1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating Efficiency of Electric Vehicle

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    Power = Force x Velocity
    Energy = Power x Time

    3. 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!
     
  2. jcsd
  3. Nov 9, 2014 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.)
     
  4. Nov 9, 2014 #3
    Thanks for the quick reply!

    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
     
  5. Nov 9, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Remember, 1 watt = 1 J/s.

    If the car is travelling 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.
     
  6. Nov 9, 2014 #5
    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!
     
  7. Nov 9, 2014 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    1 watt = 1 J/s The '/' means you divide by the time. Always check your units.
     
  8. Nov 9, 2014 #7
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculating Efficiency of Electric Vehicle
Loading...