Wednesday, April 06, 2011

ENGR 224 - HW #2

This HW covers CB chapters 2, 4 and parts of 6 or LT chapters 3 & 4.

The homework consists of 16 problems for a total of 57 pts.

Please begin your question with the problem number you are asking about.

Cengel & Boles: Ch 3:

3.26(2pts) -
Use either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in.
Use the default reference state for both the NIST and TFT.

3.29E(2pts) -
Use either the NIST Webbook or the Thermal/Fluids Toolbox (TFT) Excel plug-in.
Use the default reference state for both the NIST and TFT.

Cengel & Boles: Ch 4:

4.8(6pts) : W ~ -22 kJ
4.42(5pts) : Q ~ 12750 kJ
4.59E(4pts) : See textbook
4.60(4pts) : ΔU ~ 6200 Btu/lbm

Special Problems

WB-1(5pts) : ΔHvap ~ 575 to 600 Btu/lbm
WB-2(8pts) : ΔH14 ~ 9900 J
WB-3(6pts): Q23 ~ -85 Btu , Wcycle ~ -20 Btu
WB-4(3pts) : q ~ -97 Btu/h-ft2
WB-5(3pts) : Q ~ 3500 W
WB-6(3pts) : Lins ~ 6 cm
WB-7(3pts) : Wb ~ 350 kJ

Cengel & Boles: Ch 6: 6.23(2pts), 6.41(2pts), 6.55(2pts). These problems should be pretty easy.

56 comments:

Jayson said...

Just a quick question regarding integration. Today in class when you did an integration of A+Bt+Ct^2+Dt^3 and you said that it would be A(t2-t1)+B/2(t2^2-t1^2)+C/3(t2^3-t1^3)+D/4(t2^4-t1^4). This confused me because I have never seen it like this, I would think that the integration for that polynomial would be simple At+(Bt^2)/2+(Ct^3)/3+(Dt^4)/4. I am fine with using your method, I am just unaware of how it came about and what the difference is between the two. In other words if I integrate using the method I knew would I get the wrong answer?

Jayson said...

Was that integration after applying the fundamental theorem of calculus? Like the integral evaluated at t2- the integration at t1? I think this is the case?

Jayson said...

Your solution for example 3-C1 does not match the problem

Anonymous said...

When calculating work on 4-8 part b which pressure do you use for P since it changes from the initial phase to final phase?

Dr. B said...

Anon, 6:41pm, 4.8
Good question.
The answer should be very clear if you mentally carry out the process and then draw the process path on a PV Diagram. Hint: this process is really a two-step process. Do you see why ?

I hope this helps.

Dr. B said...

Jayson, 5:42pm, 3-C1
I am sorry, bud, but I don't know what you mean when you say 3C-1. Is that really part of HW#2 ?

I am going to guess that you mean the link to the solutions for Lecture #3 (on the "Solutions" page of the course website) does not work. I just figured that out and I will fix it when I get home today.

Thanks for the heads-up.

Dr. B said...

Jayson 4:53pm and 5:01pm
Yes, I did two steps at once by integrating and applying the limits of integration. I hope this clear things up.

Riley said...

On problem 4-60 for part a it asks you to use the polynomial equation in table A-2c to find delta u over the range of 200-800K. However, the range the polynomial is valid is give to be 273-1800K. Will this approximation still work even though it is out of the valid range?

Dr. B said...

4.60, Riley, 2:31pm
Whoa ! Good eye. I didn't even notice that, even though I put the applicable range in my solution. I just did not think about it. Let's go ahead and use the polynomial anyway for practice. It is most likely STILL better than the methods you are instructed to use in parts (a) and (b). Thank you for pointing this issue out.

Dr. B said...

6.23 Hint
This problem is a bit confusing and I did not realize it until just now. ASSUME that 1.878e12 kW-h is the shaft work produced by turbines in electric power generation facilities in the US.

