Calculate the Motor Efficiency when?

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SUMMARY

The discussion focuses on calculating the efficiencies of an electric motor and generator system. The 2hp motor, drawing 15A from a 120V DC source, has been calculated to have an efficiency of 82.85% based on input and output power. The input power is determined as 1800W using the formula P = I.V, while the output power is calculated as 1491.4W from the motor's horsepower rating. The conversation also emphasizes the assumption of no losses in the mechanical connection between the motor and generator for accurate efficiency calculations.

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  • Understanding of electric motor and generator principles
  • Knowledge of power calculations using P = I.V
  • Familiarity with efficiency formulas, specifically n = Po/Pin
  • Basic concepts of electrical resistance and Ohm's Law
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  • Explore the impact of mechanical losses in motor-generator systems
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An electric motor drives an electric generator.The 2hp motor draws 15A from a 120V D.C source and the generator supplies 5A to a 48Ohm LOAD. It is given that 1hp= 745.7W



Calculate the motor efficiency.
Calculate the generator efficiency.
Calculate overall efficiency.



Ok soo I went forward and did this.
Since 1hp = 745.7Watt I assumed that the motor would be going at 2x745.7W = 1491.4W .
So if you take the P = I.V and you insert 15A.120V = 1800W
So using n = Po/Pin for Efficiency 1491.4/1800 *100 = 82.85% efficiency.
Not sure if this is correct but now?
I can say P = I^2 / R and get the Watt but to what comparison? Which Po and which Pin ??

Please help
 
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Totter said:
An electric motor drives an electric generator.The 2hp motor draws 15A from a 120V D.C source and the generator supplies 5A to a 48Ohm LOAD. It is given that 1hp= 745.7W



Calculate the motor efficiency.
Calculate the generator efficiency.
Calculate overall efficiency.



Ok soo I went forward and did this.
Since 1hp = 745.7Watt I assumed that the motor would be going at 2x745.7W = 1491.4W .
So if you take the P = I.V and you insert 15A.120V = 1800W
So using n = Po/Pin for Efficiency 1491.4/1800 *100 = 82.85% efficiency.
Ignoring any rounding differences/errors, yes, that looks correct to me. :approve:
Totter said:
I can say P = I^2 / R and get the Watt but to what comparison? Which Po and which Pin ??
To calculate the generator's efficiency, just stick with the generator. What's the input power of the generator?

(Hint: the problem statement says, "An electric motor drives an electric generator." I'm pretty sure you should assume no losses between the output of the motor and the input to the generator in this case. Assume the power is transferred between them via a frictionless rotating shaft. The only losses are within the motor and within the generator themselves.)

(Continuing the hint: Keep in mind that this is not the case where a generator drives a motor. In that case, there could be I2R losses in the cables/wires connecting them. But in this problem, the motor and generator are connected mechanically [not electrically], and I suspect you should take the connection as loss-less.)
 

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