss per meter length

of the tube.

Given

Fluid: Saturated steam

Fluid: Air

Surrounding surfaces:

Tube:

Insulation: Rockwool

Cladding: Painted metal

Find

q/L (heat loss per meter length of tube)

Assumptions

Figure 5.1.2 Conductive,

convective and radiative heat

transfer from an insulated steam

Tsat 200oC=

T 25oC=

Ts 25oC=

2ri 5 cm=

d 2ro= 15 cm=

0.8=

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(8) The condensation heat transfer coefficient is large and the convective resistance on the steam

side is negligible. Therefore, the tube surface is at the steam saturation temperature of 200 oC.

(9) As the tube and the metal cladding are thin, their conductive resistances are negligible.

(10) The room surface is much larger than the insulation outer surface.

(11) Quiescent air.

Solution

Heat transfer is by conduction across the insulation and by convection and radiation from theouter surface to the air and surrounding surfaces, respectively, as shown in Figure 5.1.2. From

an energy balance on the outer surface,

[1:2.1]

Substituting the expressions for the heat transfer rates and rearranging, we obtain

[1:2.2]

qk qc qR+=

qk

2kinsL Ti To( )

ro ri( )ln-----------------------------------------= qc h2roL To T( )= qR 2roL To

4 Ts4( )=

kins Ti To( )ro ro ri( )ln------------------------------ h To T( ) To

4 Ts4( )+=

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1.9

(1) Assume .

(2) Properties of atmospheric air at the film temperature of 42.5 oC and at , are

After solving Equation [1:2.2] for , we obtain the heat transfer rate per meter length by

conduction (or as the sum of the convective and radiative heat transfer rates from the outer

surface). To solve for we need h. As the air is quiescent, we assume the convective heat

transfer is by natural convection. The natural convective heat transfer coefficient depends on ,

which is determined by the following iteration scheme.

(1) Assume a value of .

(2) Compute h.

(3) Compute from Equation [1:2.2].

(4) Using the computed value of , repeat steps 2 and 3 until two consecutive values ofh are veryclose to each other.

To

To

To

To

To 60oC=

T

1.118 kg/m3= 1.93 510 Ns/m2= k 0.0268 W/m K=

cp 1007 J/kg K= 3.354310 K 1= Pr 0.7248=

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Employing Equation (4.4.20),

(3) Compute from Equation [1:2.2]. With kins for rockwool (Table A3) = 0.04 W/m K

As the computed value of is much lower than the value used for computing h, recompute the

values starting with 33.1 oC for . The results are

One more iteration starting with for computing h, yields a value of 34.5 oC for ,

which is very close to the starting value for this step. Employing ,

Rad

g2 To T( )d3Pr

2

------------------------------------------------ 9.461610= =

Nud 0.60.387Rad

1 6

1 0.559 Pr( )9 16+[ ]8 27--------------------------------------------------------------+

2

27.88= =

h

Nudk

d------------ 4.98 W/m2 K= =

To

kins Ti To( )

ro ro ri( )ln------------------------------ h To T( ) To

4 Ts4( ) To+ 33.11

oC= =

To

To

h 3.28 W/m2

K= To 34.6o

C=To 34.6

oC= To

To 34.5oC= h 3.4 W/m2 oC=

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1.11

Comments

(1) Note that the radiative heat transfer rate is greater than the convective heat transfer rate.(2) The solution to Equation [1:2.2] can also be obtained by employing the radiative heat

transfer coefficient.

___________________________________________________________________________


qc

L

----- h2ro To T( ) 15.4 W/m= =qR

L

------ 2ro To4 Ts

4( ) 22.5 W= =

qk

L-----

2k Ti To( )

ro ri( )ln------------------------------- 37.9 W/m= =

qc

L-----

qR

L------+ 37.9 W/m=

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Multimode

1.12

Reconsider Example 7.4.7. The furnace arch

is constructed with 40-cm thick firebricks and

there is no radiation shield (Figure 5.1.3). The

inside surface temperature of the furnace is

1000 oC. Surrounding air and wall surfaces are

at 35 oC. Modeling the arch as a 6-m wide, 20-

m long rectangular surface with an emissivity

of 0.9, determine the heat transfer rate from

the arch to the surroundings

(3) If convection is neglected.

(4) If convection is included.

Given

Material: Firebrick

L = 0.4 m Width = 6 m

Length = 20 m = 0.9

Find

(a) q neglecting convection

Figure 5.1.3 A furnace arch heat transfer by

conduction, convection and radiation

T1 1000oC=

T Ts 35oC= =

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1.13

(b) q including convection

Assumptions(5) Steady state.

(6) Air does not participate in radiation.

(7) Constant properties.

(8) One-dimensional temperature distribution in the arch.

(9) Surrounding surface area is much larger than the furnace top surface area.

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Solution

Heat transfer is by conduction across the arch and by convection and radiation from the outer

surface to the air and surrounding surfaces respectively (Figure 5.1.3). Denoting the thermal

conductivity of firebrick by kb From an energy balance on the outer surface,

[1:3.1]

whereA is the cross-sectional (also surface) area. Substituting the expressions for the heat

transfer rates and rearranging, we obtain

[1:3.2]

After solving Equation [1:3.2] for , we obtain the heat transfer rate by conduction (or as the

sum of the convective and radiative heat transfer rate from the outer surface). To solve for we

need h. As the air is quiescent, we assume the convective heat transfer is by natural convection.

The natural convective heat transfer coefficient depends on , which is determined by the

following iteration scheme.

(1) Assume a value of .(2) Compute h.

(3) Compute from Equation [1:3.2].Using the computed value of , repeat steps 2 and 3 until two consecutive values

qk qc qR+=

qkkbA T1 T2( )

L--------------------------------= qc hA T2 T( )= qR A T2

4 Ts4( )=

kb T1 T2( )L

---------------------------- h T2 T( ) T24 Ts

4( )+=

T2

T2

T2

T2

T2 T2

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1.15

[1:3.3]

[1:3.4]

Equation [1:3.4] can be solved for directly [see Spiegel (1968)] or by numerical methods. Here,

we solve it by the bisection method. To solve Equation [1:3.4], as cannot be higher than or

lower than , we set and . From Table A3, for firebricks, kb = 1 W/m K.

Substituting the values into Equation [1:3.4], and dividing it by 108, we obtain

[1:3.5]

Setting and , we get a sequence of values, some of which are given

below.

K K

K

3.082 12.732 7.907 -473 3582 < 0

kb T1 T2( )

L---------------------------- T2

4 Ts4( )=

T24

kb

L----------T2

kb

L----------T1 Ts

4+ 0=

T2

T2 T1

Ts TH T1= TL Ts=

T2100---------

4 48.99 T2

100---------

714+ 0=

TH 1273.2 K= TL 308.2 K=

TL 100 TH 100 T 100 TL( ) T( ) TL( ) f T( )

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Multimode

1.16

From the table it is clear that the value of is close to 469.1 K. From Equation [1:3.1] the heattransfer rate is given by

An alternate method of finding is to recast Equation [1:3.4] in the form

[1:3.6]

3.082 7.907 5.494 -473 466 < 0

3.082 5.494 4.288 -473 -166 > 0

4.288 5.494 4.891 -166 97 < 0

4.691 4.693 4.692 -0.047 0.42 < 0

4.691 4.692 4.6915 -0.047 0.18 < 0

K K

K

TL 100 TH 100 T 100 TL( ) T( ) TL( ) f T( )

T2

q kbAT1 T2

L----------------- 1 6 20

1273.2 469.1

0.4------------------------------------ 241 200 W= = =

T2

x 714 48.99 x( )[ ]1 4= x T2 100=

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1.17

Starting with any reasonable value ofx on the right-hand side, a new value ofx is computed. This is

then used on the right-hand side again and the process repeated. When the absolute value of the

difference between two successive values ofx is within a pre-assigned convergence value, a good

approximation to the correct value is obtained. Starting with orx = 12.73, the

following sequence of values is obtained.

