A pipe of cross-sectional area A= 0.01m^2 and total length 5.5m is used for siph

Question

A pipe of cross-sectional area A= 0.01m^2 and total length 5.5m is used for siphoning water from a tank. The discharge from the siphon is 1.0m below the level of the water in the tank. At its highest point, the pipe rises 1.5m above the level in the tank.

(a) What is the water velocity v (m/s) in the pipe?

(b) What is the lowest pressure in bar (gauge), and where does it occur? Neglect pipe friction.

(c) Are your answers reasonable?

(d) If the siphon reaches virtually all the way to the bottom of the tank (but is not blocked off), is the time taken to drain the tank equal to t=V/vA, where V= initial volume of water in the tank, and v= velocity computed above when the tank is full?

EXPLAIN YOUR ANSWER please. :)

Explanation / Answer

use bernoulli theorem
simply for 2 places one at inside the tank(1) and other at discharge(2)

P1/ g + v12 /2g + z1 = P2/ g + v22 /2g + z2+ losses

in this case no loss due to friction so losses =0

hence taking discharge line as centre line =0

then

P1=P2= atm pressure hence cancels out

v1 is negligble

v2 is considerable

z1= 1m z2=0 m

hence

1= v22 /2g

v2 = 1*2*1000*10= 141.42 m/s

b) lowest pressure is at max hight since bernolli P/ g + v2 /2g + z = constant

z is max and considerable velocity hence pressure will reduce

P1/ g + v12 /2g + z1 = P2/ g + v22 /2g + z2

now p2 is discharge where p2 is 1 atm while v2=141.42 m/s

now z1=2.5 m while z2 =0

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