Solids and Fluids: Siphoning Gas

[huybinhs]

Homework Statement

A siphon tube is filled with gasoline and closed at each end. One end is inserted into a gasoline tank 0.20 m below the surface of the gasoline. The outlet is placed outside the tank at a distance 0.50 m below the surface of the gasoline. The tube has an inner cross-sectional area of 4.2 × 10-4 m2. The density of gasoline is 680 kg/m3. Ignoring viscous effects, what is the velocity of the gasoline in the tube shortly after the tube is opened?
What is the corresponding rate of flow of the gasoline?

2. The attempt at a solution

a) v = sqrt(2gh) = sqrt[2*9.8*(0.5-0.2)] = 2.42 m/s => incorrect.

b) Q = Av = 4.2*10^-4 * 2.42 = 0.001 m^3/s => incorrect.

Please helps! Thanks!

 

[anathema]

it is important to approach problems like this systematically and use the appropriate equations and units. In this case, the Bernoulli's equation can be used to solve for the velocity of the gasoline in the siphon tube.

First, we need to convert all units to SI units:
- The distance between the gasoline tank and the outlet is 0.5 m
- The height difference between the surface of the gasoline and the outlet is 0.3 m (0.5 m - 0.2 m)
- The inner cross-sectional area of the tube is 4.2 x 10^-4 m^2
- The density of gasoline is 680 kg/m^3

Next, we can plug these values into the Bernoulli's equation:
P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
where:
- P1 and P2 are the pressures at the two points (in this case, they can be assumed to be equal since the tube is closed at both ends)
- ρ is the density of the fluid (in this case, gasoline)
- v1 and v2 are the velocities at the two points (v1 = 0 since the tube is initially closed)
- g is the acceleration due to gravity
- h1 and h2 are the heights of the fluid at the two points (h1 = 0.2 m, h2 = 0.5 m)

Solving for v2, we get:
v2 = sqrt(2gh2) = sqrt(2*9.8*0.3) = 1.72 m/s

To calculate the rate of flow, we can use the equation:
Q = Av
where:
- Q is the volumetric flow rate (in m^3/s)
- A is the cross-sectional area of the tube (in m^2)
- v is the velocity of the fluid (in m/s)

Plugging in the values, we get:
Q = 4.2 x 10^-4 * 1.72 = 7.2 x 10^-4 m^3/s

Therefore, the velocity of the gasoline in the tube shortly after it is opened is 1.72 m/s and the corresponding rate of flow is 7.2 x 10^-4 m