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

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

The Isentropic efficiency of the turbine is 80.76%

Step-by-step 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)`

=>
`((2000)/(2.083)+(2610-3660))/(300 K)+(7.91-7.17)`

=>
`(960.153+(-1050))/(300 K)+(0.74)`

=>
`(-89.847)/(300 K)+(0.74)`

=> -0.299+(0.74)

=>0.441 kJ/kgK

Isentrophic efficeny of the turbine

=>
`\eta=(h_(1)-h_(2))/(h_(1)-h_(2 s))`

=>
`(3660-2610)/(3660-2360)`

=>
`(1050)/(1300)`

=>0.8076

=>80.76%