What is the estimated battery life?

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Discussion Overview

The discussion revolves around estimating the battery life of an electric motor system, given specific parameters such as power rating, voltage, and energy capacity of the battery. Participants explore the relationship between power consumption and battery capacity, focusing on calculations to determine how long the battery can sustain the motor's operation.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents initial calculations to convert battery capacity from kW*h to Joules, arriving at 2.7 x 10^7 Joules.
  • Another participant clarifies that batteries are rated in Amp*hours or kW*h, emphasizing the importance of understanding energy versus power in the context of battery life.
  • A participant introduces additional parameters such as mass, velocity, and energy usage fraction to refine the power requirement for maintaining the motor.
  • A later reply provides a calculation for average energy usage, resulting in an estimated battery life of approximately 5.3 hours based on the given parameters.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the relationship between power, energy, and battery life. While some calculations are presented, there is no consensus on the final estimation of battery life, and different approaches are discussed without resolution.

Contextual Notes

Participants rely on specific assumptions regarding the energy usage fraction and the parameters provided by the teacher, which may affect the accuracy of their calculations. The discussion does not resolve the potential implications of these assumptions on the overall estimates.

Pete789
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Homework Statement


Electric motor in put power rating is 17.5kw
Voltage is 48V
Battery is 7.5kw/h
Average W needed to maintain motor per sec is 1446.7W

Homework Equations


Motor: W=I*V > I = 364.58Amps



The Attempt at a Solution


My attempt: If 1kw = 3.6 x10^6J > 7.5 * 3.6 x 10^6 = 2.7 x10^7
Then the battery is 2.7 x10^7 Jouls

I can't go any further than this. Can anyone help please?

P789
 
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Pete789 said:

Homework Statement


Electric motor in put power rating is 17.5kw
Voltage is 48V
Battery is 7.5kw/h
Average W needed to maintain motor per sec is 1446.7W

Homework Equations


Motor: W=I*V > I = 364.58Amps



The Attempt at a Solution


My attempt: If 1kw = 3.6 x10^6J > 7.5 * 3.6 x 10^6 = 2.7 x10^7
Then the battery is 2.7 x10^7 Jouls

I can't go any further than this. Can anyone help please?

P789

Batteries are rated in Amp*hours, or in this case in kW*h (not kW/h). Power multiplied by time is energy, and the battery initially holds some amount of energy.

So to work problems like this, you are given the average power supplied to the load, and asked how long the battery will last...
 
Thanks Berkeman, However, the only other information my teacher gave me was
The average power required to maintain the battery = M*V^2*fr = W
M= 176kg
V=13.8ms
Fr=energy usage fraction= 12%

Any ideas?
P789
 
Ok this is my answer.
Aver E = 170kg*8.3^2*0.12=1405.35w or 1.405kw
The Battery has 7.5kwh so 1.405/7.5= 5.3 hrs of use

Thanks Peter
 

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