Question

Steam enters an adiabatic turbine operating at steady state at 2MPa, 400 C with velocity of 50 m/s and exit at 15 KPasquality of 90% with velocity or 180 m/s) The elevation of inlet is47m higher than at the exit and the mass flow rte is 5 kg/s. Let g = 9.81 m/s. determine the powerdeveloped by the turbine.

Answer 1

Answer:

P=4833.45 KW

Explanation:

At inlet:

Properties of steam at 2 MPa and 400°C

$h_1=3248.22\frac{KJ}{Kg}$

$V_1=50 m/s$

At exit:

Properties of steam at 15 KPa

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

So at 15 KPa

$h_f= 33.61\frac{KJ}{Kg} ,h_g=2515.2\frac{KJ}{Kg}$

$V_2=180m/s$

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=33.61+0.9(2515.2-33.61)\frac{KJ}{Kg}$  $h_2=2267.04\frac{KJ}{Kg}$

Now from first law of thermodynamics for open system

$h_1+\dfrac{V_1^2}{2}+Z_1g+Q=h_2+\dfrac{V_2^2}{2}+Z_2g+w$

$Z_1=47 m,Z_2=0$

Here given that turbine is adiabatic so Q=0

Now put the all values

$3248.22+\dfrac{50^2}{2000}+47\times 9.81\times 10^{-3}+0=2267.04+\dfrac{180^2}{2000}+W$

$W=966.69\frac{KJ}{Kg}$

$So power developed P=m\times W$

 $P=5\times966.69$                                            

P=4833.45 KW