Energy Problem: Calculating Mean Engine Power & Battery Range

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of mean engine power and battery range for a diesel car. Participants explore the energy consumption of the car, its efficiency, and how to determine the energy required for a specific range based on given parameters.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the mean engine power based on the energy content of diesel and the car's operational hours, arriving at 53.75 kW.
  • Another participant suggests a method for calculating the energy required for the batteries, emphasizing the need to consider the efficiencies of both the engine and the batteries separately.
  • A later reply provides an alternative calculation for the energy supplied to the batteries, indicating a total of 84.7 MWh needed for a 300-mile range, but questions the correctness of the energy per distance calculation.
  • One participant critiques the method used to calculate energy for 300 miles, suggesting that it should be based on energy per mile rather than total power multiplied by distance.
  • Another participant expresses confusion and seeks clarification on whether their lengthy calculation approach is valid, indicating a struggle with the problem.
  • Further responses encourage a simpler approach by calculating energy per distance based on the car's fuel efficiency, suggesting that this method may yield clearer results.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate energy based on efficiency and distance, but there are differing opinions on the best method to approach the problem. Some calculations are challenged, and no consensus on the final answer is reached.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations, particularly concerning the efficiency of the engine and batteries, as well as the interpretation of energy per distance versus total energy calculations.

boyblair
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Hi everyone!

I am having a bit of a problem solving the following question, and would be very grateful for any advice given.

1. 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 per year. A litre of diesel cost £1.10 and holds 38.7MJ of primary energy.

a) Calculate the mean engine power?

b) 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?




3. The Attempt at a Solution

a) energy in 2000litres of diesel 38.7MJ * 2000litres = 77400000000MJ
number of seconds in 400hours 3600*400 = 1440000secs
power = energy/time = 77400000000/1440000 = 53750Watts or 53.75kW

b) Really stuck with this part
 
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Welcome to PF!

Hi boyblair! Welcome to PF! :smile:

boyblair said:
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 per year. A litre of diesel cost £1.10 and holds 38.7MJ of primary energy.

a) Calculate the mean engine power?
…
a) energy in 2000litres of diesel 38.7MJ * 2000litres = 77400000000MJ
number of seconds in 400hours 3600*400 = 1440000secs
power = energy/time = 77400000000/1440000 = 53750Watts or 53.75kW

Yes, that's ok :smile: … but if you write all those zeros, you're very likely to make a mistake so you really ought to write the whole thing as a fraction, and then do a bit of cancelling, rather than doing it in stages. :wink:
b) 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?
…
b) Really stuck with this part

efficiency is energy output divided by energy input …

so do that for the batteries (95%) and the engine (20%) separately, then combine them, so as to find how much of the energy put into the batteries wil come out of the engine. :smile:
 
Hey, thanks for the quick response.
I have attempted part b) and would be grateful for any feedback.

Primary energy consumes in 1 year: 300h*53.75kw=16.1MWh
Energy into 20% efficient motor: 21.5MWh/20%=80.5MWh
Energy supplied to 95% efficient batteries: 80.5MWh\95%=84.7MWh

Many thanks
 
boyblair said:
Primary energy consumes in 1 year: 300h*53.75kw=16.1MWh
Energy into 20% efficient motor: 21.5MWh/20%=80.5MWh
Energy supplied to 95% efficient batteries: 80.5MWh\95%=84.7MWh

The efficiency calculations are correct

(though, as before, doing the whole thing together, instead of in stages, would be safer and would look better: 16.1 x 100/20 x 100/95 = 84.7 :wink:).

However, your energy per 300 miles is wrong …

you have multiplied the miles by the power, which is energy per time, instead of energy per mile. :redface:
 
Thanks for your help tiny-tim.

I have made another attempt at the problem:

Primary energy consumed in 1 year: 400h*53.75kw=21.5MWh
Energy supplied to 95% efficient batteries: 21.5 x 100/20 x 100/95 = 113.2MWh
For 300 miles: 300/10000*113.2=3.4MWh
 
boyblair said:
Primary energy consumed in 1 year: 400h*53.75kw=21.5MWh
Energy supplied to 95% efficient batteries: 21.5 x 100/20 x 100/95 = 113.2MWh
For 300 miles: 300/10000*113.2=3.4MWh

eugh!

why so long-winded? :rolleyes:

do it the easy way … you want the energy for 300 miles, so read the question carefully, and you'll see it goes 10,000 miles on 2,000 litres, so that's 5 miles a litre … carry on from there :smile:
 
tiny-tim said:
eugh!

why so long-winded? :rolleyes:

do it the easy way … you want the energy for 300 miles, so read the question carefully, and you'll see it goes 10,000 miles on 2,000 litres, so that's 5 miles a litre … carry on from there :smile:

Tiny Tim

Hows it going?

I've found your guidance on this thread and need help with the exact same question funnily enough...
I got to where BoyBlair got to (the long-winded response) - is this the correct answer for this problem, just the long way round?
I've got severe brainblock after 7 days of constant studying and don't even feel like i can count to 10 anymore...

Your response would be greatly appreciated

Ta!
:confused: + :zzz: + :cry: = me!
 
welcome to pf!

hi jacstar! welcome to pf! :wink:

(just got up :zzz: …)

as Zryn says in the thread you've started on this problem (https://www.physicsforums.com/showthread.php?t=423021"), you need to calculate the energy per distance

as i said before, it goes 10,000 miles on 2,000 litres, so that's 5 miles a litre …

so try it that way, and show us what you get :smile:
 
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