1 Consider a steam turbine operating at steady-state. There is one inlet and one
ID: 1861820 • Letter: 1
Question
1
Consider a steam turbine operating at steady-state. There is one inlet and one outlet, and adiabatic conditions may be assumed. Apply the first law for a control-volume to determine the power output. Inlet conditions are 125 bars, 450 degree C, and average flow speed of 100 m/s. The inlet is area 100 cm2 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: the mass flow rate in kg/s the change in specific enthalpy in kJ/kg the change in kinetic energy per unit mass in kJ/kg the change in potential energy per unit mass in kJ/kg the shaft work in kJ/kg the exit area in cm2 the power output of the turbine in kW the ( plusminus ) percent power output due to changes in enthalpy the ( plusminus ) percent power output due to changes in kinetic energy the ( plusminus ) percent power output due to changes in potential energyExplanation / 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%
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