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Understanding Electrical Units

Submitted: Tuesday, Oct 04, 2005 at 16:08

Mike DiD

It’s much harder to understand something that’s invisible, so people use something like the flow of a liquid to explain the flow of electricity. Rather than just explain in theory, here I’ll use a practical example - trying to understand poor performance under full load.

The first sentence in each paragraph below talks about the flow of fuel from the tank with internal pump to the engine after a turbocharger has been added. The second sentence in each talks about the electrical equivalent where electricity flows from the battery to an upgraded compressor in the rear.

a) The problem is that the engine hesitates under full load, so should you upgrade the pump, or is something else causing the problem - some measurements will help you decide.
b) The problem is that the compressor slows down under full load - should you upgrade the battery, or is something else causing the problem - some measurements will help you decide.

a) When the engine is off, the pressure at the fueltank is 12.7 psi and the pressure at the engine is 12.7 psi.
b) When the compressor is off, the voltage at the battery is 12.7 volts and the voltage at the compressor is 12.7 volts.

a) When the engine is at idle, the pressure at the fueltank is 12.6 psi and the pressure at the engine is 12.1 psi. There is a 0.5 psi drop across the fuel line when there is a flow of 1 litre-per-hour.
b) When the compressor has no load, the voltage at the battery is 12.6 volts and the voltage at the engine is 12.1 volts. There is a 0.5 volt drop across the cable when there is a flow of 1 amp.

a) When the engine is at full load, the pressure at the fueltank is 12.2 psi and the pressure at the engine is 10.2 psi. There is a 2.0 psi pressure-drop across the fuel line when there is a flow of 4 litre-per-hour.
b) When the compressor is at full load, the voltage at the battery is 12.2 volts and the voltage at the compressor is 10.2 volts. There is a 2.0 volt drop across the fuel line when there is a flow of 4 amps.

a) Obviously the pump can supply 4 litres-per-hour, but the fuel line is too small and is causing too much pressure drop at the needed flow rate.
b) Obviously the battery can supply 4 amps, but the wire is too small and is causing too much voltage drop at the needed current.

a) How long will the 80 litre fueltank last at full load ? At 4 litre-per-hour it will last 20 hours (80/4).
b) How long will the 80 amphour battery last at full load ? At 4 amps it will last 20 hours (80/4).

a) How many days will the 80 litre fueltank last if I run the engine at full load for 5 hours a day ? At 4 litre-per-hour it will last 4 days (80/4/5).
b) How many days will the 80 amphour battery last if I run the compressor at full load for 5 hours a day ? At 4 amps it will last 4 days (80/4/5).

a) Since it is not a hydraulic power system. no-one calculates the power in moving the fuel
b) The power the compressor draws at full load from the battery is 48.8 watts (12.2 x 4) and the compressor is using 40.8 watts (10.2 x 4) - 8 watts is lost in the cable as heat.
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AnswerID: 133025   Submitted: Tuesday, Oct 04, 2005 at 18:56

ACDC replied:

You have got to much time on your hands..^_^..
Reply 1 of 3
AnswerID: 133039   Submitted: Tuesday, Oct 04, 2005 at 20:30

VK3CAT replied:

Assuming a typical 4 x4 compressor. The motor is not rated for a full duty cycle so it will probably die well before the 4 days! :-)
Also note that current consumpion will increase as voltage decreases.
Extreme example, in a short circuit condition you will have close to 0 volts potential difference between the positive & negative battery terminals but will have maximum current flow. It is impossible to achieve maximum voltage and current at the same time.
Similarly, as the compressor continues to run, the battery volts will decrease and the current flow will increase meaning the wattage (volts x amps) will remain the same. Thus the "available running time will be decreased from the ideal 4 days.

Nice comparison though.
Cheers Tony
Reply 2 of 3
FollowupID: 387296   Submitted: Tuesday, Oct 04, 2005 at 23:02

Wok posted:

Tony,
Can't follow the logic,in Mike's example the load[impedance] is assumed constant ie [voltage is directly proportional to current]...hence lower battery volts results in lower current........or am I missing something?

eng
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FollowupID: 387351   Submitted: Wednesday, Oct 05, 2005 at 13:21

Mike Harding posted:

>Also note that current consumpion will increase as voltage decreases.

No it won't.

Except in very special circumstances in devices which exhibit negative resistance - a tunnel diode is one and I have observed the phenomenon in a small neon.

Mike Harding

FollowUp 2 of 3
FollowupID: 387392   Submitted: Wednesday, Oct 05, 2005 at 20:23

VK3CAT posted:

Current flow in a device (ie motor) will increase when it is running at low voltage.
I am assuming that the motor is driving a fixed mechanical load (pump or compressor). A certain amount of power is required to drive that load.
That driving power can be calculated in watts (volts x amps in a DC circuit).
Decreasing the voltage will subsequently increase the current in order to provide the power to drive the load.
This is why many aircompressors used on building sites burn out their motors - due to voltage drop caused by running off long extension leads. Note that for other reasons (magnetic saturation) over voltage will also increase current flow.

motor info

Mike, can't say that I have any practical experience with tunnel diodes except for some works in electronic motor control at TAFE in 1979. I think that they work at low voltages (less than 500mv) and that an increase in voltage in this region will cause the current to decrease.

Eng. I believe that Mike has just simplified things as calculations and the like get too difficult when you are dealing with variables rather than constants. Mikes basic calculations are based directly on ohms law where I (current) = E (voltage) divided by R (resistance). I=E/R.

Anyway, enough of that & cheers to all, Tony

FollowUp 3 of 3
AnswerID: 133048   Submitted: Tuesday, Oct 04, 2005 at 21:10

porl replied:

uhm, wasn't mikes point that u should use decent wire for your fridge or your battery will die and your beer will be cold, or alternatively get a decent sized fuel line or you won't gain the max potential of your new blower ?
Reply 3 of 3
FollowupID: 387290   Submitted: Tuesday, Oct 04, 2005 at 22:22

VK3CAT posted:

Yes, but once all the beer is gone, what is the point of continuing to use the fridge!
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