The equivalent thickness of air layers in millimeters is to be calculated. Concept Introduction: The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow. Q = − k A ( T 2 − T 1 ) x ..... (1) Here, x is the thickness of the material, T 1 is the temperature of the hot side of the wall, T 2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material. For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become: Q = − A ( T 2 − T 1 ) ∑ i N x i k i ..... (2)

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Answer 1

Question

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

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Answer 2

Question

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

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Answer 3

Question

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

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Answer 4

Question

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

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Answer 5

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

Answer 6

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

Answer 7

Chapter 10, Problem 10.1P

Interpretation Introduction

Interpretation:

The equivalent thickness of air layers in millimeters is to be calculated.

Concept Introduction:

The law of heat conduction states that the heat transfer rate through any material is proportional directly to the negative of the temperature gradient and the area of the heat flow.

Q=−kA(T2−T1)x..... (1)

Here, x is the thickness of the material, T1 is the temperature of the hot side of the wall, T2 is the temperature of the cold side of the wall, A is the area of the heat conduction, and k is the thermal conductivity of the material.

For N materials present in series in a wall, the rate of heat transfer is taken same throughout the wall. The heat transfer equation will now become:

Q=−A(T2−T1)∑iNxiki..... (2)

Answer 8

Expert Solution & Answer

Answer to Problem 10.1P

The air layer thickness is calculated as, xair=298 mm .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

Thickness of the refractory fireclay brick, xA=200 mm

Thickness of the kaolin brick, xB=100 mm

Thickness of the steel pipe, xC=6 mm

Thin layers of air are between brick and steel.

Fire side refractory temperature, T1=1150∘C

Temperature of the outside of steel side, T2=30∘C

Heat loss from the wall per meter square of the wall, Q=300 Wm2

For the appendix 11, take the thermal conductivity of fireclay brick, kaolin brick, steel plate, and air as:

kA=1.7 Wm⋅∘CkB=0.11 Wm⋅∘CkC=45.34Wm⋅∘Ckair=0.11Wm⋅∘C

Let the thickness of air layer be xair . Use equation (2) for 1 m² of wall area to calculate the air layer thickness as:

Q=−(T2−T1)xAkA+xBkB+xCkC+xairkair300 Wm2=−(30∘C−1150∘C)200 mm×m1000 mm1.7 Wm⋅∘C+100 mm×m1000 mm0.11 Wm⋅∘C+6 mm×m1000 mm45.34Wm⋅∘C+xair0.11Wm⋅∘Cxair=0.298 mxair=298 mm

Answer 12

Ch. 10 - Prob. 10.1P Ch. 10 - Prob. 10.2P Ch. 10 - Prob. 10.3P Ch. 10 - Prob. 10.4P Ch. 10 - Prob. 10.9P Ch. 10 - Prob. 10.10P Ch. 10 - Prob. 10.11P Ch. 10 - Prob. 10.12P Ch. 10 - Prob. 10.13P Ch. 10 - Prob. 10.14P

Answer to Problem 10.1P

Explanation of Solution

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Chapter 10 Solutions