Steam Turbine (SST-9000)

According to the trusty Steam Tables, here on the desk, 290degC/75barg is saturated steam. 277degC 9barg will be high superheat, and therefore a good feedstock for a turbine.

The output of the turbine depends upon flowrate at that pressure, and therefore the size of the steam plant feeding it. 1700MW is a large turbine.

What's the real problem - the absence of the turbine-maker's telephone number, perhaps?

Thank you to your reply. My doubt is how they are producing this much power with lower parameter without compromising the efficiency? As per your reply, HP is running with saturated steam and the LP is running With super heated steam. I want to know , whether pressure is a criteria to have large power with higher efficiency or not!!

The answer lies within a detailed study of Thermodynamics, a topic for third-year undergraduate students on Chemical Engineering courses, among others.

They must be maintaining very high flow rate of steam. The turbine might consist of HP,IP and a number of LP turbines to take care of expansion of steam in later stages. Also, there might be a number of condensers.

Dear Mr.dhinesht,

Considering the Steam Parameters, the STEAM FLOW SHOULD BE HIGH. But you have NOT MENTIONED about the exhaust Steam Conditions - to know or calculate about the Heat Drop/Kg. of Steam.

To develop this level of Power of 1700 MW, 860/HEAT DROP per Kg. of Steam, will be the Specific Steam Consumption and this Figure multiplied by 1700 x 1000 will be the STEAM FLOW in Kg/Hr. You can work out this.

DHAYANANDHAN.S

Thank you to all your replies.

Mr.Dhayanandhan,

How has you derived this values. I haven't understand that. Can you please explain me little bit more.

Thanks in advance.

Dear Mr. dhinesht,

It is very simple. This purely from THERMO DYNAMICS SUBJECT. One KwHr. = 860 K.Calories of HEAT ENERGY or 3600 KJoules of HEAT ENERGY. Hence we have 860 Kcalories of Heat = One KWHr.

We can find out the Total Heat in the Steam at Inlet and Outlet of Turbine - from STEAM TABLES or MOLLIER CHART. The Difference is the HEAT DROP per Kg. OF STEAM.

860/Heat Drop per Kg. of Steam = Steam Required per KWHr. You know the total output and multipy this output (in KWHr.) by Steam per KWHr. You get the Steam Flow.

First READ THERMODYNAMICS - TURBINE and STEAM CHAPTERS.

Thanks,

DHAYANNANDHAN.S