12.73, 3.083, 4.871, 4.669, 4.693, 4.691, 4.691

In this case a rapidly converging solution is obtained. There is no assurance that the convergence

would be as rapid in all cases or even that convergence would be attained. For example, if instead of

casting Equation [1:3.4] in the form of Equation [1:3.6] it is cast as , a

diverging sequence of values is obtained.

(b) Including convection, simplifying Equation [1:3.2],

[1:3.7]

Equation [1:3.7] is solved for by recasting it in the form,

[1:3.8]

T2 1273 K=

x 714 x4( ) 48.99=

T24

kb

L----------

h

------+

T2kbT1

L-----------

hT

----------+

T4++ 0=

T2

T2

kbT1

L-----------

hT

----------+

T4+

kb

L----------

h

------+

T2

1 4

=

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Multimode

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The convective heat transfer coefficient, which depends on the is found by the same algorithm as

in part (a).Start computation with an assumed value of . Find h by natural convection from a

horizontal surface with the heated surface facing up, employing Equation (4.4.12) or Equation

(4.4.13). Properties of atmospheric air at the reference temperature of

and at , are

With W(width) as the characteristic dimension for computing Ra and Nu,

The value of is greater than the upper limit for Equation (4.4.13). However, we will use thecorrelation with the assumption that correlation was developed with experimental values of up to

, that it is valid even at and that the value ofh is a good

approximation to the actual value.

T2

T2 100oC=

T2 0.25 T2 T( )[ ] 83.8 oC= T 0.25 T2 T( )+[ ] 324.5 K=

0.9891 kg/m3= 2.112 5 Ns/m2= cp 1008 J/kg K=

k 0.02979 W/m K= 3.082 310 K 1= Pr 0.7146=

RaW

g2 T2 T( )W3Pr

2-------------------------------------------------- 6.652

1110= =

RaW

RaW 1011= RaW 6.652

1110=

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1.19

Employing Equation [1:3.8], the value of after the first iteration is 189.8 oC. Repeating the

computations with and continuing the iteration process we obtain,

. The heat transfer rates are

Comments

(1) With the inclusion of convection the surface temperature goes down from 469.1 K to 430.8 K butthe heat transfer rate increases slightly from 241 200 W to 252 913 W, an increase of 4.9%.

(2) The radiative heat transfer rate is greater than the convective heat transfer rate.

(3) Natural convection correlation for still air is used, but with a furnace there will be some air

movement and the actual heat transfer coefficient will be higher than the one obtained. If the con-vective heat transfer coefficient is 12 W/m2 oC, the values are and q = 258 770 W.

Thus with an increase in the convective heat transfer coefficient there is a decrease in the outer

surface temperature but a slightly higher (2.3%) heat transfer rate.

NuW 0.13RaW1 3 1135= = h

NuWk

W-------------- 5.64 W/m2 K= =

T2

T2 189.8oC=

T2 157.6oC, h 6.61 W/m2 K= =

qc hA T2 T( ) 97247 W= = qR A T24 Ts

4( ) 155665 W= =

q qc qR+ 252913 W= =

T2 135oC=

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(4) Because of the simultaneous changes in h and while iterating Equation [1:3.8], the number of

iterations is significantly greater than the number of iterations with Equation [1:3.6] without con-

vection.

_____________________________________________________________________________


The temperature of air flowing in a duct ismeasured by a thermocouple. The

thermocouple junction is attached to the

bottom surface of the thermometer well

(Figure 5.1.4). The brass thermocouple

well is a hollow cylinder with inner andouter diameters of 2.5 mm and 5 mm,

respectively, and a length of 4 cm. The air

flows with a velocity of 10 m/s at a

pressure of 400 kPa. The thermocouple

indicates a temperature of 395 K. Thetemperature of the well at the duct wall

( ) is 375 K. Estimate the true

temperature of the air.

T2

Figure 5.1.4 Thermocouple in a well to measure airtemperature

TB

Ad d H T f

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1.21

Given

Fluid: Air

Thermometer well: Brass

Find

Fluid temperature

Assumptions

(1) Negligible heat transfer from the inner surface of the well.

(2) Duct wall is at the same temperature as the air, i.e., .

(3) Duct surface is much larger than the well surface.

(4) The thermocouple junction is at the same temperature as the end surface of the well.

(5) One-dimensional temperature distribution in the well.

Solution

The temperature of the end-surface , is less than the air temperature due to conduction along the

wall of the. The temperature distribution in the wall is found from an energy balance on an element

shown shaded in the figure.

[1:4.1]

U 10 m/s= p 400 kPa=

di 2.5 mm= do 5 mm=

L 40 mm= TA 395 K= TB 375 K=

T

Tw T=

TA

qx qx x+ qs 0=

M lti d

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where , is the sum of the convective and radiative heat transfer rates to the air and

the duct walls, respectively.

The duct surface is much larger than the well surface; the radiative heat transfer rate is given by

Substituting into Equation [1:4.1], dividing it by x and taking the limit

as , we obtain

[1:4.2]

Boundary conditions:

[1:4.3]

To solve the nonlinear Equation [1:4.2], linearize it by employing the radiative heat transfercoefficient, . Denoting h = hc + hR, = T - Tand m2 = hP/kAx,

Equation [1:4.2] and the associated boundary condtions are

qs qc qR+=

qc hcP T T( ) x= P do=

qR P T4 T

4( ) x=

c x +

x 0

xd

dqx hcP T T( ) P T

4 T4( ) 0=

Tx 0=

TA= Tx L=TB=

hR T2 T

2+( ) T T+( )=


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1.23

[1:4.4]

[1:4.5]

The solution to Equation [1:4.4] with the associated boundary conditions given by Equation [1:4.5]

is

[1:4.6]

Equation [1:4.6] contains the unknown fluid temperature T

We ned an additional condition to

determine it. Such a condition is obtained from an energy balance on the surface atx = 0.

or

[1:4.7]

Solving for Tfin Equation [1:3.7]

x2

2

d

d m2 0=

x 0=

A TA T= = x L= B TB T= =

A mx( )coshB A mL( )cosh

mL( )sinh------------------------------------------- mx( )sinh+=

kxd

d

x 0=

hx 0=

=

km

B A mL( )cosh

mL( )sinh------------------------------------------- hA=

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[1:4.8]

Both h and m (which depends on h) on the right hand side of Equation [1:3.8] depend on the value of

Tf, which is found by the following iteration scheme.

(1) Assume a reasonable value ofT.

(2) Determine hc, hR, and h.(3) Determine T

from Equation [1:4.8].

Employing the computed value ofT

, repeat steps 2 and 3 until two consecutive values are

sufficiently close to each other

(1) Assume

(2) Both and depend on the wall surface temperature, which varies from 395 K atx = 0 to 375

K atx = L. We will evaluate them at the mean surface temperature Tm = 385 K. The properties of

air at 400 kPa and the film temperature of 392.5 K are

T

h mL( )sinh km mL( )cosh+[ ]TA TB

mL( )cosh 1 h mL( )sinh km+---------------------------------------------------------------------------------------------=

T 400 K 126.8 oC= =

hc hR

3.551 kg/m3= cp 1012 J/kg K= 2.261510 Ns/m2=

k 0.03231 W/m K= Pr 0.7075= Red

Udo

------------- 7852= =


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1.25

Employing Equation (4.2.19)

From Table A1 the thermal conductivity of brass kb is 110 W/m K and from Table A9, the emissivity

of oxidized brass is 0.66. The value of at a surface temperature of 385 K and a duct wall

temperature of 400 K is

Employing the thermal conductivity of brass

(3) Substituting the values into Equation [1:4.8], we obtain .

