Which Turbine Work Done Maximum?

Compounding of steam turbines - Wikipedia, the free encyclopedia

Obviously, LP turbine will do maximum work.

Dear Mr.Irshad,

If you refer the Mollier Chart, as the pressure drops the area under the Saturation Curve of in the Mollier chart increases, WHICH REPRESENTS THE POWER DEVELOPED. This will be correct where there is no RE-HEATING of Steam is involved and Steam Expansion in the Turbine is straight with out withdrawing and re-heating.

But the original post indicates that there is REHEATING of Steam, with Moisture Seperator.

Hence the Power developed is to be precisely calculated. Pl. refer my comment - Post No. 4 here. The moisture seperator will remove the moisture and accordingly this quantity to be deducted from the steam flow.

DHAYANANDHAN.S

there's some tricky thing to tuning the turbines (seek guangzhou power news -- for instance - there're ? were old dudes knew about) - some wave matching for the sys - no 2-ries? or some special Fq.-up-path = better performance - the wireing . . . i donno http://english.giec.cas.cn/rh/rp/201210/t20121017_92055.html some start

Dear Mr.awaiscnpp

Power developed depends upon HOW MUCH HEAT DROP IS ACHIEVED per KG of the Steam x STEAM FLOW/Hr. is the deciding criterion. Use Steam Tables or Mollier Chart to find Heat Drop/Kg. of Steam and multiply by Steam Flow/Hr.

In simple words POWER DEVELOPED = (Steam Flow in Kgs/Hr) x (Heat Drop/Kg. of Steam)/860 KWHr.

Since the RE-HEATING of steam is involved the heat content is increased and therefore the heat drop will increase in the Turbines.

If you want I will work out and post the details. By the way, you have not indicated the Steam Flow/Hr.

DHAYANANDHAN.S