Steam flows steadily through a turbine at a rate of 45,000 lbm/h, entering at 1000 psia and 900°F and leaving at 5 psia as saturated vapor. If the power generated by the turbine is 4.1 MW, determine the rate of heat loss from the steam. The enthalpies are h1 = 1448.6 Btu/lbm and h2 = 1130.7 Btu/lbm.

Q_out = 182.0(Btu/s)

Explanation:

Given:

m = 4500 Ibm/h

P1 = 1000 psiu

T1 = 900°

P2 = 5 psiu

Wout = 4.1 MW

h1 = 1448.6

h2 = 1130.7

We need to find the energy balance first.

Energy balance;

;

Therefore;

Since we are to find Qout, let's make it the subject of the formula

Substituting figures into the equation, we have:

Qout= -(45000/3600 ibm/s)(1130.7 - 1448.6)Btu/ibm - 4100kj/s(1btu/1.055kj)

Qout = 182.0(Btu/s)

Therefore, rate of heat loss of system = 182.0(Btu/s)

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