Heat Engines and Efficiency
If three identical samples of an ideal gas are taken from initial state I to final state F along the paths IAF, IF, and IBF as shown in the PV‑diagram above, which of the following must be true? (A) The work done by the gas is the same for all three paths. (B) The heat absorbed by the gas is the same for all three paths. (C) The change in internal energy of the gas is the same for all three paths. (D) The expansion along path IF is adiabatic. (E) The expansion along path IF is isothermal.
If three identical samples of an ideal gas are taken from initial state I to final state F along the paths IAF, IF, and IBF as shown in the PV‑diagram above, which of the following must be true? (A) The work done by the gas is the same for all three paths. (B) The heat absorbed by the gas is the same for all three paths. (C) The change in internal energy of the gas is the same for all three paths. (D) The expansion along path IF is adiabatic. (E) The expansion along path IF is isothermal.
State Variables vs Processes P, V, T, and U are all state variables. They depend only on the state of the gas, and are well-defined at each point on the PV diagram. Q and W are processes. They bring the gas from one state to another, and the path taken along the PV diagram matters!
ΔU = (3/2)nR(Tf-Ti) No matter which path is taken, Although there are many different ways to get from state I to state F that involve many different Q’s and W’s… No matter which path is taken, ΔU = (3/2)nR(Tf-Ti) ΔU depends only on the initial and final states
Always return to the same state at which they started! Cyclic processes Always return to the same state at which they started! These are common in engines, when the same process is repeated in order to constant have work done by the gas.
TATBTCTDTA For a cyclic process, the gas will always return to the same state at which it started. Therefore, ΔTcycle = 0 K! Therefore, ΔUcycle = 0 J!
Gas in a chamber passes through the cycle ABCA as shown in the diagram above. In the process AB, 12 J of heat is added to the gas. In the process BC, no heat is exchanged with the gas. For the complete cycle ABCA, the work done by the gas is 8 J. How much heat is added to or removed from the gas during process CA? (A) 20 J is removed. (B) 4 J is removed. (C) 4 J is added. (D) 20 J is added. (E) No heat is added to or removed from the gas.
Therefore, Qcycle = -Wcycle For a cyclic process, the gas will always return to its initial state. ΔUcycle = 0 J! Therefore, Qcycle = -Wcycle However much energy is transferred to the gas by one method, must be transferred out of the gas by the other method! This is another statement of the Law of Conservation of Energy.
A heat engine is a machine that takes thermal energy from an external source (Qin), and turn it into useful work done by a gas (Wnet), and release some excess thermal energy (Qout). Qin Qout heat engine (+Q) (-Q) Wnet (-W)
The most common example is the modern combustion engine. Let’s use our knowledge of thermodynamics to analyze what is going on in one of the engine’s cylinders!
Start by setting up your whiteboards with the following PV diagram and a chart to keep track of each step of the cycle. Q W ΔT ΔU AB BC CA Cycle
We’ll start with the gas (a gasoline/air mixture) at state A, with the gas at its lowest volume, ready to be ignited. n = 0.8 moles
A B The spark plug ignites the gasoline/air mixture, and the gas starts to burn! A B Assume that the temperature of the gas triples, and that the piston doesn’t have enough time to move significantly during this part of the process. Draw the process on your graph, and fill in the first row of the chart.
n = 0.8 moles Q W ΔT ΔU AB 6 kJ 600 K BC CA
As the gasoline/air mixture continues to burn, it expands isobarically, pushing the piston and turning the crank shaft! C B Assume that the volume of the gas multiplies by 2.5 during this expansion. Draw the process on your graph, and fill in the second row of the chart.
-W n = 0.8 moles Useful! Q W ΔT ΔU AB BC 22.5 kJ -9 kJ 1,350 K (Work done by gas) Useful! Q W ΔT ΔU AB BC 22.5 kJ -9 kJ 1,350 K 13.5 kJ CA
A B C So, what has happened so far? Thermal energy has been transferred to the gas (+Q) during both processes. This thermal energy actually came from chemical bonds being broken in the gasoline molecules when the gas ignites! Chemical potential energy transformed into thermal energy As a result, the gas expands and we get some useful work (-W) done by the gas!
