A siphon tube is filled with gasoline and closed at each end. One end is inserte

Question

A siphon tube is filled with gasoline and closed at each end. One end is inserted into a gasoline tank 0.25 m below the surface of the gasoline. The outlet is placed outside the tank at a distance 0.45 m below the surface of the gasoline. The tube has an inner cross-sectional area of 3.3 × 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?

Explanation / Answer

since change in atmospheric pressure in this case with the difference in height 0.2 m is negligible, at the immersed end of gas, pressure = P = density*g*h = 680*9.8*0.25 in going up the pipe and down outwards, further increase in pressure : P' = density*g*h = 680*(0.45 -0.25)*9.8 so, net difference in pressure = 680*0.45*9.8 = 1/2*density*v*v so, v = v[2*9.8*0.45] =2.9698 m/s

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