According to Carnot efficiency formula $\eta=1-T_C/T_H$, can we say that the engines of cars are more efficient on cold days where $T_C$ (the temperature of the surroundings) is less than on hot days?
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4Is it actually reasonable or true to assume that the car's exhaust temperature is the outside temperature? AKA, why not touch the exhaust manifold? – dmckee --- ex-moderator kitten Jan 28 '15 at 13:54
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3In fact they're less efficient, as determined by experimental measuring of mpg. In summer I average around 72-74 mpg (Honda Insight hybrid), in winter only 66-68. – jamesqf Jan 28 '15 at 17:48
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4Efficiency aside, the cold air is denser and hence contains more oxygen per unit of volume. On engine without turbo compressor, more oxygen swallowed by the piston mean you can inject more fuel per cycle (and still get it all to react with the oxygen). This allows you to get more "oompf" from your engine at colder temperature, but as the answers here tends to demonstrate, it is probably at the expense of a lesser thermodynamic efficiency. – Hoki Jan 28 '15 at 18:04
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3To make @dmckee's point explicit: the $T_C$ in your formula is the temperature of the gas exhaust when it leaves the piston-- not the temperature of the outside air – pentane Jan 28 '15 at 20:03
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8@jamesqf The recipe for making gasoline varies seasonally so your comparison can't be made directly. If you're interested there is a decently interesting article in popular mechanic about it as well as other sources. – jkeuhlen Jan 28 '15 at 20:21
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Within reason, the engine is most efficient for the operating conditions for which it was designed and tuned. In general, that means that your best efficiency is in "temperate" conditions. – Hot Licks Jan 28 '15 at 21:34
6 Answers
The actual efficiency of the engine is most likely not driven by the "outlet temperature" but by the viscosity of the lubricants. On a cold day an engine has to work a lot harder to move the oil, and this will definitely affect (negatively) the efficiency - at least until the engine heats up.
Next, the temperature at the hot part of the cycle will be lower since the same amount of fuel heats the intake air to a lower final temperature (it starts at a lower temperature). This is the point that Kieran made.
Finally, the engine doesn't "give off heat" at the end of the cycle - rather, it releases the exhaust gases at the end of the compression stroke, when they still have a significant pressure. The change in volume is the same, but the initial pressure is lower (since the gas started out cooler). This means that the work done per stroke will be less. I think this is actually the most important aspect in this case.
I conclude that an ordinary internal combustion engine would be less efficient at lower temperatures. And where I live, I certainly get significantly less mileage from my car in winter (but there are many other factors contributing - not just thermodynamics).
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9It's likely the increased drag force due to larger air densities is why you get less MPG in the winter, rather than any differences in engine efficiency. Also, gasoline in the US in the winter is different than in the summer with winter blends being more volatile which will also influence the efficiency. – tpg2114 Jan 28 '15 at 21:17
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2@tpg2114 - those are the "other factors contributing" that I mentioned in my last paragraph. I suspect that the density and fuel blend are indeed the most significant. – Floris Jan 28 '15 at 21:49
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1lower density means more mass of air is ingested and more fuel can be burnt. This most definitely means the work extracted per stroke is more. See my answer for more details, including in particular the practice of installing intercoolers on turbocharged engines. The effect on efficiency is more difficult to predict, but theoretical analysis of the otto cycle predicts it should be higher at cold temperatures. For short journeys a cold engine may well be less efficient but once properly warmed up, the lubricant temperature is whatever the designer wants it to be and is therefore a non-issue. – Level River St Jan 29 '15 at 09:19
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Your reasoning that the initial pressure is lower due too a lower temperature is missing the density component. The pressure before combustion should be atmospheric and at a lower temperature, so it should have a higher density. If combustion adds a constant amount of heat then the temperature would be increased by a proportionally larger amount, which would increase the pressure by a proportionally larger amount. Thus the pressure after combustion should be higher not lower... – Rick Jan 29 '15 at 13:58
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1+1, but if you end up editing in any additional factors (I think it would be nice but it's your answer) here are a couple: Winter traffic is often worse as people are more inclined to drive in bad weather; Any time you're running the engine to de-ice/demist while not moving your fuel consumption is 0 mpg. – Chris H Jan 29 '15 at 14:50
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@rick if I add the same amount of heat to a larger mass of air the temperature rise will be lower not higher? – Floris Jan 29 '15 at 15:06
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@steveverrill I had not considered the possibility to burn more fuel per stroke - clearly that increases the efficiency (more work done for same friction). There is more to this than meets the eye... – Floris Jan 29 '15 at 15:10
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@Floris good point. I'd forgotten that the heat would also be distributed over additional mass. If we had $\frac9{10}$ of the starting temperature we'd have $\frac{10}9$ more mass so our temp would go up $\frac9{10}$ as much so then everything would be proportionate and we'd get the same power for the same heat. Thus, second order effects like fuel air mixture and complete combustion would dominate the change in efficiency. I could see the efficiency increasing for highly loaded engines and decreasing for lightly loaded engines. Maybe I'll write up an answer. – Rick Jan 29 '15 at 15:30
There are a lot factors involved here, but as internal combustion engines follow the Otto or Diesel cycle, those cycles (which are less efficient than the Carnot cycle) are the right model to follow, rather than the Carnot cycle itself.
