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

Answer:

The power produced by the turbine is 23309.1856 kW

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 =

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

Which gives;

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

= -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.