Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lbf/in^2, T2 = 700R the mass flow rate of air entering the turbine is 11 lb/s, and 65000 Btu/h of energy is rejected by heat transfer.

a. Neglecting kinetic and potential energy effects, determine the power developed in hp.

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = - 65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m (h1) - m (h2) = 0

W = Q - m (h2 - h1)

W = (-65000 Btu/h) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h) - (-1928.3) Btu/s

W = (-65000 Btu/h) * {1hr / (60*60) s} - (-1928.3) Btu/s

W = - 18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp