Question

Water at 20 bar, 400°C enters a turbine operating at steady state and exits at 1.5 bar. Stray heat transfer and kinetic and potential energy effects are negligible. A hard-to-read datasheet indicates that the quality at the turbine exit is 98%. Can this quality value be correct? If no, explain. If yes, determine the power developed by the turbine, in kJ per kg of water flowing

Answer 1

Answer 2

3-solution:

Given : water at inlet condition of turbine

P= 20 bar, T = 400°C

water at exit of turbine

P = 1.5 bar

quality of steam = •98

Since exit Temperture is not mentioned.

we can assume either water is in dry saturated or in wet steam at the exit of turbine

From steam table we can get following value of water :

At P = 20 bar , and Temperature 400°C ( water is at at superheated state)

enthalpy (h1)= 2945kj/kg , entropy (S1)= 7.127 KJ/kg-K

At P = 1.5 bar Temperature = 111.4 °C (assumed as it is not mentioned)

enthalpy (hf) = 467.1KJ/Kg , enthalpy at dry saturated (h2) = 2694KJ/Kg , entropy ( Sf)= 1.437Kj/kg-K , S2 = 7.223 KJ/Kg-K

let us assume that steam is expanded isentropically , so entropy at inlet condition will be ewual to entopy at the exit condition.

using relation: S1=Sf+x(S2-Sf) ( here x is the quality of steam)

7.127= 1.437+ x( 7.223-1.434)

On solving

x = 0.982 which is almost equal to 98% . Hence we can say that quality of steam estimated is correct.

Using steady flow energy eauation and ignoring kinetic aand potential effects negligible we have below relation

Change in enthalpy (inlet to exit) = work done / sec ( power) by the turbine

Now using relation : h*= hf+ x( h2- hf) for getting enthalpy at exit of turbine

h = 467.1+0.98*( 2694-467.1)

= 2649.462KJ/Kg

now power developed/ kg = h1- h*

= 3248- 2649.462

= 598.538KJ/Kg