Whether the prescribed temperature distribution possible or not along with an explanation. The temperature distribution is not possible as the conduction and convection heat transfer is not possible. Given: The given diagram is shown in Figure 1. Figure 1 The thermal conductivity of the wall is k = 4.5 W / m ⋅ K . The steady state temperature of the wall is T ∞ = 20 ° C . The thickness of the wall is L = 0.018 m . The convection heat transfer of the wall is h = 30 W / m 2 ⋅ K . Formula Used: The expression for the energy balance equation for the steady flow is given by, q c o n d − q c o n v = 0 The expression for the conduction heat transfer per unit area is given by, q c o n d = − k ( T L − T 0 ) L Here, T L is the thickness of the wall at complete thickness of the wall and T 0 is the temperature of the wall at 0 m . The expression for the convection heat transfer per unit area is given by, q c o n v = k ( T L − T ∞ ) Calculation: The conduction heat transfer per unit area is calculated as, q c o n d = − k ( T L − T ∞ ) L = − ( 4.5 W / m ⋅ K ) ( 120 ° C − 0 ° C ) 0.18 m = − 3000 W / m 2 The convection heat transfer per unit area is calculated as, q c o n v = k ( T L − T ∞ ) = 30 W / m 2 ⋅ K ( 120 ° C − 20 ° C ) = 3000 W / m 2 The energy balance equation is calculated as, q c o n d − q c o n v = 0 − 3000 W / m 2 − 3000 W / m 2 = 0 − 6000 W / m 2 ≠ 0 The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal. Conclusion: Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

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Answer 1

Question

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Expert Solution & Answer

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Answer 2

Question

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Expert Solution & Answer

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See solution

Answer 3

Question

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Answer 6

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Answer 7

Chapter 2, Problem 2.41P

(a)

To determine

Whether the prescribed temperature distribution possible or not along with an explanation.

The temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Given:

The given diagram is shown in Figure 1.

Figure 1

The thermal conductivity of the wall is k=4.5 W/m⋅K .

The steady state temperature of the wall is T∞=20°C .

The thickness of the wall is L=0.018 m .

The convection heat transfer of the wall is h=30 W/m2⋅K .

Formula Used:

The expression for the energy balance equation for the steady flow is given by,

qcond−qconv=0

The expression for the conduction heat transfer per unit area is given by,

qcond=−k(TL−T0)L

Here, TL is the thickness of the wall at complete thickness of the wall and T0 is the temperature of the wall at 0 m .

The expression for the convection heat transfer per unit area is given by,

qconv=k(TL−T∞)

Calculation:

The conduction heat transfer per unit area is calculated as,

qcond=−k( T L − T ∞ )L=−(4.5 W/m⋅K)( 120°C−0°C)0.18 m=−3000 W/m2

The convection heat transfer per unit area is calculated as,

qconv=k(TL−T∞)=30 W/m2⋅K(120°C−20°C)=3000 W/m2

The energy balance equation is calculated as,

qcond−qconv=0−3000 W/m2−3000 W/m2=0−6000 W/m2≠0

The temperature distribution is not possible as the value of conduction and convection heat transfer is not equal.

Conclusion:

Therefore, the temperature distribution is not possible as the conduction and convection heat transfer is not possible.

Answer 8

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Answer 9

(b)

To determine

The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

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Answer 13

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Solution Summary: The author explains that the temperature distribution is not possible as the conduction and convection heat transfer are not equal.

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The computation and the plot of the temperature at x=L , T(L) as a function of h for 10≤h≤100 W/m2⋅K along with an explanation of the results.

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