Why do I think the problem statement is confusing? Well, CB seem to say 1.878e12 kW-h is the amount of electricity produced. This makes the problem difficult to solve because the shaft work from the turbine is used to drive a generator. The issue I have with the problem statement is that not all of the shaft work the turbine produces gets converted into electrical work because generators do not have 100% efficiency !

So, just make the assumption above and carry on !

Anonymous said...

I have a question for number 3-29E. I don't know why, but the NIST webbook keep showing a large number of enthalpy which is completely different from the book. That's why I had a negative number for x. Can you explain why?

Thanks

Dr. B said...

3.29 , Anon , 2:51pm
Which part of this problem is giving you trouble? Unless you cut-and-paste something here, it is impossible for me to tell for sure what went wrong. One possibility is that the DEFAULT reference state for NIST is not the same reference state used in the table in your textbook. This causes several of the answers to this problem to be DIFFERENT than we got in HW#1 using the tables in the book.
OK, here is a good hint. In the 1st row in the table, part (a), quality is undefined because the system is a subcooled liquid.

I hope this helps !

Anonymous said...

How do I calculate the heat transfer in 4-9 part b?

I know that we can use Qin-Wb,out=m(u2-u1). How do I find u1?

Anonymous said...

Dr. B 3-43 PM

Yeah, I got a trouble on part (a) because I had x=-0.243. In my previous HW1, I had x is approximately 0.556 using the table in the book.

Thank you very much!

Dr. B said...

4.8b , Anon , 3:56pm
You know both T1 and P1, so U1 should come from the thermodynamic tables. So, I am guessing you really want to know U2b. Read my comment "Anon, 6:41pm, 4.8" and you will see that you know P2b and V2b and with these 2 intensive properties, you can determine the others (like U2b !).

I hope this helps.

Anonymous said...

hey, i want to ask if the homework is due thursday or friday as in the calendar it is due on thursday whereas in the homework screen, it is due on friday

Jayson said...

In problem 4-8 would it be correct to say that steam is a diatomic gas? Also it says that the stops are at 60% of the initial volume, so would it be a sound assumption that the final volume will be 60% of initial volume since it is cooling? or is that merely a minimum volume?

Dr. B said...

HW Due date , Anon , 11:30am
HW #2 is due on Th, 4/14 at 9 AM.
I will fix the HW page typo.

Dr. B said...

4.8 , Jayson , 3:21 pm
No, steam is not a diatomic gas. Daitomic gases have 2 identical atoms: O2, N2, Cl2, H2, etc.
H2O is not diatomic.

Anonymous said...

how do you get A B and C for the Antoine equation on NIST?

Dr. B said...

WB-2 , Anon , 7:28pm
Follow the 2nd link on the course website:"Search for Species Data by Chemical Name"
or go directly to
http://webbook.nist.gov/chemistry/name-ser.html
Type in the chemical name you are interested in. Then, check the "Phase Change" box and press the Search button.
Scroll down the results page until you find the Antoine parameters.

I hope this helps.

Anonymous said...

How do I do #WB-1?

Dr. B said...

WB-1 , Anon , 12:02am
I think the easiest way to learn how to do this problem would be to study Lesson 3E on the LearnThermo.com website. Alternatively, you could look at pages 668-671 of your textbook. I did not assign this reading from the textbook because the author did not handle the material very well. In any case, I can help you with this problem in my office hours or in our problem session in class on Wed.

I hope this helps.

Anonymous said...

How do you find the constant C in Clausius-Clapeyron Equation for WB-1?

And do we use T in Rankine for both equations?

Thanks

jimak said...

I just figured out part a) Problem 60 Chapter 4 [delta U] = 8668.6 kJ/kg....
It's quite different to the answers from part b) and c)
Does anyone have similar answer with me, or I am wrong somewhere.
I used empirical equation. (almost same as the problem 59E Chapter 4)

Anonymous said...

