Question

Water vapor at 6 MPa, 500°C enters a turbine operating at steady state and expands to 20 kPa. The mass flow rate is 3 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible.
Determine:

(a) the isentropic turbine efficiency and

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

Answer 1

Answer:

a)75.8%

b)2.517KW/K

Explanation:

Hello!

To solve this problem follow the steps below

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

1. use thermodynamic tables to find the following variables.

a.enthalpy and entropy at the turbine entrance

h1=Enthalpy(Water;T=500;P=6000)

=3422KJ/kg

s1=Entropy(Water;T=500;P=6000)

=6.881KJ/kgK

b. enthalpy and ideal entropy at the turbine outlet

h2i=Enthalpy(Water;s=6.881;P=20)

=2267KJ/kg

s2i=s1=6.881KJ/kgK

2. uses the output power and the first law of thermodynamics to find the real enthalpy at the turbine's output

W=m(h1-h2)

h2=h1-W/m

h2r=3422-2626/3=2546.6KJ/kg

3.

find efficiency with the following equation

$eficiency=\frac{h1-h2r}{h1-h2i}$

$\frac{h1-h2r}{h1-h2i}=\frac{3422-2546.6}{3422-2267} =0.758=75.8%$

4.

find the real entropy at the turbine exit

s2=Entropy(Water;h=2546,6;P=20)=7.72KJ/kgK

5.Finally find the entropy generated, using the following equation

ΔS=m(s2-s1)=(3kg/s)(7.72  KJ/kgK-6.881 KJ/kgK)=2.517KW/K