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Carnot cycle

by aaaa202
Tags: carnot, cycle
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aaaa202
#1
Apr1-12, 06:43 PM
P: 1,005
I have some trouble understanding the carnot cycle and why you have to make it so complicated, i.e. this process involves 2 isothermal and 2 adiabatic processes.

If you want an engine with maximum efficiency why can't you just have a system whose temperature is just a tiny inch above your hot resevoir and then absorbs heat at a rate such that it continually expands isothermally? - by this I mean that all the heat goes to producing work and keeping the proces isothermal.

Edit: I also have a second question. The efficiency of an engine is given by:

e = 1- Tc/Th , where Tc and Th are the temperatures of the hot and cold resevoir. You can see that the smaller temperature for the cold resevoir the higher efficiency - why is that? Then the entropy expelled to the environment is bigger but how does that make you lose less energy?
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Jolb
#2
Apr1-12, 11:05 PM
P: 419
Quote Quote by aaaa202 View Post
I have some trouble understanding the carnot cycle and why you have to make it so complicated, i.e. this process involves 2 isothermal and 2 adiabatic processes.

If you want an engine with maximum efficiency why can't you just have a system whose temperature is just a tiny inch above your hot reservoir and then absorbs heat at a rate such that it continually expands isothermally? - by this I mean that all the heat goes to producing work and keeping the proces isothermal.
You actually can design something that converts heat to work of this sort: for example, a steam turbine basically causes water to isothermally expand into steam forever. The problem with efficiency here is that in addition to the heat you supply to the system, you also are supplying chemical energy in the form of more water (which has a nonzero chemical potential.) The heat engines that we study in elementary thermodynamics are considered to have a constant number of particles N.

The short answer is that an engine of this sort, with constant N, violates the Zeroth Law of thermodynamics. Basically what you're saying is: put the reservoir and the system, initially at different temperatures, in thermal contact. Now we let them exchange heat. Instead of reaching equillibrium (both temperatures equal), they both stay isothermal--the system, which receives heat, just expands isothermally forever, and they never reach thermal equilibrium. Not only does this violate the Zeroth law (everything comes to equilibrium eventually), but it's physically nonsense--how could anything expand infinitely?

If the engine's substance is an ideal gas (let's say one mole), p=RT/V so the work done going from V1 to V2 is
∫p dV = ∫ RT/V dV = RT ln (V2/V1)
If the gas expands at a constant rate, V2/V1 → 1 as t→∞, so the work done is 0. So in order to do work at a constant rate, the medium must expand exponentially faster and faster. Even more unphysical.


Edit: I also have a second question. The efficiency of an engine is given by:

e = 1- Tc/Th , where Tc and Th are the temperatures of the hot and cold resevoir. You can see that the smaller temperature for the cold resevoir the higher efficiency - why is that? Then the entropy expelled to the environment is bigger but how does that make you lose less energy?
The nicer way to look at this is in terms of heat and work. We define Efficiency η = W/Qh

Qh-Ql=W by conservation of energy.
=> η=(Qh-Ql)/Qh, so Ql must be zero for maximum efficiency. Remember now that Ql happens during an isothermal leg--the substance is already in thermal equilibrium with the Tl reservoir at the end of the adiabatic leg. The heat transferred along an isothermal leg is given by -RT ln (Vi/Vf), and this is zero only when T=0 (since the volume is changing). So you want your low-temperature isotherm leg to not lose any heat Ql, and the only time this is possible in an expansion/compression is when the gas is at T=0, where it has no heat to give off.


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