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# Heat Engines & PV Diagrams

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Sections
Heat Engines and the Second Law of Thermodynamics
Heat Engines & PV Diagrams
The Otto Cycle
The Carnot Cycle
Refrigerators
Entropy and the Second Law of Thermodynamics
Entropy Equations for Special Processes
Statistical Interpretation of Entropy

Concept #1: Calculating Work in Heat Engines Using PV Diagrams

Example #1: Efficiency of a Four-Step Engine

Transcript

Hey guys, let's do an example, a proposed four step engine is produced n moles of a monoatomic ideal gas undergo the process shown in the following PV diagram what would the efficiency of engine be? The efficiency is given by the work output by the engine divided by the heat input into the engine remember there's two heats in a cyclic process in an engine there's going to be an amount of heat inputted which is going to be positive heat and an amount of heat output which is going to be negative heat. The work is easy to calculate the work is just the area contained by the cycle with the appropriate sign now first could this even be an engine, yes it can it's a clockwise cycle so it's absolutely an engine, the area is going to be the base times height because it's just a rectangle this is going to be 2Vnot this is going to be one half Pnot so this is 2vnot one half P not write that 2 and one half cancel this is just Pnot Vnot and this is absolute going to be positive it's going to be positive right energy released by engine so that's just the work done by the engine now in order to find how much heat is input we have to analyze the steps individual. I'm going to call this step one this step two this step three and this step four now something important to know about P.V diagrams and you can show this for yourself if you want is that when the process is going up into the right. The heat is always input into the system so process one and process two heat is going to be added into the system so Q is going to go in. When the process is down or to the left heat is always leaving the system. All we need to know is how much heat enters the system right this how much heat enters the system so we only have to look at steps one and two you can use the same method the same process that I'm using here to analyze steps three and four and show that the heat is always going to be leaving in those steps for step one first of all the work is always going to be 0 because this is an isochoric process this means that the first law of thermodynamics which says the change in internal energy is the heat transferred plus the work done just means ops sorry wrong color, it just means that the heat transferred is the change in internal energy for step one. Now the internal energy of any ideal gas is F over 2 NRT right an equation that we used a bunch of times because this is a monoatomic ideal gas the number of degrees of freedom is simply 3 one for each translational direction.

Now the change in internal energy is what we're interested in and this is going to be three halves NR delta T. We're talking about n moles simply changing its state there's no heat coming into the sorry there's no gas coming into the system no gas leaving the system so the only thing that changes when the internal energy changes is the temperature so the temperature is the only thing that gets this delta, the question is what is the change in temperature. Well PV equals N R T the ideal gas equation is going to tell us how the temperature changes as the gas changes state. This is going to tell us because the volume is a constant this is an isochoric process that V delta P is N R delta T the change in the left sides due entirely to a change in pressure the change on the right side is due entirely to a change in temperature so delta T is V delta P over N R now plugging that in to delta U so I can say delta T1 is equal to this, this equals three halves and N R times V1 delta P1 over N R notice that the N R in the numerator and the N R in the denominator cancel so this is three halves V1 delta P1 which is three halves what's the volume for step one it's just a Vnot what's the change in pressure in step one right from Pnot it's from one half Pnot to Pnot so this is just one half Pnot so this becomes three fourths Pnot Vnot. That's the change in internal energy for step one and by the first law of thermodynamics therefore the heat exchanged in step one is which is delta U1 is going to be positive three fourths Pnot Vnot. It's very important that the heat is positive because we're only looking for heat input into the gas we are not looking for heat output by the gas for step two the pressure doesn't change for step two delta P is 0, but Delta U is still equal to Q plus W. However we do know that at constant pressure W is equal to negative P delta V so this is for step two what's the pressure at step two it's just Pnot right what's the change in volume it's three Vnot minus 2Vnot sorry minus Vnot which is going to be positive 2Vnot so this is negative to Pnot Vnot right this is positive 2Vnot. So that's how much work is done during step two now what's the change in internal energy during step two well it's still three halves NR delta T during step two, what's delta T during step two? Well P.V equals N R T pressure does not change only the volume changes so we have P delta V equals N R delta T so P2 delta V2 equals N R delta T2, NR delta T2 is this whole right portion so delta U2 is three halves P2 delta V2 which is three halves Pnot right the pressure for step two times once again the same change in pressure that we found here. So this is just 3Pnot Vnot so by the first law of thermodynamics Q during step two is just delta U 2 minus W which is three Pnot Vnot minus negative 2Pnot Vnot which is 5Pnot Vnot. So for the cycle sorry the heat the total heat input during the cycle is just the heat input during steps one and steps two because steps three and step four release heat so the amount of heat input is just during step one and step two, step one had three fourths Pnot Vnot and step two had five Pnot Vnot if we find the least common denominator this is three fourths Pnot Vnot plus this common denominator is four this is twenty over four Pnot Vnot so that's 23 over 4Pnot Vnot and what's the work done by the cycle we calculated it to be Pnot Vnot this means that the efficiency which is just the work over the amount of heat input is Pnot Vnot over 23 over 4 Pnot Vnot. Those Pnots Vnots cancel and finally.The efficiency is 4 over 23 now 4 over 23 is an exact answer and you can leave it like that if you want or you can approximate a number 0.174 which is 17.4 % so this engine actually has an incredibly low efficiency only 17.4 % other engines can have much much much higher efficiency. I showed a case in an earlier problem where a car not engine had an 80 % efficiency when it was placed between a reservoir of a 1000 Kelvin and a reservoir of 100 Kelvin this seventeen and a half percent efficiency sort of leaves a lot to be desired but this is an application of the first law of thermodynamics on heat engines and how to find the efficiency using a P.V diagram which can be very useful. Alright guys that wraps up this problem thanks for watching.