Question

A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-vapor mixture with a quality of 0.94 at a pressure of 325 kPa and a velocity of 15 m/s. In addition, there is heat transfer from the turbine to the surroundings for 560 kW. Find the power produced by the turbine and express it in kW?

Answer 1

The power produced by the turbine is 23309.1856 kW

Step-by-step explanation:

h₁ = 3755.39

s₁ = 7.0955

s₂ = sf + x₂sfg =

Interpolating fot the pressure at 3.25 bar gives;

570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175

2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345

h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg

Power output of the turbine formula =

`Q - \dot{W } = \dot{m}\left [ \left (h_(2)-h_(1) \right )+(v_(2)^(2)- v_(1)^(2))/(2) + g(z_(2)-z_(1))\right ]`

Which gives;

`560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+(15^(2)- 60^(2))/(2) \right ]`

= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856

`- \dot{W }` = -22749.1856 - 560 = -23309.1856 kJ

`\dot{W }` = 23309.1856 kJ

Power produced by the turbine = Work done per second = 23309.1856 kW.