A piston-and-cylinder device contains 5 kg of steam at 100°C with a quality of 50 percent. This steam undergoes two processes as follows:
1-2 Heat is transferred to the steam in a reversible manner while the temperature is held constant until the steam exists as a saturated vapor.
2-3 The steam expands in an adiabatic, reversible process until the pressure is 15 kPa.
a.) Sketch these processes with respect to the saturation lines on a single TS diagram
b.) Determine the heat added to the steam in process 1-2, in kJ
c.) Determine the work done by the steam in process 2-3, in kJ
17 comments:
Im getting confused on the first part, I think we have an isothermal internally reversable process so delta s = Q/T. I know T as 100+273.15 K and I can find entropy by looking at the tables which i find them to be S2 = 7.3541 and S1= 4.33065 using the quality given and the fact that at step two the entopy is all due to the saturated vapor But this gives me the wrong answer.
Anon 5:52
I am not sure, but I think you did the problem correctly. The answer in my hints is wrong. I apologize. I do not know where those numbers came from. I just fixed the hints and I will send an email to my class. The correct answers are:
b.) 1128 kJ/kg
c.) 1290 kJ
graham 5:56 PM
You are correct. See my response to anonymous.
Now for part b I tried two methods the first method was that I know Wb is (Uhat1-Uhat2) and I know that the precess is isentropic I looked in the tables for steam at saturation for sat vap s and i got s1 =7.3541 yet when I go to look for s2, and i flip to the superheated vapor page, an s2 at 7.3541 is off the charts and when I nist it, the readings jump from like .00 something to 8 so I have no idea how to find Uhat 2 if I cant find the corresponding entropy value of 7.3541 under a pressure of .015MPa
Anon 10:03 PM
I think you have confused the steps in the process. Draw the TS diagram and label the points for each of the three states in this process.
In part (b), step 1-2, is isobaric, so the Wb combines with deltaU to become deltaH. The Wb is not zero.
S2 = S3 = 7.3541 but S1 is not 7.3541.
Use x1 & T1 to get H1. Use x2 & T2 to get H2. Then, use the 1st law to get Q12.
Something is just not working here. For part c. the 1st law boils down to W = -delH. You know that S2 = S3 so that gives you the quality. H1 is the sat vapor at 100deg, H2 is a sat mix and .015MPa and at the quality found using S, subtract and times by 5kg should give the right answer....what am I doing wrong?
Katie,
Not to take your power dr. b, but I think I can help. In part c. the process is adibatic, deltaQ is zero, so the first law term, deltaQ - deltaW = deltaU, turns into deltaW = -deltaU. Hope it helps.
Katie and Adam
Adam correctly diagnosed the problem with your approach, Katie. Thank you Adam.
But never, never, ever write or think deltaQ or deltaW ! The integral of dQ is Q not deltaQ and the integral of dW is W not deltaW. This is true because Q and W are PATH variables.
So, the correct form of the 1st Law for (c) is Q - W = deltaU. Since the process is adiabatic, Q = 0 and the 1st Law simplifies to W = -deltaU.
I'm totally failing on calculating U3 so I can find the boundary work. I assume that the process is isentropic but I am unsure how to incorporate this information to find U3. I tried interpolating between values at the pressure 0.15 kPa and I get a negative value close to 0. Can anyone point me in the correct direction?
Lisa 3:08 PM
OK, if step 2-3 is isentropic, then S3hat = S2hat. S2hat is not hard to get because it is a sat'd vapor at 100 degC.
So, now we know both S3hat and P3. This is enough info to allow us to use the saturation pressure tables to evaluate x3. Once we have that we use it to calculate U3hat.
Let me know if this doesn't get you going in the right direction.
So im confused. I understand that you can calculate Q from the entropy equation, but shouldnt you also be able to use the 1st law and say that Q is equal to deltaH plus W? I guess I'm confused about how we're supposed to know how to solve certain problems
Anon 4:36 PM
I don't think you are confused. The 1st Law always applies and is almost always useful (at least in this class :)
Q and deltaS are only easily related through the definition of entropy for internally reversible processes. So, usually you cannot relate Q and deltaS in this way.
In this problem (and many others) you can reliably use the 1st Law. The only question is which form of the 1st law is easiest to use.
These two processes occur in a closed system, so the best form of the 1st Law to choose is: Q - Wtotal = deltaU. In this problem, the only form of work is Wb. Apply this form of the 1st Law to steps 1-2 and then 2-3 to determine Q12 and Wb,23.
Because step 1-2 occurs at constant P in a closed system, Wb = PdeltaV = delta(PV), P = constant. When you combine this fact with the form of the 1st law that I mentioned above you get Q = deltaU + Wb = deltaU + delta(PV) = deltaH !
Oooh wow, for some reason I thought it was a vapor at the third stage? Oh well, now I get it. Thank you for your help Dr. B!
Lisa 6:07 PM
Yep, a lot of people thought that.
Draw the TS diagram. State 2 is Sat'd Vap. Now, we take some of the weights off the back of the piston, so the sat'd vapor expands. So, when you think of a PV diagram, you think you have a superheated vapor.
But here the gas expands adiabatically. So the T drops ! On the TS Diagram you can see that as the T of a sat'd vapor drops isentropically the state of the systems slides straight down into the 2-phase envelope ! Kinda cool. (geek humor).
How do you calculate u3 by only using P3 and S3, i see that you describe that it is possible to find the x by the given properties.
Btw great blog!
Got it ! (S3-Sf@15kpa)/Sfg@15kpa => x = 0.9099
Good work!
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