Posted by gburns on April 12, 2001 at 13:03:19:
In Reply to: specific steam consumption of turbines posted by gopi krishna on April 04, 2001 at 04:15:46:
On the steam list I sent you a correction saying you would produce only 1,041 KW with your conditions. I have just looked at what I told you and realized that I was incorrect and the model is right to predict 1,200 KW. I tried to make one simple equation and I made an error. The 85% efficiency reflects how close to an isentropic process (constant entropy) it is and not a 15% loss as my calculation showed. I'm sorry for using the English system below but these calculations should give you a rough idea how you can estimate the power production.
You first determine the enthalpy and entropy of you existing conditions:
924.48 psia at 914 deg F = Enthalpy of 1459.22 btu/lb with an entropy of 1.62867 btu/lb/deg F
Next with a 100% efficient turbine the exit entropy would be the same and you have the discharge pressure so you can use an isentropic curve or steam software to determine the Btu value.
Entropy of 1.62867 btu/lb/deg F with a pressure of 142.48 psia = Btu of 1240.03 Btu/lb.
You then subtract this from the original btu value and multiply by the turbine efficiency (85%) to give you the amount of heat lost per lb of steam.
(1459.22 btu/lb - 1240.03 btu/lb) * 0.85 = 186.13 btu/lb
Next multiply this number times the steam flow rate to get total btu/hr then convert to kilowatts.
186.12 btu/lb * 23369 lbs/hr * 0.00029306 KW/(btu/hr) = 1275.9 KW
Lastly multiply this number times the generator efficiency (96%)
1275.9 KW * 0.96 = 1225 KW.
You can calculate the projected outlet temperature by subtracting the 186.12 btu/lb from the original 1459.22 btu/lb and knowing the pressure you can determine temperature from the steam tables.
Sorry for the incorrect information I sent you. I am sure this is the last correction I make as I will not try to make calculations simple in the future.
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