Question

A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump.

Answer 1

a) 69,630KW

b) 203 KW

Step-by-step explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

`h_(1) = h_(f,@20KPa) = 251.42 KJ/kg\\v_(1) = v_(f,@20KPa) = 0.001017 KJ/kgK\\\\w_(p,in) = v_(1) * (P_(2) - P_(1))\\w_(p,in) = (0.001017)*(4000-20)\\\\w_(p,in) = 4.05 KJ/kg\\\\h_(2) = h_(1) - w_(p,in) \\h_(2) = 251.42 + 4.05\\\\h_(2) = 255.47KJ/kg\\\\P_(3) = 4000KPa\\T_(3) = 700 C\\s_(3) = 7.6214 KJ/kgK\\\\h_(3) = 3906.3 KJ/kg\\\\P_(4) = 20 KPa\\s_(3) = s_(4) = 7.6214KJ/kgK\\s_(f) = 0.8320 KJ/kgK\\s_(fg) = 7.0752 KJ/kgK\\\\`

`x_(4) = (s_(4) - s_(f) )/(s_(fg) ) \\\\x_(4) = (7.6214-0.8320)/(7.0752) = 0.9596\\\\h_(f) = 251.42KJ/kg \\h_(fg) = 2357.5KJ/kg \\\\h_(4) = h_(f) + x_(4)*h_(fg) = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\`

The power produced and consumed by turbine and pump respectively are:

`W_(T,out) = flow(m) *(h_(3) - h_(4)) \\W_(T,out) = 50 *(3906.3-2513.7)\\\\W_(T,out) = 69,630 KW\\\\W_(p,in) = flow(m) *w_(p,in) = 50*4.05 = 203 KW`