Question

A well-insulated turbine operating at steady state develops 25 MW of power for a steam flow rate of 50 kg/s. The steam enters at 15 bar with a velocity of 61 m/s and exits as saturated vapor at 0.06 bar with a velocity of 130 m/s. Neglecting potential energy effects, determine the inlet temperature, in °C.

Answer 1

Answer:

T=332.26°C.

Explanation:

Power=25 MW  

Steam flow rate =50 kg/s

Inlet pressure at turbine=15 bar,Inlet velocity=61 m/s

Exits pressure=0.06 bar (at saturated vapor) ,exits velocity=130 m/s.

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here given that turbine is adiabatic so Q=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}+w$

$h_1-h_2=\dfrac{V_2^2}{2}-\dfrac{V_1^2}{2}+w$

$W=\frac{25000}{50}$ KJ/kg

$h_1-h_2=\dfrac{130^2}{2000}-\dfrac{61^2}{2000}+500$

$h_1-h_2=506.58$ kj/kg

Now from steam table

Properties of saturated steam at 0.06 bar

$h_g=2566.2\frac{KJ}{Kg}$

$h_g=h_2$

So $h_1=2566.2+506.58\frac{KJ}{Kg}$

$h_1=3072.78\frac{KJ}{Kg}$

 Properties of saturated steam at 15 bar

$h_g=2791.5\frac{KJ}{Kg}$ ,saturation temperature=198.32°C

If super heated steam will treat as ideal gas

$C_p=2.1\frac{KJ}{kg-K}$

$h_1=h_g+C_pΔT$

3072.78=2791.5+2.1(T-198.32)

T=332.26°C

So inlet temperature to turbine  T=332.26°C.