As the new value of is almost equal to the assumed value, no further is needed. Therefore, the

correct value is

Nud 0.3

0.62Red1 2 Pr1 3

1 0.4 Pr( )2 3+[ ]1 4----------------------------------------------------+ 43.3= = hcNudk

do------------ 279.8 W/m2 K= =

hR

hR T2 T

2+( ) T T+( ) 9.055 W/m2 K= =

h hc hR+ 288.9 W/m2 K= =

m hP k bAx( )1 2 288.9 0.005

110 0.0052 0.00252( ) 4------------------------------------------------------------------------------ 52.93 m 1= = =

mL 52.93 0.04 2.117= =T

400.8 K=

T

T 400.8 K=

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Comments

(1) There is a 5.8 K difference between the indicated value and the true value of . Althoughmany simplifying assumptions were made in the solution, the computed value is much closer

to the true value than the indicated value.

(2) With forced convection the radiative heat transfer coefficient is much less than the

convective heat transfer coefficient. Therefore, in this case, neglecting the radiative heat

transfer rate will give a very good approximation to the true value of , 407.7 K.

___________________________________________________________________________

Example 5.1.5 Conduction and Convection

Cooling water enters a steam condenser (Figure 5.1.5) at 20 oC with a flow rate of 0.1 kg/s. Sat-

urated steam at 100 oC condenses on the outer surface of the tube. To measure the temperature of thecooling water at inlet, a thermocouple is attached to the outer surface of the 9-mm I.D., 12.5-mm

O.D. copper tube. The outer surface of the thermocouple junction is insulated. If the thermocouple is

placed too close to the inlet flange of the condenser, the indicated temperature will be affected by

heat transfer along the tube from the steam. How far away from the inlet flange should the thermo-

couple be placed if the indicated temperature is to be within 0.1 oC of the cooling water inlet temper-

ature?

Given

Cooling water inlet temperature = 20 oC

T

T


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1.27

Tube material: Copper

Fluid inside tube: Water

Find

The distancexoso that the temperature To should not be greater than 20.1oC.

Assumptions

(1) The temperature of the tube at the flange is affected by the temperature of the condensing steam

in the condenser. Because of the high heat transfer coefficient expected with condensing steam,

we assume that the temperature of the tube at the flange equals the saturation temperature of the

steam, i.e., .

(2) The air side convective heat transfer coefficient is much less than the convective heat transfercoefficient on the water side. Compared with the expected much higher heat transfer rate to the

water, the heat transfer rate to the air is negligible.

di 0.009 m= do 0.0125 m=

m

0.1 kg/s=

Tbase 100oC=

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1.28

(3) One-dimensional temperature distribution in the tube.

(4) Water temperature is uniform in the tube up to the flange. This means that the heat transfer ratefrom the tube to the cooling water (outside the condenser) is negligible so far as the cooling

water temperature is concerned but is significant in determining the temperature distribution in

the tube.

(5) The tube is an infinite fin with its base at 100 oC.

Solution

For an infinite fin, the temperature distribution is given by

[1:5.1]

Figure 5.1.5 Thermocouple attached to cooling water

inlet tube of a condenser

T Tf=

bemx=


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1.29

where is the thermal conductivity of the copper tube and Tb is the temperature of the flange.

To find the value ofxo where the temperature is 20.1 oC, i.e., we need the

value ofm. To find the value of on the water side, from Table A5 the properties of water at 20 oC

Employing Equation (4.3.5)

From Table A1, the thermal conductivity of copper, .

From Equation [1:5.1],

m hP kAx 4hdo kcu do2 di

2( )[ ]= = b Tb Tf=

kcu

o b 0.1 80=

hw

998.2 kg/m2= 1.0023

10 Ns/m2= cp 4183 J/kg K=

k 0.5984 W/m K= Pr 7.006= Red4m

di----------- 14117= =

Nud 0.023Red0.8Pr0.4 104.6= = h

Nudk

di------------ 6958 W/m2 K= =

kcu 401 W/m K=

m 4hdo kcu do2 di2( )[ ] 106.7 m 1= =

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The thermocouple junction must be located at a minimum distance of 6.2 cm from the flange.

Considering the uncertainty in the value of the convective heat transfer coefficient, to assure satisfy-

ing the requirement, it may be placed 10 cm from the flange.

One of the assumptions made in the analysis was that the bulk temperature of the water was not

affected by the heat transfer from the tube to the water before it reaches the flange. The heat transferrate from the tube to the incoming water is given by

Denoting the temperature of the incoming water by and the water temperature at the flange

by , we have . The temperature of the water

increases from 20 oC to 20.5 oC, a change, which is negligible.

Comments

(1) The tube surface temperature at the flange will be less than 100 oC due to the heat transfer from

the tube to the incoming water. The assumption that the tube surface at the flange is at 100 oC will

lead to a value ofx slightly higher than the actual value.

xo1

m----

0

b

-----ln1

106.7-------------

0.1

80-------ln 0.062 m 6.2 cm= = = =

q kcuAx xd

d

x 0=

kAxmb 202.3 W= = =

Tf

Tx 0= q m cp Tx 0= Tf( )= Tx 0= 20.5

oC=


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(2) Computing the total convective heat transfer coefficient, h, by assuming that has a range of

values from 4 W/m2 oC to 12 W/m2 oC, has a negligible effect on the value ofx.

_____________________________________________________________________________


The roof of a small shed is inclined at 45o to the vertical plane (Figure 5.1.6). It is modeled as a

10-mm thick plywood sheet with an emissivity of 0.9. The solar insolation is 800 W/m2 and the

ambient air is at 40 oC. The attic space is well ventilated and the attic air is also at 40 oC. The outside

convective heat transfer coefficient is 10W/m2 oC. The equivalent surrounding temperature Te, is 20

oC. Determine

(a) The temperature of the surface of the roof exposed to the sun.

(b) The heat flux across the roof.

Given

Roof: 10-mm thick plywood sheet inclined at = 45o to the horizontal, = 0.9

Find

(a) (b)q''

ha

G 800 W/m2= T 40oC= h1 10 W/m

2 oC=

Te 20 oC=

T1

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Solution

From an energy balance on the upper surface of the roof,

[1:6.1]

Assume that the area of the attic surfaces other than surface exposed to the sun is much greater.

From an energy balance on the lower surface,

[1:6.2]

is the temperature of the inner surfaces of the attic, which are not directly affected by the solar

radiation.

Substituting the expressions for the heat fluxes into Equations [1:6.2] [1:6.1] and [1:6.2] we

obtain

[1:6.3]

[1:6.4]

Solutions to Equations [1:6.3] and [1:6.4] will give the values of , and .

G q1c'' q1R'' qk''=

qk

'' q2c

'' q2R

''+=

q1R'' T1

4 Te4( )= q1c

'' h1 T1 T( )= qk'' kw T1 T2( ) t=

2 2 = 2 Ts2

G h1 T1 T( ) T14 Te

4( ) kw T1 T2( ) t=

kw T1 T2( ) t h2 T2 T( ) T24 Ts2

4( )+=

T1 T2 qk''


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As the attic is well insulated, the temperatures of the air and the surfaces not exposed to the sun

are assumed to be a little higher than , i.e., . Assuming that the convective

heat transfer from the inner surface of the roof is by natural convection, find for an inclined sur-

face. Thus, the only parameter required for the solution is , which depends on the yet unknown .

and are determined by iteration.

