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How to calculate required battery charge for 300 miles... |
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| Aug18-10, 03:43 PM | #1 |
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How to calculate required battery charge for 300 miles...
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:
1. The problem statement, all variables and given/known data 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? 2. Relevant equations n/a 3. 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 + + = me1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution |
| Aug18-10, 09:48 PM | #2 |
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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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