Water flows through a 30 cm radius pipe at the rate of 0.20 m^3/s. The pressure

Question

Water flows through a 30 cm radius pipe at the rate of 0.20 m^3/s. The pressure in the pipe is atmospheric. The pipe slants downhill and feeds into a second pipe of radius 15 cm, positioned 60 cm lower. What is the gauge pressure in the lower pipe?
30cm=.3m
60cm=.6m
15cm=.15m

P = Patm + pgh alone didn't solve it.
KE=PE
v= sqrt 2gh
v= 3.43 m/s

Bernoulli's
P + 1/2 pv^2 + pgy =1.13 x 10^5 Pa
should have been 2.1 x 10^3 Pa. I'm obviously doing something wrong. Thanks for the help. It's an online class and I get faster feedback here than from the class tutor.

Explanation / Answer

according to Bernoulli's Equation,

P1 + (1/2)*v12 + gh1 = P2 + (1/2)*v22 + gh2 + loss due sudden contraction in a pipe

30 cm of pipe connected to 15 cm pipe. therefore , contraction loss will take place.

head loss = 0.5v22/2g

assuming lower pipe having 15 cm dia is reference postion.above this pipe, consider elevation head as well.

pipe having dia is 30 cm , flow = 0.2 m3/s

consider lower pipe having dia 15 cm as 'pipe 2' and upper pipe having dia 30 cm as 'pipe 1' .

for pipe 1, cross-sectional area, A1 = *0.302/4 = 0.0707 m2

for pipe 2, cross-sectional area, A2 = *0.152/4 = 0.0177 m2

according to continuity theory,

A*v = constant = Q

v1 = Q/A1 = 0.2/0.0707 = 2.83 m/s

and v2 = Q/A2 = 0.2/0.0177 = 11.3 m/s

head loss = 0.5* 11.32/(2*9.81) = 3.25 m

(atmoshpheric pressure) + (1/2)*(1000*2.832 ) + 1000*9.81*0.6 = (atmoshpheric pressure + gauge pressure) +(1/2)*(1000*11.32 ) + 1000*9.81*0 + 3.25*1000*9.81

solving above equation, we get,

gauge pressure = - 85837.05 N/m2

or 85.84 KN/m2 or 85.84 kPa

actually gauge pressure is relative pressure. this is with respect to some value.

here, negative sign means gauge pressure is lesser with respect to upper one.

this solution is based on fluid dynamics. (fluid flowing through pipes.)

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