Question

what is the velocity head at the inlet pump?

Find the value of head loss (consider entrance, friction, elbow and valve losses). Also, find pressure at the inlet pump

-115 m E Pump Globe valve fully open DN 65 Schedule 40 steel pipe 1.40 m 0.75 m Standard elbows (2)

water is at 80 degrees C

Water is being pumped from the tank at a rate of 475 L/min

Answer 1

We will first figure out the various losses that have to be accounted for, in our analysis.

1. Major loss (Frictional loss)
2. Minor loss (Entry losses, elbow losses and valve losses).

We will first calculate Minor losses.

1. Elbow Loss - 90 (deg) flanged elbow smooth bend - Loss coefficient = 0.3
2. Entrance loss - Reentrant - Loss Coefficient = 0.8
3. Globe Valve (fully open) - Loss Coefficient = 10

Also, Velocity in the pipe = ?

Velocity = Areg

where

• Q = Volumetric Flow rate = 475 L/min = 7.9166e-3 m3/sec
• Area = $\frac{\prod dia^{2}}{4}=\frac{\prod 0.06268^{2}}{4}=3.0856e-3\: m^{2}$
  • Inner dia = 62.68 mm = 0.06268 m (Given pipe DN65 SCH 40 pipe - referred the data from the internet)

Thus, $Velocity=\frac{Q}{Area}=\frac{7.9166e-3}{3.0856e-3}=2.5656\: m/s$

Thus, total Minor Losses = $\left (\sum k \right )\times \frac{V^{2}}{2g}$

where

• k =
  • 0.3 (elbow loss)
  • 0.8 (entrance loss)
  • 10 (valve loss)
• V = Velocity = 2.5656 m/s
• g = 9.81 m/s2

Minor Losses = $\left (\sum k \right )\times \frac{V^{2}}{2g}=(0.3+0.8+10)\times \frac{2.5656^{2}}{2\times 9.81}=3.7239\: m$

Now, we will calculate Major Loss.

Given material - Steel (since no other information is given about the material of the pipe, we will consider the pipes to be Stainless Steel)

Absolute roughness value = k = 4.5e-5 m (from the internet)

Inner dia = 0.06268 m

We will now find the Reynold's number, to be able to figure out whether the flow is laminar or turbulent:

$Re=\frac{VD}{\nu }$

where

• $\nu$ = kinmeatic viscosity of water at 80 C = 3.639e-7 m2/s

Therefore, $Re=\frac{VD}{\nu }=\frac{2.5656\times 0.06268}{3.639e-7}=441912.086$

FLOW IS TURBULENT!!

We will now have to calculate the friction factor using Colebrook Equation,

$\frac{1}{\sqrt{f}}=-2log\left (\frac{k}{3.71D}+\frac{2.51}{Re\sqrt{f}} \right )$

Therefore, we have, $\frac{1}{\sqrt{f}}=-2log\left (\frac{4.5e-5}{3.71\times 0.06268}+\frac{2.51}{441912.086\sqrt{f}} \right )$

Using Engineering Equations Solver to solve this equation, we get, f = 0.019

Now, we have the major loss formula as Major Loss = $f\times \frac{L}{D}\times \frac{V^{2}}{2g}$

where

• L = total length of the pipe = 1.4 + 11.5 m = 12.9 m
• D = inner dia = 0.06268 m
• V = 2.5656 m/s
• f = 0.019

Thus, Major loss = $f\times \frac{L}{D}\times \frac{V^{2}}{2g}=0.019\times \frac{12.9}{0.06268}\times \frac{2.5656^{2}}{2\times 9.81}=1.312\: m$

Thus, total loss of head = major loss + minor loss = 1.312 + 3.7239 = 5.0357 m of water

[Here, the minor losses are greater than major losses]

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Velocity head of water = $\frac{V^{2}}{2g}=\frac{2.5656^{2}}{2\times 9.81}=\mathbf{0.3355\: m}$

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For finding the pressure at the inlet of the pump, we will apply Bernoulli's equation between the 2 points as follows:

11 sm Pump Globe valve fully open 2 Flow DN 65 Schedule 40 steel pipe Standard elbows (2)

$\frac{P_{1}}{\omega }+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\omega }+\frac{V_{2}^{2}}{2g}+z_{2}+H_{loss}$

where

• P1 = 0 (gauge pressure at datum is 0)
• V1 = 0 (no flow at datum)
• Z1 = 0 (Considered as datum)
• P2 = ?
• V2 = 2.5656 m/s
• Z2 = 1.4 - 0.75 = 0.65 m (from the diagram)
• H(loss) = 5.0357 m
• $\omega$ = 9810 N/m3 (for water - known to us)

$\frac{P_{1}}{\omega }+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\omega }+\frac{V_{2}^{2}}{2g}+z_{2}+H_{loss}$

$0+0+0=\frac{P_{2}}{9810}+\frac{2.5656^{2}}{2\times 9.81}+0.65+5.0357=>\mathbf{P_{2}=-59067.972\: Pa}$

$\mathbf{P_{2}=59.068\: kPa(Vacuum\: gauge )}$

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