[PDF] Question Known: Find: Assumptions: PDF

[PDF] Question Known: Find: Assumptions:

Heat transfer rate through the roof Assumptions: Steady state condition Inner wall temperature is constant throughout the length

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[PDF] Question Known: Find: Assumptions:

Each term is a function of the surface temperature Radiation The heat transfer due to radiation was given in the problem: qrad = qradA = A (375 − εσT4

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127940_3tutorial9_s16.pdf Forced Convection ECE309 Introduction to Thermodynamics and Heat TransferTutorial 9

Question

A truck, pulling a refrigerated trailer, travels at 105 km/hr. Solar radiation heats the roof of the truck, with the net radiatvie flux being q ??rad = 375-εσT 4o ?W m 2 ? whereε=0.5 andT o is the temperature of the roof. The roof is composed of urethane foam (t 2 =50mm,k foam = 0 .

026W/m·K) that is sandwiched between two sheets of

aluminum ( t 1 =5mm,k al = 180W/m·K) and the cool- ing system of the trailer keeps the inner surface of the roof at a temperature of-10 ◦

C. The air temperature

isT ∞ =20 ◦

C. If the length of the trailer is 10m and

the area of the roof is 35 m 2 , find the heat transfer rate through the roof. Assume that the surface of the roof is isothermal. 20 C

105km/h°L

Radiation

t1=5mm t2=50mm t3=5mmAluminum

Aluminum

Urethane foam

Known:

?Trailer roof ?Length,L=10m ?Area,A=35m 2 ?Aluminum conductivity,k al = 180W/m·K ?Urethane foam,k foam =0.026W/m·K ?Inner wall temperature,T i =-10 ◦

C= 263K

?Wall thickness ?Outer aluminum,t 1 =510 - 3 m ?Urethane foam,t 2 =5010 - 3 m ?Inner aluminum,t 3 =510 - 3 m ?Radiation q ??rad = 375-εσT 4o ?W m 2 ?

ε=0.5

?Air ?Velocity,u ∞ = 105 km/h=29.167m/s ?Temperature,T ∞ =20 ◦

C= 293K

Find: ?Heat transfer rate through the roof

Assumptions:

?Steady state condition ?Inner wall temperature is constant throughout the length ?Outer wall temperature is constant throughout the length 1 Forced Convection ECE309 Introduction to Thermodynamics and Heat TransferTutorial 9

Schematic:

?The control volume is the top surface of the roof ? ?T o T i ?q cond ?????q rad ???q conv ?T ∞ u ∞ ??L

Analysis:

An energy balance on the roof surface results in:

q rad +q conv =q cond

The heat tranfer through the roof isq

cond . Each term is a function of the surface temperature.

Radiation

The heat transfer due to radiation was given in the problem: q rad =q ??rad A =A?375-εσT 4o ?, whereσ=5.6710 - 8 W/m 2 ·K 4 andε=0.5.

Conduction

The heat transfer through the wall can be found using a resistance network: ? T o ??? ?????? ???????? ?????? ???????? ?????? ?????T i ?q cond R al,1 R foam R al,2 q cond =T o -T i R eq , R eq =R al,1 +R foam +R al,2 . 2 Forced Convection ECE309 Introduction to Thermodynamics and Heat TransferTutorial 9

The individual resistance are

R al,1 =t 1 k al A = 510
- 3 m (180W/m·K)(35m 2 ) =7.936510 - 7 K/W R foam =t 2 k foam A = 5010
- 3 m (0 .

026W/m·K)(35m

2 ) =0.054945K/W R al,2 =R al,1 =7.936510 - 7 K/W

The equivalent resistance becomes:

R eq =R al,1 +R foam +R al,2 =7.936510 - 7

K/W+0.054945K/W+7.936510

- 7 K/W =0.054947K/W

Convection

The plate is assumed to be constant temperature; thus, the heat transfer due to convection is q conv =¯hA(T infty -T o ) , where

¯his the overall heat transfer coefficient. The heat transfer coefficient depends on the fluid velocity, length

of the plate, and the properties of the air. The properties of the air must be found at the film temperature, which

is defined as T f =T o +T∞ 2.

However,T

o is not known. Therefore, the solution will be iterative and an initial guess forT f is required. A good first estimate is assume thatT f ≂=T ∞ = 293 K. From Table A-19 (assuming that the air is at 1atmpressure),

ν=1.5110

- 5 m 2 /s, k air =0.02554W/m·K,

Pr=0.714.

Using these values, the Reynolds number for the plate can be calculated: Re L =u ∞ L

ν=(29.167m/s)(10m)1.5110

- 5 m 2 /s=1.9310 7

SinceRe

L >510 5 , theNu L number correlation must be valid for turbulent flow. Equation (10-27) is valid for 510 5 ?Re L ?10 7 ,0.6?Pr?60 : ¯ Nu L =? 0 . 037
Re 4 / 5 L -871? Pr 1 / 3 .

For the assumedT

f = 293 ,Nu L = 21514.4.The convection coefficient becomes ¯ h= Nu L k air L = (21514 .

4)(0.02554W/m·K)

10 m =54.95W/m 2 ·K 3 Forced Convection ECE309 Introduction to Thermodynamics and Heat TransferTutorial 9

Final Solution

The energy balance on the surface of the roof can be expressed as A ?375-εσT 4o ?+¯hA(T ∞ -T o )-T o -T i R eq =0,

13125-(9.92510

- 7 ) T 4o +5.635110 5 -1923.25T o -18.199T o + 4786.3=0.

Solving the equation results inT

o = 295 .

58. The¯hvalue was calculated assuming thatT

o =T ∞ = 293 K; therefore,¯hshould be recalculated using a new film temperature, T f =293 + 295.58

2= 294.3K

Get the properties from Table A-19 again,

ν=1.5310

- 5 m 2 /s, k air =0.02574W/m·K,

Pr=0.713.

The iterations are summarized in the table below.

it#T f [ K]Re L Pr Nu L ¯ h [W/m2

·K]T

o [ K]

1 293 1

. 9310
7

0.714 21514.4 54.95 295.6

2 294.3 1

. 9110
7

0.713 21271.0 54.75 295.6

Thus, the surface temperature isT

o = 295 . 6 K. The heat transfer rate through the wall becomes q cond =T o -T i R eq =295.6K-263K 0 .

05494K/W

= 593

[PDF] Question Known: Find: Assumptions:

Question

026W/m·K) that is sandwiched between two sheets of

C. The air temperature

C. If the length of the trailer is 10m and

105km/h°L

Radiation

Aluminum

Urethane foam

Known:

C= 263K

ε=0.5

C= 293K

Assumptions:

Schematic:

Analysis:

An energy balance on the roof surface results in:

The heat tranfer through the roof isq

Radiation

Conduction

The individual resistance are

026W/m·K)(35m

The equivalent resistance becomes:

K/W+0.054945K/W+7.936510

Convection

However,T

ν=1.5110

Pr=0.714.

ν=(29.167m/s)(10m)1.5110

SinceRe

For the assumedT

4)(0.02554W/m·K)

Final Solution

13125-(9.92510

Solving the equation results inT

58. The¯hvalue was calculated assuming thatT

2= 294.3K

Get the properties from Table A-19 again,

ν=1.5310

Pr=0.713.

The iterations are summarized in the table below.

·K]T

1 293 1

0.714 21514.4 54.95 295.6

2 294.3 1

0.713 21271.0 54.75 295.6

Thus, the surface temperature isT

05494K/W

W←-Answer

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