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# Efficiency of a Carnot engine

Definition of efficiency for a heat engine. Efficiency of a Carnot Engine. Created by Sal Khan.

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• How similar are the engines we use in our cars to the "ideal" Carnot engine? •  Usually 25% of the gasoline is converted to work. Cars with diesel engines however can convert up to 40% to work.
• what happened to the work done from points B to C, then D to A in the cycle? this seems to be totally neglected? • The work that the system does from B to C is counterbalanced by the work that is done to the system from D to A.

Visualize the Carnot cycle illustrated on a PV diagram. We know that internal energy is a state variable, which means the internal energy at point A will always be the same regardless of the path that leads to point A. Therefore, there is no change in internal energy when a Carnot cycle is completed, because we started at point A and finished at point A.

We know that there is no change in internal energy from A to B, because there is no change in temperature from point A to B. The same is true from C to D. We know that there is a decrease in internal energy from B to C, because there is a decrease in temperature. However, this decrease in internal energy is counterbalanced by an increase in internal energy from D to A, because the internal energy at the end of the cycle must equal the internal energy at the start of the cycle.

Finally, we know the thermodynamic processes from B to C and from D to A are adiabatic, which means the system did not gain or lose heat during these processes. We know that "change in internal energy" equals "heat gained" minus "work done by the system" (or plus "work done to the system"). Since no heat was gained or lost from B to C, the "change in internal energy" from B to C equals negative ( - ) "work done by the system" (the gas did work to increase the volume). Since no heat was gained or lost from D to A, the "change in internal energy" from D to A equals "work done to the system" (work was done to the gas to decrease the volume). We've already established that the change in internal energy from B to C counterbalances the change in internal energy from D to A. Therefore, the work done by the system from B to C counterbalances the work done to the system from D to A. This is why Sal neglected the work involved in these two adiabatic processes.
• how much will the engine's efficiency be if both the temperatures are kept at 0 degrees? • 0/0 does not simply equal 1. However, if we start at a temperature of 1 for the hot reservoir and for the cold reservoir, the efficiency will be 1 - 1/1 =0. If we halve the temperatures: 1 - .5/.5= 0. Halved again; 1 - .25/.25=0 and again: 1 - .125/.125=0... and so on. You'll never get to zero, but you can get infinity close, and as you approach zero, you will still observe that the efficiency is zero.
Still unconvinced? 0 degrees Celsius is the same temperature as 273 Kelvin. Because you are dividing one temperature by another, units will cancel out. Therefore, they do not matter, as long as they are the same. 1 - 273/273 = 0, so after converting back to Celsius, we can safely say that the efficiency is 0 at 0 degrees Celsius.
Hope this helped
• Could a Carnot engine actually be built in real life? • why isn't the Q2 integrated from Vc to Vd? • You see, when we are talking about Q2 we assume it is positive because that's better to understand. Its definition should be the heat given off by the system. If you integrate from Vc to Vd you derive a heat given to the system, which is negative. (Now I'm also a novice on thermodynamics I may be wrong... But the general idea is probably like that...)
(1 vote)
• I can't seem to figure out where this "net work" ends up going. Does the piston end up higher than when it started, even if the thermodynamic macrostates all go back to where they started? If someone could explain where the net positive work is actually doing I would greatly appreciate it. • Can some please explain to me the difference between the efficiency of the Carnot cycle and a normal engine? I had a conceptual question and the correct answer was "The efficiency of this engine is greater than the ideal Carnot cycle efficiency." However, when i tried to calculate the efficiency of it i ended up with the same number for both the theoretical (carnot) and real engine efficiency. Used sals equation e = 1 - (Tc/Th) for the Carnot cycle and e = w/Qh for the real one. Any advice? • At Sal says that the efficiency equation 1- Q_2/Q_1 applies to ANY engine, but this equation was derived using the assumption that all net heat added is converted to Work. This would not be true in a non-Carnot engine in which some of the heat added is lost to friction. Please explain. • How do you calculate the efficiency of a Carnot refrigerator? Is it Qc/(work input)?

I tried it and came up with eff = 1/(1+(Th/Tc)) where Th is the hot side temp, and Tc the low side temperature. Is this correct?
If so, then maximum efficiency would be 0.5 (when Th = Tc). That seems very low, I suspect that the efficiency should be 1 (for a Carnot refrigerator) when the hot and cold side temperatures are the same.
(1 vote) • Your first equation is correct, but your second one isn't. For the Carnot engine, the equation for efficiency with respect to temperature is [eff = 1+(Tc/Th)].

To get the efficiency of a Carnot refrigerator, you would have to start from your first equation [eff = (Qc/W)], substitute W with (Qh-Qc), and then do the same procedure as Sal where [Qc = nRTc(ln(VA/VB))] and [Qh = nRTh(ln(VD/VC))].
After substituting and simplifying, you should come to [eff = Tc/(Th-Tc)] 