- #1

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So this is what I was thinking:

1) change deg. C to deg K

2)interpolate for h

3)Wcv=Qcv+[mi1(hi1+Vi1^2/2)+mi2(hi2+Vi2^2/2)]-me(he+Ve^2/2)

However: I think I should find the entropy change 1st but, I'm unsure how to

- Thread starter CRich
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- #1

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So this is what I was thinking:

1) change deg. C to deg K

2)interpolate for h

3)Wcv=Qcv+[mi1(hi1+Vi1^2/2)+mi2(hi2+Vi2^2/2)]-me(he+Ve^2/2)

However: I think I should find the entropy change 1st but, I'm unsure how to

- #2

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howver the answer my prof gave me is 307.4K .... why isn't my equation correct?

- #3

- 12,121

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Your equation would be correct if no energy is added to the air. If energy

- #4

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steady state means m1=m2 but in this case can you say m1+m2=m3?

- #5

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I know:

T1, T2, and I found T3

P1, P2, P3

cp=1 kJ/kgK

if m1+m2=m3 then m3=50kg/min ... right?

- #6

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Yes, steady state meanssteady state means m1=m2 but in this case can you say m1+m2=m3?

m_{inlets} = m_{outlets}

- #7

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I ment "how do I find Wcv"?

- #8

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also how do I relate d S = δq/T to my problem and my knowns or can/should I?

- #9

Q_Goest

Science Advisor

Homework Helper

Gold Member

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Normally, the first law is written:

dU = dQ + dW

For a control volume with flow in and/or out, heat in and/or out, work in and/or out, and the change in internal energy being only that change internal to the control volume, the first law can be rewritten:

dU = dQ + dW + dH

Where dH = H

and H=mh (mass flow rate times specific enthalpy)

This assumes energy entering the volume is positive and energy leaving the volume is negative.

For you problem then, the first law can be rewritten:

dU = Win + H1

For a steady state process, dU = 0 so you can now solve for work. Note that to do this, you need a way of determining h.

Also, you can neglect velocity on the inlet and outlet. Although it is more correct to include it, there isn't any information that allows it to be calculated and even if there was, the error imposed is generally less than the error in the values for enthalpy in your database.

- #10

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- #11

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okay so my T equations is correct except for the addition of work...

In order to calculate work I tried :

m1h1+m2h2+m3h3=0

m1h1+m2h2+(m1+m2)h3=0

m1h1+m2h2+m1h3+m2h3=0

m1(h1-h3)+m2(h1-h3)=0

dh=CpdT

m1Cp(T1-T3)+m2Cp(T2-T3)=0

inserting my values

m1=30kg/min

m2=20kg/min

Cp=1kJ/kgK

T1=-50C=223.15K

T2=150C=423.15

T3=m1T1+m2T2/m3=303.15

30*1*(223.15-303.15)+20*1*(423.15-303.15)=0 .... But according to my prof W=-3.67kJ

So what am I doing wrong???????????

Also how do I find δ? What equation do I use? How do I use the 2nd Law to find δ?

- #12

Q_Goest

Science Advisor

Homework Helper

Gold Member

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Hi CRich. That's a good shot at it. What you've done however is to simply calculate this as an isenthalpic process. There's no calculation to see what pressure should be assuming the low pressure stream is isentropically compressed.

I've been playing around with this for a while now and I still haven't nailed it down. However, here's what I'd suggest.

The best that one could hope for is to take the work that comes from the isentropic expansion of flowstream 1 from 200 kPa to outlet pressure (190 kPa), and put that work into an isentropic compression of flowstream 2.* The pressure in flowstream 2 must then come up to some higher pressure. If that's not high enough to bring that flow stream up to the outlet pressure (190 kPa), then additional work must be put in. You can assume that work is put in at 100% isentropic efficiency as eluded to in the problem statement.

Once you've brought flowstream 2 up to outlet pressure, the rest is just the mixing of two flow streams which you've already done (enthalpy in = enthalpy out). This gives you the final outlet temperature.

I appologize for not being more helpful. I'll have another look tomorrow and see if I can finish solving this but I've run out of time tonight. Good luck on your exam.

*Edit: Actually, if you were to transfer as much heat as possible from flowstream 2 to flowstream 1 prior to expanding to the lower pressure, you could get more work out of flowstream 1 and it would take less work to compress flowstream 2. That could be done using a perfect counterflow heat exchanger. The part I struggle with is how to determine that this is theoretically, the best possible process. I suspect it is but can't prove it.

I've been playing around with this for a while now and I still haven't nailed it down. However, here's what I'd suggest.

The best that one could hope for is to take the work that comes from the isentropic expansion of flowstream 1 from 200 kPa to outlet pressure (190 kPa), and put that work into an isentropic compression of flowstream 2.* The pressure in flowstream 2 must then come up to some higher pressure. If that's not high enough to bring that flow stream up to the outlet pressure (190 kPa), then additional work must be put in. You can assume that work is put in at 100% isentropic efficiency as eluded to in the problem statement.

Once you've brought flowstream 2 up to outlet pressure, the rest is just the mixing of two flow streams which you've already done (enthalpy in = enthalpy out). This gives you the final outlet temperature.

I appologize for not being more helpful. I'll have another look tomorrow and see if I can finish solving this but I've run out of time tonight. Good luck on your exam.

*Edit: Actually, if you were to transfer as much heat as possible from flowstream 2 to flowstream 1 prior to expanding to the lower pressure, you could get more work out of flowstream 1 and it would take less work to compress flowstream 2. That could be done using a perfect counterflow heat exchanger. The part I struggle with is how to determine that this is theoretically, the best possible process. I suspect it is but can't prove it.

Last edited:

- #13

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Also how do I find δ? What equation do I use? How do I use the 2nd Law to find δ?

- #14

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I tried

-entropy production=m1(cp*ln(T3/T1)-R*ln(P3/P1)+m2(cp*ln(T3/T2)-R*ln(P3/P2)

but I don't get the answer my prof gave ... and even if I did how would I use my 'knowns' to calculate work?

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