Learning undergraduate engineering thermodynamics might be less painful with a blog. I hope that students, faculty and interested observers will share their thoughts on the laws of thermodynamics, phase and chemical equilibrium and many related topics.

I have a question on The example # 1 exam from 2001. On question number 1 there is a polytropic process, but there is no polytropic equation given. Are you supposed to make one up or what?

Jayson , 12:03 pm Yes, sort of. A polytropic process follws a path of the form: PV^d=C. Since you know P1, V1, P2, V2, you can determine d. This allows you to use the eqn I gave in LT and in my PPT for Wb for polytropic processes with d not= 1.

Oh, okay so I set P1V1^d = P2V2^d? Then do I need to take the natural log of both sides to bring down the d? And if I do that does the pressures go in the front of the natural log or inside? for instance After taking the Ln of both sides would it be d*P1LnV1 or would it be d*LnP1V1?

nevermind it makes senses disregard that last one again I was talking crazy talk. And you were right it was too hard to study on a sunny Saturday I had to get out of the house.

Unfortunately I got stuck again cause have Wb = -16.945kJ/(1-d). The problem is finding the boundary work and I know it should be positive cause it is work by the system on the surroundings, so d will be larger than 1, but I still have two unknowns and only one equation and I cannot figure out the other equation. Please help!!

Professor, Test #1 Problem 3 - Heat and Work for a Cycle Executed in a Closed System Containing R-134a 26-Oct-04

I have a question about step 1-2 in his problem, if we have P1<P(sat@-10C) 200kPa<200.74kPa( value from our textbook), then R-134 must be in saturated mixture phase, but solution says it is superheated, Please explain this to me, am I missing something here?

Jayson 7:49pm Well, d > 1 is true, so Wb>0 too. All is well. Just solve for d. I found my solution for this ancient test. It is not an ideal gas, so I used the generalized compressibility chart. You can use any EOS you like, just not the Ideal Gas EOS. I got V1 = 0.00987 m3/kg and V2 = 0.01588 m3/kg. These led me to d = 1.07 and Wb = 115 kJ +/- 5 kJ because choice of ES affects V1 and V2.

Andrei 8:38pm Imagine that you have a piston and cylinder device filled with saturated R-134a vapor: -10C and 200.74kPa. This P is maintained by a small stack of weights on the back of the piston. Now, imagine taking some of the weights off until the P inside the cylinder drops to 200 kPa. What happens and what is the phase of the R-134a in the cylinder when it reaches equilibrium at -10C and 200 kPa ? Draw yourself a nice PV diagram. Put the 2-phase envelope on it and 2 or 3 isotherms. Now, draw the process path for the process I just described. I think you will see that at -10C and 200 kPa the R-134a is indeed a superheated vapor.

I hope this helps. Keep working. I am sure it will pay off on Monday.

After some thinking and "imagination" I understand now. It is easy to understand superheated substance based on Temp, then Pressure. Thank you for your help.

On problem 3 of test 2 of 2004, it asks for x2 and I thought that x was vapor kg/ total kg, so I thought I could just take 7.48727kg/153.504 kg, (7.48727 is the mass of the air in the tank)because only liquid leaves the tank, but this doesn't work and I have to use the x = (v2 -vsatliq)/(vsatvap- vsatliq) I was just wondering why this is so

Also I saw isentropic efficiency on some of the problems, will that be a possible topic on the test. I dont remember reviewing it and it wasnt on the summary sheet, but I just wanna make sure, and also I see on the summary sheet that it only goes up to chapter 4, is there no chapter 5? no flow work shananigans?

Andrei , 9:19pm Yes, it is easier to imagine that superheat is only related to pressure. Similarly, you only think about liquids boiling as you heat their temperature increases. But they boil just the same as pressure decreases at constant T !

Camden , 6:05pm & 6:08pm The test tomorrow covers Ch 1-5 in CB & LT. The old tst #1's only covered Ch 1-4. You should look at the old test #2's for some Ch 5 test problems.

No, isentropic efficiency is NOT part of our test tomorrow. It is not in Ch 1-5 in either CB or LT.

I think you stumbled ontoisentropic efficiency in some old test #2 problems because those tests covered LT Ch 5-8. If it is not in Ch 1-5 (CB & LT), it is not on the test tomorrow.

Camden , 5:33pm , 2004 test2, prob3 Quality, x, is the mass of vapor divided by the total mass of liquid and vapor in the system.

But 7.48 kg is the mass of R-134a vapor in the tank initially. There is NO AIR in the system. You do not know the mass of R-134a vapor in the system in the final state. So, you must determine it "when half of the total mass has been removed from the tank". Not when half of the VAPOR in the tank has been removed, but when half of the total mass of R-134a in the tank has been removed.

