What is the estimated battery life?

AI Thread Summary
The discussion focuses on estimating the battery life for an electric motor with a power rating of 17.5 kW and a 7.5 kWh battery. The average power needed to maintain the motor is calculated to be approximately 1.405 kW, based on the given parameters of mass, velocity, and energy usage fraction. Using this average power, the estimated battery life is determined to be around 5.3 hours. Participants clarify the importance of understanding energy in terms of Joules and the correct units for battery capacity. The conversation highlights the process of calculating battery life based on power consumption and energy storage.
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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