Two hundred kg/min of steam enters a steam turbine at 350°C and 40 bar through a 7.5-cm diameter ...

a) T =?
P = 5 bar ---------> table, appendix B.6        Tsaturation = 151.8oC
The discharge is all liquid at 75oC

b) Energy transfer? (Neglect ΔEp but not ΔEk)

T = 350oC at inlet, P = 40 bar -------> table, appendix B.7  Ĥin = 3095 kJ/kg, Vin = 0.0665 m3/kg


T = 75oC at outlet, P = 5 bar -------> table, appendix B.7  Ĥout = 314.3 kJ/kg, Vout = 1.03 x 10-3  m3/kg

(Velocity), uin = (volume), Vin/ (Area), Ain, where, A = πD2/4

(Velocity), uin = (200kg/ min) x (1min/ 60s) x (0.0665 m3/kg/ π (0.075)2/4 m2) = 50.18 m/s



(Velocity), uout = (volume), Vout/ (Area), Aout = = (200kg/ min) x (1min/ 60s) x (0.00103 m3/kg/ π (0.05)2/4 m2) = 1.75 m/s


Energy balance: Q - Ws = ΔH - ΔEk = ṁ (Ĥ2 - Ĥ1) + m/2 (u22 - u11)


Q - Ws = (200kg/ min) x (1min/ 60s) x (314 - 3095)kJ/kg + (200 kg/2min) x (1min/ 60s) x (1.752 - 50.182 m2/s2) = - 13460 kw

∴ 13460 kw transferred from the turbine (since the answer was in negative then it's transferred from)

At the last line, 1st part's unit is kJ/kg whereas 2nd part's unit is J/kg. To do the addition, both parts unit should be same. Actual answer is -9274 kW