Carnot Cycle

Concept: The Carnot Cycle and Maximum Theoretical Efficiency

9m
Video Transcript

Hey guys, in this video we're going to cover the carnot cycle and the maximum possible theoretical efficiency for an engine. Alright let's get to it remember no heat engine can convert all of the heat from a hot reservoir into usable work so we have some hot reservoir we have some engine we have some cold reservoir whatever heat flows into the engine from the hot reservoir leads to the production of work but it cannot all be converted into work some heat has to be released into the cold reservoir in order for heat to flow some of that heat has to be returned to the cold reservoir otherwise the engine is not possible the efficiency of any heat engine remember guys is a measurements of how much usable work right how much of this, you can pull out of the heat that's input right out of this and the more efficient an engine is the more usable work for a given heat input the carnot cycle or the carnot engine is theoretically the most efficient engine possible between two reservoirs so if you have some hot reservoir and you have some cold reservoir the most efficient engine you can put between them is going to be a carnot engine that's going to get the most work out of any pair of reservoirs. Let me minimise myself the carnot cycle which powers a carnot engine is composed of four steps, step one right here is an adiabatic expansion at the temperature of the hot reservoir right at this temperature, step two right here is an adiabatic expansion so we have two expansions from the temperature of the hot reservoir to the temperature of the cold reservoir. This is how we're switching temperatures, step three right here is an isothermal compression so we're going backwards now we're compressing the gas at the temperature of that cold reservoir and finally step four is another compression this time adiabatic returning us from the temperature of the cold reservoir to the temperature of the hot reservoir, so we get back to our starting point because this is a cyclic process.

In a carnot cycle heat is transferred during only two of the steps right two of the steps are adiabatic. So heat absolutely cannot flow during those steps but the other two steps the isothermal steps heat can flow in step one heat is added into the system. That's how the gas expanded it needed some energy to increase its volume and that was provided by the heat added to the system. In step three heat is removed from the system and dumped into the cold reservoir right that's this heat that has to exist carnot cycle can have that heat equal to zero and as I said steps two and four are adiabatic so heat is not transferred in those steps alright the carnot engine is the theoretically most efficient engine you can ever place between any pair of hot and cold reservoirs and that efficiency is 1 minus the temperature of the cold reservoir divided by the temperature of the hot reservoir right and this is the maximum possible efficiency for any engine between a hot reservoir at T.H and a cold reservoir at T.C so no matter what type you use if you put it between the same pair of reservoirs the carnot engine is always going to output more work than any other possible engine. A car engine is designing a car engines design is aimed at getting as close to the theoretical maximum efficiency as possible say the engine is designed to operate when the engine compartment is cooled to room temperature 27 degree Celsius by the radiator inside the engine compartment. What is the least percentage of heat wasted or wasted heat the designers could hope to get? The least amount of wasted heat note that gasoline ignites at around 260 degrees Celsius so if we want the least wasted. What does that mean that means we need the most efficient.Right the designers were targeting the theoretical maximum efficiency which is just the carnot efficiency

I don't remember how to spell efficiency, there we go sorry about that. The most efficient possible engine is the carnot engine and the highest the largest efficiency possible is the carnot efficiency so that maximum efficiency is going to be 1 minus the temperature of the cold reservoir divided by the temperature of the hot reservoir well N an engine right in a car's engine what's the hot reservoir and what is the cold reservoir at least what's those temperatures the cold reservoir is the temperature of the environment that that engine sits in that's what's releasing the heat into and we were told that that environment inside the hood of the car is 27 degrees Celsius Now this is not a difference in temperature remember this is one temperature divided by another you can only use Celsius and Kelvin interchangeably when it's a difference in temperature since this is not a difference we have to convert this into Kelvin so we add 273 now what is the temperature of a hot reservoir well the hottest that engine's ever going to get is the temperature of the gases ignited at and that happens to be 260 degrees Celsius so the hot reservoirs is going to be 260 Celsius but we have to convert it to Kelvin so we have to add another 273 alright plugging this into a calculator we get 0.437 so that maximum efficiency is 43.7% so what does that mean about the least percentage of heat wasted. The minimum wasted percentage is just going to be 100% minus the efficiency which is 56.3 % so even theoretically the minimum amount of heat that's wasted in this engine is over 56% that means that no engine designed could ever be better than 56% wasted energy as the minimum. That wraps up this discussion on the carnot cycle and the theoretical minimum sorry the theoretical maximum efficiency. Thanks for watching guys.

Carnot Cycle Additional Practice Problems

You want to increase the efficiency of a Carnot engine by altering the temperature of either of the reservoirs by a temperature of ΔT. Which of the following will produce the greatest increase in efficiency?

A) Increasing the temperature of the hot reservoir by ΔT

B) Increasing the temperature of the cold reservoir by ΔT

C) Decreasing the temperature of the hot reservoir by ΔT

D) Decreasing the temperature of the cold reservoir by ΔT

E) Option A or D, because the temperature change ΔT is the same for either case

F) Option B or C, because the temperature change ΔT is the same for either case

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A Carnot heat engine operates between a high-temperature reservoir at T= 800 K and a low-temperature reservoir at T= 200 K. In one cycle the engine rejects 400 J of heat energy to the low temperature reservoir. How much work does the engine perform in one cycle?

(a) 900 J

(b) 1200 J

(c) 225 J

(d) 300 J

(e) 1500 J

(f) 2000 J

(g) none of the above answers

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In one cycle, a Carnot heat engine absorbs 800 J of heat and performs 600 J of work. If the temperature of the high-temperature reservoir is 560 K, what is the temperature of the low temperature reservoir?

(a) 140 K

(b) 240 K

(c) 320 K

(d) 420 K

(e) none of the above answers

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The high temperature reservoir of a Carnot engine has a temperature of 227°C and the temperature of the low-temperature reservoir is 17°C. If the engine rejects an amount of heat of magnitude 320 J each cycle, how much mechanical work does the engine perform each cycle?

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In each cycle a Carnot heat engine takes in 620 J of heat from the high temperature reservoir that is at 500 K and performs 390 J of mechanical work.

a) What is the thermal efficiency of the engine?

b) What is the temperature of the low temperature reservoir?

c) During one cycle how much heat is rejected into the low temperature reservoir?

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