Q W ΔT ΔU AB 6 kJ 600 K BC 22.5 kJ -9 kJ 1,350 K 13.5 kJ CA So far, we have put in 28.5 kJ of thermal energy (Qin), and gotten out 9 kJ of work done by the gas This seems very wasteful, but having some excess thermal energy is inevitable in every heat engine!
Draw the process on your graph, and fill in the last row of the chart. Finally, as the gas cools down by dissipating thermal energy into the surroundings, it is compressed back to its initial state. C A Assume that this process is a straight line on the PV diagram (for the sake of simplicity) Draw the process on your graph, and fill in the last row of the chart.
(Work done to bring gas back to its initial state) n = 0.8 moles (Work done to bring gas back to its initial state) Not useful! +W Q W ΔT ΔU AB BC CA -25.5 kJ 6 kJ -1,950 K -19.5 kJ
During the last part of the process, 6 kJ of work is done on the gas to bring it back to its initial state (this is not useful work!) At the same time, 25.5 kJ of thermal energy is dissipated by the gas into the environment (-Q) as the gas cools off. This is why your engine feels so hot! It dissipates tons of thermal energy into the surroundings! This is also wasted energy, but it is impossible to avoid at least some of it being dissipated!
Now you can complete the last row of the chart! To finish things off, complete the diagram below to summarize the way that this heat engine works. Qin = ? Qout = ? heat engine Wnet = ?
Wnet by gas = 3 kJ Qin = 28.5 kJ Qout = 25.5 kJ heat engine Q W ΔT ΔU AB 6 kJ 600 K BC 22.5 kJ -9 kJ 1,350 K 13.5 kJ CA -25.5 kJ -1,950 K -19.5 kJ Cycle 3 kJ -3 kJ 0 K 0 kJ Qin = 28.5 kJ Qout = 25.5 kJ heat engine Wnet by gas = 3 kJ
The efficiency of a heat engine is defined as Efficiency is how much net work you get out divided by how much energy you put in! The more net work that you get out of the process, the closer the efficiency will be to 1. 0 < e < 1
What is the efficiency of the combustion engine that we have been analyzing? Q W ΔT ΔU AB 6 kJ 600 K BC 22.5 kJ -9 kJ 1,350 K 13.5 kJ CA -25.5 kJ -1,950 K -19.5 kJ Cycle 3 kJ -3 kJ 0 K 0 kJ e = (3 kJ/28.5kJ) = only 10.5%! Although this sounds low, the average efficiency of a modern car engine is only about 15 - 20%!
Efficient Whiteboard! B C A Calculate the efficiency of this engine! 1 mole of an ideal gas in a heat engine undergoes the process shown below. The work done on the gas from state A to B is 6 kJ. B C Calculate the efficiency of this engine! A
B C A Q W ΔT ΔU AB -6 kJ 6 kJ 0 K 0 kJ BC 30 kJ -12 kJ 1,444 K 18 kJ Cycle
= 20% Q W ΔT ΔU AB -6 kJ 6 kJ 0 K 0 kJ BC 30 kJ -12 kJ 1,444 K 18 kJ CA -18 kJ -1,444 K Cycle = 20%
This is the most efficient possible cycle for a heat engine. AP Test Side Note! Sometimes on the AP test, there will be a question about the Carnot Cycle. This is the most efficient possible cycle for a heat engine. We will not be covering this, but should you encounter a question on this (where it says “maximum efficiency” or “Carnot efficiency”)… They are basically just asking you to plug into the equation for the efficiency of a Carnot engine.
This will not be on your test, but you may see it on the AP. Tcold is the temperature of the cold reservoir of the engine. Thot is the temperature of the hot reservoir of the engine. This will not be on your test, but you may see it on the AP. It is easy to check your work, because you should always get an efficiency that is between 0 and 1! Make a note to yourself to memorize this before May 13th. I am not teaching it because it would take four days to actually do it justice and have it make sense, and the AP people literally just want you to plug into one equation :-/