Why is your Clausius-Clapeyron equation totally different than the one in the book? The book has a P1/P2 term and a 1/T1-1/T2 term which are totally different. What is the constant C supposed to be in your equation?

Anonymous said...

Jimak- my answer for part a of 4-60 is very close to my answers for parts b and c. You probably have an error somewhere

Dr. B said...

WB-1 , Anon , 4:12 pm
You don't need to determine the value of C in the Clausius-Clapeyron Equation. It is just the y-intercept on a plot of Ln{P*} vs 1/T. We are really only interested in the slope for determining the heat of vaporization.

Yes, use T in Rankine.

Dr. B said...

4.60a , Jimak , 9:24 pm
In part (a) you need to integrate the polynomial for the ideal gas heat capacity to get delta-H and then use it to get delta-U. I got about 6200 kJ/kg for delta-U by this method. This is similar to what I got in (b) and (c), so I have to agree with Anon at 9:39pm that you made an error somewhere.

I hope this helps.

Dr. B said...

WB-1 , Anon , 9:35 pm
The Clausius-Clapeyron Equations in LT and in my PPT and in your textbook are equivalent. The eqn in the book is not an eqn of a line but is only a statement of the fact that the SLOPE of the line = Ln{P*2/P*1} / (1/T2-1/T1) is equal to the negative of the latent heat of vaporization divided by the universal gas constant, R. I presented the complete C-C Eqn which looks like y = m x + b, a linear eqn in which y = Ln{P*}, and x = 1/T. The slope is m = -DHvap/R and the y-intercept is the not-so-important constant, C.

I hope this helps.

Jayson said...

I know this seems crazy but i am still stuck on 4.8 on how to get the X value and then using that to get the H final value for part a).

Andrei said...

in problem 3.59, how do you go from kJ/kg*K to Btu/lbm. specifically in part (a) i got an answer for specific heat in kJ/kmol*K and don't know how to go to Btu/lbm as problem asks.
Thanks

Dr. B said...

4.8 , Jayson , 3:45pm
In part (a), P = 1000 kPa. Tsat at 1000kPa is about 180 degC. T1 and T2a are BOTH greater than 180 degC. So, in part (a), there is no liquid present and there is no need to try to calculate the quality. The water is a superheated vapor and you don't even need to interpolate ! So, I think you are good-to-go !

I hope this helps.

Dr. B said...

3.59 , Andrei , 3:59pm
You used the wrong heat capacity table. You used Cp in SI units on pages 909-911. You SHOULD use the Cp tables in AE units on pages 959-961.

I hope this helps.

Andrei said...

I think I just realized that after integration we get "delta"(h) in kJ/kg which is easily converting to Btu/lbm! right?
Andrei

Andrei said...

thank you Professor...and American units tables

Andrei said...

Dr.B,
in 4.8(a) i've got confused,
Q-W(boundary)=mass*"delta"U
the prblem asks to find W(boundary) and Q. I can look up "delta U's but I don't know how to find Q and/or W. so I have 1 equation and 2 unknown (Q and W)?

Andre said...

isobaric W(boundary)=IntgralPdV , and then by finding Wboundary we can find Q. right?

Dr. B said...

4.8a , Andrei , 6:24 pm & 6:35pm
Yes, you should use Wb = INT{P dV} to evaluate Wb and then use the 1st Law to evaluate Q.

Andrei said...

yes it is clear to me and I've solved it up to part where I have to figure out work (b). I understand that, the Work is the area under the graph P-V.Then we have W=("delta"V)*"delta P, but What "delta"P we have to consider 1MPA-0.5MPa or 1MPA-0Mpa...? and to find temperature I just have to look up in the table(sat water at 500kPa(table A-5 p.916)?
Thank you

Dr. B said...

4.8 , Andrei , 7:30pm
Draw the PV diagram that represents what happens in this process and you will have a much better understanding of the process and probably be able to solve the problem.

