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% Engine Load vs. % Power
3

% Engine Load vs. % Power

% Engine Load vs. % Power

(OP)
Is it safe to assume that an engine at X% load will make X% of the rate power at that RPM?

Example: an engine produces 100 HP at 5000 RPM at 100% load. If I reduce the load to 60%, does the engine only produce 60 HP?

RE: % Engine Load vs. % Power

An engine produces exactly as much power as the load requires or it else will slow down and stop. Throttles regulate engine power to keep the rpm at the desired level.  

RE: % Engine Load vs. % Power

(OP)
I understand that. It's simply reaching equilibrium. If the engine produces any more or less power, then it will speed up or slow down, respectively.

My question is, is the HP produced linearly proportional to the load reduction. I'm almost certain that it's not, but just want someone to explain why.

RE: % Engine Load vs. % Power

The correct answer to your question, within limits, is yes. Torque (if that is what you mean by "load") multiplied by engine speed is proportional to the engine power output.  If, by load, you mean the torque at the engine tailshaft, then cutting load in half at the same speed cuts the power output in half. That doesn't mean exactly half the fuel consumption at that operating point, which is probably why folks are reluctant to say yes to your question, even though it comes straight out of an engines-101 equation.

If you describe load in terms of %, then you can't say the same thing about engine speed (i.e. cut in 1/2 then cut power in 1/2) since the same % load at different engine speeds will be a different torque output value.

RE: % Engine Load vs. % Power

Quote:


an engine produces 100 HP at 5000 RPM at 100% load. If I reduce the load to 60%, does the engine only produce 60 HP?

This is not right how you have it.
For one, acceleration must be accounted for. Secondly, it's force that propels you forward not HP. HP is a measure of torque at certain engine speeds.

Quote:


Is it safe to assume that an engine at X% load will make X% of the rate power at that RPM?

Nope. This is also not right. Engines cannot produce any torque at any RPM.
It is a highly nonlinear process. There are many more factors to account for.

 

peace
Fe

RE: % Engine Load vs. % Power

I thought "percentage load" was by definition "percentage of maximum available torque output" under given conditions.

In that case, 60% load means 60% of available torque (i.e. power, if RPM is held constant) because that's how the term "load" is defined.

Fuel consumption will, of course, not be in direct proportion due to various secondary effects.

Powertrain engineers speak in terms of "brake mean effective pressure" to avoid the uncertainties associated with other ways of defining how much load an engine is under.

RE: % Engine Load vs. % Power

The original question is both specific and vague.  Can the OP perhaps describe the problem he/she is trying to solve to put things into context?

Otherwise, BrianPetersen has it licked.

- Steve
 

RE: % Engine Load vs. % Power

(OP)
All HP, torque, and fuel consumption curves are given at 100% engine load (measure by brake mean effective pressure, or BMEP). It is obvious that I can keep the same engine speed, and vary the load at that speed.

Example: A car driving at constant velocity uphill at 3000 RPM WOT vs drive a car driving at constant velocity downhill with the throttle shut, with the car at 3000 RPM. The engine is producing less torque (which is linearly proportional to BMEP, and therefore load) while moving downhill. It must be producing less torque because there is less force to overcome, e.g. gravity. Equilibrium dictates that the engine cannot be producing the same force on the downhill or the car would be accelerating.

My question comes from the fact that BSFC is not scalable. As people have replied, I can't halve the torque value at constant engine speed and expect half the fuel consumption. From what I've read, brake specific fuel consumption often, if not always, increases as load is decreased.

Hopefully that clears up any confusion in the question. Again, my ultimate goal is to accurately estimate the BSFC at a reduced engine load, if I don't have the fuel consumption map.
 

RE: % Engine Load vs. % Power

2
are you really trying to estimate BSFC, or are you trying to estimate fuel consumption?

I agree with Brian, and Jsteve has an important point about 100% load being rpm-specific (for a variable-speed engine).


 

RE: % Engine Load vs. % Power

(OP)
I'm trying to calculate fuel consumption (lbs/hr) at a 60% load. A fuel consumption curve is given at 100% load. If RPM is constant, then an engine at 60% load will produce 60% of the power of the 100% load. BSFC is not scalable in the same way. I need to find the BSFC at 60% load so that I can multiply it by the 60% power in order to gain the fuel consumption in lbs/hr.

Ex:

Engine makes 100HP @ 5000 RPM under 100% load. It's BSFC is 0.40 under these conditions. If I reduce the load to 60%, then at the same engine speed it will make 60HP. The new BSFC will be some other value, probably higher, than 0.40.

Is there a logical assumption to find the new BSFC value?

*Sorry to all if I haven't been clear.

RE: % Engine Load vs. % Power

I think that between this thread and the former, you have the info you need to make your estimate.  Estimate the BSFC, then multiply by 60% power (at 60% load) to get fuel consumption.

 

RE: % Engine Load vs. % Power

Since I think we've gotten this thread to the point the OP needed, let me pick some bones elsewhere:

 

Quote:

(FEX32)  This is not right how you have it. For one, acceleration must be accounted for. Secondly, it's force that propels you forward not HP. HP is a measure of torque at certain engine speeds.

- based on the context, I thought it was clear enough that this question was w/r/t steady state operation (so acceleration is zero)
- Power is absolutely required to move something down the road, and is a key factor determining how fast somethign accelerates and its maximum speed.  Power is transferred via force, but force without velocity is not useful for propulsion.  Be that as it may, none of that is relevant to this discussion.
- Power is the product of force and velocity, or of torque and angular velocity, regardless of speed.  It's not true that there are only certain speeds where that mathematical relationship holds.


 

Quote:

(FEX32) This is also not right. Engines cannot produce any torque at any RPM. It is a highly nonlinear process. There are many more factors to account for.
actually, the OP pretty much had the %load vs. power output correct, perhaps the only bone to pick was the use of the word "rated" to describe max power at a given RPM (rated power is often the max power at the rated RPM - but sometimes rated power is lower than the max power, even at rated rpm)

 

RE: % Engine Load vs. % Power

Your are right ivymike.
Sorry, I must have read the OP wrongly. Not sure why I said that.

Yea, seems it's too simple. smile

peace
Fe

RE: % Engine Load vs. % Power

"Power is the product of force and velocity, or of torque and angular velocity, regardless of speed.  It's not true that there are only certain speeds where that mathematical relationship holds."

I am well aware of this. I never meant to contradict the fundamental relationship which I have known since high school.
Guess I was too tired to think last night. smile

peace
Fe

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