The efficiency of a vehicle engine

[heavenjai2007]

Homework Statement

A vehicle engine has a power output of 6.2 KW and uses fuel which releases 45 MJ per kilogram when burned. At a speed of 30ms ^-1 on a level road, the fuel usage of the vehicle is 18km per kilogram. Calculate: the useful energy supplied by the engine in this time.

Homework Equations

efficiency= (useful energy) / (energy supplied)

The Attempt at a Solution

Well, first i know that the time taken by the vehicle to travel 18 km at 30ms^-1 is 600s using s=d/t. i also know that (useful energy) is equal to (mg x change in height). the problem is that I am not given the value of mg(N) therefore i have to work it out. so i know F=mxa but again, we're not given the value of m and a. i might have confused my self.

 

[Andrew Mason]

The efficiency is also the rate of work output / rate of energy input. That is easier to work with here. You are given the rate of work output. All you need to do is figure out how much energy it consumes per unit time (denominator).

To calculate the denominator, first work out the number of kilograms of fuel it consumes per second. Then multiply it by the amount of energy per kilogram.

AM

 

[Phrak]

1.Calculate: the useful energy supplied by the engine in this time.

i also know that (useful energy) is equal to (mg x change in height).

That's an interesting definition of "useful energy". The change in height is zero isn't it? In that case you should be able to work out the useful energy.

 

[Redbelly98]

Homework Statement

A vehicle engine has a power output of 6.2 KW and uses fuel which releases 45 MJ per kilogram when burned. At a speed of 30ms ^-1 on a level road, the fuel usage of the vehicle is 18km per kilogram. Calculate: the useful energy supplied by the engine in this time.

Welcome to Physics Forums.

What is the problem actually asking for? The useful energy supplied (as asked in the above post), or the efficiency of the engine (as mentioned in the thread title)?

And if the question is to find the useful energy supplied "in this time", that is troublesome because there is no time interval specified. Nor is it possible to calculate a time interval; one would need more information such as the distance traveled or the available mass of fuel.

 

[rwisz]

I would suggest approaching this problem by first considering the efficiency of the engine. Efficiency is defined as the ratio of useful energy output to the total energy input. In this case, the useful energy output is the power output of the engine (6.2 KW) and the total energy input is the energy released by the fuel (45 MJ per kilogram) multiplied by the amount of fuel used (18 km per kilogram).

So, the efficiency can be calculated as:

efficiency = (6.2 KW) / (45 MJ/kg x 18 km/kg) = 0.0000037 or 0.00037%

This means that only a very small percentage of the energy released by the fuel is actually being used to power the vehicle.

To calculate the useful energy supplied by the engine, we can use the formula for power (P = W/t) and rearrange it to solve for work (W):

W = P x t

Substituting in the values we have:

W = (6.2 KW) x (600 s) = 3720 kJ

This is the useful energy supplied by the engine in 600 seconds (the time it takes to travel 18 km at 30ms^-1).

In conclusion, the efficiency of the engine is very low, with only 0.00037% of the energy released by the fuel being used to power the vehicle. This highlights the importance of developing more efficient engines and alternative energy sources for vehicles in order to reduce our reliance on fossil fuels and decrease emissions.