I have an exam on monday and the prof gaves us some sample practice problems.... I have no idea where to start or how to approach the following; any help would be much apprecieated. I google'd the hell out of it... no luck.

A Carnot engine operating on air accepts 50 kJ/kg of heat and rejects 20 kJ/kG. Calculate the high and low reservior temperatures if the maximum specific volume is 10m^3/kg and the pressure after the isothermal expansion is 200kPa.

I know that thers has to be some way to use P2/P1=(v1/v2)^k and that T1=T2, but not sure where to go from there. Thanks.

8/3/2007 12:51:22 PM

just thought to use T2/T1=(v1/v2)^(k-1) where T2/T1=1 and k =1.4 .... then I can solve for v2. with v2 I can solve for P1. is My reasoning correct? is the "max specific volume indeed v2? If so, then what are the properties of points 3 and 4?

8/3/2007 12:54:44 PM

no, that notion will not work b/c it will imply that 1=v1/v2 which would imply that P2/P1=(1)^k=1 and P2 definitaly does not equal P1 in a Carnot engine.....

8/3/2007 1:14:13 PM

are u assuming that its an isentropic process for an ideal gas? if you can find Cp/Cv and use that as k, i would assume you can use that to solve...

8/3/2007 2:18:32 PM

yes, air is an IG, thus I am using 1.4 for K as given in the steam tables.

8/3/2007 2:44:58 PM

I actually enjoyed thermo 1 and 2 but its been a while, it's late but I'd probably try to utilize a combination of the ideal gas law and some of the 3 isentropic relations of T,V, and P like you've already done. Drawing a P-v and/or T-s chart sometimes is the only way to recognize and proceed through the steps, if you're taking Boles he should have that drilled into your head by now. Draw the bloody chart, everytime. Each process from 1 to 2, 2 to 3, 3 to 4, etc will have it's own little equation, property, or constant to let you go through each process and find what you need.

I think there was a proportional equation relating the heat rejected/gained and the reservoir temps in the carnot cycle iirc. The specific volume and the pressure is probably given to use in the ideal gas law to get your temps.

Kinda vague but hope all this helps. If a problem this simple is giving you trouble your theory,reasoning, or process path might need work as a whole.

8/4/2007 3:02:46 AM

bah i should remember this. about the only thing i remember about carnot is that it was the most efficient and that efficiency = 1- T2/T1 and Q1/T1 = Q2/T2 (and from these two the efficiency could also be written as 1-QL/QH of course)

but im sure you already have all those equations

8/6/2007 9:42:01 AM

good luck dude....thermo blows....not to mention you rarely ever use it again after taking the course.....unless of course you're some nerdlinger in a lab somewhere that nobody cares about....

8/7/2007 2:34:55 PM

i disagree.

-more than half the jobs you would end up doing in the power industry end up dealing with some sort of thermo.

-i deal with thermo on an everyday basis working with HVAC and water systems for pharm. companies

furthermore HVAC and power jobs make up a huge percentage of jobs people take when they graduate with a Mechanical Eng. degree

8/7/2007 2:40:30 PM

well i'm not in engineering so

8/7/2007 2:47:03 PM

haha well that would explain it

8/7/2007 2:47:50 PM

I'm in chemistry and other than in the class, I dont use thermo much....


I use a whole hell of a lot more organic than thermo

8/7/2007 2:49:04 PM

well yea that makes sense.

8/7/2007 2:55:11 PM