# Using Bernoulli's Equation to find Power Requirement

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1. Apr 24, 2017

### GSXR-750

1. The problem statement, all variables and given/known data

Calculate head required for the pump and then its power requirement assuming 70% efficiency.
The lower storage vessel is vented to atmosphere (assume 1 bar pressure) .

I have the following given information:
Pipe Area = 0.00636m^3.
Flow(Q)= 0.01m^3/s
Average Velocity = 1.57m/s
Density of Fluid = 960kg/m^-3.
Liquid viscosity = 0.081 Pa/s

I have calculated the head losses of the whole system, using firstly the Equivalent Head (Hm) to be 1.03m and Number Velocity head (Hf) to be 0.54m.

2. Relevant equations

Bernoullis

3. The attempt at a solution

Pump inlet
$\frac {0} { 960 * 9.81 } + \frac {1.57^2} { 2 * 9.81 \ } + 0 = \frac {P2 }{\ 960 * 9.81 \ } + \frac {1.57^2} { \ 2 * 9.81 \ } + 4 + 0.38$

P2 = -41249 Pa
= -0.41Bar

Pump Outlet

$\frac {P1} { 960 * 9.81 } + \frac {1.57^2} { 2 * 9.81 \ } = \frac {200000 }{\ 960 * 9.81 \ } + \frac {1.57^2} { \ 2 * 9.81 \ } + 15 + 1.19$

P1= 352470 Pa
P1 = 3.5 Bar + 2 Bar(Pressurised Cylinder)

Power = qpgh
= 0.01 x 960 x 9.81 x 19.57
= 1.84 kW
Assuming 70% = $\frac { 1.84 } { \ 0.7 \ }$
= 2.63 kW

Is my method correct so I can say the pump required is an 2.63 kW to overcome the 5.5 bar.

Any help would be great

Thanks

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Last edited: Apr 24, 2017
2. Apr 29, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Oct 11, 2017

### Ltowley

I have the pump pressure at 2.56 Barg and the power at 70 % efficiency 2.5 kW

4. Oct 12, 2017

### Ltowley

Equivalent length;
Re=1.572*0.09*960/0.081
Re=1676.8 thus streamlined flow

Friction factor=16/1676.8 =9.54*10^-3

Hf/L =4 (9.54*10^-3 * 1.572^2)/2*9.81*0.09
=0.0534
L = 32 m (total length of pipe)

Hf = 1.7088 m

Le/d = 213 (4 x 90 bends, 1 x entry, 1 x exit)

Cf=16/Re = 9.542*10^-3 (Cf)

Hm=2*Cf*Um^2/g *Le/d
Hm=426*Cf*1.572/9.81
Hm=1.024 (equivalent length method)

1.572^2/2*9.81 + 0 + 101325/960*9.81 + 1.572^2/2*9.81 + 2 + Pp/960*9.81 = 1.572^2/2*9.81 + 16 + 200000/960*9.81 + 1.0788 + 1.024

=13.011 + Pp/960*9.81=39.970

Pp=26.959*(960*9.81 )

Pp=253889 Pa (2.54 Barg)

H=3+15-1+1.7088+1.024
H=19.733 m

Power required = 0.01*9.81*960*19.733
=1.858 kW

At 70% 1.858/0.7= 2.654 kW

5. Oct 12, 2017

### Staff: Mentor

I didn't check all your arithmetic, but your approach is correct. It looks like the OP omitted the frictional and joints resistances, and only used Bernoulli.

6. Oct 12, 2017

### Ltowley

Thanks for taking the time to look at that for me

7. Oct 21, 2017 at 5:11 AM

### Babadag

8. Oct 21, 2017 at 5:21 AM

### Babadag

If Re<2000 it is a laminar flow then friction factor =64/Re. I'm only an electrical engineer and I saw this equation in a few articles.Why here
f=16/Re?

9. Oct 21, 2017 at 5:28 AM

### Nidum

This explanation is given in many fluid mechanics textbooks :

Friction Factors: Fanning and Darcy

There are two common friction factors in use, the Darcy and Fanning friction factors. The Darcy friction factor is also known as the Darcy–Weisbach friction factor or the Moody friction factor. It is important to understand which friction factor is being described in an equation or chart to prevent error in pressure loss, or fluid flow calculation results.

The difference between the two friction factors is that the value of the Darcy friction factor is 4 times that of the Fanning friction factor. In all other aspects they are identical, and by applying the conversion factor of 4 the friction factors may be used interchangeably.

10. Oct 21, 2017 at 5:32 AM

### Babadag

Thank you, Nidum.

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