Question

Water vapor enters a turbine operating at steady state at 500 degrees C, 40 bar, with a velocity of 200 m/s, and expands adiabatically to the exit, where it is saturated vapor at 0.8 bar, with a velocity of 150 m/s and a volumetric flow rate of 9.48 cubic meters per second.

The power developed by the turbine, in kW, is approximately:

A. 3500

B. 3540

C. 3580

D. 7470

Answer 1

Answer:

correct option is  C. 3580

Explanation:

given data

temperature = 500 degrees C

steady state at = 40 bar

velocity =  200 m/s

saturated vapor = 0.8 bar

velocity = 150 m/s

volumetric flow rate = 9.48 cubic meters per second

solution

we use here steam tables for temp 500 deg C, 40 bar

we get h1 = 3445300 J/kg and 0.8 bar

we get hg = h2 = 2665800 J/kg

and vg = v2 = 2.087 m³/kg

and

we know power that is express as

Power = w × (Q ÷ v2)    ....................1

and

here

$h1 + \frac{V1^2}{2} = h2 + \frac{V2^2}{2} + w$

so

w = $(h1 - h2) + \frac{V1^2-V2^2}{2}$      

put here value and we get

w = (3445300 - 2665800) + $\frac{200^2 - 150^2}{2}$    

w = 788250 J/kg

so put value in equation 1 we get  

Power =  $788250\times \frac{9.48}{2.087}$

Power = 3580551 W

Power = 3580 kW