From Table A3, for plywood, kw = 0.115 W/m K. Start with an assumed value of. The characteristic length for determining the Rayleigh and Nusselt

number is . With

Properties of atmospheric air at are

T

Ts2

Ti

45oC= =

2

2 T2

T2 h2

T2 60oC 333.2 K= =

L 4 45ocos 5.66 m= = Ti 45oC=

Tf T2 Ti+( ) 2 52.5 oC= =

1.084 kg/m3= 1.974 510 Ns/m2= cp 1006 J/kg K=

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For a downward facing heated surface from Equation (4.4.4) with g replaced by g cos ,

Figure 5.1.6 Heat transfer by different

k 0.0275 W/m K= (at Ti

45 oC) 3.143310 K 1==

Pr 0.721=

RaL

g ( )cos 2 T2 T( )L3Pr

2------------------------------------------------------------------ 1.287

1110= =

NuL 0.8250.387RaL

1 6

1 0.437 Pr( )9 16+[ ]8 27--------------------------------------------------------------+

2

581= =


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Substituting the known values into Equations [1:6.3] and [1:6.4] and solving for and by

a numerical method or by linearizing the radiative heat transfer terms with radiative heat transfer

coefficient, we obtain

The assumed value of is slightly higher than the computed value of 331.8 K. No

further iteration is needed. Thus, we obtain the correct values ofT1, T2, and q as

___________________________________________________________________________

5.2 SOLAR COLLECTORS

A solar collector is a good example of the application of the principles of heat transfer involving multimode heat transfer and therefore, it is considered in

some detail in this section. Here, we consider one type of solar collector, the flat plate collector, utilized extensively for residential use for space heating and hot

water supply.

The sun is a star in our galaxy. It is considered a sphere of hot gases. Some of the data on the sun are given in Table 5.2.1.

Table 5.2.1 Some data of the sun

Diameter of the sun 1.39109

m

h2c

NuLk

L------------- 2.83 W/m2 K= =

T1 T2

T1 343.4 K= T2 331.8 K (58.6oC )=

T2 333.2 K=

T1 343.4 K= T2 331.8 K= qk''

kw T1 T2( )

t

----------------------------- 134 W/m2= =

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The solar constant is the rate of incident solar energy flux on a surface perpendicular to the solar beam in the outer reaches of the atmosphere. As the solar

energy passes through the atmosphere, it is attenuated by scattering by the air molecules (Raleigh scattering), absorption by dust, water vapor, carbon dioxide, and

ozone; and reflection and absorption in cloud layers. The extent of the attenuation depends on the height above the sea level, the position of the sun (time of day and

time of the year), and dust and gas concentrations in the atmosphere. On a clear day approximately 70% of the of the solar constant, 1000 W/m2, reaches the surface

of the earth. About 99% of the solar energy is in the wave band of 0.2 m - 4 m and 94% in the band 0.2 m - 2 m.

Nomenclature: In this section the following terms are used.

Mean distance between the sun and

the earth

1.49510

11 mEffective blackbody surface temper-

ature of the sun

5777 K

Ratio of the mass of the sun to the

mass of the earth

332 488

Total power emitted by the sun 3.831026

W

Solar power intercepted by the earth 1.71017

W

Solar constant 1367 W/

m2

Incident solar energy flux (W/m2)

Table 5.2.1 Some data of the sun

Gs


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A solar collector is a heat exchanger in which a fluid is heated by a surface that absorbs incident solar energy. A tube, exposed to the sun, in which a fluid iscirculated, is a collector in its simplest form. In a flat plate collector, a flat surface is used to absorb solar energy and transmit it to a fluid adjacent to it.

There are two types of solar collectors - the concentrating collector and the flat plate collector. In a concentrating collector the solar beam is focussed on a

much smaller area by suitably shaped mirrors or lenses. There is no such concentration in a flat plate collector.

Two types of flat plate collectors for space heating, with air or water as the working fluid, are shown in Figure 5.2.1. A flat plate collector consists of an

absorber plate covered with one or more glass cover plates and an insulated frame. The absorber plate is heated by the absorbed solar energy. The plate transfers a

Spectral solar insolation per unit area per unit

wavelength (W/m2m)

Blackbody emissive power at the temperature of

the sun, 5777 K (W/m2)

Blackbody monochromatic emissive power at

the temperature of the sun, 5777 K (W/m2m)

Solar absorptivity (absorptivity for solar radia-

tion)

Infrared emissivity - emissivity in the range 300K - 400 K

Cover plate (glass) transmissivity for solar radia-tion

Cover plate infrared transmissivity in the range300 K - 400

Gs

Ebs

Ebs

s

s

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part of the energy to air flowing between the plate and the adjacent glass cover plate (Figure 5.2.1a) or to water flowing in tubes attached to the plate (Figure 5.2.1b).

Most of the solar energy is transmitted through the glass plates which are essentially transparent to the incident solar energy.

However, the glass plates are almost opaque to infrared radiation. Therefore, the radiant energy

emitted by the absorber plate, usually maintained at around 350 K - 380 K, is blocked.

The space between the glass cover plates is occupied by air, which is essentially stagnant and serves to reduce the heat loss from the collector to the

surroundings.

Figure 5.2.1 Two types of flat plate collectors (a)


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Figure 5.2.2 shows a simple system for the supply of hot water to a residential building. Water is circulated through the

collector mounted on the roof. The heated water passes through a coil in the holding tank and heats the cold water from the

main cold water line. The cooled water from the coil is forced back into the collector. The warm water from the holding tank is

fed into a hot water heater, where the water is further heated as needed, before being supplied to the various consuming points.

To analyze the operation of a collector, a knowledge of the radiation characteristics of the absorber plate and the glass

cover plates is needed. A brief discussion of some of the characteristics follows.

Solar absorptivity and infrared emissivity The collector plate receives solar energy from the sun radiating as a black

body at 5777 K. It generally operates at temperatures less than 400 K. A good absorber plate should absorb all the incident

solar energy but should emit very little energy. Therefore, the absorber plate should have a high absorptivity at short

wavelengths (99% of the solar energy is in the 0.2 m - 4 m range) and low infrared emissivity. The solar absorptivity isdefined by Equation (8.1.15). As the spectral distribution of solar energy is fixed, the solar absorptivity has a unique value foreach surface except for minor variation due to the temperature dependence of the spectral absorptivity. Similarly, the infrared

emissivity is defined by Equation (8.1.10) with the monochromatic blackbody emissive power in the 300 K - 400 K temperate

range.

Even if the spectral emissivity of a surface is temperature independent, its total infrared emissivity is temperature

dependent, as the monochromatic emissive power depends on the temperature. However, if the temperature range is not large,

the variation of the infrared emissivity is small. When an infrared emissivity value is given without indicating the temperature at which it is evaluated, it is usually in

the 300 K - 400 K range. The solar absorptivities and infrared emissivities of a few special surfaces [Howll et al. (1982)] are given in Table 5.2.2.

Table 5.2.2 Solar absorptivity and infrared emissivities of a few special surfaces

Surface

Black nickel 0.87-0.96 0.07-0.12 7.3-13.7

Copperoxide

0.81-0.93 0.11-0.17 4.8-8.5

Lead sulfide

(crystals)

0.89 0.20 4.45

Figure 5.2.2 Schem

s s

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Example 5.2.1A special opaque surface has an emissivity of 0.9 for 0 < < 3 m and 0.1 for > 3 m. Compute its

(a) Solar absorptivity.

(b) Infrared emissivities at 300 K, 400 K, and 500 K.

