Friday, May 20, 2011

ENGR 224 - Test #2 , May 24, 2011

Please post any questions relating to the second test as comments on this blog post.

Test 2 will focus on CB chapters 6 & 7 or LT chapters 6 - 8. Material from earlier chapters will also be part of this test, although it will not be the focus.

The test will be available for you to take between 8AM and 7PM in the Testing Center on Tuesday, 5/24/11. It is a 2-hour, closed book test. You will be allowed to use TWO 8.5"x11" cheat sheets. You can write or print on both sides of your cheat sheets. You will be penalized ONE point for every TWO minutes over 2 hours that you have the test in your possession. Do NOT forget the time stamps !

We will have an optional test-prep class on Mon, 5/23, and NO CLASS MEETING on Tue, 5/24. There is a QUIZ on 5/25. It is not my fault. The testing center situation messed up our schedule.

Best of luck to you !

42 comments:

Jayson said...

So I was making my Cheat Sheet for the test, and I was putting the table that is on LT CH 8B page 20 and I noticed for polytropic process were delta equals one the equation used starts with -delta/delta-1. However if delta = 1 that means 1/0 which is infinity... what does that mean and it that an error? should it be when delta is greater than 1?

Jayson said...

IF there was a compressor with more than 2 stages, what would be added to the equation for the 2 stage compressor?

Anonymous said...

Would it be a good idea to put the constants for the polynomial on p. 911 on our cheat sheet or would that just be wasted space?

Dr. B said...

Jayson 12:17pm
Yes, this is an error ! The third rwo in the table applies when delta is NOT equal to 1. The equation is correct. Thank you for pointing this out !

Dr. B said...

Jayson 1:18pm
I have never worked out a big equation for the total work in a 3-stage compressor (comparable to the one on page 8C-17. I have also never derived an equation for the intermediate pressures in a 3-stage compressor. Having said that, here is what I THINK.

I think the eqn on page 8C-17 for each compressor would still apply for each compressor in the 3-stage system. I also think that the optimal compression ratio for EACH stage would be the same. For example, let's say that you want to use a 3-stage compressor to increase the pressure from 100 kPa to 2700 kPa. Ii thin kthe optimal outlet pressures for each of the 3 compressor stages would be 300 kPa, 900 kPa and 2700 kPa. So, the compression ratio for each stage would be the same, in this case 3.

Does that help ?

Dr. B said...

Anon 3:03pm
I think that would be a waste of space. If you need any constants for a problem on the test, I will supply them.

Camden said...

Im having a problem with the 2001 exam 2 problem 1. How do I find du for step 12, and 34. I thought I would use cv(t2-t1) but the temperature is constant. maybe its just zero but that didn't seem right. Also I dont know how to find Wb34 because I dont have the pressures for the equation RT1 ln(p1/p2) and I don't see how ideal gas would help. How would I get those or do I use a different equation?

Jayson said...

If a system is closed, even if it is a cycle that indicates no flow work and therefore you have the first law with dU instead of dH?

Jayson said...

The practice tests that are on the site; were they designed to be taken in 1 hour or 2?

Dr. B said...

Hi Jayson,
It has been a few years, but those tests were MOST LIKELY designed for 2 hours. Except the final exams were 3-hours long.

I hope that is a relief !

Dr. B said...

2001, ex2, p#1, Camden

Steps 1-2 and 3-4 are isothermal and deltaU = 0 because U = fxn(T) only for ideal gases.

As we covered in class today, you can get P3 from P2 and P4 from P1 using: TP^((1-gamma)/gamma) = constant. I checked and this is correct in LT. Check out the derivation leading up to this result on page 7E-6.

Once you know P3 and P4 you are in good shape.

I hope this helps.

Dr. B said...

Jayson 11:14pm
If a system is closed, it will usually be easier to use the form of the 1st law that includes deltaU.
Q-Wtotal = deltaU +deltaEkin + deltaEpot

Camden said...

I can find p3 for practice test 2001 #2 but i cant seem to find P4. How would I get p4?

Camden said...

On practice test 2001 #2 problem 2 I get up to Q = Ws +H2-H1, and look up h2 and h1, but I then I dont know where I need to go from there. When it asked about the relation between the heat transfer between the compressor and the surroundings I assumed it was Q in the equation above and whether that was positve or negative would give me the relation. Is this correct or is it asking something else

Jayson said...

When you use a function for a polytropic process involving temperature such as
TP^(1-delta/delta)=C does the temperature need to be in Kelvin and not in Celsius for things to go smoothly?

Jayson said...

Hey Camden when finding P4 in that problem you do the same Process as you did to find P3. T3P3^(1-gamma/gamma)=T4P4^(1-gamma/gamma) and since you already found p3 the only unknown is P4.

Camden said...

thanks for answering. but aren't t3 and t4 the same? and wouldn't that result in p3= p4 which isn't possible? correct me if im wrong. and in response to your last question yes im pretty sure it has to be in kelvin because if you have t1 = 1c and t2 = 50c, then p2 would be essentially multiplied by 50, but if there in kelvin then its 278k for t1 and t2 is 328k which would only multiply p2 by 1.3 ish.

Jayson said...

Actually Camden that process does not work because the process is isothermal. So instead I think you can use process 4-1 to find P4. In other words same thing but T1P1^(1-gamma/gamma)=T4P4^(1-gamma/gamma). I think that will work... I am trying right now.

Jayson said...

In Sample exam 2001 number 1 (the one we were doing in class today. I am calculating Wbhat12 and using the equation for isothermal Wb12=(R/MW)T(Ln(Vhat2/Vhat1). In the table there is no Vhat values, however there is relative specific volume. Are we supposed to know the relationship between Vr and Vhat in order to find Vhat?

