Mass of fuel burned in car over distance with changing mass

[Belginator]

Homework Statement

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 $v$, and assume there is a drag force $D = 0.5 \rho v^2 C_D h w$ and a frictional force μ. We know that the fuel has an effective energy content of $E_f = 15kWh/kg$. 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: $P_{req} = (μ(m+M)g + D)v$. I think the average power of the fuel can be defined as: $P_{fuel} = \frac{E_f m v}{R}$ 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: $P_{fuel} = \frac{E_f dm v}{dR}$. 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:

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

$\frac{E_f}{\mu g (m+M)+D} dm = dR$

After integrating and solving for fuel mass, m:

$m = \frac{e^{\frac{\mu g R}{E_f}} - D}{\mu g} - M + C_0$

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 $C_0$ I get:

$m = \frac{e^{\frac{\mu g R}{E_f}} - 1}{\mu g}$

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 don't know. Again, any help is appreciated, thanks.

 

[LawrenceC]

If the frictional force is mu as stated in your problem statement, why are you concerned with the weight of the vehicle?

 

[Belginator]

μ is actually stated as a percent of the instantaneous weight. So the frictional force: $F_f = \mu mg$ 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]

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.

 

[paranormal]

Hello, it seems like you have made some good progress in your attempt to solve this problem. However, there are some aspects that need to be clarified.

Firstly, your equation for power required to keep the car moving at a constant velocity is correct, but the units are not. The units of power are Joules per second (J/s) or Watts (W), so the equation should be P_req = (μ(m+M)g + D)v.

Secondly, your equation for average power of the fuel is also correct, but the units are not. The units of average power are Joules (J) per distance (dR), so the equation should be P_fuel = E_fmv/R.

Now, when setting these two equations equal to each other, you are essentially equating power with power. This is not the correct approach, as the power required to keep the car moving is not equal to the power provided by the fuel. The power required is the power needed to overcome the forces (drag and friction) and maintain a constant velocity, while the power provided by the fuel is the energy released per distance traveled. These are two different quantities and cannot be equated.

To solve this problem, you need to use the concept of work. The work done by the forces (drag and friction) is equal to the change in kinetic energy of the car. This work can also be expressed as the force multiplied by the distance traveled. On the other hand, the work done by the fuel is equal to the energy released per distance traveled multiplied by the distance traveled. Setting these two works equal to each other and solving for the fuel mass will give you the correct result.

I hope this helps. Good luck with your homework!