Given

Find

(a) (b) at 300 K, 400 K, and 500 K

Solution

(a) From Kirchoffs law the spectral emissivity equals the spectral absorptivity. Denoting the quanti-

ties in the wave band 0 to 3 m by subscript 1 and those in the wave band 3 m to by subscript2, from Equation (8.1.15) the solar absorptivity is given by

Black paints 0.90-0.98 0.90-0.98 ~ 1

Black

chrome

0.90-0.95 0.15-0.25 3.6-6.3

Table 5.2.2 Solar absorptivity and infrared emissivities of a few special surfaces

Surface s s

0 3 m< < 0.9= 3 m < 0.1=

s


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[2:1.1]

is to be evaluated at T= 5777 K. From Table 7.2.1, for

(b) T= 300 K For , . From Equation [2:1.1]

Similarly, at

T= 400 K, , , and

T= 500 K, , , and

Comment

(a) For the surface, the ratio of varies from 8.83 at 300 K to 8.0 at 500 K.

s

1 Gs d0

3

Gs----------------------------

2 Gs d3

Gs-----------------------------+ 1f0 1T 2 1 f0 1T( )+= =

0 1

T

1T 3 5777 17331 m K= = 0 1

T 0.9789=

s 0.9 0.9789 0.1 1 0.9789( )+ 0.883= =

1T 3 300 900 m K= = 0 1

T 8.704510=

0.9 8.7045

10 0.1 1 8.7045

10( )+ 0.1= =

1T 1200 m K= 0 1

T 2.134310= 0.102=

1T 1500 m K= 0 1T 1.2852

10= 0.11=

s

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(b) The infrared emissivities at 300 K and 400 K are almost equal but the emissivity at 500 K is

slightly higher.

_____________________________________________________________________________

Example 5.2.2

The solar insolation on an absorber plate attached to a space ship is 800 W/m2. The underside of the plate is perfectly insulated. The absorber plate transfers

energy to a cooling fluid flowing in tubes attached to it. the Assuming a sky temperature of 0 K, determine

(a) The equilibrium temperature of the plate (without heat transfer to the cooling fluid) if the

surface is (1) black and (2) black nickel.

(b) The heat flux to the cooling fluid in the tubes attached to

the plate for the two surfaces. The cooling fluid maintains

the plate at 320 K in both cases.

Given Absorber plate surface: Black or black nickel,

Find

(a) The equilibrium temperature with q = 0 for each case.

(b) q''for each case if = 320 K.

Assumption

Figure 5.2.3 Flat plate

Gs 800 W/m2=

Ts


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As the plate is located in the outer reaches of the atmosphere, the heat transfer is by radiation

only.

Solution

From an energy balance on the surface,

energy reaching the surface = energy leaving the surface or

Energy reaching the surface from the sun

= Reflected solar energy + emitted energy from the surface + Heat transfer to the coolantThere is no heat transfer to the coolant. With the absorbed energy equal to the incident energy less

the reflected energy, we have

Absorbed energy flux = Emitted energy flux (emissive power)

(a)(1) For a black surface ( = = 1). Absorbed energy flux = incident energy flux, and

(2) For the black nickel surface: Absorbed energy flux = , Emissive power =

To find the absorbed energy, the solar absorptivity is used. As the plate temperature is expected

to be in the 300 K - 400 K range the infrared emissivity is used to find the emissive power.

From Table 5.2.2, for a black nickel surface, and .

Gs Ts4= Ts

Gs

------

1 4344.6 K= =

sGs Ts4

s 0.94= 0.12=

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(b) From the energy balance on the plate including the heat transfer to the coolant, Absorbed solar energy flux = emissive power of the surface + Heat flux to the coolant (q'')

q''= Absorbed solar energy flux - emissive power of the surface =

(1) For the black surface,

(2) For the black nickel surface with and

Comments

(1) The equilibrium temperature of the black nickel surface is 576.6 K, which is above the 300 K -

400 K range for which the infrared emissivities are applicable. However, no appreciable increase

in the emissivity at 576 K is expected and the computed equilibrium temperature is a good

approximation.

(2) Though the black surface absorbs all the energy it also emits the highest amount of energy at a

given temperature. At the same temperature, the black nickel surface absorbs 94% of the incident

s----

. 8

----------------------------------------- =

sGs Ts4

q '' Gs Ts4 800 5.67 810 3204 205 W/m2= = =

s 0.94= 0.12=

q '' sGs Ts4 0.94 800 0.12 5.67 810 3204 681 W/m2= = =


b i l 12% f h i d b h bl k f Th f h h fl

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energy but emits only 12% of the energy emitted by the black surface. Therefore, the heat flux to

the coolant increases from 205 W/m2 for a black surface to 681 W/m2 for the black nickel surface.

(3) If the efficiency of the collector is defined as the ratio of the heat flux to the coolant to the inci-

dent energy flux, the efficiency of the black surface is 25.7% and that of the black nickel surface is

85%. The special surface has a much higher efficiency than the black surface, but such special

surfaces are expensive and many of them lose their selective properties with time.

_____________________________________________________________________________Solar Transmissivity A flat plate collector is usually provided with one or two (and rarely three) glass cover plates. The cover plates transmit most of the

solar energy while reducing both the transmission of the emitted energy from the absorber surface, and the convective heat transfer from the collector to the

surroundings.

The transmissivity of a glass plate depends on the material, the thickness of the plate, and the orientation of the plate to the incident solar energy. When radiant

energy strikes a glass plate, a part of it is absorbed, a part reflected and the rest transmitted. The reflected part is affected by the angle between the incident beam and

the normal to the plate. Further, with more than one plate, there is interaction between the two plates - the reflected energy from the second plate reaching the first

plate and so on.

The solar transmissivity ( ) and infrared transmissivity ( - evaluated at a surface temperature in the range 300 K - 400 K) of a cover plate are defined as

(5.2.1)

where and are evaluated at the absorber temperature of 300 K - 400 K.

Most glass plates used in solar collectors are nearly opaque for radiation in wavelengths greater than 3 m.

s

s

Gs d0

Gs----------------------------=

Eb d0

Eb---------------------------=

Eb Eb

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_____________________________________________________________________________

Example 5.2.3

The spectral transmissivity of a 13-mm thick fused-silica glass is shown in Figure 5.2.4. Determine its

(a) Solar transmissivity.

(b) Infrared transmissivity at 300 K and 400 K.

Solution

(a) The solar transmissivity is calculated from Equation (5.2.1). The spectral transmissivity given inFigure 5.2.4 is approximated as

Figure 5.2.4 Typical

0.17 m 2.5 m< < 0.88=


0 17 d 2 5 0

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< 0.17 m and > 2.5 m

Equation (5.2.1) reduces to

With ,

, , ,

(b) The infrared transmissivity is given by Equation (5.2.1), evaluated at .

With for 0 < < 0.17 m and for > 2.5 m, and for 0.17 m < < 2.5 m

where 1 = 0.17 m and 2 = 2.5 m. Evaluate the expressions at 300

K or 400 K.

0=

s

0.88Gs d0.17

2.5

Gs

-------------------------------------- 0.88Gs

Gs--------- d

0

2.5

Gs

Gs--------- d

0

0.17

= =

Gs

Gs

Eb

Eb

5777 K

f0 T 5777 K

= =

2.5 m= 0 T 0.9658= 0.17 m= 0 T 2.595410=

s 0.88 0.9658 2.595410( ) 0.85= =

Ts

Eb0

Eb----------------------=

0=

0.88=

0.88 f0 2

T f0 1

T( )=

Multimode

At 300 K 0 0 0

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At 300 K:

At 400 K,

Comment

From the calculated values, the glass has a high solar transmissivity and is practically opaque to

infrared emission.

_____________________________________________________________________________

Glass Cover Plates From Figure 5.2.4 we observe that fused-silica glass has a high transmissivity in the wave band of 0.17 m to 2.5 m. When more thanone glass cover plate is used in a collector, the combined transmissivity is approximately given by the product of the individual transmissivities, i.e.,

. The transmissivity of a glass plate is a function of the angle of incidence of the radiant beam, but is approximately constant

for angles up to 60o between the normal to the surface and the beam. Most collectors operate at angles less than 60o, and the effect of the incident angle may be

neglected.

