A pipe with a diameter of D1 = 8 cm rises vertically a height H = 3.2 m and expa

Question

A pipe with a diameter of D1 = 8 cm rises vertically a height H = 3.2 m and expands to a pipe of D2 = 36 cm in diameter. The flow rate of the fluid through the pipe is 0.14 cubic meters per second. The pressure in the lower end of the pipe is 1.4?10^5 Pascals. The fluid has a density of 1 kg/m3 ..

Find the velocity of the water in the lower section of the pipe. Answer in m/s

Find the difference in hydrostatic pressure from the high section to the low section. Answer in Pascals

Find the pressure in the upper section of the pipe. Answer in Pascals

Explanation / Answer

Velocity = Flow rate/area

V1 = Q/A1 = 0.14/(3.14/4*0.08^2) = 27.9 m/s

V2 = Q/A2 = 0.14/(3.14/4*0.36^2) = 1.38 m/s

P1 + 1/2**V1^2 + *g*Z1 = P2 + 1/2**V2^2 + *g*Z2

(P1 - P2) = 1/2**(V2^2 - V1^2) + *g*(z2 - z1)

(P1 - P2) = 1/2*1*(1.38^2 - 27.9^2) + 1*9.81*3.2

= -356.86 Pa

Pressure in upper section P2 = P1 + 356.86 = 1.4*10^5 + 356.86 = 140356.86 Pa

Mass flow rate = density*vol. flow rate

= 1*0.14 = 0.14 kg/s

Weight of the fluid in 14.5 minutes = 0.14*14.5*60*9.81 = 1194.9 N

(Density of fluid = 1 kg/m3 looks too low to me. Just check I believe if it is water, it should be 1000 kg/m3.We need to redo the calculations in that case.)

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