Question

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negigible.

Determine :

(a) the isentropic turbine efficiency

(b) the rate of entropy production within the turbine, in kW/K.

Answer 1

Step-by-step explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

`h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k`

Obtain the following properties at 10kPa from the table "saturated water"

`h_(f2)=191.81KJ/Kg.K\\h_(fg2)=2392.1KJ/Kg\\s_(f2)=0.6492KJ/Kg.K\\s_(fg2)=7.4996KJ/Kg.K`

Calculate the enthalpy at exit of the turbine using the energy balance equation.

`(dE)/(dt)=Q-W+m(h_1-h_2)`

Since, the process is isentropic process
`Q=0`

`0=0-W+m(h_1-h_2)\\h_2=h_1-(W)/(m)\\\\h_2=3658.8-(2626)/(2)\\\\=2345.8kJ/kg`

Use the isentropic relations:

`s_1=s_(2s)\\s_1=s_(f2)+x_(2s)s_(fg2)\\7.1693=6492+x_(2s)(7.4996)\\x_(2s)=87`

Calculate the enthalpy at isentropic state 2s.

`h_(2s)=h_(f2)+x_(2s).h_(fg2)\\=191.81+0.87(2392.1)\\=2272.937kJ/kg`

a.)

Calculate the isentropic turbine efficiency.

`\eta_(turbine)=(h_1-h_2)/(h_1-h_(2s))\\\\=(3658.8-2345.8)/(3658.8-2272.937)=0.947=94.7%`

b.)

Find the quality of the water at state 2

since
`h_f` at 10KPa <
`h_2`<
`h_g` at 10KPa

Therefore, state 2 is in two-phase region.

`h_2=h_(f2)+x_2(h_(fg2))\\2345.8=191.81+x_2(2392.1)\\x_2=0.9`

Calculate the entropy at state 2.

`s_2=s_(f2)+x_2.s_(fg2)\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K`

Calculate the rate of entropy production.

`S=(Q)/(T)+m(s_2-s_1)`

since, Q = 0

`S=m(s_2-s_1)\\=2(kg)/(s)(7.398-7.1693)kJ/kg\\=0.4574kW/k`