So, it will take a bit more thought and work to determine x2 and that is what I did in part (b) of my solution.

I hope this helps. Blog again if this is still not clear.

kay thanks i get it now, but I got another question. On 2001 problem 2 I get the -Wb12 = delta u and Q23 = delta u and on wb12 I find T2 using the ideal gas law to be 641.983 K and I attempt to set up cv = cp-R = (3.355+.575*10^-3*T-1600/T^2)*8.314+8.314 and assuming T is 641.983 , i dont know if thats right, i get cv = 1.35325 kj/kg K after the conversions and using that T value, which leaves me with W = 93.1482 kj kg which isnt the 115 kj im looking for but its close. Do you know what I'm doing wrong?

Camden , 8:24pm & 8:28pm Good work ! Yes, I do see where you went wrong, but it was right near the end ! Everything was fine until you tried to USE Cv to evaluate deltaU. What you did was plug values into the polynomial to get a single constant value for Cv and that is not correct. You need to INTEGRATE the polynomial from T1 to T2 (300K to 642K). Doing this, I got W12 = -256.6 kJ/kg.

Problem 1 test 1 2001-well then after I take the Ln of both sides, I get 5*d*ln(.o5045)= 3*d*ln(.078435) so how do I solve for d from that? I think I must have done it wrong cause the way I have it written, if you divide by d it goes away...

Jayson , 8:45pm Try setting up this eqn: P1 V1^d = P2 V2^2 and doing some algebra before taking the LN of both sides. P1/P2=(V2/V1)^d LN(P1/P2)=d LN(V2/V1) ...

Best of luck

I am not sure practicing algebra right now is the best way to use your time. Just a thought. Gnite.

Is there going to be something like 2001 problem 3 on the test tommorrow, cause that Qe and Qs, and We and Wp is kinda throwing me for a loop. would the Qe and Qs just be Q or does that energy balance tie in somehow?

## 27 comments:

I have a question on The example # 1 exam from 2001. On question number 1 there is a polytropic process, but there is no polytropic equation given. Are you supposed to make one up or what?

Jayson , 12:03 pm

Yes, sort of. A polytropic process follws a path of the form: PV^d=C. Since you know P1, V1, P2, V2, you can determine d. This allows you to use the eqn I gave in LT and in my PPT for Wb for polytropic processes with d not= 1.

Hard to study on a sunny Sat. Best of luck !

Oh, okay so I set P1V1^d = P2V2^d? Then do I need to take the natural log of both sides to bring down the d? And if I do that does the pressures go in the front of the natural log or inside? for instance After taking the Ln of both sides would it be d*P1LnV1 or would it be d*LnP1V1?

On our cheat sheet do we get front and back for our notes?

Camden 6:02pm

Yes, front and back.

nevermind it makes senses disregard that last one again I was talking crazy talk. And you were right it was too hard to study on a sunny Saturday I had to get out of the house.

Unfortunately I got stuck again cause have Wb = -16.945kJ/(1-d). The problem is finding the boundary work and I know it should be positive cause it is work by the system on the surroundings, so d will be larger than 1, but I still have two unknowns and only one equation and I cannot figure out the other equation. Please help!!

Thanks!

Professor,

Test #1

Problem 3 - Heat and Work for a Cycle Executed in a Closed System Containing R-134a 26-Oct-04

I have a question about step 1-2 in his problem, if we have P1<P(sat@-10C) 200kPa<200.74kPa( value from our textbook), then R-134 must be in saturated mixture phase, but solution says it is superheated, Please explain this to me, am I missing something here?

Jayson 1:46pm

Yes, take LN of both sides and solve for d.

Jayson 7:29

Good. I hope you had a good day. It was far too beautiful for school work.

Jayson 7:49pm

Well, d > 1 is true, so Wb>0 too. All is well. Just solve for d. I found my solution for this ancient test. It is not an ideal gas, so I used the generalized compressibility chart. You can use any EOS you like, just not the Ideal Gas EOS. I got V1 = 0.00987 m3/kg and V2 = 0.01588 m3/kg. These led me to d = 1.07 and Wb = 115 kJ +/- 5 kJ because choice of ES affects V1 and V2.

It will rain tomorrow, for better and for worse.

Andrei 8:38pm

Imagine that you have a piston and cylinder device filled with saturated R-134a vapor: -10C and 200.74kPa. This P is maintained by a small stack of weights on the back of the piston. Now, imagine taking some of the weights off until the P inside the cylinder drops to 200 kPa. What happens and what is the phase of the R-134a in the cylinder when it reaches equilibrium at -10C and 200 kPa ? Draw yourself a nice PV diagram. Put the 2-phase envelope on it and 2 or 3 isotherms. Now, draw the process path for the process I just described. I think you will see that at -10C and 200 kPa the R-134a is indeed a superheated vapor.

