At steady state steam with a mass flow rate of 10 lbs enters

At steady state, steam with a mass flow rate of 10 lb/s enters a turbine at 800 F and 600 lbf/in^2
and expands to 60 lbf/in^2 The power developed by the turbine is 2852 horsepower. The steam
then passes through a counterflow heat exchanger with a negligible change in pressure, exiting
at 800 F. Air enters the heat exchanger in a separate stream at 1.1 atm, 1020 F and exits at 1
atm, 620 F. Kinetic and potential energy changes can be ignored and there is no significant heat
transfer between either component and its surroundings. Determine

(a) the mass flow rate of air, in lb/s.
(b) the rate of entropy production in the turbine, in Btu/s·R.
(c) the rate of entropy production in the heat exchanger, in Btu/s·R.

Solution

a)
From steam properties at 800 F and 600 lbf/in^2, we get h1 = 1408.9 Btu/lb, s1 = 1.6359 Btu/lb-R
2852 hp = 120945.4 Btu/min = 120945.4 / 60 Btu/s = 2015.76 Btu/s

Q - W = m*(h2 - h1)
0 - 2015.76 = 10*(h2 - 1408.9)
h2 = 1207.3 Btu/lb

From steam properties at P2 = 60 psi and h2 = 1207.3 Btu/lb we get, T2 = 346 F, s2 = 1.682 Btu/lb-R
From steam properties at P3 = P2 = 60 psi and T3 = 800 F we get, h3 = 1432.4 Btu/lb, s3 = 1.9039 Btu/lb-R

For air, at 1020 F, we have Cp = 0.263 Btu/lb-R
For air at 620 F, we have Cp = 0.25 Btu/lb-R

Average Cp = 0.2565 Btu/lb-R

Heat gained by steam = heat lost by air
m_steam*(h3 - h2) = m_air*Cp*(Tout - Tin)
10*(1432.4 - 1207.3) = m_air*0.2565*(1020-620)
= 21.94 lb/s

b)
Entropy production in turb = m*(s2 - s1)
= 10*(1.682 - 1.6359)
= 0.461 Btu/R