Gas consumption to overcome drag force

[warfreak131]

Homework Statement

A car moving with speed $v$ is subject to a drag force $F_{d}=0.5\rho A c_{d}v^{2}$, where $\rho$ is the density of air ($1.25 kg/m^{3}$ at STP), A is the cross sectional area of the car, and $c_{d}$ is the drag coefficient.

A Honda Civic has $A=1.9 m^{2}$ and $c_{d}=0.36$.

Assuming gasoline contains 125 MJ/gallon of energy, and that the car's engine is 25% efficient, calculate the gasoline consumption (in miles per gallon) needed to overcome this drag force if the car is traveling 75 mi/h (=33.3 m/s). Hint, 1 mile = 1600 m.

B. What is the power output of the engine needed to overcome this drag force at 33.3 m/s? Hint, 1 horsepower = 750 watts.

Homework Equations

W=Fd
P=W/t=Fv

The Attempt at a Solution

From the given information, I know that the drag force is equal to 474 N (using MKS units), and the power needed to overcome the drag is approximately 15,800 W.

From unit analysis, I have to somehow convert MJ/gal into mi/gal, which would mean multiplying by mi/MJ, but I am not sure how to get that.

I know from the efficiency of the engine, that you can only extract a maximum of 31 MJ/gal of gas.

 

[milesyoung]

If I might make a suggestion, try solving it symbolically first using, for instance:
$$\eta = \frac{P_{out}}{P_{in}} \,, P_{in} = u Q$$
where:
- η is the efficiency of the car engine in converting fuel to mechanical work.
- P_out is the mechanical power supplied by the engine.
- P_in is the chemical power supplied by the fuel.
- u is the energy density of the fuel.
- Q is the volumetric flow rate of fuel into the engine.

 

[warfreak131]

But how would the volumetric flow rate factor into miles/MJ? VFR would give me m^3 /s or gallons/s

 

[milesyoung]

But how would the volumetric flow rate factor into miles/MJ? VFR would give me m^3 /s or gallons/s

If you know the velocity of the car (assume it's constant), you know how far it travels in some period of time.
If you know the flow rate of fuel, you know how much fuel it'll use in that same period of time.

Edit:
To clarify, can you use this to find an expression that relates the fuel flow rate [m³/s], the fuel consumption [m/m³] and car velocity [m/s]?

You can use the dimensions of the units to give you a hint. Try coming up with a symbol for the fuel consumption and relate it to the fuel flow rate Q and car velocity v.

 

[Nickie]

Therefore, I can calculate the maximum distance the car can travel with 1 gal of gas, which is approximately 248 mi.

To calculate the gasoline consumption (in miles per gallon) needed to overcome the drag force, I would need to divide the distance traveled by the car by the distance traveled with 1 gal of gas, which gives a value of approximately 0.3 miles per gallon.

To calculate the power output of the engine, I would need to use the equation P=W/t, where W is the work done (474 N x 1600 m = 758,400 J) and t is the time it takes to travel 1600 m at a speed of 33.3 m/s, which is approximately 0.048 s. This gives a power output of approximately 15,800 W, which is consistent with the initial calculation.

In conclusion, the gasoline consumption needed to overcome the drag force at a speed of 33.3 m/s is approximately 0.3 miles per gallon and the power output of the engine needed is approximately 15,800 W. This highlights the importance of reducing drag in order to increase fuel efficiency and decrease gasoline consumption.