Wednesday, May 10, 2006

HW #8, P#8 - Entropy Generation and Lost Work in an Evaporator - 10 pts

Air enters the evaporator section of a window air conditioner at 100 kPa and 37°C with a volumetric flow rate of 10 m3/min. The refrigerant, R134a at 150 kPa with a quality of 0.3 enters the evaporator at a rate of 0.8 kg/min and leaves as saturated vapor at the same pressure. Determine the exit temperature of the air and the rate of entropy generation for this process ...
a.) Assuming the outer surfaces of the air conditioner are perfectly insulated
b.) Assuming heat is transferred to the evaporator of the air conditioner from the surrounding medium at 37 °C at a rate of 30 kJ/min.
c.) The lost work in kW, for both parts (a) and (b).

24 comments:

Anonymous said...

I'm lost on calculating Sgen! I thought for part a it would be mdot(deltaS) = Sgen. So I assumed mdot was the sum of mdot1 and mdot2, and deltaS was the sum of deltaS12 and deltaS34. For deltaS34 I just found the change in entropy using the NIST table. For delta S12 I found the new volume using Vr, found the new pressure, and used the ideal gas Entropy function to get the change in entropy (I interopolated to get the Sknot for T1 and T2). My answer is off by a factor of 100, but I think I have a concept error rather than a math error. Any guidance would be really appreciated!

Dr. B said...

Lisa 11:58 AM
Your intuition is correct. There are some conceptual problems in your approach.

You are on the right track about how to calculate Sgen for part (a) but this is a MIMO system. Sgen = Sum{mi Si}out - Sum{mi Si}in = mref*deltaSref + mair deltaSair.

Your method for calculating deltaSref is correct. I think you also deterined deltaSair correctly, but I am not completely sure. I think for this process, it would be easier to use Gibbs 2nd Eqn and the ideal gas entropy function: S^o. The air pressure is the same at the inlet and at the outlet.

Does this mean 1-7 are done ? If so, whoohoo !

Anonymous said...

Ooh, okay, my problem was that I was finding a new pressure for the air using Vr and those relative property equations. Thank you so much!

And yep, all done now. Not like this is a common event or anything o_o

Dr. B said...

Lisa 11:39 AM

Congratulations ! Enjoy the rest of the week !

Anonymous said...

Could you remind me how to convert the volumetric flow rate of air to the mass flow rate, pretty please?

Anonymous said...

So i calculated H2 of the air by setting mdot(deltaH)R134 = -mdot(H2-H1)air, which i looked up values for H1, H2 R134,NIST and H1 of Air in the back tables. Then I cannot look up T2 with H2 in the back becuase there is nothing to interpolate with and there is no "air" chemical in NIST. how are you supposed to find T2?

Dr. B said...

Julie 4:09 PM
Use the IG EOS: PV=nRT
or P Vdot = ndot R T
n = m / MW
or: ndot = mdot / MW
Putting all of this together, we get:
P Vdot = (mdot R T)/MW
Just solve for mdot !

Dr. B said...

Anon 6:12 PM
Your method sounds correct to me. I think you must have made a calculation error in determining H2air. You are supposed to determine H2air in the way you described and then go back to the ideal gas property tables and use H2 to interpolate and determine T2. What value did yo get for H2 ? I got about 87 kJ/kg.

Anonymous said...

Im getting lost at the entropy part of a, I know my m air which is 11.24 kg/min which works beacuse it works for part one.. well i shouldent say it works but it gives me the right answer. Then I know the m of air which is given as .8kg/min and I know delta s for the refridge cause I can use nist and I get .5788kj/kg. Then I know the delta s for the air beacue i used the ideal gas property tables to find my T2 and then just looked over at the column for H and i got -.035kj/kg. I plug this all in and bam..close but not close enough!! around .7ish when it should be 1.2. ideas??

Anonymous said...

i got 1.4 for Sgen for a... then used same method on b except with Q/T factor in and got 4.4... why would i be off by such a factor, when in calculating Sgen the only number that changes in the Sumation balance equation is S2 of Air and the added -Q/T number?

Dr. B said...