Jayson said...

Yes Camden thanks I realized that Kelvin is a must and I decided when in doubt GO KELVIN! Cause Celsius just causes problems.

Camden said...

since its an ideal gas you can substitute vhat for rt/p and since t is constant and the r's are the same your (v2/v1) turns into (p1/p2).

Camden said...

and that last comment was for jaysons 5:21 comment

Dr. B said...

2001 test #2, p1, Camden, 4:22pm
You must be talking about problem #1. You know T1, T2, T3 and T4, right ?
T1*P1^((1-gamma)/gamma) = T4*P4^((1-gamma)/gamma). The only unknown is P4. Solve for it !

Maybe I don't understand your question ?

Dr. B said...

2001 test #2, p2, Camden 4:25pm
Sketchon the TS USING the Shat values you find for states 1 and 2.
The 1st Law is: Q - Ws = H2 - H1 (all hats).
We know Ws < 0 for a compressor.
You looked up H2 and H1, which is greater ?
What can you conclude about the SIGN of Q ? This will tell you the direction of heat transfer.

I hope this helps.

Dr. B said...
This comment has been removed by the author.
Dr. B said...

Jayson, 4:29pm
YES ! T in Kelvin ! Or Rankine.

Dr. B said...

Jayson 4:55pm
No. The equation only applies to isentropic processes and step 3-4 is isothermal, not isentropic.

Dr. B said...

Camden, 4:58.

All correct. T in Kelvin or Rankine and that eqn does not apply to step 3-4, just 2-3 and 4-1.

Dr. B said...

Jayson, 5:06pm
Yes ! Precisely.

Dr. B said...

Jayson, 5:21pm
You might be better off to never look at or even consider the Pr and Vr columns in the ideal gas property tables.
Just use the IG EOS: PV=nRT and all will be well.

Dr. B said...

Camden, 5:35pm
Yes. PV=nRT or PV^=RT/MW or V^ = RT/(P*MW). So, V2^/V1^ = P1/P2.

Camden said...

is Sgen for a cycle only have to consider the boiler and condenser, because the system has a ds of 0, but the entropy of the reservoirs change? so for a common four step cycle, or any cycle really with just two reservoirs, and all the other components are adiabatic, Sgen,cycle = Qcond/t +Qboiler/t and theres no ds right?and the t is the temperature at which the heat transfer occurs, but is that given every time?

Jayson said...

Ok thanks Camden and Dr B. I was using the ideal gas table along with the definition of entropy to find the Qs in that question but since you can replace vhats with pressure all is well.

Thanks for the help!

Jayson said...

During an isothermal expansion does Q go out and therefore this equation would take on this form: S2-S1 = -Q/Tres + Sgen. All with hats. Also would that be Qh or Qc? That last part might be confusing sorry.

Jayson said...

Also on problem one of that exam when I solved for W23 (the isentropic one), I ended up using the first law for turbines Wb=-(H2-H1) with hats. But it got kinda nasty cause you have to interpolate to get both H2 and H1. I know you can use Wb=-Cvnaught*(T2-T1), however I was unsure on how to determine what to plug in for Cvnaught. Basically is the way I did it the quickest way or is there a quicker alternative?

Camden said...

Also theres several other things I got questions on. for an incompressible liquid or solid does ds = cpln(T2/T1). Also what is the difference between cop(the o is supposed to be above the p) and the cp and does polytropic mean it has a constant heat capacity or is that just isentropic?

Dr. B said...

Camden 7:09pm
As long as no heat transfer occurs anywhere else in the cycle, then
Sgen,total,cycle = Qcond/TC +Qboiler/TH
Where Tc is the cold reservoir T in Kelvin or Rankine and TH is the hot reservoir T in Kelvin or Rankine.

Dr. B said...

Jayson 8:19pm
Think about an expansion of an IG. The T would tend to drop. So, to keep the process isothermal, heat must be transferred INTO the system.
S2-S1 = Q/Tsys + Sgen,int
or
S2-S1 = Q/Tres + Sgen,total
If this is part of a power cycle, then this Q is QH because it goes INTO the system. But if it is a refrigeration cycle, then this Q is QC.

I hope this helps.

Dr. B said...

Jayson 8:37pm
If you have an IG and Cp & Cv are NOT constants, then it is quickest and best to use the ideal gas property table. I CP is constant, then you can use the formula for isothermal processes (polytropic, delta=1, IG) to quickly calculate work.
Open systems use the deltaH form of the 1st Law. Closed systems, use the deltaU form of the 1st Law.

Best of luck !

Dr. B said...

Camden 8:39pm
For incompressible liquids and solids with constant Cp, yes deltaS = Cp Ln(T2/T1).

The superscript "o" means that the ideal gas heat capacity. This Cp is technically only useful when the ideal gas assumption is valid. ALL of the gas phase Cp we use in this course are really ideal gas heat capacities: CoP.

Polytropic ONLY means that a process follows a process path that is described by the equation: PV^delta=constant. Any type of fluid can follow such a path. IF IF IF the fluid is an IG with constant Cp, Cv and gamma, then an isentropic process follows a polytropic path with delta=gamma.

Polytropic does NOT imply IG OR constnat Cp. It only means PV^delta=constant.

Best of luck!

Tuan said...

Problem 2, part e)
when I calculate S12gen,int, does the sign of Qdot12 matter because I think Qdot12 = -76kW

Calvin said...

I got a question on test 2 part d, so to find the COP of refrigeration cycle are we going to use the equation 1/((QH/Qc)-1)? and to find the QH/Q34 i used the first law,
Q34-W34=mdot(H4-H3)..is the work done in step 34 is 0? i am not sure with this.