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Steam at 400°C and 40 bar flows steadily through an adiabatic turbine at a volumetric flowrate...

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.

Solutions

Expert Solution

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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