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Mass of fuel burned in car over distance with changing mass

by Belginator
Tags: burned, distance, fuel, mass
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Belginator
#1
Feb21-12, 09:42 AM
P: 11
1. The problem statement, all variables and given/known data

Hi! I seem to be having some difficulty with this one, any help would be appreciated. There is a car, assume it for simplicity to be a rectangular prism of dimensions l,w and h. The car is moving at a constant velocity [itex] v [/itex], and assume there is a drag force [itex] D = 0.5 \rho v^2 C_D h w [/itex] and a frictional force μ. We know that the fuel has an effective energy content of [itex] E_f = 15kWh/kg [/itex]. The total mass of the car is defined as M + m, where m is the fuel mass and M everything else. R is the distance the car travels. Essentially how much fuel is needed to go a certain distance R, knowing that over the distance the car gets lighter as fuel is burned, which means the force required to keep the constant velocity decreases.

2. The attempt at a solution

My thought process is that the Power required to keep the car moving at the constant velocity is the amount of power needed from the fuel. The power required: [itex] P_{req} = (μ(m+M)g + D)v [/itex]. I think the average power of the fuel can be defined as: [itex] P_{fuel} = \frac{E_f m v}{R} [/itex] I want everything in terms of distance rather than time that's why I have v and R in that equation rather than time.

I believe fuel power is average since it's over a distance, R, so to make it instantaneous power I say: [itex] P_{fuel} = \frac{E_f dm v}{dR} [/itex]. The power required by the car is already instantaneous since m is instantaneous and changes over R (or time, same thing, as fuel is burned).

I set the power required and the fuel power equal to each other as follows:

[itex] \frac{E_f dm v}{dR} = (\mu (m+M)g + D) v [/itex], the v's cancel. Seperation of variables:

[itex] \frac{E_f}{\mu g (m+M)+D} dm = dR [/itex]

After integrating and solving for fuel mass, m:

[itex] m = \frac{e^{\frac{\mu g R}{E_f}} - D}{\mu g} - M + C_0 [/itex]

It seems to make sense that the mass of fuel m needed to drive a distance, R of 0 is 0. Using that as the initial condition to solve for [itex] C_0 [/itex] I get:

[itex] m = \frac{e^{\frac{\mu g R}{E_f}} - 1}{\mu g} [/itex]

That doesn't make logical sense however, it should be a function of drag and velocity as well, I would think, because if you have two cars, one with a higher velocity than the other, but constant, the higher velocity car should burn more fuel since it has more drag. Perhaps my setup is completely wrong, I dont know. Again, any help is appreciated, thanks.
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LawrenceC
#2
Feb21-12, 10:26 AM
P: 1,195
If the frictional force is mu as stated in your problem statement, why are you concerned with the weight of the vehicle?
Belginator
#3
Feb21-12, 10:35 AM
P: 11
μ is actually stated as a percent of the instantaneous weight. So the frictional force: [itex] F_f = \mu mg [/itex] changes as mass changes which changes the power required to keep the velocity constant. The power required to keep the constant velocity, I believe, would be exponentially decreasing, as less and less fuel is required to maintain the velocity due to a lighter and lighter vehicle.

LawrenceC
#4
Feb21-12, 12:53 PM
P: 1,195
Mass of fuel burned in car over distance with changing mass

Can you write an equation that relates the power input to the power expended? You basically have it with your first equation:

P = (mu(m+M)g+D)v

The power input can be written in terms of fuel consumption

P = (dm/dt)E

If you solve the differential equation with time as the independent variable, you get an expression for m in terms of t. The t variable can always be related to R, the range, by dividing R by v.


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