does the fuel pump really draw 20 amps?
Thread Starter
mad scientist
Joined: Feb 2001
Posts: 2,665
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From: Savannah, GA
does the fuel pump really draw 20 amps?
I was curious how much power the fuel pump draws, so I measured the resistance of 2 pumps(.6 ohms), and calculated the current draw at 12V. I came up with 20 amps. Does the pump really draw that much power?
Thread Starter
mad scientist
Joined: Feb 2001
Posts: 2,665
Likes: 2
From: Savannah, GA
I thought that was kind of strange also. The fuse is only rated for 15 amps, but using Ohms law, current / resistance. 12/.6 = 20 amps. I measured 2 pumps, and both read .6 ohms.
I'm a boost creep...
Joined: Jan 2002
Posts: 15,608
Likes: 8
From: Auckland, New Zealand
OK, I asked an electrical guru at work about this, and he found the answer in a book on motors:
It goes on to explain how to calculate CEMF, and it all works out right in the end!
The fuel pump probably draws no more than 5A.
Counterelectromotive Force
A 240V, 10hp motor turning at its rated rpm will draw ~38A at full load. The DC resistance of the armature is only 0.06ohm. Based on Ohm’s Law 204V/0.06ohm = 4000A current calculated in the circuit. The difference between the actual and calculated current is explained by the presence of counterelectromotive force (CEMF).
As the armature of the rotor rotates in the magnetic field of the stator, a voltage is induced into it (generator action). This voltage opposes the applied voltage, and reduces the amount of current required by the motor to develop its full power.
A 240V, 10hp motor turning at its rated rpm will draw ~38A at full load. The DC resistance of the armature is only 0.06ohm. Based on Ohm’s Law 204V/0.06ohm = 4000A current calculated in the circuit. The difference between the actual and calculated current is explained by the presence of counterelectromotive force (CEMF).
As the armature of the rotor rotates in the magnetic field of the stator, a voltage is induced into it (generator action). This voltage opposes the applied voltage, and reduces the amount of current required by the motor to develop its full power.
The fuel pump probably draws no more than 5A.
Full Member
Joined: May 2002
Posts: 209
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From: B.C.
Hey guys!
Is this something I should be worrying about?
Also, could you guys help me out on my other thread?
https://www.rx7club.com/forum/showth...threadid=81104
Thanks in advance!
Is this something I should be worrying about?
Also, could you guys help me out on my other thread?
https://www.rx7club.com/forum/showth...threadid=81104
Thanks in advance!
Thread Starter
mad scientist
Joined: Feb 2001
Posts: 2,665
Likes: 2
From: Savannah, GA
Thanks. How much work is it to find out the CEMF? Do you think you could post it if it isnt too long? I would really like to know how much power the fuel pump draws. I want to make an adjustable voltage regulator for my fuel pump. My car is running too rich, and I dont have $400 to spend on an S-afc.
I'm a boost creep...
Joined: Jan 2002
Posts: 15,608
Likes: 8
From: Auckland, New Zealand
Here's the rest of it:
Really, it would be easiest just to measure the current going through the pump. 
The CEMF can be calculated by subtracting the actual voltage drop across the motor from the applied voltage.
CEMF = Ea – Ira = 240 – 2.28 = 2.37.72
Using these new values, we find that Ohm’s Law does apply. The voltage across the armature is 2.28V instead of 240V. 2.28V / 0.06ohm (the armature’s resistance) gives the 38A the motor draws under full load conditions.
CEMF = Ea – Ira = 240 – 2.28 = 2.37.72
Using these new values, we find that Ohm’s Law does apply. The voltage across the armature is 2.28V instead of 240V. 2.28V / 0.06ohm (the armature’s resistance) gives the 38A the motor draws under full load conditions.
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