This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Pa. Chapter 7 - 62 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-63 From the steam tables at state 1 kJ h1 = 3698.6 P = 1 MPa kg 1 kJ T1 = 600o C s1 = 8.0311 kg ⋅ K At the end of the isentropic expansion process, see Example 7-6. kJ h2 s = 2545.6 kg kJ s2 s = s1 = 8.0311 kg ⋅ K x2 s = 0.984 The actual turbine work per unit mass flow is (see Example 7-6) P2 = 0.01 MPa wa = h1 − h2 a h2 a = h1 − wa kJ = (3698.6 − 1037.7) kg kJ = 2660.9 kg
For the actual turbine exit state 2a, the computer software gives P2 = 0.01 MPa h2 a URT = 86.85 C | kJ VS = 2660.9 |TSuperheated kg W
Chapter 7 - 63 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-64 A second method for finding the actual state 2 comes directly from the expression for the turbine isentropic efficiency. Solve for h2a. h2 a = h1 − ηT ( h1 − h2 s ) kJ kJ = 3698.6 − (0.9)(3698.6 − 2545.6) kg kg kJ = 2660.9 kg
Then P2 and h2a give T2a = 86.85oC. Example 7-15 Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27oC, to a final state of 0.5 MPa. Find the work done on the air for a compressor isentropic efficiency of 80 percent. System: The compressor control volume
T1 P1 WC Compressor or pump T2 P2 Property Relation: Ideal gas equations, assume constant properties.
Chapter 7 - 64 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-65 T 2s 2a P2 1 P1 s Process and Process Diagram: First, assume isentropic, steady-flow and then apply the compressor isentropic efficiency to find the actual work. Conservation Principles: For the isentropic case, Qnet = 0. Assuming steady-state, steady-flow, and neglecting changes in kinetic and potential energies for one entrance, one exit, the first law is Ein = E out m1h1 + WCs = m2 h2 s
The conservation of mass gives m1 = m2 = m
The conservation of energy reduces to Chapter 7 - 65 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-66 WCs = m(h2 s − h1 ) wCs = WCs = (h2 s − h1 ) m Using the ideal gas assumption with constant specific heats, the isentropic work per unit mass flow is wCs = C p (T2 s − T1 )
The isentropic temperature at state 2 is found from the isentropic relation T2 s F 0.5 MPa IJ = (27 + 273) K G H 01MPa K .
= 475.4 K
The conservation of energy becomes FG P IJ =T HPK
1 2 1 ( k −1)/ k (1.4 −1)/1.4 wCs = C p (T2 s − T1 ) kJ (475.4 − 300) K kg ⋅ K kJ = 176.0 kg . = 1005
Chapter 7 - 66 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-67 The compressor isentropic efficiency is defined as ηC =
wCa = ws wa ηC wcs kJ kJ kg = 220 kg 0.8 176 =
Example 7-16 Nitrogen expands in a nozzle from a temperature of 500 K while its pressure decreases by factor of two. What is the exit velocity of the nitrogen when the nozzle isentropic efficiency is 95 percent? System: The nozzle control volume. T
T1 P1 V1 Nozzle T2 P2 V2 1 P1 2s 2a P2 s Property Relation: The ideal gas e...
View Full Document
This note was uploaded on 09/17/2009 for the course MAE 301 taught by Professor Hassan during the Fall '08 term at N.C. State.
- Fall '08