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EXAMPLE 5–7 Siphoning Out Gasoline from a Fuel Tank During a trip to the beach (Pam = 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. 5–40). 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 4 mm, and frictional losses in

EXAMPLE 5–7 Siphoning Out Gasoline from a Fuel Tank During a trip to the beach (Pam = 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. 5–40). 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 4 mm, and frictional losses in

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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EXAMPLE 5–7 Siphoning Out Gasoline from a Fuel Tank During a trip to the beach (Patm = 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. 5-40). 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 4 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to with- Gasoline siphoning tube 2 m Gas tank 0.75 m draw 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. SOLUTION 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 esti- mate. 3 The change in the gasoline surface level inside the tank is negligible compared to elevations z, and z, 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, = Patm (open to the atmosphere), V = 0 (the tank is large relative to the tube diameter), and z, = 0 (point 2 is taken as the reference level). Also, P, = Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to Gas can FIGURE 5-40 Schematic for Example 5–7. + z1 = 2g + z2 2g Solving for V, and substituting, V = V2gz, = V2(9.81 m/s³)(0.75 m) = 3.84 m/s
Transcribed Image Text:EXAMPLE 5–7 Siphoning Out Gasoline from a Fuel Tank During a trip to the beach (Patm = 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. 5-40). 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 4 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to with- Gasoline siphoning tube 2 m Gas tank 0.75 m draw 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. SOLUTION 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 esti- mate. 3 The change in the gasoline surface level inside the tank is negligible compared to elevations z, and z, 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, = Patm (open to the atmosphere), V = 0 (the tank is large relative to the tube diameter), and z, = 0 (point 2 is taken as the reference level). Also, P, = Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to Gas can FIGURE 5-40 Schematic for Example 5–7. + z1 = 2g + z2 2g Solving for V, and substituting, V = V2gz, = V2(9.81 m/s³)(0.75 m) = 3.84 m/s
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