Let's say we reduce the inlet air temperature from 300 Kelvin to 270 Kelvin (90%) while keeping compression ratio and RPM the same.
Now, the engine ingests the same volume of air, but greater mass. This means that a greater mass of fuel will be needed and greater power will be produced.
The compression step remains the same, that is, if before it doubled the temperature from 300K to 600K, it will now double the temperature from 270K to 540K. The combustion chamber pressure before ignition will also be the same.
The burning of the fuel will now increase the temperature by a fixed amount, so in the first case it will reach 600+2000=2600K and in the second case it will reach 540+2000=2540K.
Expansion will now half the temperature (assuming reversibility and neglecting the minor change in composition of the gas during combustion) so the exhaust temperatures are 1300K and 1270K.
What we can see from this theoretical internal combustion engine cycle is that the efficiency of both compression and expansion is determined only by the compression ratio. But also note that a significant proportion of the energy generated in the expansion is needed to drive the compression. The proportion is marginally less at lower temperature (due to the fixed temperature increase provided by combustion) so the engine will theoretically run slightly more efficiently at lower temperature.
For identical efficiency, we would need the fuel to increase the temperature by 2000K in the 300K ambient case and 1800K in the 270K ambient case. But as the temperature increase from combustion is independent of the ambient temperature, the lower ambient temperature gives higher efficiency, at least in theory.
There are many practical factors that may affect this analysis. The first is that my exhaust temperatures are extremely high. I can assure you that a temperature increase of 2000K is typical for complete combustion of hydrocarbon in air (I am a boiler engineer, but here's a table for those who like to check http://en.wikipedia.org/wiki/Adiabatic_flame_temperature.) Reciprocating internal combustion engines do run at close to complete combustion of all the air ingested, so it would appear that there are significant losses to the cooling system during the combustion process.
My own experience is more with industrial gas turbines, which have proportionally much less loss to cooling systems, and an exhaust temperature arount 800K. Gas turbines never burn sufficient fuel to use all the oxygen in the inlet air, because the expansion turbine would melt. If suitable materials were available they would use all the oxygen, which would make them a lot more efficient.
The next thing to consider is the valve timing. It will be apparent that if we have the same ratio on the compression and expansion stroke, we have effectively the same mass of gas entering the engine at 270-300K and leaving it at a higher temperature. The higher temperature means there is a higher pressure at the end of the expansion stroke, and if the exhaust valve opens too early, about 2-3 bar absolute (1-2 bar gauge) of pressure will be wasted. Although there are from time to time suggestions of ways to improve this situation, including closing the inlet valve late (to reduce compression) and opening the exhaust valve late (to increase the expansion) this is not a common pattern of operation. One reason is that when there is only 1-2 bar gauge pressure left, the amount of friction means that not much shaft power can be usefully extracted. However modern engines are often able to vary their valve timing and this complicates real-world analysis considerably.
Thirdly, as per the above analysis an engine will produce more power in colder conditions due to the greater mass of air ingested. If more power is produced and friction stays the same, we would expect a small gain in engine efficiency, too.
