Finding the cross sectional area given fuel efficiency. Your car's fuel efficiency is 35 miles per gallon. Assume that your car's fuel pump pulls fuel out of the fuel tank at a constant rate through a tube. If the speed of the fuel through the tube is equal to the speed of the car, what must be the cross-sectional area of the tube in square inches?
I am not sure where to start. I know that the cross sectional area is equal to the area of a circle $A=\pi r^2$. What I don't get is how I would get the radius of the pipe using 35 miles per gallon. Could someone give me a hint on what I am supposed to do?
 A: Let the speed of the car be $v$ miles per hour. Then we use fuel at the rate of $\dfrac{v}{35}$ gallons per hour. 
Let $K$ be the number of cubic inches in a  gallon. You can look up the value of $K$.
So we use fuel at the rate of $\dfrac{Kv}{35}$ cubic inches per hour. 
Let $A$ be the cross-sectional of the tube in square inches. Then since the speed of the fuel in the tube is $v$, the amount of fuel that flows per hour is $Av$. 
We conclude that
$$\frac{Kv}{35}=Av.\tag{1}$$
We can now use Equation (1) to determine $A$.
A: Let's examine the quantities you have, using dimensional analysis.


*

*Efficiency: miles per gallon, or $\mathcal{L}/\mathcal{L}^3$;

*Cross-sectional area: $\mathcal{L}^2$;

*Vehicle speed: $\mathcal{L}/\mathcal{T}$;

*Flow velocity: $\mathcal{L}/\mathcal{T}$.


Here, $\mathcal{L}$ represents a length quantity (miles, centimeters, whatever...), and $\mathcal{T}$ represents a time quantity. So $\mathcal{L}^3$ represents length cubed, or volume.
Now, we have that flow rate is equal to flow velocity times cross-sectional area. We know this because it pumps constantly. So,
$$\textrm{flow rate} \propto \frac{\mathcal{L}}{\mathcal{T}} \cdot \mathcal{L}^2 = \frac{\mathcal{L}^3}{\mathcal{T}}.$$
This means that the flow rate of the fuel is proportional to the fuel flow velocity times the cross sectional area; in other words, fuel flow rate is some constant times the product of the tube area and the flow velocity. What that constant is doesn't really matter, yet.
Now, if you spend 35 miles per gallon, and you go 70 miles per hour, then in 1 hour you spend 2 gallons worth of gas, right? Dimensionally, this can be represented as
$$\textrm{fuel usage rate} \propto \frac{\mathcal{L}}{\mathcal{T}} / \frac{\mathcal{L}}{\mathcal{L^3}} = \frac{\mathcal{L}^3}{\mathcal{T}}.$$
Now, the flow rate and fuel usage rate should be equal, by conservation of mass.
Therefore, using this dimensional analysis approach
$$\frac{\textrm{vehicle velocity}}{\textrm{fuel economy}} = \textrm{fuel flow velocity}\cdot \textrm{cross sectional area}.$$
Since $\textrm{fuel flow velocity} = \textrm{vehicle velocity}$, you're done.