Chris Lee 11:18 PM
Your mair is correct.
I am not sure what you mean by deltaS for the refridge. I am guess from the numbers that this is deltaS for the R-134a. Am I right about that ?
Yes the next step is to lookup S^o for T1 and T2. Here you said H= -0.035 kJ/kg. I think you meant deltaS air = -0.035 kJ/kg-K. Am I correct about this ? You got about 0.7 but for what ? I need more info here and some units too. My answer is in W/K. Can you tell me how calculated Sgen ?

Dr. B said...

Anon 1:16 AM
I cannot tell what went wrong for you in part (a) from what you told me.

In part (b), the temperature of the air leaving the evaporator is not the same as it is in part (a). You need to apply the MIMO form of the 1st Law again with Q = 0.5 kW and solve for H2^o and interpolate for T2air again.

Anonymous said...

i am having some problems with the entropy generation for part a. Sgen = mref*deltaSref + mair deltaSair.

Mref = 0.8 kg/s
Mair = 0.187 kg/s

I am solving for deltaSref from the saturation R124a tables. I am looking at P = 150kpa, and then finding the S values for x = 0.3, and x= 1.0.

I get deltaSref = 0.668 Kj/kgK

I am solving for deltaSair simply from the Air IG table. Using 37*C and 26.5*C.

I get deltaSair = 0.0327Kj/Kg-K.

then I can calculate Sgen to be 0.539 KW/K..

can anyone tell me what concept I have completely wrong. Thank you

Anonymous said...

For the Work lost for part b, I think I'm slightly off. I'm coming up with 399.1 W, not 390. I'm just multitling Sgen * Tsurr, so 1.29 * 310.15. Is there some other equation I should be using?

Anonymous said...

From Lan

answers to anonymous: Note that the exit temp for air is unknown and needs to be calculated! It's not necessarily the surrouding temp.

Anonymous said...

Lan.. I do not quite understand what you are saying. How do you solve for T(exit).. I was using T(exit) = 26*c. How is this wrong

Anonymous said...

So i got part A right. But i can't find s gen for part b.

DeltaSref is the same as in part A, correct?

I recalculate deltaSair with the new final temperature. I get -0.026. I think this is where I am going wrong. S1 equals 0.0389 like in part a and I interpolate to S2 at T = 302K to get a S2 equal to 0.013.
Where am I going wrong?

Dr. B said...

Anon 4:39 PM
Your approach is correct, but Mref = 0.8 kg/min not kg/s.
I also got a different value for deltaSref. I think you used x1 = 0 instead of S1 = 0.3.
Your deltaSair is close to my value, but it should be negative because the Tair drops at constant P.
I don't think you have any concepts wrong. Just a few minor problems you can quickly fix.

Dr. B said...

Rachel 5:23 PM
I think you did the problme correctly and this is a round-off issue. My Sgen was 1.26 W/K. I think you have no worries here.

Dr. B said...

Mark 3:35 AM
Yes, deltaSref is the same in part (b) as in part (a).
Your deltaSair is the same as mine to 3 digits.
What makes you think something is wrong ?

Anonymous said...

I don't understand how to solve for the T(exit). I tried using the equation Mdot(air)*deltaHhat(air)=Mdot(134)*deltaHhat(134) I used the shomate equation to get deltaHhat(air), but when I solved for deltaT I didn't get the correct answer

Lan said:

"T(exit) for air should be calculated using the energy balance, i.e., Q(in)=delta(134a)+delta(air) from where you solve for T(exit) for air."

Could someone clear up for me what delta(134a) and delta(air) are?

Dr. B said...

Greenpepper 7 2:56 PM
Your method appears to be correct. Perhaps you made a calc error when integrating the Shomate Eqn ? It is not very easy at all to determine an unknown T when using the Shomate Eqn.

You will find it much easier to use the IG Property Tables on page 323 in the TCD Notebook. You solve for Hair,out and then interpolate on the H column of the table to determine Tair,out.

Charissa said...

Cool!

Dr. B said...

Hello Charissa,
I am glad you like my old blog. I hope you find it useful.
Dr. B