How Fast Does the Siphon Flow? [Challenge Solved]

With gas prices soaring, several friends decide to save money by siphoning gas from other people's vehicles. They need to siphon 40L of gas in order for the group to have enough for their mopeds for the week. Sunday night, they were siphoning gas out of a large truck at the local market. They inserted a 1cm ID tube into the gas tank and started the siphon. The gasoline level in the truck was about 1 m from ground level. The flow of fuel from this hose can be estimated as 3/4 of the flow of a frictionless hose. Assume SG of gas to be 0.8. After 5 minutes, the owner of the truck hits the keyless entry button and the group takes off.

Did they siphon enough gas out of the truck?

ANSWER

The first correct answer received was provided by Smriti Gupta of India's National Institute of Technology, Trichy. Congratulations! And a special mention to those who were observant enough to note that the truck pictured is fueled by diesel, while the mopeds in the challenge question are undoubtedly fueled by gasoline!

Smriti's Winning Response

Here Bernoulli's equation may be applied to derive the flow-rate and velocity of the fluid through the siphon. Bernoulli's equation: (v^2)/2 + gy + P/rho = constant We consider a large truck hence, the height of the falling fluid is nearly constant. The velocity at the upper surface of the truck tank may be set to zero. Furthermore, the pressure at both the surface and the exit point is equal to atmospheric pressure. (P = 1 atm) Considering an energy balance between the upper surface of the tank and the outlet of the tube we have: 0 + g*0 + 1/rho = (v^2)/2 - gh + 1/rho Solving for velocity at the outlet of the tube: v = sqrt(2g) = 4.4286 m/s Area of cross section of the tube = (pi/4)*(0.01)^2 = 0.7853 sq.cm density = 0.8*1000 = 800 kg/cu.m So, the mass flow rate = 800*4.4286*0.7853*(10)^(-4)=0.278257 kg/s volumetric flow rate q= (0.278257/800) cu.m/s total volume of gasoline siphoned out of the truck in 5 minutes= (0.278257/800)*5*60 cu.m=0.104348 cu.m=104.348 L Since the Actual flow of the gas is 0.75 times the flow in a frictionless hose, the actual volume of the gas siphoned =104.348*0.75=78.261 L So they siphoned enough gas out of the truck (38.261 L extra!!)