a.) The compressor power requirement in kW and the mass flow rate of the cooling water in kg/s.
b.) The rate of entropy generation in kW/K and the rate at which work is lost in kW for the compressor. Assume the temperature of the surroundings is 300 K.
c.) The rate of entropy generation in kW/K and the rate at which work is lost in kW for the heat exchanger. Assume the temperature of the surroundings is 300 K.
14 comments:
I don't understand how the water stream flow rate is the same as the ideal gas flow rate (0.5 kg/s). Can you help me understand?
Oh crap. That was not my flow rate either.
I did get the same answer as part C though using another flow rate??
-
Dr. B, for part B, I don't understand how Ws,lost = 4.94. I was thinking along the lines that we use the "big result" equation utilizing Tsurr.
ThermoSoulja4Lyfe:
Good name !
I think I messed something up when I posted my answers. The mass flow rate of cooling water should be 0.405 kg/s as opposed to 0.50 kg/s for the air. I am sorry I caused this confusion.
Parwiz:
I am not sure what the 1st 2 parts of your comment mean. Read my reply to Soulja.
Sadly, this is a typo on my part.Wlost = 5.94 kW. You do use the "Big Result" Wlost = Tsurr*Sgen.
Sorry for the confusion.
Yes sir, that was what I meant. I got the same answer for part C using 0.405 kg/s :)
Parwiz:
Excellent. You are done ! Whoohoo.
hello professor....
i can't seem to find Sgen for b or c. is there a trick or something that you think i'm not picking up on? or do you have any suggestions?
Harry P:
I used the ideal gas property tables to get deltaShat for the air. Then multiply by m-dot-air. I used this to calc Sgen for the compressor in kW/K. Then Wlost is just Tsurr * Sgen.
In part (c), you can treat water as an incompressible fluid and calculate deltaS as in the TCD notebook.
Either way, you calculate Sgen from its definition (the equality form of the Clausius Inequality).
I hope this helps.
For part a) & b), I got wrong answers but I don't know what I did wrong.
For part a),
First, I got the Vhat from the table to get (m)dot by doing (V)dot/Vhat, and because m1=m2=m3 due to steat state assumption. Then use the heat exchanger equation - (m)c*delta(Hhat)c=-(m)H*delta(Hhat)H but finally i got it wrong. What did i do wrong?
For part b),
since we neglect heat exxhange,
Sgen=(m)dot*delta(S hat),right?
I think the problem is the (m)dot air from part a) because everything related to that.
But i think it is simple...
m dot = V dot/ Vhat
I am still lost...
Help:
It sounds like you did it correctly. Did you use the ideal gas property tables in the back of the book ? What values did you get for each variable ?
The good news is that I think you learned what you were supposed to from this problem. So, don't worry too much about it. I will post the solution later tonight.
In example problem number 8C-1 in the Thermo-CD, part D, you use a compressor efficiency n=W(S,iso-T)/W(S,act). I looked through the book, but couldn't find that definition anywhere else. Where did this come from?
In example problem 8D1 in the Thermo-CD notebook, you state that the lost work is the difference between the work the HE puts out and the work required to run the HP. Why isn't the lost work simply defined as the work that the HE puts out?
Ethan:
I copied this post to the new Test #2 blog.
The lost work is the difference between the work output of the HE and the work requirement of the HP. This system must result in exactly the same heat transfer from the hot reservoir at 1000K and to the cold reservoir at 320K as the original heat "leak". This way we are comparing apples to apples. The hot and cold reservoirs cannot tell the difference between the heat leak and the HE/HP system. The only change on the universe is that we end up with a "free" net amount of work = WHE - WHP at zero cost because the net heat transfer from the dead state (surroundings at 298 in this problem) has no value and therefore no cost either.
Post a Comment