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.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

100%

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

A vessel contains 10 kg of water at 10 x last digit kPa (if last digit is 2 then 10 x 2 = 20 kPa). Determine specific and total internal energy, specific and total enthalpy, and specific and total entropy. Consider different cases/states: I. T = Tsat and x = 0II. T = Tsat and x = 0.5III. T = Tsat and x = 1IV. T = 300 ⁰C (Plot these states on the Tv diagram and label magnitudes. Refer property tables.) very important.

(B) 2.4 kg of air at a temperature of 523K and a pressure of 650 kPa expand according to the processes (a) isothermal process , (b) polytropic process of index n=1.2 , (c) adiabatic process, to a pressure of 100 kPa. Find the change in entropy in each case. Take y = 1.41 and cy =0.708 kJ/kg.K

Five cubic meter of nitrogen are heated at constant volume so that the pressure changes from 276 kPaa to 1375 kPa gage. For nitrogen, R = 0.296 kJ/kg-K and Cp = 1.039 kJ/kg-K. Find the change of entropy.

Answer 2

Textbook Question

100%

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

100%

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

100%

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

100%

Answer 6

Textbook Question

100%

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

Ch. 12 - Air is expanded in a steady flow process through a...Ch. 12 - Five kilograms of air is cooled in a closed tank...Ch. 12 - Air is contained in a piston-cylinder device. The...Ch. 12 - Calculate the power delivered by the turbine per...Ch. 12 - If hydrogen flows as a perfect gas without...Ch. 12 - A 1-m3 tank contains air at 0.1 MPa absolute and...Ch. 12 - Air enters a turbine in steady flow at 0.5 kg/s...Ch. 12 - Natural gas, with the thermodynamic properties of...Ch. 12 - Carbon dioxide flows at a speed of 10 m/s in a...Ch. 12 - In an isothermal process, 0.1 cubic feet of...

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.

Concept explainers

Videos

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

Additional Engineering Textbook Solutions