Calculate the R-value for the following materials:
a. 4 in. thick brick
b. 10 cm thick brick
c. 12 in. thick concrete slab
d. 20 cm thick concrete slab
e. 1 cm thick human fat layer
(a)
Find the R-value for 4 in thick brick.
Answer to Problem 9P
The R-value for 4 in thick brick is 0.10 m2⋅KW.
Explanation of Solution
Given data:
The thickness of the brick, L=4 in.
From table 11.3 in the textbook, the thermal conductivity of the brick, k=1.0 Wm⋅k.
Formula used:
The formula to find R-value is,
R=Lk (1)
Here,
L is the thickness of the material, and
k is the thermal conductivity of the material.
Calculation:
Substitute 4 in for L, and 1.0 Wm⋅k for k in equation (1) to find R,
R=4 in1.0 Wm⋅k [∴ 1 in=0.0254 m]=(4 in)(0.0254 m1 in)1.0 Wm⋅k=(4×0.0254) m1.0 Wm⋅kR=0.10 m2⋅KW
Therefore, R-value for 4 in thick brick is 0.10 m2⋅KW.
Conclusion:
Hence, the R-value for 4 in thick brick is 0.10 m2⋅KW.
(b)
Find the R-value for 10 cm thick brick.
Answer to Problem 9P
The R-value for 10 cm thick brick is 0.10 m2⋅KW.
Explanation of Solution
Given data:
The thickness of the brick, L=10 cm.
Calculation:
Substitute 10 cm for L, and 1.0 Wm⋅k for k in equation (1) to find R,
R=10 cm1.0 Wm⋅k [∴ 1 m=100 cm]=(10 cm)(1 m100 cm)1.0 Wm⋅k=0.1 m1.0 Wm⋅kR=0.10 m2⋅KW
Therefore, R-value for 10 cm thick brick is 0.10 m2⋅KW.
Conclusion:
Hence, the R-value for 10 cm thick brick is 0.10 m2⋅KW.
(c)
Find the R-value for 12 in thick concrete slab.
Answer to Problem 9P
The R-value for 12 in thick concrete slab is 0.22 m2⋅KW.
Explanation of Solution
Given data:
The thickness of the concrete slab, L=12 in.
From Table 11.3 in the textbook, the thermal conductivity of the concrete slab, k=1.4 Wm⋅k.
Calculation:
Substitute 12 in for L, and 1.4 Wm⋅k for k in equation (1) to find R,
R=12 in1.4 Wm⋅k [∴1 in=0.0254 m]=(12 in)(0.0254 m1 in)1.4 Wm⋅k=(12×0.0254) m1.4 Wm⋅kR=0.22 m2⋅KW
Therefore, R-value for 12 in thick concrete slab is 0.22 m2⋅KW.
Conclusion:
Hence, the R-value for 12 in thick concrete slab is 0.22 m2⋅KW.
(d)
Find the R-value for 20 cm thick concrete slab.
Answer to Problem 9P
The R-value for 20 cm thick concrete slab is 0.14 m2⋅KW.
Explanation of Solution
Given data:
The thickness of the concrete slab, L=20 cm.
Calculation:
Substitute 20 cm for L, and 1.4 Wm⋅k for k in equation (1) to find R,
R=20 cm1.4 Wm⋅k [∴ 1 m=100 cm]=20 cm×1 m100 cm1.4 Wm⋅k=0.2 m1.4 Wm⋅kR=0.14 m2⋅KW
Therefore, R-value for 20 cm thick concrete slab is 0.14 m2⋅KW.
Conclusion:
Hence, the R-value for 20 cm thick concrete slab is 0.14 m2⋅KW.
(e)
Find the R-value for 1 cm thick human fat layer.
Answer to Problem 9P
The R-value for 1 cm thick human fat layer is 0.05 m2⋅KW.
Explanation of Solution
Given data:
The thickness of the human fat layer, L=1 cm.
From table 11.3 in the textbook, the thermal conductivity of the human fat layer, k=0.2 Wm⋅k.
Calculation:
Substitute 1 cm for L, and 0.2 Wm⋅k for k in equation (1) to find R,
R=1 cm0.2 Wm⋅k [∴ 1 m=100 cm]=(1 cm)(1 m100 cm)0.2 Wm⋅k=0.01 m0.2 Wm⋅kR=0.05 m2⋅KW
Therefore, R-value for 1 cm thick human fat layer is 0.05 m2⋅KW.
Conclusion:
Hence, the R-value for 1 cm thick human fat layer is 0.05 m2⋅KW.
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Chapter 11 Solutions
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