The process 1 => 2b takes place in 2 steps. The first step ends when the piston just comes down to touch the stops. This process is isobaric and you can fairly easily evaluate Wb = INT{P dV}. In the second step, heat is removed, pressure drops, but volume remains constant. Wb = 0 here.

I hope this helps.

Andrei said...

Thank you, I understand now.
in WB1 part a, I am trying to find "delta"V(specific volume), NIST gives me only Vvap and V(liq) and I don't know how to find V"hat"
in part b, what is the value of C in clausius-clapeyron equation?
part c, having trouble to find the right table in NIST for latent heat vaporization @-10F

Thank you in advance

Camden said...

On problem 4.60 you have to vind cv instead of cp, and since they only have the equation for cp for part a of 4.60, i assume you have to find cp using the a, b, c, and c constants and convert it to cv, but heres where I have a problem. Since its not a incompressible gas cp = cv + R, but how do I include the R in the integration. Do I simply subtract it at the end or integrate with it? I'm kinda fuzzy with integral stuff or i would probably know this

Anonymous said...

Is the W=PdV equation always applicable, or is it conditional, like the Q=DH, Q=DU equations? On 4-42, if i calculate work as the area under a P-V curve, i get a different answer than if i use the Q-W=M(delta(U-hat)). I found the Q utilizing an intermediate state and the Q=DH, Q=DU equations.

Anonymous said...

Camden-
For 4-60(b) i found Cv values from the table on pg 910, and used those in dU=CvdT

Andre said...

hey guys! do you know how to solve WB-2 change in enthalpy for 1-2 path. I am puzzled! I used Antoine equation and found Psat, but don't know what to do next.
thnks

T said...

@Andrei: use slope that we talked in the class today, then multiply with R, you got delta H -wiggle.
Multiply it with mole, you will get delta H. I got around 7340J.

Hope it helps!

Dr. B said...

Andrei 8:26 pm
delta-V-vap = Vsatvap - Vsatliq

In the Clausius-Clapeyron Eqn, C is the not-so-important y-intercept. It is a constant.

Click the "Search for Species Data by Chemical Name" link on the front page of the course website, enter "heptane", choose "Gas Phase" checkbox under the Thermodynamic Properties category and press the search button. Scroll down to the Cp data table.

Dr. B said...

4.60 Camden 8:52pm
DH~ = DU~ + D(PV~)
Assume the gas is ideal and PV = RT
So, use Cp to calculate DH~ and then subtract R*DT to get DU~.

This is equivalent to what you wanted to do. Cp = Cv+R
DH~ = INT{Cp dT} = INT{(Cv + R) dT} = INT{Cv dT} + R*DT = DU~ + R*DT.

Either way, it comes out the same !

I hope this helps.

Dr. B said...

4.42 , Anon , 9:20
Wb = INT{P dV} as long as the process is quasi-equilibrium. It does not depend on whether the process is isobaric, adiabatic or whatever.

I am not sure which part you are talking about, but you cannot determine Q for the whole process without determining Wb first and to do that, you need to use Wb = INT{P dV}. Maybe we can discuss this in class tomorrow or in my office hours.

Dr. B said...

4.60 , Anon , 9:37 pm
Yes, you can use dU = Cv dT and integrate instead, but only in parts (b) and (c). In part (a), you must use the Cp polynomial. See my comment from 10:07 for more information about the relationship between Cp and Cv and how to use them.

Dr. B said...

WB-2 , Andrei , 9:46pm
Use Clausius-Clapeyron just like you did in WB-1.

Best of luck!

Dr. B said...

WB-2 , T , 9:55 pm
Thank you for helping.

Andre said...

Thank you guys for your help!
WB-3 What is P2 for 1-2process in Work(1-2)?
Work(1-2)=(P2V2-P1V1)/(1-1.4);P2-?!

Camden said...

It seems like it should be a simple problem but I just cant seem to figure it out. does anyone know how to solve WB-4?

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