|May15-12, 07:03 AM||#1|
IC Engine performance test
I came across an engine performance graph as attached.
I have basically two questions looking at that:
1.To do any engine testing we keep some parameter constant - May be throttle position or applied load (torque) or speed or power. What is kept constant for this test?
2.In the torque Vs. engine speed curve, till 1750 rpm (approx.) the speed appears to be rising. I am little confused here. If torque rises, it means the load on the engine climbs up. Thus, should it not result in drop of the speed instead of rising?
I have these questions unresolved for long. Pl help. Thanks.
|May15-12, 11:05 AM||#2|
It tricky to answer as although it seems simple, it's a very broad question.
A dyno run is done at full (wide open - WOT) throttle, it's a measure of the output across the cars operating range. This is nothing braking the engine apart from the engine itsself and the measuring device.
Brake dynos can apply a braking torque, they can be used to do part throttle testing, and hold a car at a specific rpm, by appliyng a brakeing torque equal to the engine torque output.
Hence the term: brake horsepower.
At WOT @ 1750 rpm. The torque output of the engine is 400 Nm.
At WOT @ 4000 rpm. The torque outout of the engine is 290 Nm.
Torque output, is indicitave how the engine breathes.
At very low RPM, the cylinders don't take in all the air they can.
As you climb in RPM and get to peak torque, the engine is sucking in peak amount of air and making the biggest 'bang' it can.
After peak torque, as RPM gets high the air is restricted by the inlet, so torque begins to drop. However power will still increase becuase although you are getting a smaller 'bang' per cycle, you are doing more bangs per unit time.
There is a point where torque will drop off enough that no increase in rpm will compensate. At this point power output will drop, and you have reached the maximum useful rpm of the engine.
|May16-12, 01:37 AM||#3|
I was comparing this graph with the tests I have conducted on engines as a college student. There, I used to measure torque using loading the engine. There used to be a drum connected to the crank shaft. I used to apply load on it. Thus the product of load and drum radius becomes the torque. In that case, torque in the graph is directly proportional to the load applied on the engine. That was the basis of beginning this thread.
In this particular test we are discussing on, can you please explain me how torque is meausred? I mean the working principle of the dyno used here. Also, looking at the Power Vs Speed curve, you can identify easily that power is found out by multiplying the speed and torque of all the instances from the Torque Vs Speed curve. Is it applicable for all type of dyno test? Thanks.
|May16-12, 03:03 AM||#4|
IC Engine performance test
You have a fixed load (a weighted drum) that you know the polar moment of inertia of (angular mass) and are using engine output to accelerate it. By measuring the acceleration of the drum, you know the applied torque.
The linear analogue, is someone pushing a block. The force appied can be measured by measuring the acceleration of the known mass (F=ma). Changing the block will change the acceleration, but the force can still be calculated.
As there is no braking force to react the applied torque, you can't steady state test. So all tests are typically done at Wide open throttle. Then the rig is left to coast down, the rate of deceleration measures the losses in the engine.
Just like the block will slow down due to friction, losses in the engine will act like a braking torque on the drum.
Brake Dyno: You actively apply a torque that acts as a brake on the drum. So you can steadily apply a load to decelerate the engine. When you apply a load that is equal to engine output, you hold at a steady rpm.
This can be used to do steady state testing.
It's a fundamental relation between the two, so it always applies.
|May16-12, 04:34 AM||#5|
Thanks again Chris.
I guess I have got the point clearly now. Just for confirmation, I rephrase what you have explained.
Torque (T) = Polar mass moment of inertia (I) x angular acceleration(α)
In this experiment, polar mass moment of ineria (flywheel) is known.
This flywheel is accelerated by the engine when speeds & time taken are noted. Thus acceleration is known.
Then by using the above mentioned equation, torque Vs speed curve can be plotted.
Initially the acceleration increases, flattens and drops down. But because its still ACCELERATION, the speed never drops.
Is my summary right?
I am sorry for asking a couple more questions. Why are there no points after 4000 rpm? At a point there will be a steady speed which means zero acceleration. That is no torque condition. Why is that point ignored? Thanks in advance.
|May16-12, 12:46 PM||#6|
Maximum test rpm is somewhat arbitrary, generally speaking. It's certainly possible to continue testing an engine at higher and higher speeds. At some point, the engine will reach its maximum possible air and/or fuel flow capability and not accelerate any further. Far more likely, the engine will experience catastrophic failure of one or more components before reaching its "breathing" limit. (Not uncommon when testing new designs.)
|May16-12, 11:39 PM||#7|
"Judging by the graph, the engine appears to peak at about 4000 rpm, so there is no reason to run it higher. Also, I'd guess it's a diesel engine, and they operate at lower speeds than spark-ignition engines"
Thanks, Pantaz. :)
|Similar Threads for: IC Engine performance test|
|air flow in inlet manifold affecting engine performance||Mechanical Engineering||6|
|Performance Envelope of a Turbojet Engine||Aerospace Engineering||0|
|Engine performance variables||Mechanical Engineering||2|
|Can it hurt an engine's performance by having to big of a throttle body?||Mechanical Engineering||10|
|Engine Performance(Finding Power)||Introductory Physics Homework||3|