A) The work input and entropy generation in an adiabatic pump are;
w_p = 1.10 kJ/kg and s_gen,p = 0.0004 kJ/kg.k
B) The work input and entropy generation in an adiabatic compressor if the inlet state is saturated liquid are; w_p = 610.1 kJ/kg and s_gen,p = 0.1385 kJ/kg.k
We are given;
- Initial Steam Pressure; P₁ = 100 kPa
- Final Steam Pressure; P₂ = 1 MPa
- Isentropic efficiency; η = 85% = 0.85
A) For an adiabatic pump;
- From the first steam table attached;
At P₁ = 100 kPa;
Enthalpy of saturation liquid; h_f = h₁ = 417.51 kJ/kg
Entropy of saturation liquid; s_f = s₁ = 1.3028 kJ/kg.k
Also, from steam tables at P₂ = 1 MPa and s_2s = 1.3028 kJ/kg.k, we have;
h_2s = 418.45 kJ/kg
- Formula for actual specific enthalpy at pump exit is;
h₂ = h₁ + ((h_2s - h₁)/η)
Thus;
h₂ = 417.51 + ((418.45 - 417.51)/0.85)
h₂ = 418.61 kJ/kg
Also, from steam tables at P₂ = 1 MPa and s_2s = 1.3028 kJ/kg.k, we have;
Actual specific entropy; s₂ = 1.3032 kJ/kg.k
- The formula for the work input is;
w_p = h₂ - h₁
w_p = 418.61 - 417.51
w_p = 1.10 kJ/kg
- Formula for the entropy generation is;
s_gen,p = s₂ - s₁
s_gen,p = 1.3032 - 1.3028
s_gen,p = 0.0004 kJ/kg.k
B) For an adiabatic compressor;
From the first steam table attached;
At P₁ = 100 kPa;
Enthalpy of saturation vapour; h_g = h₁ = 2675 kJ/kg
Entropy of saturation vapour; h_f = s₁ = 7.3589 kJ/kg.k
Also, from steam tables at P₂ = 1 MPa and s_2s = 7.3589 kJ/kg.k, we have;
h_2s = 3193.6 kJ/kg
Formula for actual specific enthalpy at pump exit is;
h₂ = h₁ + ((h_2s - h₁)/η)
Thus;
h₂ = 2675 + ((3193.6 - 2675)/0.85)
h₂ = 3285.1 kJ/kg
Also, from steam tables at P₂ = 1 MPa and s_2s = 7.3589 kJ/kg.k, we have;
Actual specific entropy; s₂ = 7.4974 kJ/kg.k
The formula for the work input is;
w_p = h₂ - h₁
w_p = 3285.1 - 2675
w_p = 610.1 kJ/kg
Formula for the entropy generation is;
s_gen,p = s₂ - s₁
s_gen,p = 7.4974 - 7.3589
s_gen,p = 0.1385 kJ/kg.k
Read more about entropy generation at; https://brainly.com/question/16014998
