Question

Answer: 14.8 kW, Explain?

Liquid water is to be compressed by a pump whose isentropic efficiency is 81 percent from 0.2 MPa to 5 MPa at a rate of 0.15 m3/min. What is the required power input to this pump? Solve using appropriate software. Multiple Choice 12.0 kW 14.8 kW 8.5 kW 15.3 kW 10.2 kW

Answer 1

Given P 0.2 MPa = 200 kPa P2 = 5 MPa = 5000 kPa Flow rate, Q = 0.15 m3/min = (0.15/60) m3/s = 0.0025 m3/s Isentropic efficiency, Nis = 81 % = 0.81 Applying the first law of thermodynamics (energy conservation) to the compression process 0 + (, ti + 92) = n( n + i + 922) + (Note: in the above equation heat transfer to the system and work done by the system is considered positive) Neglecting the heat transfer and kinetic - potential energy changes mh = mh, + w Or W = -m(hy-hy) = -mAh The above expression will return a negative value for compressor (since, in a compressor the work is done on the system. The isentropic power input to the pump can be written as Wis = mAhis (Note: in the above equation we have omitted the sign; we are only considering the magnitude of power input) Now from the thermodynamic property relations dh = Tds + vdP Where, v is the specific volume

For an isentropic process ds = 0. Therefore dhis = vdP Expressing this in finite difference form Ahis = v(P2 - P.) Substituting this Wis = mv (P2 - P) Or Wis (Q)()<P> (P2 - P) = (P2 - P) = Q(P2 - P) Where we have used the relations 1 V = m = pQ Substituting all the known values, power input for an isentropic pump corresponding to the given pressure change can be calculated as Wis = 0.0025(5000 – 200) = 12 kW Now from the definition nis = Wis Wact Where, act is the actual power input to the compressor. On rearranging Wact = Wis mis Substituting the known values

Wact 12 0.81 14.81 kW Power input to the compressor = 14.8 kW

MatLab code to compute the power

1 - clear all close all clc coo) un WN 5 - Q = 0.15/60; P2 = 5*10^3; P1 = 0.2*10^3; eta = 0.81; flow rate, m^3/s $ Final pressure, kPa $ Initial pressure, kPa Isentri=opic efficiecy 8 - 9 -- rho =1000; Density 10 11 - v = 1/rho; Specific volume, m3/kg 12 13 14 - m = rho*Q; Mass flow rate, kg/s 15 16 - Wis = m*v* (P2-Pl); *Isenropic Power input 17 18 - Wact = Wis/eta; $ Actual power input 19 20 - fprintf('Power input to the compressor: $.3 kW \n', Wact); $Display the $ Power input 21

MatLab output

Command Window Power input to the compressor: 14.815 kW