Question

Steam at 400°C and 40 bar flows steadily through an adiabatic turbine at a volumetric
flowrate of 5,000 m3/h. The steam leaving the turbine at 1 bar is then cooled at constant
pressure in a condenser to 25°C. The rate of transfer from the condenser is 50 MW.
Calculate the power output generated by the turbine (MW). Clearly state assumptions (if
any) and reference state.

Answer 1

Initial steam temperature = 400°C= 673 K

P_{​​​​​​1} = 40 bar

V_{​​​​​​1} = 5000 m^{​​​​​​3}/h

n = PV/(RT)

n = (40×10⁵×5000)/(8.314×673)

n = 3574414.484 mol/h = 3574.414 Kmol/h

= 0.992892 Kmol/s

The condenser outlet = 25°C and 1 bar

Q_{​​​​​​cool} = 50 MW

Water at 1 bar boils at 100°C

From handbook

Cp of steam is computed

Q = nCp (T-25)+ nCp(100-25) + n(latent heat)

50(10⁶)=

0.992892(1000)((Cp(T-25))+ latent heat + (Cp(100-25)))

The cooling involves condensation of vapor and subcooling it to 25°C

From handbook Cp(liquid) is taken at average temperature of (25+100)/2 = 62.5°C and latent heat at T = 100°C

Cp(liquid)= 75.5 KJ/Kmol°C

Latent heat = 40650 KJ/kmol

50(10³)= 0.992892((75.5)(100-25)+40650+ 0.992892Cp(T-100))

4016.68925 = 0.992893 C_{​​​​​​pgas}(T-100)

Trial 1

At T = 200°C

From handbook Cp(gas) at average temperature of (100+200)/2 = 150 = 34.463 KJ/kmol°C

By calculating we get T = 218°C

Trial 2

T = 217°C

Average temperature = (100+217)/2= 158.5°C

Cp(gas) at 158.5 °C = 34.54 KJ/Kmol°C

By calculating we get T = 217°C

So turbine exit temperature = 217°C

From steam tables

At P = 40 bar and T = 400°C

H = 3214.5 KJ/Kg

Exit turbine conditions

T = 217°C, P = 1 bar

H' = 2909.06 KJ/Kg

H - H' = W = 3214.5-2909.06 = 305.44 KJ/Kg

Moles of steam = 3574.414 Kmol/h

M.W of steam = 18 Kg/Kmol

Kg of steam = 18(3574.414)= 64339.452 Kg/h

Work given out by turbine = 64339.452(305.44)= 19651842.22 KJ/h = 5458.845 KW = 5.4588 MW

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