Calculating the Efficiency of a Heat engine

[grandpa2390]

Essentially, yes. We are trying to determine the relationship between P, V (or T) and heat flow (Q). A T-S diagram will show the relationship between temperature and heat flow. A T-S diagram will be a vertical line for the adiabatic portion and a curve on the right side (looking like a D shape overall) for the expansion portion. So if you can draw a T-S diagram for the expansion, you look for the point where S is maximum (i.e. the point farthest to the right). The area below the lower part of the curve from at that point is Qc and the area below the upper part of the curve is Qh. The area inside the curve is Qh-Qc. We are trying to find the point at which S is maximum, ie. where dS/dT = 0. This should also be where dQ/dV = 0 (in which case dU/dV = -P).

U = C_v(PV/R)
Work out dU/dV and set it equal to -P. The solution will give you the point on the line where dQ/dV is zero. Before that point, heat flow is in. After, heat flow is out.

By the product rule:
dU/dV = (C_v/R)(P + VdP/dV)
Work out dP/dV from the relationship between P and V that you derived.

AM

$\frac{dU/dV}{C_V/R} - VdP/dV = P$

I don't know what to do with this. thanks for trying, but my assignment is due. Time to embrace my F. I wasted an entire week trying to figure out this one problem, my time would have been better spent studying for other classes :(
it's why i pleaded to be directed towards a video lecture or something that could walk me through to understand the concepts I need to solve this problem.
I do not have the knowledge I need to follow hints. I need instruction.

 

[Andrew Mason]

$\frac{dU/dV}{C_V/R} - VdP/dV = P$

Set dU/dV = -P to determine where heat flow direction changes (dQ/dV = 0 or dS/dT = 0)
ie. $dU/dV = \frac{Cv}{R}(P + V\frac{dP}{dV}) = -P$. Substitute dP/dV from the equation for your expansion line (hint: it is a line of constant, negative slope and dP/dV is the slope).

$\frac{Cv}{R}(P + mV) = -P$

3) $V = \frac{P}{m}\left(\frac{Cv}{R} + 1\right)$

Since the expansion line is of the form: P = mV + b substitute mV + b for P in 3). That will give you the value of V where dQ/dV = 0. I didn't say it was easy.

I don't know what to do with this. thanks for trying, but my assignment is due. Time to embrace my F. I wasted an entire week trying to figure out this one problem, my time would have been better spent studying for other classes :(

This is a very difficult problem. I doubt that you were expected to solve it. Give it your best shot. Look at it as a good learning experience.

AM

 

[grandpa2390]

Set dU/dV = -P to determine where heat flow direction changes (dQ/dV = 0 or dS/dT = 0)
ie. $dU/dV = \frac{Cv}{R}(P + V\frac{dP}{dV}) = -P$. Substitute dP/dV from the equation for your expansion line (hint: it is a line of constant, negative slope and dP/dV is the slope).

$\frac{Cv}{R}(P + mV) = -P$

3) $V = \frac{P}{m}\left(\frac{Cv}{R} + 1\right)$

Since the expansion line is of the form: P = mV + b substitute mV + b for P in 3). That will give you the value of V where dQ/dV = 0. I didn't say it was easy.
This is a very difficult problem. I doubt that you were expected to solve it. Give it your best shot. Look at it as a good learning experience.

AM

I don't know. We took a test and I didn't do so well. Apparently nobody else must have either because he sent this problem out for us to do for an extra three points if correctly done.

And he said it was OK to ask for help from people and so forth otherwise I wouldn't of asked. But it was due today but now it's too late. I wrote down all the information I found it figured out. And explained the problem I ran into. I probably failed. But maybe I'll get enough partial credit to pass

I'm already having to do my best to teach myself on the book and these problems are not like anything in the homework or otherwise

Take home questions ought to be difficult. But to ask questions that require concepts not taught... :(