# Energy problem

1. Jun 26, 2009

### boyblair

Hi everyone!

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

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 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

2. Jun 26, 2009

### tiny-tim

Welcome to PF!

Hi boyblair! Welcome to PF!

Yes, that's ok … 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.
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.

3. Jun 28, 2009

### boyblair

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 in to 20% efficient motor: 21.5MWh/20%=80.5MWh
Energy supplied to 95% efficient batteries: 80.5MWh\95%=84.7MWh

Many thanks

4. Jun 28, 2009

### tiny-tim

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 ).

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.

5. Jun 29, 2009

### boyblair

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

6. Jun 29, 2009

### tiny-tim

eugh! :yuck:

why so long-winded?

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

7. Aug 18, 2010

### jacstar

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!
+ :zzz: + = me!

8. Aug 19, 2010

### tiny-tim

welcome to pf!

hi jacstar! welcome to pf!

(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

Last edited by a moderator: Apr 25, 2017