Consider a steam turbine operating at steady-state. There is one inlet and one o
ID: 1861821 • Letter: C
Question
Consider a steam turbine operating at steady-state. There is one inlet and one outlet, tand adiabatic conditions may be assumed. Apply the first law for a control-volume to determine the power output. Inlet conditions are 125 bars, 420 degrees C, and average flow speed of 100 m/s. The inlet is area 100 cm^2 at a height of 2 m above the datum. The exit quality is 90% at 1 bar and average flow speed of 50 m/s, at a height of 50 cm above the datum. Determine the following:
a) the mass flow rate in kg/s
b) the change in specific enthalpy in kJ/kg
c) the change in kinetic energy per unit mass in kJ/kg
d) the change in potential energy per unit mass in kJ/kg
e) the shaft work in kJ/kg
f) the exit area in cm^2
g) the power output of the turbine in kW
h) the percent power output due to changes in enthalpy
i) the percent power output due to changes in kinetic energy
j) the percent power output due to changes in potential energy
Explanation / Answer
At Inlet P = 12500 KPa and T = 450 Deg
v1 = 0.02299 m^3/Kg
h1 = 3199.78 KJ/Kg
these values are from steam tables
Density at inlet(p1) = 1/v1 = 1/0.02299 = 43.497 Kg/m^3
At exit P= 100Kpa and x =0.9
vf = 0.001043 and vg= 1.694
v2 = (1-x)*vf + x*vg = 1.5247 m^3/Kg
Density at exit(p2) = 1/v2 = 0.656 Kg/m^3
hf = 417.44 and hg = 2675.46
h2 = (1-x)*hf + x*hg = 2449.658 KJ/Kg
a)Mass flow Rate = p1*V1*A1 = 43.497*100*100*10^-4 = 43.497 Kg/s
b) Change in specific enthalpy = h1-h2 = 3199.78 - 2449.658 = 750.12 KJ/Kg
c) Change in KE = V1^2 - V2^2 = (100^2 - 50^2) = 7.5 KJ/Kg
d) Change in PE = g(Z1-Z2) = 9.8*(2-0.5) = 0.0147 KJ/Kg
e) shaft work = Change in specific enthalpy + Change in KE + Change in PE
Shaft Work = 750.12 + 7.5 + 0.0147 = 757.637 KJ/Kg
f) mass flow is constant
p2*V2*A2 = 43.497
0.656*50*A2 = 3.497
A2 = 1066.375 cm^2
g)Percent Power Due to enthalpy = (Change in specific enthalpy/Shaft Work)*100 = (750.12/757.637)*100 = 99 percent
h)Percent Power Due to KE = (7.5/757.637)*100 = 0.989%
i)Percent Power Due to PE = (0.0147/757.637)*100 = 0.0019%