Change in volume of the system at given pressure and temperature. Concept introduction: Thermodynamic property tables: Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition. The change in volume ( Δ V ) . Δ V = M ( V ^ 2 − V ^ 1 ) Here, the mass of steam is M , volume of the final state is V ^ 2 and volume of the initial state is V ^ 1 .

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Answer 1

Question

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (2)

Substitute 100 kg for M, 0.001398 m3/kg for V^1, and 0.523 m3/kg for V^2 in Equation (2).

ΔV=M(V^2−V^1)=100 kg(0.523 m3/kg−0.001398 m3/kg)=52.16 m3

Thus, the change in volume (ΔV) is 52.16 m3_.

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Answer 2

Question

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (2)

Substitute 100 kg for M, 0.001398 m3/kg for V^1, and 0.523 m3/kg for V^2 in Equation (2).

ΔV=M(V^2−V^1)=100 kg(0.523 m3/kg−0.001398 m3/kg)=52.16 m3

Thus, the change in volume (ΔV) is 52.16 m3_.

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Answer 3

Question

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (2)

Substitute 100 kg for M, 0.001398 m3/kg for V^1, and 0.523 m3/kg for V^2 in Equation (2).

ΔV=M(V^2−V^1)=100 kg(0.523 m3/kg−0.001398 m3/kg)=52.16 m3

Thus, the change in volume (ΔV) is 52.16 m3_.

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Answer 4

Question

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (2)

Substitute 100 kg for M, 0.001398 m3/kg for V^1, and 0.523 m3/kg for V^2 in Equation (2).

ΔV=M(V^2−V^1)=100 kg(0.523 m3/kg−0.001398 m3/kg)=52.16 m3

Thus, the change in volume (ΔV) is 52.16 m3_.

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Answer 5

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (2)

Substitute 100 kg for M, 0.001398 m3/kg for V^1, and 0.523 m3/kg for V^2 in Equation (2).

ΔV=M(V^2−V^1)=100 kg(0.523 m3/kg−0.001398 m3/kg)=52.16 m3

Thus, the change in volume (ΔV) is 52.16 m3_.

Answer 6

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

Answer 7

Chapter 1.6, Problem 9E

A)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

Answer 8

A)

Expert Solution

Explanation of Solution

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at initial pressure and temperature as 2 bar and 200°C respectively.

The specific volume for initial state is V^1@200°C=1.080 m3/kg

Refer the Appendix A-3, “Superheated steam table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 350°C respectively.

The specific volume for final state is V^2@350°C=0.570 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (1)

Here, the mass of steam is M.

Substitute 50 kg for M, 1.080 m3/kg for V^1, and 0.570 m3/kg for V^2 in Equation (1).

ΔV=50 kg(0.570 m3/kg−1.080 m3/kg)=−25.5 m3

Here, negative sign indicates that the volume is leaving the system.

Thus, the change in volume (ΔV) is −25.5 m3_.

Answer 9

A)

Expert Solution

Answer 10

A)

Expert Solution

Answer 11

Expert Solution

Answer 12

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.

Calculate the change in volume (ΔV).

ΔV=M(V^2−V^1) (2)

Substitute 100 kg for M, 0.001398 m3/kg for V^1, and 0.523 m3/kg for V^2 in Equation (2).

ΔV=M(V^2−V^1)=100 kg(0.523 m3/kg−0.001398 m3/kg)=52.16 m3

Thus, the change in volume (ΔV) is 52.16 m3_.

Answer 13

B)

Interpretation Introduction

Interpretation:

Change in volume of the system at given pressure and temperature.

Concept introduction:

Thermodynamic property tables:

Thermodynamic property tables is provide the data about pressure, temperature, , specific enthalpy and entropy and specific volume at particular saturation pressure or saturation temperature or both are combined in superheat condition.

The change in volume (ΔV).

ΔV=M(V^2−V^1)

Here, the mass of steam is M, volume of the final state is V^2 and volume of the initial state is V^1.

Answer 14

B)

Expert Solution

Explanation of Solution

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at initial pressure and temperature as 100 bar and 300°C respectively.

The volume at the initial state is V^1@300°C=0.001398 m3/kg.

Refer the Appendix A-4, “Compressed Liquid table”, obtain the following properties of the steam at final pressure and temperature as 5 bar and 300°C respectively.

The volume at the final state is V^2@300°C=0.523 m3/kg.