Answer:
(a) The net power output is 6081 kW
(b) The back work ratio is 0.536
(c) The thermal efficiency is 37.4%
Explanation:
Process 1-2 is compression
Obtain the properties from EES software.
Enthalpy is function of Temperature for an ideal gas whereas the entropy is a function of both temperature and pressure.
T₁ = 40°C, P₁ = 100 KPa
Enthalpy, h₁ = 313.6 kJ/kg
Enthropy, S₁ = 5.749 kJ/kg.k
Entropy at state 2, s₂ = s₁ = 5.749 kJ/kg.k
P₂ = 1600 kPa and Entropy s₂ = 5.749 kJ/kg.k
Saturated Enthalpy, [tex]h_{2s}[/tex] = 691.9 kJ/kg
The Compressor efficiency is,
[tex]n_c = \frac{h_{2s} -h_1}{h_2 - h_1}[/tex]
[tex]0.85 = \frac{691.9 - 313.6}{h_2 - 313.6}[/tex]
h₂ = 758.6 kJ/kg
Process 3 - 4 is expansion
T₄ = 650°C, specific enthalpy, h₄ = 959.2 kJ/kg
The turbine efficiency is,
[tex]n_t = \frac{h_3 - h_4}{h_3 - h_{4s}}[/tex]
Enthalpy cannot be found directly at state 3.
Using the EES software together with isentropic efficiency relation, we find
Temperature, T₃ = 1353°C
Enthalpy, h₃ = 1790 kJ/kg
Entropy, s₃ = 6.75 kJ/kg.k
Note that: h₃ = enthalpy(Air, T=T₃)
s₃ = entropy(Air, T=T₃, P=P₂)
[tex]h_{4s}[/tex] = enthalpy(Air, P=P₁, s=s₃)
The mass flow rate is determined from
m = P₁V₁ ÷ RT₁
= (100 kPa) (850 m³/min) ÷ (0.287) (40 + 273)
= 15.77 kg/s
The work input to the compressor is:
[tex]W_{c,in} = m (h_2 - h_1)[/tex]
= (15.77 kg/s) (758.6 kJ/kg - 313.6 kJ/kg)
= 7017 kW
The work output from the turbine is:
[tex]W_{T,out} = m (h_3 - h_4)[/tex]
= (15.77 kg/s) ( 1790 kJ/kg - 959.2 kJ/kg)
= 13098 kW
(A) the net power output,
[tex]W_{net} = W_{T, out} - W_{C,in}[/tex]
= 13098 kW - 7017 kW
= 6081 kW
(B) The back work ratio is
[tex]r_{bw} = \frac{W_{C, in}}{W_{T, out}}[/tex]
= 7017 kW / 13098 kW
= 0.536
(C) The thermal efficiency is
[tex]n_{th} = \frac{W_{net}}{Q_{in}}[/tex]
= 6081 kW ÷ [m(h₃ - h₂)]
= 6081 kW ÷ [15.77 kg/s (1790 kJ/kg - 758.6 kJ/kg)]
= 37.4%
