A siphon tube of cross=sectional area 3X10 -4 m2 is used to drain a tank of wate
ID: 1913619 • Letter: A
Question
A siphon tube of cross=sectional area 3X10 -4 m2 is used to drain a tank of water. The tube is initially filled with water, and with the ends closed, one end is placed in the tank 0.25 m below the water surface. The other end is outside the tank and at a distance of 0.5 m below the immersed end. a) What is the velocity of the water flowing out of the tube shortly after the ends are opened? b) is the flow continuous? C) What is the velocity of the water when the water surface in the tank is 0.1 m above the immersed end?Explanation / Answer
the velocity at the immersed end,v is given by v=0.5*g*t^2 the velocity at the tube's other end is V.....from the equations of motion applied to both ends of tube we have V^2-v^2=2*g*0.5 a)initially t=0 V^2=2*9.8*0.5 V=3.13 m/sec b)For a fluid flow to be continuous, we require that the velocity be a finite and continuous function of and t Therefore we can say that the flow here is continuous c)here 0.15m of tank has drained out velocity at the surface is u=2*g*0.15 =2.94m/sec velocity at immersed end v^2=2.94^2+2*g*0.1 v=3.256m/sec velocity at tubes other end V^2=3.256^2+2*9.8*0.5 V=4.517 m/sec