There is one further complication. With one cover plate, the energy flux incident on the absorber plate is . The absorber plate absorbs and the

balance is reflected, which reaches the glass plate and the process is repeated an infinite number of times. The total absorbed energy is then

given by

0 2

T 0= 0 1

T 0= 0=

0 2T 34

10= 0 1T 0= 0

1 2 3 n=

Gs Gs 1 Gs( )

Absorbed energy Gs1 1 ( )d-------------------------------- ( )effGs= =

( )effAbsorbed energy

Incident solar energy--------------------------------------------------

1 1 ( )d--------------------------------= =


where is the diffuse reflectivity of the glass plate The value of is approximately 0 9 for a single

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where is the diffuse reflectivity of the glass plate. The value of is approximately 0.9 for a single

cover plate, 0.8 for two cover plates, and 0.7 for three cover plates. The infrared emissivity of glass

is approximately 0.88.

5.2.1 Performance of Flat Plate Collectors

The heat transfer rate to the working fluid in a solar collector is given by

where is the mass rate of flow of the fluid, and and are the inlet and exit temperatures of the

fluid, respectively. The heat transfer rate and the fluid exit temperature depend on the absorber plate

temperature and the convective heat transfer coefficient associated with the working fluid. The

absorber temperature, in turn, depends on the absorbed radiant energy, and the heat transfer rate to

the working fluid and the surroundings.

The efficiency of a collector is defined as the ratio of the heat transfer rate to the working fluid to the incident radiant energy on the absorber plate area. It is

usually computed for defined values of the temperatures of the absorber plate and the surroundings.

d

q m cp

Te

Ti

( )=

m Ti Te

Multimode

Consider a flat plate collector with one glass cover plate (Figure 5.2.5) with water as the working fluid.

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p g p ( g ) g

Water flows in the tubes attached to the absorber plate. The temperature of the surroundings is for computing

both the convective and radiative heat transfer rates. Given the solar irradiation, the heat transfer rate to the water

is to be determined.

In analyzing the performance of the collector the following assumptions are made.

(1) Steady state: Although solar insolation is time dependent, on

clear days, the variation is slow enough that a quasi-steady-

state assumption is a good approximation. However, if rapid

changes occur due to changes in cloud cover, the steady-state

assumption is not valid.

(2) Uniform absorber plate temperature: Depending on the construction of the collector, there is

some variation in the absorber plate temperature both in the direction of the flow of the water

and between the tubes. Such variations are neglected.(3) Negligible temperature variation across the cover plate.

(4) The glass cover plate is opaque to infrared radiation.

(5) The solar absorptivity of the cover plate is negligible in so far as it affects the heat transfer to

the surroundings. Solar absorptivity is significant in determining the effective transmissivity

but its effect on the temperature of the plate is negligible. The effect of the infrared radiationfrom the absorber plate is more significant than that due to solar absorption.

(6) The temperature distribution in the back insulation of the absorber plate is one-dimensional.

Figure 5.2.5 Schematic of

Ta


(7) As the differences between the temperatures of the absorber plate, the glass cover plate and

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(7) As the differences between the temperatures of the absorber plate, the glass cover plate and

the surroundings are not expected to be large, the radiative heat transfer coefficient can be

used to determine the radiative heat transfer rates.From the above assumptions, the heat transfer in the collector can be analyzed by employing the thermal circuit shown in Figure 5.2.6. The nomenclature is

A Tube surface area Solar insolation on

the cover plate

Solar insolationabsorbed by the

absorber plate

Convective heattransfer coefficient

associated with the

back insulation

Convective heat trans-

fer coefficient associ-ated with the cooling

water

Radiative heat

transfer coefficient back insulation

surface

Convective heat trans-

fer coefficient betweenthe absorber and cover

plates

Convective heat

transfer coefficientbetween the cover

plate and the sur-

rounding air

Gs

G hb

h hrb

hpc hca

Multimode

Radiative heat transfer Radiative heath h

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Radiative heat transfer

coefficient cover

plate and surroundingair

Radiative heat

transfer coefficient

between the coverplate and absorber

plate

k Insulation material

thermal conductivity

Cover plate temper-

aturet Insulation thickness Working fluid inlet

temperature

Mass flow rate of water Working fluid exit

temperature

Net heat flux to the

absorber plate

Absorber plate tem-

perature

q Heat transfer rate to the

working fluid

Temperature of

insulation surface

exposed to air

hrca hrpc

Tc

Ti

m Te

q '' Tp

Ts


Figure 5.2.6 shows the individual resistances associated with the conductive, convective, and radiative

h f b d i f f h b b l N i h h l i i

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heat transfers based on unit surface area of the absorber plate. Notice that there are several resistances in

parallel. Using the radiative heat transfer coefficient permits the radiative heat flux to be related to the

temperature difference. (If the radiative heat flux is computed exactly, it would be related to the difference in

the fourth power of the temperatures and the convective heat flux to the difference in the temperatures; in sucha case a thermal circuit like the one shown cannot be used.) In Figure 5.2.6b the various resistances have been

replaced by the equivalent resistances at the top surface of the absorber plate and at the bottom

surface.

The reciprocals of the resistances for a unit area and are termed the bottom and

topheat loss coefficients, and respectively.

From Figure 5.2.6b, we have

(5.2.2)

Recast Equation (5.2.2) as

Figure 5.2.6 (a) Thermal

circuit for the collector

RT'' RB

''

RB''

R2''R3

''

R2'' R3

''+------------------- R1''

+=

RT''

R4''R5

''

R4'' R5

''+-------------------

R6''R7

''

R6'' R7

''+-------------------+=

1 RB'' 1 RT''UB UT

( )GsT

p

Ta

RB''-----------------

Tp

Ta

RT''----------------- q ''+ +=

Multimode

(5 2 3)( )G

U U+( ) T T( ) q ''+

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(5.2.3)

If the temperature of the absorber plate is known, can be computed. Usually, only the inlet temperature of the working fluid and the solar insolation,

, are known. All other quantities are computed by iteration to satisfy the following relations with the assumption of uniform absorber surface temperature:

(5.2.4)

(5.2.5)

The different convective heat transfer coefficients are evaluated from appropriate correlations. In terms of the temperatures, the radiative heat transfer

coefficients are:

where and are the infrared emissivities of the absorber plate and glass cover plate respectively,

and is the emissivity of the insulation surface.

( )Gs UB UT+( ) Tp Ta( )= q+

q ''

Gs

q m cp Te Ti( )=

Tp Te Tp Ti( )hAm cp--------- exp=

hrb i Ts2

Ta2

+( ) Ts Ta+( )= hrca g Tc2

Ta2

+( ) Tc Ta+( )=

hrpc

Tp2 Tc

2+( ) Tp Tc+( )

1 p 1 g 1+--------------------------------------------------=

p gi


With fi d l f d b th d d d t l lth h th i d hi h f d b it tiG

T U U T T T

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With fixed values of and , both and are dependent on only , although they require and , which are found by iteration.

and can be computed for different values of . With these values of and as functions of , we find . One possible iteration

scheme is

(1) Compute h.

(2) Compute and for different values of .

(3) Guess a value of .

(4) Find from Equation (5.2.5) and q from Equations (5.2.4) and (5.2.3). If the two values of

q are close to each other, the values are correct. Otherwise, repeat steps 2 through step 4 with a

different value of . The computations are illustrated in Example 10.2.4.

_____________________________________________________________________________

Example 5.2.4The specifications of a solar collector with one glass cover plate (Figure 5.2.7) are given in the following table.

Working fluid Water

Solar insolation 800W/m2

Gap between absorber plate and glass

cover plate (L)

30 mm

Gs Ta UB UT Tp Tc Ts

UB

UT

Tp

UB

UT

Tp

q

UB UT Tp

Tp

Te

Tp

Multimode

Solar transmissivity of glass cover 0.88

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Solar transmissivity of glass cover

plate

0.88

Absorber plate emissivity ( ) 0.9

Glass infrared emissivity ( ) 0.9

Air and surroundings Temperatures

(Ta)

0 oC

Absorber plate temperature (Tp) 90oC

Angle of tilt of collector from hori-

zontal ()50o

Convective heat transfer coefficients

( and )

9 W/m2 oC

Insulation thickness (t) 75 mm

Insulation Fiber-

glass

Collector height (H) 1 m

Insulation covering - thin metal sheet

p

g

hca hb


Find

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(1) The top heat loss coefficient

(2) The bottom heat loss coefficient(3) The heat transfer to the water per unit area of the

absorber plate.

