Electrical question how to solve?

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Danny89

Member
Location
Indian Land
Question:
What is the Amp load of a single phase motor at 240 volts with an efficiency rating of 83%?

How do I solve this and with what formulas do I use? Thanks.


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Dennis Alwon

Dennis Alwon

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Staff member
Location
Chapel Hill, NC
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Retired Electrical Contractor
Danny89 said:
Question:
What is the Amp load of a single phase motor at 240 volts with an efficiency rating of 83%?

How do I solve this and with what formulas do I use? Thanks.


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Click to expand...
What is the Horsepower of the motor and is it single phase or 3 phase
 
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David Castor

Senior Member
Location
Washington, USA
Occupation
Electrical Engineer
You also need the horsepower and power factor, at least in the real world. Convert hp to watts. That provides the shaft power output. Then divide that by the power factor and efficiency to get the input kVA. Then divide that input kVA by the voltage (assuming single-phase) to get the current.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
230128-2112 EST

What is missing is the mechanical load on the motor.

With mechanical load, then you can use use efficiency to get input power. Then calculate input current.

.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
230129-1136 EST

Danny89:

When you know power output, and efficiency, then you can calculate power input.

To get input current you will need the power factor at that operating point.

.
 
kwired

kwired

Electron manager
Location
NE Nebraska
Most these types of questions that are hypothetical in nature will be presuming the driven load is the output rating of the motor.

Real world the driven load might be a varying load that is often less than motor rating but may intermittently be more than the motor rating, or it might be a steady load but is probably close to or below motor rating but probably higher than the next lower standard size motor. This is kind of typical for induction type motors of 1700-3600 RPM speeds anyway.
 
H

Hv&Lv

Senior Member
Location
-
Occupation
Engineer/Technician
Danny89 said:
Question:
What is the Amp load of a single phase motor at 240 volts with an efficiency rating of 83%?

How do I solve this and with what formulas do I use? Thanks.


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Click to expand...
Formula
I = P(out) * 746 / V* eff * pf. (if its rating is in hp. Use 1000 if its in kW)

See how much information your missing?
 
T

Tulsa Electrician

Senior Member
Location
Tulsa
Occupation
Electrician
Ok I will do my best.
I have limited math skills so.
I went with was is known and solve for x.
Since it ask for I Load based on efficiency. Efficiency is based of of unity or 100%
So I solve x, using X *.83= unity
.83 × 1.204819277= 100
So load amps = LA
LA( unity, 100%) = 1.204819277 × .83

(100/.83= 120.4819277)/100
Plus I just guessed.
The math for x works
 
ptonsparky

ptonsparky

Tom
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
Tulsa Electrician said:
Ok I will do my best.
I have limited math skills so.
I went with was is known and solve for x.
Since it ask for I Load based on efficiency. Efficiency is based of of unity or 100%
So I solve x, using X *.83= unity
.83 × 1.204819277= 100
So load amps = LA
LA( unity, 100%) = 1.204819277 × .83

(100/.83= 120.4819277)/100
Plus I just guessed.
The math for x works
Click to expand...
So that works for a 100 HP or a 1 HP?
 
T

Tulsa Electrician

Senior Member
Location
Tulsa
Occupation
Electrician
A comment on question.
Sometimes when I read these types of questions it's not the ansawer as much as it causes one to pause and think. The goal is to find what's missing in order to make since. Then during an actual test it would cause an automatic response.

Just like the basic triangle E I R.
So we need to find X or what's missing. Now when testing we can weed out un-needed information.
For basic test we need HP or 746 for the above question.
Now we have what we need to solve for I (LA) based on basic info.
HP ( in watts 1= 746)
EFF ( in %)
Voltage ( applied)

For 1 HP @ 240v and 83% EFF.
(746*.83)/240 = 2.57991
746/240= 3.1083
3.1083*.83 = 2.57991( LA)
1/.83= 1.20841×(X)

2.57991*(1/.83) = 3.1083
Now we take each ansawer and convert to HP.

In the test I would bet it will give you LA,voltage and EFF ask for HP.
I see this a lot in the psi stuff.
This why I added a comment.
I like to think about it this way,Look for the need underneath the question.
 
T

Tulsa Electrician

Senior Member
Location
Tulsa
Occupation
Electrician
A comment on question.
Sometimes when I read these types of questions it's not the ansawer as much as it causes one to pause and think. The goal is to find what's missing in order to make since. Then during an actual test it would cause an automatic response.

Just like the basic triangle E I R.
So we need to find X or what's missing. Now when testing we can weed out un-needed information.
For basic test we need HP or 746 for the above question.
Now we have what we need to solve for I (LA) based on basic info.
HP ( in watts 1= 746)
EFF ( in %)
Voltage ( applied)

For 1 HP @ 240v and 83% EFF.
(746*.83)/240 = 2.57991
746/240= 3.1083
3.1083*.83 = 2.57991( LA)
1/.83= 1.20841×(X)

2.57991*(1/.83) = 3.1083
Now we take each ansawer and convert to HP.

In the test I would bet it will give you LA,voltage and EFF ask for HP.
I see this a lot in the psi stuff.
This why I added a comment.
Look for the need underneath the question.
 
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