I hope this helps. Keep working. I am sure it will pay off on Monday.

After some thinking and "imagination" I understand now. It is easy to understand superheated substance based on Temp, then Pressure.

Thank you for your help.

On problem 3 of test 2 of 2004, it asks for x2 and I thought that x was vapor kg/ total kg, so I thought I could just take 7.48727kg/153.504 kg, (7.48727 is the mass of the air in the tank)because only liquid leaves the tank, but this doesn't work and I have to use the x = (v2 -vsatliq)/(vsatvap- vsatliq) I was just wondering why this is so

Also I saw isentropic efficiency on some of the problems, will that be a possible topic on the test. I dont remember reviewing it and it wasnt on the summary sheet, but I just wanna make sure, and also I see on the summary sheet that it only goes up to chapter 4, is there no chapter 5? no flow work shananigans?

nevermind forget that last comment except the isentropic efficiency part, my bad.

Andrei , 9:19pm

Yes, it is easier to imagine that superheat is only related to pressure. Similarly, you only think about liquids boiling as you heat their temperature increases. But they boil just the same as pressure decreases at constant T !

Camden , 6:05pm & 6:08pm

The test tomorrow covers Ch 1-5 in CB & LT. The old tst #1's only covered Ch 1-4. You should look at the old test #2's for some Ch 5 test problems.

No, isentropic efficiency is NOT part of our test tomorrow. It is not in Ch 1-5 in either CB or LT.

I think you stumbled ontoisentropic efficiency in some old test #2 problems because those tests covered LT Ch 5-8. If it is not in Ch 1-5 (CB & LT), it is not on the test tomorrow.

Camden , 5:33pm , 2004 test2, prob3

Quality, x, is the mass of vapor divided by the total mass of liquid and vapor in the system.

But 7.48 kg is the mass of R-134a vapor in the tank initially. There is NO AIR in the system. You do not know the mass of R-134a vapor in the system in the final state. So, you must determine it "when half of the total mass has been removed from the tank". Not when half of the VAPOR in the tank has been removed, but when half of the total mass of R-134a in the tank has been removed.

So, it will take a bit more thought and work to determine x2 and that is what I did in part (b) of my solution.

I hope this helps. Blog again if this is still not clear.

kay thanks i get it now, but I got another question. On 2001 problem 2 I get the -Wb12 = delta u and Q23 = delta u and on wb12 I find T2 using the ideal gas law to be 641.983 K and I attempt to set up cv = cp-R = (3.355+.575*10^-3*T-1600/T^2)*8.314+8.314 and assuming T is 641.983 , i dont know if thats right, i get cv = 1.35325 kj/kg K after the conversions and using that T value, which leaves me with W = 93.1482 kj kg which isnt the 115 kj im looking for but its close. Do you know what I'm doing wrong?

sorry keep everything the same but the 8.314's are supposed to be .008314 and thats how i got those results, I just typed those 8.314's by mistake

sorry again i got 462.788 kj/kg after i found another mistake which is about 4 times 115, so maybe thats connected i dont know

Camden , 8:24pm & 8:28pm

Good work ! Yes, I do see where you went wrong, but it was right near the end ! Everything was fine until you tried to USE Cv to evaluate deltaU. What you did was plug values into the polynomial to get a single constant value for Cv and that is not correct. You need to INTEGRATE the polynomial from T1 to T2 (300K to 642K). Doing this, I got W12 = -256.6 kJ/kg.

Note W = 115 kJ is the answer to problem 1 !

Carry on !

Problem 1 test 1 2001-well then after I take the Ln of both sides, I get 5*d*ln(.o5045)= 3*d*ln(.078435) so how do I solve for d from that? I think I must have done it wrong cause the way I have it written, if you divide by d it goes away...

Jayson , 8:45pm

Try setting up this eqn: P1 V1^d = P2 V2^2 and doing some algebra before taking the LN of both sides.

P1/P2=(V2/V1)^d

LN(P1/P2)=d LN(V2/V1) ...

Best of luck

I am not sure practicing algebra right now is the best way to use your time. Just a thought.

Gnite.

Is there going to be something like 2001 problem 3 on the test tommorrow, cause that Qe and Qs, and We and Wp is kinda throwing me for a loop. would the Qe and Qs just be Q or does that energy balance tie in somehow?

would it be overkill to have the SRK EOS on our cheat sheet?

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