Answer to Question #306547 in Mechanical Engineering for james

Question

A reheat cycle with two stages of reheating is executed, with steam expanding initially from 90 bar and 530 deg. C. The two reheater pressures are 10 bar and 0.6 bar, and the steam leaves each reheater at 480 deg. C. Condensation occurs at 0.03 bar. For an ideal cycle, find:

a. Q_a and e

b. For the engine, ignore the pressure drop through the reheaters, let the engine operate through the same states, and compute W and e.

c. What is the steam flow rate for an engine output of 20,000 kW?

 

Answer 1

Solution;

Let subscripts 1,23... Represent various stages of the cycle;

At state 1,90bar,530°c;

$h_1=3462kJ/kg$

$s_1=6.7553kJ/kgK$

Since process 1-2 is isentropic expansion;

$s_1=s_2=6.7553kJ/kgK$

At P=10bar;

$h_2=2856.92kJ/kg$

At state 3;10bar,480°c;

$h_3=3435.8kJ/kg$

$s_3=7.7075kJ/kgK$

Process 3-4 is isentropic expansion;

$s_3=s_4=7.7075kJ/kgK$

At 0.6 bar;

$h_4=2719.06kJ/kgK$

At state 5,0.6bar,480°c;

$s_5=9.0166kJ/kgK$

$h_5=3446.6kJ/kg$

Process 5-6 is isentropic expansion;

$s_5=s_6=9.0166kJ/kgK$

At $P_6=0.04bar;$

$h_6=2745kJ/kg$

$h_7=121.40kJ/kg$

$h_8=h_7+u\Delta p$

$h_8=121.40+(90-0.04)×10^2×1.0063×10^{-3}=130.396kJ/kg$

Hence;

$Q_a=(h_1-h_8)+(h_3-h_2)+(h_5-h_6)$

Heat added;

$Q_a=(3462-130.396)+(3435.8-2856.92)+(3446.6-2719.061)=4638.024kJ/kg$

Efficiency;

$W=(h_1-h_2)+(h_3-h_4)+(h_5-h_6)$

$W=(3462-2856.92)+(3435.9-2719.06)+(3446.6-2745)=2023.22kJ/kg$

$\eta=\frac{W}{Q_a}=\frac{2023.22}{4638.024}=0.4362$

$\eta=0.4362$ or 43.62%

For output of 20,000kW;

$P=\dot mW$

$\dot m=\frac{20000}{2023.22}=9.885kg/s$