# NCEES 2008 # 512 Thermal and Fluids



## tmacier (Oct 8, 2010)

Pluggin away at these problems and in general getter better and better.

Working thru Problem 512 the answer starts with h1=h2=1,224.6 Btu/lbm (from steam table)

I cannot for the life of me figure where in the steam table they obtain this value?

I am assuming it is at 60 PSIA and I am looking under table 24.b @ 60 psia and I see H satvap @ 1178.

Can someone point me to what I am missing?

Thanks in advance for your time.

Tim


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## NerdHerd (Oct 8, 2010)

tmacier said:


> Pluggin away at these problems and in general getter better and better.
> Working thru Problem 512 the answer starts with h1=h2=1,224.6 Btu/lbm (from steam table)
> 
> I cannot for the life of me figure where in the steam table they obtain this value?
> ...


__________

Hi Tim,

Try this:

1)	You know that at state 1, P1 = 120 psia and T1 = 400 F. When you look at your steam tables by pressure at 120 psia, you see that the vaporization temperate is 341.25 F, which means you are now in the superheated region because your T1 = 400 F. Proceed to the superheated tables.

2)	Now, at the superheated tables there are values for pressures and enthalpy at 100 psia and 150 psia. Your P1 state however is in between the given values in the table, therefore you must interpolate as follows:

•	At 100 psia, h = 1227.7 btu/lbm,

•	At 150 psia, h = 1219.7 btu/lbm,

•	h(@120psia) = 1227.7 + ((120-100)/(150-100))*(1219.7-1227.7) = 1224.5

•	h(@120psia) = 1224.5 btu/lbm

The solutions in the book are not very good. They leave out a lot of important details.

I hope this helps. Good Luck!

NerdHerd


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## tmacier (Oct 9, 2010)

THank you so much for taking the time.

I spent hours yesterday working this problem, and was really focusing on the 60 PSIA.

That throttling valve thru me a curve ball!

I fully understand now.

THanks again!

Tim


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## abourne (Oct 26, 2010)

Throttling valve is Isenthalpic.

Five rules to know for Thermo:

1) Isothermal - Constant T

2) Isobaric - Constant P

3) Isentropic - Constant entropy

4) Adiabatic - no change in heat (Q)

5) Isenthalpic - no change in enthalpy (i.e. throttling device)


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## buffteya4 (Dec 20, 2012)

Hmm, this problem is ridiculous. I guess Ill look at the solution now. I got the throttling valve part, but couldnt put together where they got the equations H3=hsat-cpDeltaT and m2h2+m3h3=m4h4....


nervous for April


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## ongreystreet (Dec 20, 2012)

buffteya4 said:


> Hmm, this problem is ridiculous. I guess Ill look at the solution now. I got the throttling valve part, but couldnt put together where they got the equations H3=hsat-cpDeltaT and m2h2+m3h3=m4h4....
> 
> 
> nervous for April


I just passed the T/F, and I remember that practice question vividly, it's a pretty good mash-up of a lot of ideas that are common on the practice tests and so they could be seen often on exam.

I don't remember the specifics of the question, but h(enthalpy) = Cp*deltaT, so you've found your hsat and lowered the temp I am assuming, so also lowering the enthalpy?

Just like superheated steam is hs = hg + Cp*deltaT


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## ongreystreet (Dec 20, 2012)

buffteya4 said:


> Hmm, this problem is ridiculous. I guess Ill look at the solution now. I got the throttling valve part, but couldnt put together where they got the equations H3=hsat-cpDeltaT and m2h2+m3h3=m4h4....
> 
> 
> nervous for April


I can tell you this much, I didn't get to practicing those sort of problems until 3 or 4 weeks before exam, so you are pretty far ahead of schedule.


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