Explanation Draw the T − s diagram for the Brayton cycle with regeneration as shown in Figure (1). Write the pressure ratio relation for the process 1-2. P r 2 = P 2 P 1 P r 1 (I) Here, relative pressure at state 1 is P r 1 and relative pressure at state 2 is P r 2 . Write the pressure ratio relation for the process 3-4. P r 4 = P 4 P 3 P r 3 (II) Here, relative pressure at state 3 is P r 3 and relative pressure at state 4 is P r 4 . Write the expression for the efficiency of the compressor ( η C ) to calculate h 2 . η C = h 2 s − h 1 h 2 − h 1 h 2 = h 1 + h 2 s − h 1 η C (III) Here, enthalpy on isentropic condition 2 s is h 2 s . Write the expression for the efficiency of the turbine ( η T ) to calculate h 4 . η T = h 3 − h 4 h 3 − h 4 s h 4 = h 3 − η T ( h 3 − h 4 s ) (IV) Here, enthalpy on isentropic condition 4 s is h 4 s . Write the expression for the heat rejected by the regenerator ( q regen ) . q regen = ε ( h 4 − h 2 ) (V) Here, the effectiveness of the regenerator is ε and enthalpy at state 4 is h 4 . Write the expression to calculate the heat input to the gas turbine ( q in ) . q in = ( h 3 − h 2 ) − q regen (VI) Write the expression to calculate the net work output by the gas turbine ( w net,out ) . w net,out = w T,out − w C,in = ( h 3 − h 4 ) − ( h 2 − h 1 ) (VII) Here, the work done by the turbine is w T,out , and the work input to the compressor is w C,in . Write the thermal efficiency of the given turbine ( η t h ) as η t h = w net,out q in (VIII) Write the expression to calculate mass flow rate m ˙ . m ˙ = W ˙ net w net,out (IX) Here, mass flow rate is m ˙ , net work output is w net,out . Conclusion: From Table A-17, “Ideal-gas properties of air”, obtain the following properties at the temperature of 303 K . h 1 = 121.88 Btu / lbm P r 1 = 1.1369 Here, enthalpy at state 1 is h 1 and relative pressure at state 1 is P r 1 . From Table A-17, “Ideal-gas properties of air”, obtain the following properties at the temperature of 1561 K . h 3 = 549.35 Btu / lbm P r 3 = 230.12 Here, enthalpy at state 3 is h 3 and relative pressure at state 3 is P r 3 . Substitute 4 for P 2 P 1 , and 1.1369 for P r 1 in Equation (I). P r 2 = ( 4 ) ( 1.1369 ) = 4.548 From the Table A-17E, “Ideal-gas properties of air”. Obtain the value of enthalpy on isentropic state ( h 2 s ) at the relative pressure ( P r 2 ) of 4.548 by using interpolation method. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (X) Here, the variables denoted by x and y are relative pressure and enthalpy. Show relative pressure and enthalpy values from the Table A-17E. Relative pressure ( P r 2 ) Enthalpy ( h 2 ) , in Btu / lbm 4.193 177.23 4.548 ? 4.607 182.08 Substitute 4.193 for x 1 , 4.548 for x 2 , 4.607 for x 3 , 177.23 for y 1 , and 182.08 for y 3 in Equation (X). y 2 = ( 4.548 − 4.193 ) ( 182.08 − 177.23 ) ( 4.607 − 4.193 ) + 177.23 = 181.388 = 181.39 The value of enthalpy on isentropic state ( h 2 s ) at the relative pressure ( P r 2 ) of 4

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

See similar textbooks

Concept explainers

Laws of thermodynamics

Thermodynamics is a field of research, which needs the creation of absolute and ideal states and logical and practical examination of the actual conditions. By considering all the laws and theories of thermodynamics, the chemical reactions, the occurrenc…

Videos

Question

Chapter 9.12, Problem 103P

To determine

The thermal efficiency of the car and the mass flow rate of air for a net power output of 130 hp.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.12, Problem 102P

Chapter 9.12, Problem 104P

Students have asked these similar questions

No use chatgpt

Problem 6 (Optional, extra 6 points) 150 mm 150 mm 120 mm 80 mm 60 mm PROBLEM 18.103 A 2.5 kg homogeneous disk of radius 80 mm rotates with an angular velocity ₁ with respect to arm ABC, which is welded to a shaft DCE rotating as shown at the constant rate w212 rad/s. Friction in the bearing at A causes ₁ to decrease at the rate of 15 rad/s². Determine the dynamic reactions at D and E at a time when ₁ has decreased to 50 rad/s. Answer: 5=-22.01 +26.8} N E=-21.2-5.20Ĵ N

Problem 1. Two uniform rods AB and CE, each of weight 3 lb and length 2 ft, are welded to each other at their midpoints. Knowing that this assembly has an angular velocity of constant magnitude c = 12 rad/s, determine: (1). the magnitude and direction of the angular momentum HD of the assembly about D. (2). the dynamic reactions (ignore mg) at the bearings at A and B. 9 in. 3 in. 03 9 in. 3 in. Answers: HD = 0.162 i +0.184 j slug-ft²/s HG = 2.21 k Ay =-1.1 lb; Az = 0; By = 1.1 lb; B₂ = 0.

Chapter 9 Solutions

Thermodynamics: An Engineering Approach

Ch. 9.12 - What are the air-standard assumptions?Ch. 9.12 - What is the difference between air-standard...Ch. 9.12 - How does the thermal efficiency of an ideal cycle,...Ch. 9.12 - What does the area enclosed by the cycle represent...Ch. 9.12 - Prob. 5P Ch. 9.12 - Prob. 6P Ch. 9.12 - Can the mean effective pressure of an automobile...Ch. 9.12 - Prob. 8P Ch. 9.12 - What is the difference between spark-ignition and...Ch. 9.12 - Prob. 10P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The thermal efficiency of the car and the mass flow rate of air for a net power output of 130 hp .

Solution Summary: The author analyzes the thermal efficiency of the car and the mass flow rate of air for a net power output of 130%.

Concept explainers

Videos

The thermal efficiency of the car and the mass flow rate of air for a net power output of 130 hp.

Want to see the full answer?

Chapter 9 Solutions