Question

P2 Device 260 m DK-1 B Air 160 m т 26 m 1 Water K-0.4 325 m Determine the pressure P1 needed for flowing water at a 1001/s rate into a device at 40.000 Pa in gage pressure. It is known that the pipe is a 150mm pipe made of commercial steel and the elbows are 900 regular screwed elbows. The kinematic viscosity is given as 0.113x10-5 m/s. L. If you think data is missing in the question, you may assume it and proceed with your calculations. State such assumptions in your question.

Answer 1

Ans) ApplyBernoulli equation between point 1 and 2 loacted at water surface elevation of tank and just outside pipe exit at device respectively,  

  P1/$\gamma$ +
V1
/2g + Z1 = P2/$\gamma$ +
V22
/2g + Z2 + Hf + Hm

Given, P2 = 40000 Pa  or 40 kPa

Also, velocity at surface is negligible so V1 = 0

Velocity at point 2, V2 = Q / A2

Flow rate (Q) = 100 L/s or 0.1 $m^3$ /s

Area of pipe 2, A2 = ($\pi$/4)$(0.15)^2$ = 0.0176625 sq.m

=> V2 = 0.1/ 0.0176625 = 5.66 m/s  

Elevation, Z1 = 26 m and Z2 = 160 m

Hf is head loss due to friction

Hm is minor head loss due to bends and valves

Also,

Hf = f L $V^2$ / (2 g D)

where, f = friction factor

L = Pipe length = 325 + 160 + 260 = 745 m

D = Pipe diameter = 150 mm or 0.15 m

Since, friction factor is unknown, calculate Reynold number (Re) and reletive roughness (e/D) to determine friction factor

Re = V D / $\nu$

where, $\nu$ = Kinematic viscosity = 0.113 x $10^{-5}$ m2/s

=> Re = 5.66 x 0.15 / (0.113 x $10^{-5}$ )

=> Re = 751328

Roughness of steel pipe (e) = 0.045 mm

=> Relative roughness (e/D) = 0.045 / 150 = 0.0003

According to Moody diagram, for Re = 751328 and e/D = 0.0003 , friction factor (f) = 0.016

=> Hf = 0.016(745)($5.66^2$) / (2 x 9.81 x 0.15)

=> Hf = 129.75 m

Also,

Hm = $\sum$ K $V^2$ / 2 g

$\sum$K = sum of loss coefficients due to bends and elongation

For 90 degree elbows, K = 0.3

=> $\sum$ K = 0.4 + 2(0.3) + 1 = 2

=> Hm = 2 $(5.66^2)$ / (2 x 9.81)

=> Hm = 3.27 m

Putting values in Bernoulli equation,

=> P1/$\gamma$ + 0 + 26 = (40/9.81) + $(5.66^2)$ /(2 x 9.81) + 160 + 129.75 + 3.27

=> P1/$\gamma$ + 26 = 4.08 + 1.63 + 293.02

=> P1/$\gamma$ = 272.73 m

=> P1 = 9.81 kN/$m^3$ x 272.73 m

=> P1 = 2675.5 kPa

Hence, pressure required at point 1 is 2675.5 kPa