The efficiency of a vehicle engine

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Homework Help Overview

The discussion revolves around a vehicle engine's efficiency, specifically focusing on its power output, fuel consumption, and the useful energy supplied during operation. Participants are analyzing the relationship between energy input and output in the context of a vehicle traveling at a constant speed on a level road.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to calculate the useful energy supplied by the engine based on given parameters, while questioning the definitions and assumptions related to "useful energy" and the time interval for calculations. Some suggest focusing on the rate of work output and energy input, while others express confusion over missing information necessary for calculations.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the calculation of energy consumption per unit time, but there is no consensus on the exact requirements of the problem or the necessary information to proceed.

Contextual Notes

There are concerns regarding the lack of specified time intervals and the ambiguity in the problem statement about whether to calculate useful energy or efficiency. Participants note that additional information, such as the distance traveled or the mass of fuel, is needed for a complete analysis.

heavenjai2007
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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.
 
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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
 
heavenjai2007 said:
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.
 
heavenjai2007 said:

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.
 

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