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Old Mar 16th 2017, 06:43 PM   #1
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Thermodynamic question about open systems

A steam turbine takes in steam, with specific enthalpy 3095 kJ/kg, the steam leaves the turbine with specific enthalpy 2660 kJ/kg, work output is 0.5 MW, If heat losses from the turbine are 120 kW, determine

a) the mass flowrate of the steam.
b) the work output per kilogram , assume that kinetic-energy and potential-energy changes are negligible.
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Old Mar 17th 2017, 06:06 AM   #2
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This is a straight forward first law question.

Post your attempt for guidance.
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Old Mar 17th 2017, 06:45 AM   #3
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Hello, I had it in the exam last week and wanted to make sure I did alright, I think I did a mistake on B) though, but not sure.

I don't know how to write equations and all of that here on the internet, only in papers, but i'll try to explain it in computer writing as much as I can

for A) it wants mass flowrate which is m and dot on top of it.

I used equation of

Q-W=m(h2-h1)

cause heat is loss so its minus, so therefore the equation is like this with plugging the numbers

-120-0.5x10^3=m(2660-3095)
-620=m(2660-3095)
m=1.425 kg/s

For B) I didn't know exactly what to do, but I knew work output is given which is 0.5 MW, it's in MEga Wats, so I did convert it to KiloWats first of all, so what i did basically is i divided it to the m dot that I got from A.

so I did this basically

0.5x10^3/1.425

and cause the /second from Wats which is J/s will cancel out the seconds from Kilogram/s which is from the mass flowrate, the answer that I had was.

= 350.9 KJ/Kg

Please tell me this is right, it weights 10 marks out of the total mark of this semester. if not right tell me what the B solution is, thanks.
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Old Mar 19th 2017, 07:45 AM   #4
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Aaand no feedback
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Old Mar 19th 2017, 09:07 AM   #5
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Yes you got both parts correct, though you should brush up on your units.
You did better than the OP here.

Finding temperature before combustion

It would help you if you are able to reason out the method without needing formulae

I worked this way

Work out 500 kJ/s
Heat loss 120 kJ/s

total 620 kJ/s

This was provided by a change in internal energy soley due to a drop in specific enthalpy.

H1 - H2 = 3095 - 2660 = 435 kJ/kg

So to provide 620 kJ/sc there must be 620/435 = 1.425 kg/s flowing.

If 1.425 kg provides 500 kJ of work

So 1.0 kg provides 500/1.425 = 350.8 J of work

Does it matter how long this takes will you get more work (not power) if it takes longer?

Last edited by studiot; Mar 19th 2017 at 01:47 PM.
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Old Mar 20th 2017, 04:52 AM   #6
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Thanks for the feedback, I'm glad I did it right hopefully full mark in the exam, fingers crossed.

At the end of your sentence you had a question there? I'm not sure how to answer that, but thank you.
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