# Internal combustion engine partially loaded.

• beginner49
In summary: For the amount of power needed to maintain a certain speed, the engine will require the same amount of energy regardless of the number of cylinders. .However, the engine will require more fuel to maintain a certain speed when running at a lower rpm.
beginner49
.
About a car internal combustion engine.

Lets suppose the engine torque is maximum at 3500 RPMs when the engine gets also its maximum load, let's say, a 90%. This kind of data is usually shown at WOT.

My question is: if the throttle is not wide open, but half open, allowing the cylinders to be loaded with 45% (half of the maximum load), is the engine torque also reduced to the 50%? and so would the output power?. I mean, at the same RPMs, is torque proportional to the engine load?

Or, Does the engine efficiency change when partially loaded and then, the torque is something difficult to figure out? and so, would I need my engine BSFC map to solve this question?.

For me, your question does not make any sense as "percentage load" is by definition "percentage of maximum available torque output". So load and torque are the same thing, measured differently (or in different "units" if you will).

What I think you refer to is the "percentage load" that is estimated from intake manifold air pressure by EFI systems. In that case, I would say that it is just that, an estimation. Computer tables must be determined by comparing those inputs (intake pressure, temperature, air flow, etc) to actual torque data from dyno tests such that proper injection time and ignition timing can be chosen.

If you refer to the angle of the throttle plate, it is not directly related to the manifold pressure so even less to the torque output of the engine.

The point is that I am trying to plot a BSFC map of my small car engine according with the fuel consumption numbers I am empirically figuring out and I see that the sweet spot is drifted to the higher RPMs range (around 4500 RPM).

Most of BSFC maps (in a torque-RPM graph) I get in internet technnical pages have that sweet spot (highest efficiency) evolving a point around 70% of engine load and 40% of maximum RPMs (about 2500 RPMs).

My engine is an 2-cylinder boxer type with a maximum RPM of 6000. My question is: will this 2-cylinder engine have its sweet spot drifted to a higher rpms with respect to a normal 4-cylinder engine?. Is there any explanation that may argue this?.

This small car has a 28 HP at 5500 RPMs when in 4th gear reaches its maximum speed (110 Km/hr). Its torque peak is at 3500 RPMs (4.1 mkg).

What I want to know is, if because the fact of having 2 vice 4 cylinders, the BSFC map is drifted to a higher RPMs.

.

The number of cylinders have no real effect on fuel consumption (everything else being the same).

BSFC depends mostly on valve events, fuel enrichment and ignition timing. For example, most BSFC maps found on the net are for «everyday» cars, so they are tuned for fuel efficiency. If your engine is tuned for maximum power, it will probably have its sweet spot at a higher rpm.

Nonetheless, I would guess that this spot cannot be higher than the rpm for maximum torque as past this point, usually efficiency starts dropping. So I found suspicious that you have your best BSFC at 4500 rpm and peak torque at 3500 rpm. Maybe there is an error in your measurements.

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Thanks a lot, Jack, for your help.

I would show you all the numbers, but first, I would like to know your opinion about the following reasoning:

My car needs 18 HP to maintain a 100 kms/hr speed (accounting both rolling and drag resistances). So it needs 18 HPh or 13.2Kwh of energy to complete those 100kms.

To keep on driving at 50 kms/hr my car needs 4 HP. So it "consumes" 8 HPh or 5.9 Kwh of energy during the 2 hours it takes to complete those 100 kms.

If this is correct, it would mean that my car consumes the same amount of fuel during one hour at 100 kms/hr as during 4.5 hours driving at 50 kms/hr. In the first case, the distance would be 100 kms, while in the second case, the car will have run along 225 kms.

Up to this point, am I right?

Thank you very much in advance for spending your time helping me with my problem.

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That seems correct for the amount of energy required to propel the car. But don't forget that there are a lot of losses between the energy from the combustion to the one available at the wheels.

Thanks Jack. Let me then go step by step.

Instead of handling a bunch of numbers and for the sake of visibility, I will show you a couple of graphs.

The following is the engine curve. On green torque function, on red, the power. Both curves are considered when the throttle plate is wide open. By the way, this is an old (1975) carburated 2-cylinder boxer engine.

http://img827.imageshack.us/img827/152/2cvenginecurve1jpeg.jpg

On this other, I have include the car speed (4th geared). On blue, the power required because of drag resistance and on orange the total power to propel the car on a level road with no wind at all.

http://img32.imageshack.us/img32/6280/2cvenginecurve3jpeg.jpg

On this graph you can also see the maximum car speed, around 117 kms/hr.

Data based on car aerodinamics coefficient, really poor (0,51), and a front area=1.6 m2.

The rolling resistance (gear box is included) was obtained empirically and here, I considered it is proportional to car speed. Based as well on the car weight of 700 kgs.

Up to this, does it sound OK?

Let me know if you see anything wrong or very deviated from what you would consider another figures.

.

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Sounds OK. But this is the engine power and you need the wheel power to determine the top speed (which I estimate to 109 km/h with your numbers).