Solution

Figure 5.2.7 Flat plate collector

(a) Top Heat Loss Coefficient ( )

For the solution we use the thermal circuit in Figure 5.2.6.

[2:4.1]

UT

1

UT-------

1

hrpc hpc+------------------------

1

hrca hca+------------------------+=

Multimode

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(a) Top Heat Loss Coefficient ( )

(1) Assume .

(2) Determine and

(1) Guess a value of .

(2) Determine the heat loss coefficient .

(3) Compute from the relation

[2:4.2]

Tc

UT

Tc

UT Tp Ta( ) hrca hca+( ) Tc Ta( )=

UT

Tc 40oC=

hrpc hrc

hrpc Tp

2 Tc2+( ) Tp Tc+( )

1 p 1 g 1+--------------------------------------------------

5.67810 363.22 313.22+( ) 363.2 313.2+( )

1 0.9 1 0.9 1+----------------------------------------------------------------------------------------------------------- 7.21 W/m2 oC

=

= =

hrca g Tc2 Ta

2+( ) Tc Ta+( )

0.9 5.67810 313.22 273.22+( ) 313.2 273.2+( ) 5.17 W/m2 oC

=

= =


Determine , the convective heat transfer coefficient between the absorber plate and the glasshpc

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cover plate across the 30-mm air gap inclined at 50o to the horizontal. For the inclined cavity,

, cr= 70o. As and = cr, we employ Equation (5.1.19). Mean

temperature of the two plates = = (90 + 40)/2 = 65 oC. Properties of atmospheric air at 65 oC are:

Employing Equation (5.1.19);

p

H L 1 0.03 33.3= = H L 12>

Tf

1.044 kg/m3= 2.03 510 Ns/m2= cp 1007 J/kg K=

k 0.02845 W/m K= Pr = 0.718

(at Tf) 1 338.2 2.957310 K= =

RaL

g2 Tp Tc( )L3Pr

2----------------------------------------------- 74425= =

a 11708

RaL ( )cos-------------------------- 0.9643= = b 1

1708 1.8( )sin[ ]1.6RaL ( )cos

----------------------------------------------------- 0.9643= =

5830-------------------------- 1 =

NuL 1 1.44 a b c++ 2.362= = hpcNuLk

k------------- 2.239 W/m2 K= =

Multimode

1

U-------

1

h h------------------------

1

h h------------------------+

1

7 21 2 239------------------------------

1

5 17 9-------------------+= =

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(3) Compute : From Equation [2:4.2].

The computed value of 36 oC for is slightly less than the assumed value 40 oC in step 1.Recomputing the values with = 36 oC, we get , which is close to the assumed value. The

final values are

(a) Bottom Heat Loss Coefficient ( )

As the sheet steel insulation cladding is thin, assume that its conductive resistance is negligible.

From Figure 5.2.6b,

UT hrpc hpc+ hrca hca+ 7.21 2.239+ 5.17 9+

UT 5.67 W/m2 oC=

Tc Tc Ta

UT Tp Ta( )

hrca hca+------------------------------+ 36 oC= =

Tc

Tc 36.1=

Tc 36.1oC= hrpc 7.09 W/m

2 oC= hrca 5.06 W/m2 oC=

hpc 2.33 W/m2 oC= UT 5.64 W/m

2 oC=

UB

1

UB-------

t

kins--------

1

hb hrb+-------------------+=


The thermal conductivity of fiberglass insulation (Table A3): . To findkins 0.036 W/m2 oC= hrb

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we need , which is determined by iteration [as in part (a) to find ] such that it satisfies the

relation

[2:4.3]

Assume . Expecting the cladding to be painted, its emissivity, ,

From Equation [2:4.3]

Recomputing the values with , we obtain the final solution

(b) Heat Transfer Rate to the Working Fluid

Ts Tc

UB Tp Ta( ) hrb hb+( ) Ts Ta( )=

Ts 10oC= i 0.81=

hrb i Ts2 Ta

2+( ) Ts Ta+( )

0.81 5.67810 283.22 273.2+( ) 283.2 273.2+( ) 3.96 W/m2 oC

=

= =

1

UB-------

t

kins--------

1

hb hrb+------------------- UB 0.463 W/m2 o

C=+=

Ts Ta

UB Tp Ta( )

hb

hrb

+------------------------------+ 3.22 oC= =

Ts 3.22oC=

Ts 3.25oC= hrb 3.81 W/m

2 oC= UB 0.463 W/m2 oC=

Multimode

The heat transfer rate to the working fluid is given by Equation (5.2.3). For the collector, isapproximated as the product of the transmissivity of the glass plate and the solar absorptivity of the

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approximated as the product of the transmissivity of the glass plate and the solar absorptivity of the

absorber plate. From the specifications, it is assumed that the surface is gray with a solarabsorptivity of 0.9 so that = 0.88 0.9 = 0.792. Thus,

Comments

(1) The efficiency of the collector is: . The efficiency is low as the

collector operating temperature of 90 oC is fairly high. We may find the approximate efficiency

at other values of assuming that the heat loss coefficients remain at the same value. With a

collector temperature of 60 oC, q'' = 267 W/m2 and = 33%.(2) The bottom heat loss coefficient is less than the top heat loss coefficient. When solar

collectors are mounted on the roof of a building, there is little clearance between the underside

of the collector and the heated roof. The heat transfer rate from the collector bottom surface in

those cases is likely to be negligible.

_____________________________________________________________________________

An iterative method for determining the glass cover plate temperature with one cover plate is illustrated in Example 5.2.4. When more than one cover plate is

used, the cover plate temperatures and the loss coefficients are determined by an extension of the iterative scheme. Such a scheme would involve double iteration of

the glass temperatures in the case of two cover plates.

q '' ( )Gs UB UT+( ) Tp Ta( )0.792 800 0.463 5.64+( ) 90 0( ) 84.3 W/m2

=

= =

84.3 800 10.5%= =

Tp


The absorber temperature was assumed to be uniform. In an actual collector there is variation in the plate temperature. The non-uniformity depends on the

flow pattern of the working fluid. If air is the working fluid, the temperature variation is only in the flow direction. With water flowing in tubes attached to the plates,

the absorber temperature varies not only in the direction of flow of the water but also between the tubes

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the absorber temperature varies not only in the direction of flow of the water but also between the tubes.

___________________________________________________________________________Example 5.2.5

To estimate the variation of the absorber temperature between tubes in a collector with water as the working fluid, consider the collector (Example 5.2.4) with

the tubes 20-cm apart. The absorber plate is made of 1-mm thick copper sheet. The temperature of the sheet adjacent to the tubes is 90 oC. Determine the maximum

absorber plate temperature.

Given

Material: Copper

Find

Figure 5.2.8 Collector with

T x b=( ) Tw 90oC= = b 10 cm=

Ta 0oC=

Tp 90oC=

G 800 W/m2=

0.792=t 1 mm=

UB

0.463 W/m2 oC= UT

5.64 W/m2 oC= U UT

UB

+ 6.104 W/m K= =

Tmax

Multimode

Assumptions

(1) Steady state

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(1) Steady state

(2) The variation of the temperature of the absorber plate in the direction of flow is negligible.Solution

With the thermal conductivity of copper kcu=401 W/m K, the Biot number for the absorber plate is

As the temperature distribution is assumed to be one-dimensional, i.e., .

