Air is expanded in a steady flow process through a turbine. Initial conditions are 1300°C and 2.0 MPa absolute. Final conditions are 500°C and atmospheric pressure. Show this process on a Ts diagram. Evaluate the changes in internal energy, enthalpy, and specific entropy for this process.

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Answer 1

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Chapter 12, Problem 1P

Air is expanded in a steady flow process through a turbine. Initial conditions are 1300°C and 2.0 MPa absolute. Final conditions are 500°C and atmospheric pressure. Show this process on a Ts diagram. Evaluate the changes in internal energy, enthalpy, and specific entropy for this process.

Expert Solution & Answer

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

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Answer 2

Textbook Question

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Chapter 12, Problem 1P

Air is expanded in a steady flow process through a turbine. Initial conditions are 1300°C and 2.0 MPa absolute. Final conditions are 500°C and atmospheric pressure. Show this process on a Ts diagram. Evaluate the changes in internal energy, enthalpy, and specific entropy for this process.

Expert Solution & Answer

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

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Answer 3

Textbook Question

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Chapter 12, Problem 1P

Air is expanded in a steady flow process through a turbine. Initial conditions are 1300°C and 2.0 MPa absolute. Final conditions are 500°C and atmospheric pressure. Show this process on a Ts diagram. Evaluate the changes in internal energy, enthalpy, and specific entropy for this process.

Expert Solution & Answer

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

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Answer 4

Textbook Question

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Chapter 12, Problem 1P

Air is expanded in a steady flow process through a turbine. Initial conditions are 1300°C and 2.0 MPa absolute. Final conditions are 500°C and atmospheric pressure. Show this process on a Ts diagram. Evaluate the changes in internal energy, enthalpy, and specific entropy for this process.

Expert Solution & Answer

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

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Answer 5

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Answer 6

Textbook Question

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Answer 7

Expert Solution & Answer

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

Answer 8

Expert Solution & Answer

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The change in internal energy, enthalpy, specific entropy and show this process on a T-s diagram.

Answer 12

Explanation of Solution

Given:

Initial temperature (T1) is 1300°C.

Initial pressure (P1) is 2 MPa.

Final temperature (T2) is 500°C.

Final pressure (P2) is 1 atm.

Calculation:

Determine the change in internal energy.

dU=U2−U1=CvdT=Cv(T2−T1)=0.718 kJ/kg⋅K(773 K−1573 K)=−573.6 kJ/kg

Thus, the change in internal energy is −573.6 kJ/kg_.

Determine the change in enthalpy.

dh=h2−h1=CpdT=Cp(T2−T1)=1.004 kJ/kg⋅K(773 K−1573 K)=−803.2 kJ/kg

Thus, the change in enthalpy is −803.2 kJ/kg_.

Determine the change in specific entropy.

ds=s2−s1=Cpln(T2T1)−Rln(P2P1)=1.004 kJ/kg⋅K×ln(773 K1573 K)−1.004 kJ/kg⋅K×ln(101.325 kPa2000 kPa)=0.143 kJ/kg⋅K

Thus, the change in specific entropy is 0.143 kJ/kg⋅K_.

Draw the T-s diagram as in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 12, Problem 1P

Answer 13

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Air is expanded in a steady flow process through a turbine. Initial conditions are 1300°C and 2.0 MPa absolute. Final conditions are 500°C and atmospheric pressure. Show this process on a Ts diagram. Evaluate the changes in internal energy, enthalpy, and specific entropy for this process.

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