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Thanks again, Jack, really very kind of you.

About what wheel power is, I will have to learn something else. I thought that once we have subtracted the drag and rolling power from engine flywheel power, the result would give the real traction or wheel power. But as I said, I have to look into this deeper.

Going back to what I want to solve. From my second graph I have removed the air drag power curve and left the total power to propel the car at different speeds.

Now, my very initial question.

For plotting the engine load (or torque) needed to propel the car at those different speeds I have done the following:

First look for the percentage of power in respect to the wide open power at that particular RPM.

Let me show you, for instance, for 60 kms/hr. The necessary power is 5.7 HP, which is the 35% of the engine power (16 HP) when WOT. So, what I do is to apply this percentage of 35% to the engine torque when WOT. I then assume that to have a 35% power out the WOT power, I have to load the engine with the 35% load against the load when the throttle is wide open. In this case, at 3000 rpms WOT, the torque is 3.8 dNm. So, that 35% would be
1.35 dNm. There is where I plot with a black mark, the point of my engine load for going at 60 kms/hr.

http://img694.imageshack.us/img694/1321/2cvenginecurve4jpeg.jpg

What I mean, I do not mind if at 3000 rpms WOT the engine load is complete or only 70 or 80%. Whatever it is, what I understand is that for getting a 35% of power in respect a WOT condition, I have to load the engine with a 35% of whatever load is at WOT.

And so with the rest of engine load plot.

After this, if you agree with my method, I will get into how and where to map the engine efficiency spots in order to make them match with the real car fuel consumption.

.

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I'm following ...

Hi Jack

Here is a table with the data corresponding to the fuel consumption in litres during a 100 km trip at those different speeds.

http://img545.imageshack.us/img545/3995/2cvenginecurve5jpeg.jpg

I can see that the engine is more efficient as we go to a higher rpms and this is the point I do not find the “normal” location of the sweet spot in a “normal” BSFC map. Unless, the way I have used to compute the engine loads (black points) is totally erroneous or maybe I have a misconception about the real calculation of those numbers in the last column (energy per litre of fuel).

With these data, would you be able to map a "standard" BSFC curves evolving that sweet spot in the 2500/3000 rpm and 2.5 to 3.0 dNm area?

The only "approach" I have found is this, by duplicating the graph vertical scale: ( I have not put any number on the curves). But if this the case, my car would save more fuel if I drive those 100 kms in the 3rd gear at 50 kms/hr and 3300 rpms. I have not tried to drive in the 3rd gear for 100 kms to check the fuel consumption.

What do you think, Jack?

http://img845.imageshack.us/img845/9474/2cvenginecurve6jpeg.jpg

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I can see that the engine is more efficient as we go to a higher rpms and this is the point I do not find the “normal” location of the sweet spot in a “normal” BSFC map.

It is normal as you only have partial data, i.e. you only have the numbers where you have black dots. Put those black dots on «typical» BSFC map (like the one below) and you will find a similar progression (higher rpm --> higher efficiency):

One note: I converted the BSFC you found to g/kW/h and you obtain these numbers (from 40 km/h to 115 km/h):

0.896
0.632
0.555
0.500
0.450
0.391
0.361
0.356
0.353

The BSFC at lower rpm seem to be abnormally high. It is probably because there are HP losses that you do not take into account (the actual HP produced by the engine is higher than you think); A problem that does not reflect at higher rpm where the majority of the power goes to the wheels.

Now, if you repeat your experiment in a lower gear, the imaginary line going through your new black dots will be below the one you have right now. If you check against the typical BSFC map above, you will obtain a lower efficiency overall with the same progression as rpm increases.

If you could add a 5th gear, the imaginary line would be higher than the one you have now; Bringing you closer to the sweet spot you want to find. And you would get a better fuel efficiency.

Bottom line: Intuitively, your engine is probably most efficient when it produces approximately 15 hp @ 3000 rpm. Find the speed you can obtain at 15 hp and the gear ratio you need to match that speed to the rpm, and you will have the best fuel efficiency for you vehicle.

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Hi Jack. Thanks again for your knowledges and time.

Probably that consumption figure at 40 kms/hr is not very reliable as I only drove for half an hour at that speed to learn it and statistically this is not very correct. This is an air-cooled engine and I do not know if that has anything to do with a higher engine losses at low rpm.

What I do not see is what would happen driving in a lower gear. For keeping, say. 50 kms/hr, the engine should go at 3300 rpms, the power required would be that for 50 kms/hr, 3.8 HP, while this required power vs. the WOT power at 3300 rpms is only a 20%. That means that the engine load (as a percentage of required power vs. WOT power) is a 20% of the WOT torque (or load) at 3300 rpms, and that is about 8dNm. This would lead us to locate the corresponding (now blue) dot (50 Kms/hr-3th gear) below that of driving at the same speed but in the 4th gear.

