ENGINEERING FUNDAMENTALS
ENGINEERING FUNDAMENTALS
6th Edition
ISBN: 9781337705011
Author: MOAVENI
Publisher: CENGAGE L
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Chapter 11, Problem 28P
To determine

Calculate the reduction in heat transfer rate.

Expert Solution & Answer
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Answer to Problem 28P

The reduction in heat transfer rate is 97%.

Explanation of Solution

Given data:

Area of the concrete wall, A=1000ft2.

Thickness of the concrete wall, L1=6in.

Thickness of the insulation batt, L2=2in.

Inside temperature of concrete wall, T1=20C.

Outside temperature of concrete wall, T2=0C.

From Table 11.3 in the textbook, the thermal conductivity of the concrete wall is, k1=1.4WmK.

The thermal conductivity of the insulation batt is, k2=0.03Btuhft°F.

Formula used:

The formula for the heat transfer is,

q=kAT1T2L (1)

Here,

k is the thermal conductivity,

A is the area of cross section,

L is the thickness of the material,

T1 is the inside temperature of the material, and

T2 is the outside temperature of the material,

Calculation:

Convert the unit of area of cross section into meter.

A=1000ft2                              [1ft=0.3048m]=(1000)(0.3048 m)2=(1000)(0.0929)m2A=92.9m2

Convert the unit of thickness into meter for L1.

L1=6in                          [1in=0.0254m]=(6in)(0.0254m1in)L1=0.1524m

Convert the unit of thickness into meter for L2,

L2=2in=(2in)(0.0254m1in)L2=0.0508m

Substitute 1.4WmK for k1, 92.9m2 for A, 20C for T1, 0C for T2, and 0.1524m for L in equation (1) to find heat transfer rate of concrete wall without insulation q1.

q1=(1.4WmK)(92.9m2)((20C0C)0.1524m)q1=(1.4WmK)(92.9m2)(20C0.1524m) (2)

From the truth the temperature difference in C is equals to the temperature difference in K.

Substitute 20K for 20C in equation (2),

q1=(1.4WmK)(92.9m2)(20K0.1524m)q1=17070W

Change the unit of thermal conductivity for insulation batt,

k2=0.03Btuhft°F                           [1W=3.4123Btuh]=(0.03Btuhft°F)(1W3.4123Btuh)k2=8.792×103Wft°F

For the temperature difference (1°F=1°R). Therefore,

k2=8.792×103Wft°R           [1K=59°R]=(8.792×103Wft°R)(59°R1K)          [1m=3.28ft]=(4.885×103WftK)(3.28ft1m)k2=0.016WmK

Rearrange equation (1), we get

q=(T1T2)A(Lk) (3)

Here,

Lk is the thermal insulation.

From equation (3), the expression for the heat transfer rate of the concrete wall with insulation batt is,

q2=(T1T2)A(L1k1)+(T1T2)A(L2k2)q2=(T1T2)A[1L1k1+L2k2] (4)

Substitute 1.4WmK for k1, 0.016WmK for k2, 92.9m2 for A, 20C for T1, 0C for T2, 0.1524m for L1, and 0.0508m for L2 in equation (4) to find q2.

q2=(20C0C)(92.9m2)[10.1524m1.4WmK+0.0508m0.016WmK]=(20K)(92.9m2)[10.1524m1.4WmK+0.0508m0.016WmK]=(20K)(92.9m2)(0.3045Wm2K)q2=566W

Hence the reduction in heat transfer is,

%reductioninq=q1q2q1×100 (5)

Substitute 17070W for q1 and 566W for q2 in equation (5) to find % reduction in q,

%reductioninq=(17070W)(566W)(17070W)×100=0.97×100=97%

Therefore, the reduction in heat transfer rate is 97%.

Conclusion:

Hence, the reduction in heat transfer rate is 97%.

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Chapter 11 Solutions

ENGINEERING FUNDAMENTALS

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