Question

I have a engine and it’s connected to a generator... (Theoretical Example Below).

When the patrol engine is inputting mechanical energy( 1000Watts) and the generator is generating electricity( 900W, including the losses) and there is a load connected to the generator that is (800W) is the mechanical input able to continue doing its work with that load connected?

I know generally that there will be a major mechanical resistance (difficulty) because of the load. In some cases there has to be an increase in mechanical load. Why? It’s above the load by 200W its not equal its less...

If ma

Answer 1

If maybe I had 1000W in and a load that is like 950W that would make sense... Losses would case the mechanical load to overcome them and produce for the load. But if resistance was very very low! The mechanical load does not have to be increased.

All I'm trying to ask is if my load is less than the mechanical input why is there a need to increase it? Because I've read some example where there is mechanical difficulty I must increase input to overcome it.

In this situation could you guys answer an answer relevant to my example? Thanks!

Answer 2

1st part is the question 2nd part is the 2nd post :P

@Jemurray3 @experimentX Help me out!

Answer 3

I wanted to make sure:

do I need to increase mechanical input of the load was 800W? And the mechanical input is 1000W and the electrical output is 900W.

Will there be a difficulty that would require more mechanical input?

Answer 4

In what condition is the mechanical input has to be increased to work the load perfectly? If the load demands more?

Answer 5

I'm confused by the entire question here.

Let's say you have an engine rated at 100kW (~134HP) and you mechanically connect it to a 4-pole generator rated at 90kW. The desired output is 60hz so you calculate that the engine will have to drive the generator at 1800RPM: \[RPM=\frac{120f}{P}=\frac{120(60hz)}{4}=1800\]You start the engine with the load disconnected and ramp it up to 1800 RPM. After a warmup period you apply a 60kW load on the generator. The engine RPM will inevitably drop(~5-10RPM) for a portion of a second since the start-up currents for many types of equipment (think motors, compressors, etc) will be higher initially. The engine's governor senses the RPM dropping (various methods are used to do this) and applies more fuel to compensate for that and go back 1800 RPM. As long as the load on the generator doesn't exceed the rating of the engine, the engine can compensate for these dips. Later, as the load subsides, the governor reduces fuel as necessary to sustain 1800 RPM.

As you can see from above, a good design would require that the driving engine be rated at a somewhat higher kW than the generator. This ensures that engine should never be overloaded to the point at which it can no longer compensate and maintain rated speed (barring any mechanical problems)...even if the generator is operating at its maximum rated output.

The bottom line is that the governor's sole job in life is to continually adjust fuel to the engine to maintain the required speed...so why would you be trying to do the governor's job?

Answer 6

@Shane_B haha Im not!

In you're example you showed me that the governor applies more fuel because of the load.
Lets say that he noticed an Rpm drop and did not apply more fuel what would happen?

I mean the load is just 60kW the engine is producing about 90kW so why the need to worry about any drop? You're 30kW safe.

Answer 7

OW WAIT! Do you mean that there has to be an increase of fuel to maintain 1800RPM its the same amount used in the beginning to start it anyway right? I'm starting to get this! Its like a cycle! That is constantly repeating. So you don't shut or lose the system you need to equally distribute the mechanical input and manage the output :D

Answer 8

Im starting to get this :)

Answer 9

I hope I am xD

Answer 10

I think you're getting it :)

\[\uparrow \text {electrical load } => \space \uparrow \text{mechanical resistance} => \space \uparrow \text {fuel consumption}\]And vice versa.

Think of it in terms of driving a car at a constant speed. When going uphill you must apply more accelerator (increase fuel) and when going downhill you apply less accelerator (reduce fuel). In a car, the driver is essentially acting as the governor. In an engine-generator, the governor is a device that does these adjustments for you...just like the cruise-control feature on a car does. If the governor does not perform its job then your engine's speed and therefore, generator output frequency will fluctuate as the load goes up/down.

I think what you also need to understand is that engines (and generators) are rated based on their *maximum* HP/kW output. Just because an engine is rated at 100kW doesn't mean it's always outputting 100kW. Engine output at any given point in time is the result of the load and how much fuel the governor is sending it. The engine will only output 100kW when it's fully loaded.

Answer 11

Very good discussion :)
@Shane_B good points!
@Hope99

I think you need to compare this with the car example.

Imagine you going down a road you're applying the same input @ 100KH and when you're going up hill you need more input to stay @ 100KH :) Or you'll lose it.

If you're going in the same land level going at @ 100KH you're just applying the same equal input to stay @100 KH,

If you going down hill you don't need to apply any additional fuel at all...

In case of you're generator example I'll try to make sense:

When a generator runs due to the mechanical input it generating an output for the load.
If the load is less then the mechanical input there is need to apply more.
If you're load requires more! from the generator the generator ask more from the mechanical input! So it slows significantly because it can't power the load fully.

However, if it was of equal you just need to apply a little more mechanical input to overcome resistance.


vision.

Answer 12

Thanks @Visionary01 & Shane_B pretty much I got the idea.

Load less then input = No need to increase input.
Load greater then input = Increase of inputed needed


I got this! Thanks!

Answer 13

@Shane_B What do you think?

Answer 14

I think you got it :)

Answer 15

Thanks!
I'll just add another point:

When I use a generator and put a load to it I can feel the difficulty when turning it! Only because the load requires so much power that I can't supply.

However, if the load was less I can easily do it.

Example:

A mechanical device applied 100kW of raw power to a generator! Its generated 90kW at its maximum output (Hight efficiency) a load was connected requiring 80kW.

That 80kW is the load and 20kW went with all the losses. There is no need for more mechanical input.

Where as there is a 90kW load there will be a mechanical resistance and I have to increase it.

My conclusion is: If you have 100kW of input use a load that is less by a lot example: 80kW's of power!

Thanks again!

Answer 16

@Shane_B I hope you agree with my final post! :)

Answer 17

I thought you had it but after reading your example I'm not so sure!

Remember that the power output of an engine is dependent on the load. An engine just doesn't supply 100kW of input when there is no external load applied to it...it only generates power enough to overcome the mechanical losses incurred while running (minimal load=minimal kW output).

Once you apply a load to it the engine will be forced to provide more output power (up to it's max output) since it's actually doing more work.

Answer 18

@Shane_B Maybe the mechanical input is enough for both the mechanical losses and the load?

Answer 19

@Shane_B I understood it pretty well but I can't describe it properly thats my only problem :P

However, as you the output is dependent on the load, true.
If the load was under the input's range by a lot that means there is no worries it increase the input :)

However, if the load is EXACTLY the same as the mechanical input obviously due to the mechanical losses etc... There is a demand for more input.

Answer 20

That sounds better :)

Answer 21

good to hear :)