Question

Consider now a turbine that operates with steam with T1 = 450◦C and P1= 8.0 MPa and P2 = 30 kPa. The efficiency is 80% and the mass flow rate is 80 kg/s. You may assume the velocities are zero. You may also assume q=0.

What is the exhaust temperature and quality (if saturated) for 100% efficiency?

Answer 1

For this we will make use of steam table and obtain the value of specific enthalpy required to solve the problem

For state 1

P = 8 MPa

T =450 oC

at this pressure and temperature

H = 3273.234 KJ/Kg

s = 6.5576 KJ/KgK

Now here q is zero and hence the process is adiabatic

Now if the efficiecy is 100 % it implies that the process can be assumed to be reversible

A reversible adiabatic process is nothing but isentropic process

hence the entropy will not change

Hence we will look at State in steam table Having P = 30 KPa and S = 6.5576 KJ/KgK

From steam table

at P = 30 Kpa saturated liquid has entropy has Sv = 7.7674 KJ/Kg K and Sl = 0.94393 KJ/Kg K

Hence the outlet stream is actually saturated

The temeperature will be the saturation temperature at P =30 KPa

Tsat = 69.09 oC

We can find the quality of steam as following

S =x*Sv + (1-x)*Sl

6.5576 = x*7.7674 + (1-x)* 0.94393

x = 0.823

Hence the exhaust temperature of steam in 69.09 oC and quality of steam is 0.823 for 100 % efficiency