Finally, let us consider turbocharged engines. Here the pressure in the exhaust gas which would otherwise be wasted is used to drive a turbine, which compresses the air at the engine inlet. The primary motive for this is to increase power by increasing the mass flow into the engine. The work of compression heats the inlet air, so an intercooler is installed between the turbocharger and the engine. The purpose of this is to cool the air back down to ambient to further increase the mass flow of air into the engine.
In conclusion, colder conditions produce a measurable increase in engine power (otherwise intercoolers would not be installed on turbo engines.) According to a theoretical cycle analysis they also produce an increase in efficency, though this is likely too small to be measurable, and may be negated or reversed by other factors.
A number of people continue to believe colder temperatures lead to worse mpg. This is generally true for a variety of reasons: Cold temperatures lead to lower air pressure in tires, wet roads have increased resistance, cold temperatures lead to thicker lubricants etc. The engine is operating more efficiently (unless we choose to disbelieve 1-Tc/Th) but is working against a host of conditions, that lead to reduced mileage, that occur during the winter. Try picking a spot of road you drive every day. Put your car in neutral at a given speed at a given point. You will coast farther during the summer, with your engine disengaged.
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Why does the combustion of fuel increase the temperature by a fixed amount? Isn't it dependent from the air intake? – seldon Jan 29 '15 at 10:25
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1@mattecapu the temperature rise depends only on the mixture composition. The heat released per kg for burning x% fuel in y% air is constant, and if we assume the heat capacity is independent of temperature (which is a reasonable assumption for the purpose of this answer) the temperature rise is constant. I have assumed 2000K rise, which gives different theroretical post-combustion temperatures of 2000+600=2600K and 2000+540=2540K in the two inlet temperature cases considered. – Level River St Jan 29 '15 at 12:07
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@Floris thanks, but I'm a an industrial combustion equipment engineer (mainly furnaces, boilers, and heat recovery systems for gas turbine exhaust) so I understand combustion pretty well. In particular it's counterintuitive to many people that cold fuel/air contains more energy per unit volume than hot fuel/air. tpg2114's comment about drag on your answer is an excellent point that didnt occur to me, though I'm sure it would have done to many on aviation SE. – Level River St Jan 29 '15 at 20:54
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+1 in a car this effect is tiny, but if you play a bit with a microengine (3.5cc or so) you would experience this very clearly, both in the higher amount of fuel per cycle that requires in cold days, and in absolute performances. – DarioP Jan 29 '15 at 21:18
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I once read a paper about a five-cycle modification to an inline four-cylinder engine so that the inner two cylinders could either operate normally (when significant power was required), or could operate in tandem as an "expansion cylinder" (during times of lower energy demand). I would think such an arrangement would effectively double the expansion ratio, though I'm not sure what effect that would have on efficiency. Have you ever heard of such a thing? – supercat Jan 30 '15 at 02:03
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@supercat I assume you mean this, or a development of it: http://www.5-stroke-engine.com/ See also http://en.wikipedia.org/wiki/Six-stroke_engine . The Crower six-stroke adds (after the normal 4 strokes) a water injection which cools the engine by generating steam, followed by a steam expansion stroke and steam exhaust stroke. It's a good idea, but means carrying water as well as fuel, and the water must be pure to avoid furring up the engine. My view is, it's probably best to leave the engine as it is and add heat recovery to the exhaust: either closed cycle steam generation or thermoelectric – Level River St Jan 30 '15 at 05:48
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That's the five-stroke behavior of the engine I described, but the paper described an engine which could switch between having four combustion cylinders (and no expansion) and having two combustion cylinders and two expansion cylinders. Engines with dedicated expansion cylinders go back decades, but from what I understand they're not usually worth the weight. What was cool about the approach in the paper was that the inner two cylinders could be used for a combustion cycle in times of high power demand, but could be used for an expansion cycle, increasing efficiency, in times of lower demand. – supercat Jan 30 '15 at 13:25
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I wish I could find that paper again; does it sound like a neat concept? – supercat Jan 30 '15 at 13:26
