Chapter_7_2

# To perform the integral the pressure volume relation

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e + ∆pe 1 z 2 FG kJ IJ H kg K Turbine Since the fluid pressure drops as the fluid flows through the turbine, dP < 0, and the specific volume is always greater than zero, wrev, turbine > 0. To perform the integral, the pressure-volume relation must be known for the process. Compressor and Pump Since the fluid pressure rises as the fluid flows through the compressor or pump, dP > 0, and the specific volume is always greater than zero, wrev, in > 0, or work is supplied to the compressor or pump. To perform the integral, the pressure-volume relation must be known for the process. The term compressor is usually applied to the compression of a gas. The term pump is usually applied when increasing the pressure of a liquid. Chapter 7 - 55 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-56 Pumping an incompressible liquid For an incompressible liquid, the specific volume is approximately constant. Taking v approximately equal to v1, the specific volume of the liquid entering the pump, the work can be expressed as wrev , in = v dP + ∆ke + ∆pe 1 z 2 FG kJ IJ H kg K = v∆P + ∆ke + ∆pe For the steady-flow of an incompressible fluid through a device that involves no work interactions (such as nozzles or a pipe section), the work term is zero, and the equation above can be expressed as the well-know Bernoulli equation in fluid mechanics. v ( P2 − P1 ) + ∆ke + ∆pe = 0 Extra Assignment Using the above discussion, find the turbine and compressor work per unit mass flow for an ideal gas undergoing an isentropic process, where the pressure-volume relation is Pvk = constant, between two temperatures, T1 and T2. Compare your results with the first law analysis of Chapter 5 for control volumes. Chapter 7 - 56 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-57 Example 7-13 Saturated liquid water at 10 kPa leaves the condenser of a steam power plant and is pumped to the boiler pressure of 5 MPa. Calculate the work for an isentropic pumping process. a. From the above analysis, the work for the reversible process can be applied to the isentropic process (it is left for the student to show this is true) as WC = mv1 ( P2 − P1 ) Here at 10 kPa, v1 = vf = 0.001010 m3/kg. The work per unit mass flow is wC = WC = v1 ( P2 − P1 ) m m3 kJ = 0.001010 (5000 − 10) kPa 3 m kPa kg kJ = 5.04 kg −Wnet = m(h2 − h1 ) −(0 − WC ) = m(h2 − h1 ) b. Using the steam table data for the isentropic process, we have Chapter 7 - 57 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 7-58 From the saturation pressure table, kJ h1 = 191.81 P = 10 kPa kg 1 kJ Sat. Liquid s1 = 0.6492 kg ⋅ K Since the process is isentropic, s2 = s1. Interpolation in the compressed liquid tables gives kJ kJ h2 = 197.42 s2 = s1 = 0.6492 kg kg ⋅ K The work per unit mass flow is P2 = 5 MPa wC = WC = (h2 − h1 ) m kJ kg = (197.42 − 191.81) = 5.61 kJ kg The first method for finding the pump work is adequate for this case. Chapter 7 - 58 Student Study Guide for 5th editi...
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.

Ask a homework question - tutors are online