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matjan (Chemical) (OP)
13 Jun 06 19:58
Hi,

I would like to know how to calculate the cogeneration potential (electricity generation) that is possible to achieve with a given steam condensing turbine at some efficiency. In this case, with efficiency I mean the MWh produced per MMBTU of energy extracted from the steam.
Let's say I have 27.2 ton/h (60 klb/h) of saturated steam at 2.4 bar (35psi) going into the turbine and an outlet pressure of 0.15 bar.
Is 0.15 bar a usual outlet pressure for a steam condensing turbine?
What is a typical efficiency value (in MWh/MMBTU extracted)?
How much energy is extracted from the steam that can be converted to electricity under these circumstances?
How do I go about calculating this amount of energy?

Many thanks,
mat
Helpful Member!  rmw (Mechanical)
13 Jun 06 23:29
Go to www.katmarsoftware.com and download the turbine sizing freeware.  Plug in your values at some reasonable efficiencies for turbines in this range and iterate it until you get a reasonable answer.

rmw
Helpful Member!  timbones (Mechanical)
14 Jun 06 8:08
If you want to do it the old fashioned way, its pretty straightforward to get a basic indication.

Let's define a few terms

h1=enthalpy of sat steam
s1=entropy of saturated steam
h2=enthalpy at condenser
hs2=isentropic enthalpy at condenser
m=mass flow of steam
x=quality at condenser (isentropic)
sg=saturated gas entropy at condenser pressure
sf=satruated liquid entropy at condenser pressure
hg= saturated gas enthalpy
hf=saturated liquid enthalpy
e=isentropic efficiency (ussually in the range of 80-86%)

1. Determine quality and h2s using pressure at condenser and s1.

x=(s1-sf)/(sg-sf)
h2s=x*(hg-hf)+hf

2. Determine real enthalpy

h2=h1-e*(h1-h2s)

3. Power= m*(h1-h2)

So for your conditions:
h1=2717 kJ/kg
s1=7.053 kJ/kg-K

I would assume a condenser pressure of about 0.07 bara
hg=2572
hf=163
sg=8.274
sf=0.559

Assume an efficiency of about e=80%

therefore
x=(7.053-0.559)/(8.274-0.559)=0.84
hs2=0.84*(2572-163)+163=2191
h2=2572-0.8*(2572-2191)=2267
Power=27.2*(2572-2267)/3600=2.3 MW

EGT01 (Chemical)
16 Jun 06 2:24
timbones,

I like your "old fashion" way of determining the energy extracted but could you check your calculations for h2 and Power?  It looks like you inadvertently picked up "hg" in those calculations instead of using "h1" as indicated in your procedure.  I think that what you have outlined in the steps of your procedure is correct but I wanted to make sure.
panduru (Mechanical)
16 Jun 06 2:42
timbones would be doing a great favor in clarifying, so the method would be useful to the others too.
timbones (Mechanical)
16 Jun 06 5:36
Indeed I have picked up hg by accident.

The calculation should be

h2=2717-0.8*(2717-2191)=2296 kJ/kg
P=27.2*(2717-2296)/3.6=3.2 MW

A note about isentropic efficiencies: I am used to dealing with steam turbines in a much higher power range (30 MW to 100 MW). These machines tend to have quite high isentropic efficiencies (in the range of 83-86%). I am taking an educated guess on the efficiency of a turbine in the 2-3 MW range having an efficiency of about 80%, but I wouldn't be surprised if it was as low as 75% (although practically speaking it doesn't make a huge difference in the resulting power). Perhaps someone with more experience with turbines in this size range could "weigh in" with an answer?
panduru (Mechanical)
16 Jun 06 7:11
Thanks a lot, timbones, for the valuable post.

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