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@supercat It's nice to see people try different things, but for that last bit of expansion I think a turbine (as used in a turbocharger) is probably best. Turbines have an advantage over reciprocating expanders at low pressure differences, as they use the principle of the wheel to reduce friction: perimeter velocity is
ntimes hub velocity, so friction of the bearings feelsntimes less at the perimeter. Turbines are also smaller. The principle works with steam, too: the Titanic and her sister ships had 2 reciprocating steam engines, with the exhaust feeding into a low pressure turbine. – Level River St Jan 30 '15 at 22:04 -
@steveverrill: Turbines do have some advantages, but I would think they would "prefer" a flow of fluid which is smoother than what an engine would normally put out; expansion cylinders would have a cyclic "demand" for fluid which coincides with the main cylinder's exhaust stroke. Further, the cool thing about the approach in the white paper I wish I could find was the ability to use the expansion-harvesting cylinders as combustion cylinders during times of higher demand. Could an engine with multiple turbines use "trumpet valves" or something similar to dynamically reconfigure them? – supercat Jan 30 '15 at 22:13
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Does the greater heat loss by the engine casing and via the water-radiator cooling system have any additional benefits? – Bill Feb 09 '23 at 07:25
A car engine uses (approximately) a cycle called the Otto cycle, but this is close enough to a Carnot cycle that the same concepts apply. The hot temperature is the temperature of the gases after detonation, and the cold temperature is the temperature of the gases when they stop doing work i.e. when the exhaust valve opens and the gases exit the cylinder. It is not the external temperature. This is what I assume dmckee is alluding to in his comment, though I don't recommend touching the exhaust pipe right by the exhaust valve as the screaming noise is distracting.
In principle the external temperature has some effect because if the engine pulls in cold air the post detonation temperature will presumably be slightly lower. In practice I suspect engine management systems compensate for this. Anyhow, to a first approximation the external temperature does not affect the thermodynamic efficiency of a car engine.
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"though I don't recommend touching the exhaust pipe right by the exhaust valve as the screaming noise is distracting." Yes. Perhaps I assumed too much familiarity with engines on the part of the average reader. – dmckee --- ex-moderator kitten Jan 29 '15 at 14:09
The efficiency of the engine is limited by the ratio of the two temperatures, I.e. your ability to extract work from that temperature difference. However, most people are interested in fuel efficiency, and here you have to consider how much energy has to go into creating that temperature difference, not just how much work can be extracted from it. So on very cold days, a lot of fuel is used just to combat heat loss through conduction, as well as to bring the reagents up to combustion temperature (here the free energy is important). Lower temperatures also lead to incomplete combustion. The only saving grace is that the inlet air is denser so provides more oxygen, but I don't know how big that effect is.
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Internal combustion engines aren't Carnot engines, but they operate on a similar thermodynamic principle, so for the sake of this question, let's say they are.
$T_\textrm{H}$ is reached by adding thermal energy from the combustion to $T_\textrm{C}$, the temperature of the inflow. The energy produced by combusting a giving volume of octane is fixed (assuming you do it efficiently, it depends only on chemical properties of octane, oxygen, carbon dioxide, and water), adding $\Delta T$ to a given volume of inflow. Thus we can write $T_\textrm{H} = T_\textrm{C} + \Delta T$, and then rewrite the efficiency as: $$ \eta = 1-\frac{T_\textrm{C}}{T_\textrm{C} + \Delta T} $$
For fixed $\Delta T$, we see that $\eta$ decreases as $T_\textrm{C}$ increases - however, around standard operating temperatures, this difference will be quite small!
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For engines with oxygen sensors (Lambda sensors), the exhaust is kept at a very low oxygen concentration. As such, if the air ingested is not heated enough they pull more air (by mass) in the same volume (engine displacement) so they consume more fuel idling. Also, during winter I start my car early so have heat on windscreen (so it won't freeze on the way), and this also negatively affects the mpg. Other reasons why fuel efficiency during winter might be lower is the increased rolling resistance when running in the snow/mush, keeping engine in a lower gear to have improved engine braking, reduced overall speed. So while engines might be more efficient in winter, cars (overall) usually aren't.
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