Question

Steam at 6 MPa and 600 °C enters a turbine with a mass flow rate of 2 kg/s and exits at 10 kPa. The power Produced by the turbine is 2626 kW. Assuming the changes in kinetic and potential energies and heat loss to be negligible, determine (a) the isentropic efficiency of the turbine and (b) the rate of entropy generation within the turbine in kW/K

Answer 1

`\eta = 94.64`

entropy generation = 0.45908 kw/k

Step-by-step explanation:

Given data:

pressure of steam is 6MPa

temperature of steam is 600 degree Celcius

For temperature 600 degree Celcius

latent heat h_1 = 3658.8 kJ/kg

specific heat s_1 = 7.1693 kJ/kg K

Power produced

`P = m (h_1 -h_2)`

`2626 = 2 (3658.8 -h_2)`

solving for h_2

`h_2 = 2345.8 kJ/kg`

At 10 kPa

`h_f = 191.81 kJ/kg`

hfg = 2392.1 kJ/kg

sf = 0.6492 kJ/kg K

sfg = 7.499 kJ/kg

h = hf + x hfg

`2345.8 = 191.81 + x * 2392.1`

x = 0.9

`s_2 = sf + x* sfg`

`= 0.6492 + 0.9 * 7.4996`

`s_2 = 7.39884 kJ/kg K`

Now considering
`s_1 = s_2` to determine enthalalpy

`s_1 = s_2 = 7.1693  kJ/kg K`

s = sf + x sfg

`7.1693 = 0.6492 + x * 7.4996`

x2s = 0.869

`h2s = hf + x2s hfg`

`h2s = 191.81 + 0.869 * 2392.1`

x = 2271.4778 kJ/kg

a) isotropic efficiency

`\eta = (h_1 - h_2)/(h_1 -h_(2s)) * 100`

`\eta = (3658.8 - 2345.8)/(3658.8 -2271.4778) * 100`

`\eta = 94.64`

b) entropy generation

`= m(s_2 -s_1)`

`= 2* (7.39884 - 7.1693)`

= 0.45908 kw/k