# Calculating Efficiency of Electric Vehicle

• bnashville
In summary, to maintain highway speed, a typical electric car will need to apply a force of 165 N and generate 4583.34 watts of power.
bnashville

## Homework Statement

For the following calculations, assume a typical electric car will need to apply a constant forc of 165 N to maintain a typical highway speed of 100 km/hr (27.7778 m/s).

A: What power (watts) is needed to maintain this speed?
B: What energy (Joules) is needed to travel 100 km at 100 km/hr?
C: Given that the combustion of gasoline will produce 4.73x10^7 J /kg of gasoline consumed, and the density of gasoline is .74 kg/L, if a car can get 32 mi/gal at 100 km/hr, what power is being generated by the combustion reaction? (in watts)

There are several additional parts to answer on this question, but I think I can get going if I get these answered.

## Homework Equations

Power = Force x Velocity
Energy = Power x Time

## The Attempt at a Solution

I think I have the first 2 answers but wanted to check that they were correct:
A: 165 N x 27.7778 m/s = 4583.34 Watts
B: 4583.34 Watts x 3600s = 1.65x10^7 Joules

C: I didn't get much further than unit conversion on this one.
32.0 mi/gal = 51499 m / 3.7854 L = 13604.63888 m / L
.74 kg/L x 3.7854 L = 2.8011 kg...and this is where I start to trail off and have no grasp on what I'm doing..

Thanks for the help!

bnashville said:

## Homework Statement

For the following calculations, assume a typical electric car will need to apply a constant forc of 165 N to maintain a typical highway speed of 100 km/hr (27.7778 m/s).

A: What power (watts) is needed to maintain this speed?
B: What energy (Joules) is needed to travel 100 km at 100 km/hr?
C: Given that the combustion of gasoline will produce 4.73x10^7 J /kg of gasoline consumed, and the density of gasoline is .74 kg/L, if a car can get 32 mi/gal at 100 km/hr, what power is being generated by the combustion reaction? (in watts)

There are several additional parts to answer on this question, but I think I can get going if I get these answered.

## Homework Equations

Power = Force x Velocity
Energy = Power x Time

## The Attempt at a Solution

I think I have the first 2 answers but wanted to check that they were correct:
A: 165 N x 27.7778 m/s = 4583.34 Watts
B: 4583.34 Watts x 3600s = 1.65x10^7 Joules

C: I didn't get much further than unit conversion on this one.
32.0 mi/gal = 51499 m / 3.7854 L = 13604.63888 m / L
.74 kg/L x 3.7854 L = 2.8011 kg...and this is where I start to trail off and have no grasp on what I'm doing..

Thanks for the help!

For part C, you have calculated how many kg of gasoline would be used to travel 32 miles. This fuel consumption is based on the car traveling at a speed of 100 km/hr. The question asks you to figure out how much power is being generated from the combustion of the gasoline which is consumed at this speed. (Hint: you know how much energy in J/kg is produced from the combustion of gasoline, and you have worked out the number of kg of gasoline burned in traveling 32 miles. The speed of the car (100 km/hr) tells you how quickly the energy must be used, so you can then calculate the power derived from the combustion of the gasoline.)

So do I just take: 2.8011 kg x 4.73x10^7 J/kg x 27.7778 m/s = 3.68x10^9 Watts ?

If that is correct, I don't really understand how to units work out. How do we go from kg*((kgm^2/s^2)/kg)*(m/s) to Watts which I think is (kgm^2/s^2)/s

Remember, 1 watt = 1 J/s.

If the car is traveling at a speed of 100 km/hr, how long does it take the car to go 32 miles? That's how you find out the time in which the energy (and thus the power) is generated.

so to travel 32 miles = 51499m / 27.7778 m/s = 1853.9625 seconds
and 2.8011 kg x 4.73x10^7 J/kg x 1853.9625 s = 2.4563x10^11 Watts ?

Thanks for the help!

1 watt = 1 J/s The '/' means you divide by the time. Always check your units.

I think I see it now, I get confused if I don't break down all the units...
( 2.8011 kg x 4.73x10^7 J/kg ) / 1853.9625 seconds = 71464.24418 Watts

Thanks!

## What is the efficiency of an electric vehicle?

The efficiency of an electric vehicle refers to how well it converts its energy source (typically electricity) into motion. It is typically measured as a percentage, with higher percentages indicating greater efficiency.

## How is the efficiency of an electric vehicle calculated?

The efficiency of an electric vehicle is calculated by dividing the distance traveled by the amount of energy consumed. This can be represented by the formula Efficiency = Distance / Energy Consumed.

## What factors affect the efficiency of an electric vehicle?

There are several factors that can affect the efficiency of an electric vehicle, including the weight of the vehicle, the type and condition of the battery, the driving style, and external factors such as temperature and road conditions.

## Is it possible to improve the efficiency of an electric vehicle?

Yes, there are several ways to improve the efficiency of an electric vehicle. These include regular maintenance and upkeep of the vehicle, using eco-friendly driving techniques, and upgrading to more efficient batteries or motors.

## How does the efficiency of an electric vehicle compare to a gasoline-powered vehicle?

Electric vehicles are generally more efficient than gasoline-powered vehicles, with electric vehicles typically achieving efficiencies of 80-90%, while gasoline vehicles typically achieve efficiencies of only 20-30%. This is because electric vehicles do not have the same energy losses associated with combustion engines.

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