Gas consumption to overcome drag force

AI Thread Summary
The discussion focuses on calculating gasoline consumption and power output needed for a Honda Civic to overcome drag force while traveling at 75 mi/h. The drag force is determined to be 474 N, requiring approximately 15,800 W of power to counteract it. The efficiency of the engine is noted to be 25%, meaning only 31 MJ/gal of gasoline can be utilized effectively. Participants suggest solving the problem symbolically to relate fuel flow rate, fuel consumption, and car velocity. The conversation emphasizes the importance of unit conversions and dimensional analysis in deriving the necessary expressions for fuel efficiency.
warfreak131
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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.
 
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If I might make a suggestion, try solving it symbolically first using, for instance:
<br /> \eta = \frac{P_{out}}{P_{in}} \,, P_{in} = u Q<br />
where:
- η is the efficiency of the car engine in converting fuel to mechanical work.
- Pout is the mechanical power supplied by the engine.
- Pin 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.
 
But how would the volumetric flow rate factor into miles/MJ? VFR would give me m^3 /s or gallons/s
 
warfreak131 said:
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.
 
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