Tuesday, April 11, 2006
HW #4 - P #6 - Electrical Work and the 1st Law
A mass of 15 kg of air in a piston-and-cylinder device is heated from 25 to 77°C by passing current through a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa during the process, and a heat loss of 60 kJ occurs. Determine the electric energy supplied, in kWh.
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16 comments:
Graham 9:40 AM
You typed the 1st Law incorrectly. It should be: Q-W=deltaE. In this case, W = Welec+Wb.
Yes, evaluate deltaE by integrating heat capacity from T1 to T2.
What type of process is this ? Does this allow us to simplify the 1st Law ?
graham 3:03 PM
Your reasoning is correct. Good work !
I have two questions. First where do you get look up the values for air such as heat capacity. Second in the equation W=Welec+Wb, what is Wb?
So the answer is in kW*h, but the answer I get from the 1st law equation is in kJ. Is there sone way to convert kJ into kW*h?u
Never mind, my calculator figured it out for me.
greenpepper7 1:56 PM
The heat capacity of sir in Appendix E of the Thermo-CD notebook.
Wb = boundary work.
Kevin 2:55 & 3:00
1 kW*h = (1 kJ/s) * 3600 s = 3600 kJ
Just in case your calculator forgets.
Where can we find the different volumes of air? To find Wb, you need to integrate 300Kpa dv by the volumes, all I can find is Vr and that doesn't seem right to me.
Katie 7:08 PM
This is an isobaric process. This makes evaluating Wb = INT{P dV} easier. You can determine the volume of the air using the Ideal Gas EOS. This applies for both the initial and final states.
since we do not know the molar volume of the gas how are we suppose to know we can use the Ideal Gas formula?
I have the equation Wele = Q - Wb - deltaU.
the values i got are:
Q = -60kJ
Wb = 222kJ
deltaU = 1.361kJ
This gives me an answer of Wele = -0.079kW-h. Can someone tell me which step I have wrong. Thank You
I'm not quite sure if i'm doing it right, but when it say they we can either use the ideal gas property table or the specific heat it seems like it is significantly easier to use the table. With the specific heat you have a rather large polynomial and then have to integrate it, but the table is just pulling values. Please let me know if this is correct or if I'm missing something.
Anon 10:43
It looks like you're problem is your deltaU. The rest of it looks correct. If you integrated that big Cp polynomial, you probably made a mistake there. Try the steam tables in the Appendix D, it makes finding deltaU so much easier. Hope this helps.
Anon 10:11 PM
You know the T & P, so you can calculate the molar volume ! It is > 5, so it is OK to use the IG EOS.
Anon 10:43 PM
Your Wb is a bit off. I got about 224 kJ.
"dkress" is correct. Your value for deltaU is off. deltaU=INT{Cv dT}=INT{Cp dT}-INT{R dT} = INT{Cp dT}-R*deltaT = deltaH-R*deltaT
I hope this helps. You will need to integrate Cp dT or Cv dT on the test on Tuesday !
dkress 10:10 PM
Yes, the tables are easier to use and inthe future we will use them often.
But on the test on Tue, you will be required to integrate the Shomate Eqn. So, I suggest you take this opportunity to get some practice !
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