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gt combustion chamber problem

i am working on a small gas turbine unit, which i am going to modify to inject water into the combustion chamber to reduce the temp in there.

i'm doing some kinda preliminary calcs to work out roughly how much water i'm gonna have to inject to achieve a certain drop in temperature. my project supervisor says it will be reasonable to model it as an adiabatic mixing process.

i know the specific heat and flow rate of the combustion products, and what the temp and pressure in there is.

my question is, what value of specific heat and temperature of the water being injected should i use in the calculation. obviously it will heat up as it passes near to the CC, and vapourise pretty much as soon as it enters.

so should i use the values for water from my steam tables for the water before it gets anywhere near the injection point, since any heat gains into the water, are still losses from the cc, or should i use the values of the water/steam at the point it enters the cc.

or should i be doing this a completely different way altogether?

thanks.

Reply to
c p
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incidentally, my calculations as they stand are along the lines of:

m_dot,gas * Cp,gas * T,gas + m_dot,water * Cp,water * T,water = (m_dot,gas * Cp,gas + m_dot,water * Cp,water) * T,reduced

therefore after a bit of rearranging,

M_dot, water = [m_dot,gas * Cp,gas * ( T,reduced - T,gas )] / [ Cp,water * ( T,water- T,reduced ).

values are: m_dot,gas = 131.5e-3 kg/s Cp, gas = 1.183 kJ/kgK T,gas = 998K

T,water = 283K Cp,water = 4.193 kJ/kgK

so to reduce to 800K, m_dot,water = (approx) 0.88 L/min

which doesnt seem completely unreasonable to me, but no doubt there;s plenty wrong, so any suggestions are quite welcome,

thanks.

Reply to
c p

c p said the following on 11/12/2004 6:44 AM:

You're missing the heat of vaporization of the water which is a huge number compared to (Cp)(dT). The vaporization of the water is the part that will be doing all the temp reduction for you.

Base your calculations on enthalpy (h) instead of CpdT. It'll go a little easier.

Lance

*****
Reply to
Lance

yeah, i was pretty sure that the evaporation of the water was going to have a much bigger cooling effect than my calculations based on CpdT. Problem being, i dont really know where to begin with the enthalpy based analysis of the process. i cant really find anything relevant in my textbook (Eastop and McConkey - applied thermodynamics). could you give me a nudge in the right direction of a starting point, or to a text you think would be good for this?

cheers,

Reply to
c p

Try Google "water injection combustion" It brings up a number of articles that might be of interest. for example this one:

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Reply to
MLD

Try this approiximate approach:

m_dot.gas X Cp.gas X (T.gas - T.mix) = m_dot.water X (Cp.water) X (373 - T.water) + m_dot.water X LH.water X (Tmix - 373) + m_dot.water X Cp.steam X (T.mix - 373)

Or approximating some more.... 131.5 X 1.183 X (998 - 800) = M_dot.water X (4.193 ) X (373 - 283) + M_dot.water X 2268 X ( 800 - 373 ) + M_dot.water X 3 X ( 800 - 373)

And so...

30800 = M_dot.water X (377) + M_dot.water X 968400 + M_dot.water X 1280 = 970,000 X M_dot.water

Hence M_dot.water = 0.032 kg/sec or 32 cc/sec or

1.9 kg/min or 1.9 L/min

(Notice how my value seems to be twice yours? It's actually

1/30 as big, because you used a unit of L/min instead of L/sec)

Brian W

Reply to
Brian Whatcott

Surely the water is entering the system at a known temp and therefore a known enthalpy. If there is no chemical reaction then it all leaves as a gas (steam) at the temp of the exhaust. Its enthalpy at that condition can be looked up in your superheated steam tables. By making a few trial stabs at the exhaust temp you can balance the heat in with the heat out and Bob's yr Uncle.

Reply to
Rusty

unless i'm being really obtuse, your answer does seem to be twice as big as mine. i calculated my flow rate to be 0.014kg/s and you got yours to be

0.032 kg/s. i know my calc was fundamentally wrong, but am i missing something totally retarded in the conversion of units?!!?
Reply to
slow_graffiti

I could easily be mistaken - but I reckoned your numbers resulted in a water flow of 0.88 liters/sec Didn't they then?

Brian w

Reply to
Brian Whatcott

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