Question 5.7: Siphoning Out Gasoline from a Fuel Tank During a trip to the...
Siphoning Out Gasoline from a Fuel Tank
During a trip to the beach (P_{atm} = 1 atm = 101.3 kPa), a car runs out of gasoline, and it becomes necessary to siphon gas out of the car of a Good Samaritan (Fig. 5–41). The siphon is a small-diameter hose, and to start the siphon it is necessary to insert one siphon end in the full gas tank, fill the hose with gasoline via suction, and then place the other end in a gas can below the level of the gas tank. The difference in pressure between point 1 (at the free surface of the gasoline in the tank) and point 2 (at the outlet of the tube) causes the liquid to flow from the higher to the lower elevation. Point 2 is located 0.75 m below point 1 in this case, and point 3 is located 2 m above point 1. The siphon diameter is 5 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to withdraw 4 L of gasoline from the tank to the can and (b) the pressure at point 3. The density of gasoline is 750 kg/m³.
Gasoline is to be siphoned from a tank. The minimum time it takes to withdraw 4 L of gasoline and the pressure at the highest point in the system are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Even though the Bernoulli equation is not valid through the pipe because of frictional losses, we employ the Bernoulli equation anyway in order to obtain a best-case estimate. 3 The change in the gasoline surface level inside the tank is negligible compared to elevations z_1 and z_2 during the siphoning period.
Properties The density of gasoline is given to be 750 kg/m³.
Analysis (a) We take point 1 to be at the free surface of gasoline in the tank so that P_1 = P_{atm} (open to the atmosphere), V_1 ≅ 0 (the tank is large relative to the tube diameter), and z_2 = 0 (point 2 is taken as the reference level). Also, P_2 = P_{atm} (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to
\cancel{\frac{P_1}{\rho g} } + \overset{\approx 0}{\frac{\cancel{V^2_1} }{2g} } + z_1 = \cancel{\frac{P_2}{\rho g} } + \frac{V^2_2}{2g} +\overset{0}{\cancel{z_2} } \longrightarrow z_1 = \frac{V^2_2}{2g}
Solving for V_2 and substituting,
V_2 = \sqrt{2gz_1} = \sqrt{2(9.81 m/s^2)(0.75 m)} = 3.84 m/s
The cross-sectional area of the tube and the flow rate of gasoline are
A = \pi D^2/4 = \pi (5 \times 10^{-3} m)^2/4 = 1.96 \times 10^{-5} m^2
\dot{V} = V_2A = (3.84 m/s)( 1.96 \times 10^{-5} m^2) = 7.53 \times 10^{-5} m^3/s = 0.0753 L/s
Then the time needed to siphon 4 L of gasoline becomes
\Delta t = \frac{V}{\dot{V} } = \frac{4 L}{0.0753 L/s} = 53.1 s
(b) The pressure at point 3 is determined by writing the Bernoulli equation along a streamline between points 3 and 2. Noting that V_2 = V_3 (conservation of mass), z_2 = 0, and P_2 = P_{atm},
\frac{P_2}{\rho g} + \cancel{\frac{V^2_2}{2g} } + \overset{0}{\cancel{z_2}} = \frac{P_3}{\rho g} + \cancel{\frac{V^2_3}{2g} } + z_3 \longrightarrow \frac{P_{atm}}{\rho g} = \frac{P_3}{\rho g} + z_3
Solving for P_3 and substituting,
P_3 = P_{atm} – \rho gz_3 = 101.3 kPa – (750 kg/m^3)(9.81 m/s^2)(2.75 m) \left(\frac{1 N}{1 kg.m/s^2} \right) \left(\frac{1 kPa}{1000 N/m^2} \right)
= 81.1 kPa
Discussion The siphoning time is determined by neglecting frictional effects, and thus this is the minimum time required. In reality, the time will be longer than 53.1 s because of friction between the gasoline and the tube surface, along with other irreversible losses, as discussed in Chap. 8. Also, the pressure at point 3 is below the atmospheric pressure. If the elevation difference between points 1 and 3 is too high, the pressure at point 3 may drop below the vapor pressure of gasoline at the gasoline temperature, and some gasoline may evaporate (cavitate). The vapor then may form a pocket at the top and halt the flow of gasoline.