From an energy balance on the shaded element in Figure 5.2.8,

where Wis the length of the plate in the flow direction. Dividing the equation by x, taking the limit

as , substituting , and dividing by kWt, we obtain

where , and .

With the boundary conditions, , the solution to the differential equation is

k

-----------------------. .

401

--------------------------------- 1.= =

Bi

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With , substituting all the known values, we obtain

Comment

The plate temperature varies from a minimum of 90 oC to a maximum of 91 oC. With a characteristic temperature difference of ,

the variation of 1 oC is negligible and the assumption of uniform absorber plate temperature is valid.

Generally, we are interested in the performance of a collector during days when the solar insolation on the surface changes with time due to the changing

angle between the solar beam and the collector surface. Simulation programs that account for the variation not only during days but also during a year, the weather

data (air temperature, and wind speed and direction) for different geographical locations are now available.

_____________________________________________________________________________

Summary

Most heat transfer problems involve heat transfer by more than one mode. The methodologies to solve such problems depend on the nature of the problems,

and each case has to be considered by itself. Many of them require numerical techniques. Solutions to a few cases illustrating some of the iterative and numerical

algorithms are given. The flat plate solar collector is a good example where such multimode heat transfer occurs. A brief introduction to a flat plate collector is

given. The use of special surfaces in collectors to increase the efficiency of collectors is demonstrated. The computations of heat loss coefficients from a collector,

and the heat transfer rate to the working fluid are illustrated.

T Taq

U

-----U

mb( )cosh------------------------------ mx( )cosh=

Tmax T x 0=( ) Taq ''

U-----

Tw Taq ''

U-----

mb( )cosh------------------------------+ += =

t 6.103 401 0.001( )= =

Tmax633.6

6.103-------------

90 633.6 6.1033.9 0.1( )cosh

-------------------------------------------+ 91 oC= =

Tw

Ta

90 oC=

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Multimode

estimate the surface temperature of vegetation in those cases, consider a thin, freely suspended, black, horizontal 30-cm square plate, with the bottom surface

insulated. If the effective sky temperature is -40 oC, and the surrounding air temperature is 4 oC, determine the equilibrium temperature of the plate if

(a) The air is calm

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(a) The air is calm.

(b) The air has a velocity of 4 m/s parallel to one of the edges.5.18 Re-solve part b of Problem 5.17 if the plate is replaced by a 8-cm diameter spherical shell (simulating an orange).

5.19 A 1-mm diameter thermocouple junction placed inside a duct to measure the temperature of the air flowing in the duct indicates 400 oC with the duct walls at

380 oC. The air velocity is 10 m/s. Assuming conduction in the lead wires of the thermocouple to be negligible, estimate the true air temperature if the

emissivity of the junction is

(a) 0.9

(b) 0.1 Suggest a method to reduce the error in measuring the air temperature without changing the

emissivity of the surface.5.20 The properties of a spherical thermocouple junction are

k= 20 W/m K = 8500 kg/m3

= 0.9d= 1 mm It is placed in an air stream flowing at 10 m/s and 60 oC, in a duct whose surface is also at 60 oC.

The initial temperature of the junction is 20 oC. Estimate the time for the junction temperature to

reach 59.5 oC

(a) Neglecting radiative heat transfer.

(b) Including radiative heat transfer, making a reasonable estimate of the radiative heat transfer.5.21 A 10-cm diameter copper sphere with a surface emissivity of 0.8, initially at a uniform temperature of 100 oC, is freely suspended in a large room with air in it

at 20 oC. Determine the time for the sphere temperature to reach 25 oC.

5.22 A thermocouple junction to measure the temperature of air flowing at 4 m/s in a duct is made by soldering two wires to a thin plate at the end of a 3-mm I.D.

plain carbon-steel tube with a wall thickness of 1 mm. A 2- cm length of the tube is inserted into the duct. The temperature of the tube where it enters the duct

cp 400 J/kg K=


is 25 oC and the air in the duct is at 70 o and 400 kPa. There is concern that due to conduction in the tube, the temperature at the end of the tube may be lower

than the actual temperature of the air in the duct.

(a) Estimate the temperature that the thermocouple would indicate

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(a) Estimate the temperature that the thermocouple would indicate.

(b) If the indicated temperature is significantly different from theactual air temperature, propose methods to ensure that the

indicated temperature is close to the actual temperature.

For Problems 10.23 through Problem 5.30 on flat plate collectors

with water as the working fluid, the specifications of the

collector are:

Collector length 2.4 m

Collector width 1.2 m

Water tube inside diameter 1 cm

Pitch of water tubes 20 cm

Orientation of water tubes Parallel to the

2.4-m

dimension

Ambient air temperature 15 oC

Figure P5.22

Multimode

Air space between absorber

and cover plates

5 cm

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5.23 The convective heat transfer coefficient associated with the surface of the cover plate exposed to the surrounding air is 6 W/m2 oC. The gray absorber plate has

an emissivity of 0.9. Determine the top heat loss coefficient for an absorber plate temperature of 60

o

C.

and cover plates

Solar transmissivity of glasscover plate

0.85

Infrared emissivity of glass

cover plate

0.88

Collector tilt angle with thehorizontal

45o

Solar insolation 800 W/m2


5.24 Re-solve Problem 5.23 for an absorber plate temperature of 60 oC if the air gap is 3 cm.

5.25 The absorber plate emissivity is 0.9 for 0 < < 2 m and 0.15 for 2 m < < . Compute theinfrared emissivity of the surface at 350 K and its solar absorptivity

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infrared emissivity of the surface at 350 K and its solar absorptivity.

5.26 The emissivity of the absorber plate is 0.9 for 0 < < 2 m, 0.7 for 2 m < < 5 m, and 0.15 for5 m < < . Determine the infrared emissivity of the surface at 350 K and its solar absorptivity.

5.27 The collector is operated in a location where the air temperature ( ) and the effective sky

temperature are 15 oC. Due to a failure of the circulating water pump, the water in the collector is

stagnant. The pressure of the water in the tubes is 100 kPa and the emissivity of the gray absorber plate

is 0.9. The convective heat transfer coefficient associated with the cover plate surface exposed to the

atmospheric air is 6 W/m2 oC. The heat loss through the back insulation is negligible. Determinewhether the water in the tube boils.

5.28 Re-solve Problem 5.27 if the absorber plate has a solar absorptivity of 0.85 and infrared emissivity of

0.1, and the water in the tubes is replaced by a mixture of water and ethylene glycol having a boiling

point of 150 oC.

5.29 The collector with a gray absorber plate ( = 0.9) is used to heat water. Determine the heat transfer rateto the circulating water if the absorber plate temperature is 60 oC and that of the surrounding air is 15 oC. With a gentle breeze the air velocity parallel to the

1.2-m dimension is 3 m/s.5.30 Re-solve Problem 5.29 if the absorber plate has a solar absorptivity of 0.9 and an infrared emissivity of 0.1.

Figure P10.23 - P10-30

Ta

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5.37 An electric heater is to be fabricated by sandwiching a heating element between two plates each 72-cm long and 36-cm wide. With the plates maintained at 40oC above the surrounding air temperature, the heater is to transfer 4000 W. The ambient air flow velocity over the heater should not to exceed 5 m/s. Aluminum

and copper rods of 3-mm diameter are available for use as extended surfaces. Rod spacing should not be less than 4dand weight and volume should be low.

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Propose a design.

5.38 A package has two sources of energy generation. Each source dissipates 200 W. The temperature of the sources may be assumed to be uniform and is to be

limited to 60 oC with ambient air at 20 oC flowing at 5 m/s over the plates connecting the sources used as extended surfaces Investigate if such a heater is

feasible and if so, design the system of extended surfaces. Weight is an important consideration.

5.39 Design a 110-V (or 220 V) 1500 W electric space heater for residential use. Identify the criteria or a successful design.

Figure P5.38

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Multi Mode