I have plotted (blue dots) the theoretical engine loads vs. car speed going in 3rd gear (linked with their corresponding black dots) and the engine efficiency is even worst at all rpms.

http://img341.imageshack.us/img341/5072/2cvenginecurve8jpeg.jpg

To me, I think, I should have the throttle more depressed when driving in a lower gear. Like, for example, here I can reach 90 kms/hr in the 3rd gear with a half depressed throttle.

Jack, I think I am doing something (really big) wrong in my calculations.

Would you please be so kind of taking a look on this?

Thank you very much.

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The new graph in your last post corresponds exactly to what I said in my previous post and makes perfect sense to me:
jack action said:
Now, if you repeat your experiment in a lower gear, the imaginary line going through your new black dots [Note: blue dots in your graph] will be below the one you have right now. If you check against the typical BSFC map above, you will obtain a lower efficiency overall with the same progression as rpm increases.

In your new graph, plot a new dotted line for an imaginary 5th gear (say 0.50:1 gear ratio) and see what it will give you:
jack action said:
If you could add a 5th gear, the imaginary line would be higher than the one you have now; Bringing you closer to the sweet spot you want to find. And you would get a better fuel efficiency.

You are wrong when you said the following:
beginner49 said:
To me, I think, I should have the throttle more depressed when driving in a lower gear.

Try this experiment: If you try to move your vehicle from a complete stop in 1st gear vs 4th gear, with which gear ratio will you depress more the throttle? It will be in 4th gear (and it might not even be enough to be at full throttle and the engine will stall).

.
Thanks a lot, Jack. You are absolutely right. Now, I have extrapolated curves beyond 6000 rpms, and I learned that at 3rd gear WOT, car would not go faster than 100 kms/hr, even keeping this long is not safe for the engine.

As well, plotting a hypothetical 5th gear (1.33:1 respect to 4th) the engine would go more efficient but the car maximum speed would be 108 kms/hr.

http://img214.imageshack.us/img214/3796/2cvenginecurve9jpeg.jpg

I know that all these numbers may have an error, but I think all those are well within a +/- 5% from this car real world, and I feel happy with these results.

I would like to thank you for your invaluable help.

Anything you want to add, I really appreciate.

And now, let me show you the car we have been discussing about.

http://img834.imageshack.us/img834/7663/chicagoenginecurves.jpg

Thanks again.

.

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This chart is labeled incorrectly.
beginner49 said:
...
The following is the engine curve. On green torque function, on red, the power.
http://img827.imageshack.us/img827/152/2cvenginecurve1jpeg.jpg

Since horsepower is calculated from torque and rpm, the curves must cross at 5252 rpm.

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pantaz said:
This chart is labeled incorrectly...since horsepower is calculated from torque and rpm, the curves must cross at 5252 rpm.

I am afraid that is not right. That will be in the case of using pounds_feet as the torque unit. Apart from that, having a crossing point at a relationship constant is only an option which does not make the curve more reliable or accurate.

In this curve the torque unit is dNm and any possible crossing should be at 7120, which is far beyond the rpms range. I labeled this curve in the most comprehensive way for adding bsfc maps, different gears torques, etc...

Regards.

.

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beginner49 said:
I know that all these numbers may have an error, but I think all those are well within a +/- 5% from this car real world, and I feel happy with these results.

I would like to thank you for your invaluable help.

Anything you want to add, I really appreciate.

And now, let me show you the car we have been discussing about.

http://img834.imageshack.us/img834/7663/chicagoenginecurves.jpg

Thanks again.

Well, my help is free so I'm glad that it makes you happy and that you appreciate it. You have a nice car, which seems to be in good shape as well. The colors remind me of a car that I used to own many moons ago and that I loved dearly:

Thank you for triggering those good memories.

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beginner49 said:
In this curve the torque unit is dNm...
Sorry about that -- I should have looked at it more carefully.

## 1. What is an internal combustion engine partially loaded?

An internal combustion engine partially loaded refers to an engine that is operating at a lower load or power output than its maximum capacity. This can happen when the vehicle or equipment using the engine is not under heavy load or when the engine is idling.

## 2. What are the effects of running an internal combustion engine partially loaded?

Running an internal combustion engine partially loaded can lead to lower fuel efficiency and increased emissions. This is because the engine is not operating at its optimal level and is wasting fuel to maintain its speed.

## 3. How does a partially loaded internal combustion engine compare to a fully loaded one?

A partially loaded internal combustion engine will have lower power output and efficiency compared to a fully loaded one. This is because the engine is not able to fully utilize its capacity and is not operating at its optimal level.

## 4. Can running an internal combustion engine partially loaded damage the engine?

Running an internal combustion engine partially loaded for short periods of time should not cause significant damage to the engine. However, prolonged operation at low loads can lead to carbon buildup and potential damage to engine components.

## 5. How can the efficiency of an internal combustion engine be improved when running partially loaded?

To improve the efficiency of an internal combustion engine when running partially loaded, it is important to maintain the engine properly and use it for its intended purpose. Additionally, using high-quality fuel and avoiding prolonged idling can help improve efficiency and reduce emissions.

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