How to calculate required battery charge for 300 miles

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SUMMARY

The discussion focuses on calculating the required battery charge for a vehicle to achieve a range of 300 miles, based on its diesel consumption and engine efficiency. The car operates for 400 hours annually, consuming 2,000 liters of diesel, which translates to 77.4 billion Joules of energy. Given the engine's 20% efficiency and the battery's 95% efficiency, the total energy needed for a full charge to cover 300 miles is calculated to be 3.4 MWh.

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Need help - my brain is mush and the longer I look at it the more I don't understand it...can someone please tell me if this calculation is correct:

Homework Statement



A car is run for 400 hours per year, with a total mileage of 10,000 a year.
The car uses diesel and consumes 2,000 litres a year.
A litre of diesel costs £1.10 and holds 38.7MJ of primary energy.

If the engine is 20% efficient overall, how much energy would need to be stored in 95% efficient batteries to give a range of 300 miles on full battery charge?

Homework Equations



n/a

The Attempt at a Solution



Background Calcs:
Convert MJ to J: 38700000J
Total Joules (J x 2000 litres): 77400000000J
Convert 400 hours to seconds: 1440000s

Energy 77400000000 J
Time 1440000 seconds
= Power 53750 Watts
53.75 kW
1 watt = 0.001341022 horsepower
Convert W to hp: 72.07993734 hp

Answer (attempt!):

Primary energy (1 year): 400h*53.75kw=21.5MWh
Energy to 95% efficient batteries: 21.5 x 100/20 x 100/95 = 113.2MWh
Therefore 300 miles: 300/10000*113.2=3.4MWh

I appreciate your help!
Thanks guys

:confused: + :zzz: + :cry: = me
 
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Does the battery replace the diesel or the engine?

Either way, look at it from the point of view of a flow of units. For example, taking distance traveled & diesel used and energy in diesel, and then calculating energy per distance (MJ/mile) should give you a good grounding for calculating the energy required to travel some more distance.
 

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