Gas consumption to overcome drag force

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Homework Help Overview

The problem involves calculating gasoline consumption for a car overcoming drag force while traveling at a specific speed. The drag force is defined by a formula that includes parameters such as air density, cross-sectional area, and drag coefficient. The context includes energy content of gasoline and engine efficiency.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating drag force and power needed to overcome it, with one noting the drag force value and power output. Others suggest solving symbolically and question how to relate volumetric flow rate to fuel consumption in miles per gallon.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to relate fuel flow rate to distance traveled. Suggestions for symbolic representation and dimensional analysis are being considered, but no consensus has been reached on a specific method.

Contextual Notes

Participants are navigating the constraints of converting energy content of fuel into a distance metric, while also considering the efficiency of the engine and the need for unit conversions.

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.
 
Last edited:

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