Question

Steam enters a turbine operating at steady state at 6 MPa, 600°C with a mass flow rate 125 kg/min and exits as a saturated vapor at 20 kPa. The turbine produces energy at a rate of 2 MW. Kinetic and potential energy effects are negligible. The rate of heat loss from the turbine occurs to the air around the turbine at 27°C. a. What is the rate of entropy production for within the turbine?b. What is the isentropic efficiency of the turbine?

Answer 1

Answer:

The rate of entropy production for within the turbine is 0.441 kJ/kgK

The Isentropic efficiency of the turbine is 80.76%

Explanation:

Given:

$P_1$ = 6MPa

$T_1$ = $600^{\circ}C$

$h_1$= 3660kJ/kg

$s_1$= 7.17Kj/kgK

$p_2$= 20kPa

$x_2$ = 1

$s_2$= 7.91kJ/kgK

$h_2$ = 2610kJ/kg

Isentrophic properties of stream

$P_2$ = 20kPa

$S_{2s}$=s_1

=7.17Kj/kgK

$h_{2s}$= 2360kJ/kg

Entropy generation of the stream:

=>$S_{g e n, c v}=\frac{\dot{S}_{g e n}}{\dot{m}}$

=>$\frac{\frac{-\dot{Q}_{c v}}{\dot{m}}}{T_{b}}+\left(s_{2}-s_{1}\right)$    

=>$\frac{\frac{W_{c v}}{\dot{m}}+\left(h_{2}-h_{1}\right)}{T_{b}}+\left(s_{2}- s_{1}\right)$

=>$\frac{\frac{2000}{2.083}+(2610-3660)}{300 K}+(7.91-7.17)$

=>$\frac{960.153+(-1050)}{300 K}+(0.74)$

=>$\frac{-89.847}{300 K}+(0.74)$

=> -0.299+(0.74)

=>0.441 kJ/kgK

Isentrophic efficeny of the turbine

=>$\eta=\frac{h_{1}-h_{2}}{h_{1}-h_{2 s}}$

=>$\frac{3660-2610}{3660-2360}$

=>$\frac{1050}{1300}$

=>0.8076

=>80.76%