Calculate pump height by using the Bernoulli Equation

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  • Start date
  • #1
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Homework Statement:
A centrifugal pump has an NPSH of 4 metres. It is used to transfer hot wort from kettle, open to atm prsssure, to a whirlpool. Calculate the minimum vertical distance the pump must be positioned below the outlet of the kettle to enable effecient pump operation.

Wort flow rate is 200 hl h-1
Vapour pressure of wort at 100 degrees C = 99 kPa
Wort density = 1065 kg m-3
Pipe length from kettle to pump inlet = 8.5 m
Pipe internal diameter = 75 mm
Pipe friction factor = 0.002
Acc due to grav = 9.81 m s-2
Atm pressure = 101.3 kPa
Relevant Equations:
Darcys Friction Equation
Bernouilli Equation
Really need to help getting started with this one.

They've given NPSH = (Ps - Pvp)/qg
Where Ps = Suction pressure at pump inlet
Pvp = vapour pressure of liquid at temp of pumping
q = density of liquid
g = acc due to grav

They've also provided Darcys friction Equation

The examiners tip (its a past exam paper) said that people used equations that enable fast solutions, but expected way was Bernouilli Eq.

Bernouilli Eq:

P2/q + (V2^2)/2 + gz2 = P1/q + (V1^2)/2 + gz1

So P2 is 99,000 Pa
q = 1065 kg m-3
V2 is 200 hl h-1 = 5.5x10^-3 m3/s
Z2 = 8.5m

P1 = atmospheric pressure?
I dont have V2
Im looking for z1.

Am I on the right track here, need some guidance please.
 

Answers and Replies

  • #3
83
5
Okay I wont express in m3/s

The minimum vertical distance should be 3.94 metres.
Can someone show me how to solve please.
 
  • #4
Lnewqban
Gold Member
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710
Okay I wont express in m3/s

The minimum vertical distance should be 3.94 metres.
Can someone show me how to solve please.
I will help you, but you need to show some work, according to the rules of this forum.
Please, see the importance of understanding and learning this over just solving a problem.
We will be here for you all the way, don't quit, keep trying.

What work have you done so far?
If none, we could start by discussing what you understand about NPSH from the point of view of energy of the fluid.
Next, set a general equation for the different parts that form the NPSH, considering the signs of each.
Then, convert all the given information to units of the International System of Units (SI), so we can use them in the equations.
 
  • #5
83
5
Thanks I appreciate your help.
The NPSH is difference between suction head and liquid vapour head
Ps / yLiq + Vs^2 / 2g - Pv / yVap

Bernouilli:
P1/q + V1^2/2 + gh1 = P2/q + V2^2/2 + gh2

So like you say lets convert the flow rate to m/s
200hl / h-1
= 3.33 hl/min
= 333 L/min
1L/min = 10^-2 m^3/min
333L/min = 0.333 m^3/min
= 0.00555m^3/sec

Cross Sec Area
= 3.14(0.075/2)^2
= 0.0044

Velocity = Flow/Cross Sec Area
= 1.26 m/s
 
  • #6
83
5
NPSH is:

Ps/Unit Weight + Vs^2 / 2g - Pv / Unit Weight

101.3/1065 +1.26^2 / 2(9.81) - 99 / 1065
= 0.0831 which is > 0
 
  • #7
83
5
Darcys Friction Eq:

4(Friction Factor)(L/d)(density)(flow)^2

4 (0.002)(8.5/0.075)(1065)(1.26)^2

= 1532.94
 
  • #8
83
5
Another NPSH Equation = Abs Pressure + vertical distance - Friction Loss + Velocity - Vapour Pressure
101300 Pa + H2 - 1532.94 +1.26 - 99,000 Pa
H2 - 765.8
 
Last edited:
  • #9
Lnewqban
Gold Member
1,256
710
Sorry, it was a long day for me.
Please, see comments below:

Thanks I appreciate your help. You are welcome. :smile:

The NPSH is difference between suction head and liquid vapour head
Ps / yLiq + Vs^2 / 2g - Pv / yVap

Net positive suction head is the energy level of the liquid at the point of entrance to the impeller of the pump.
Think of it as a safety margin that the system connected to any pump should have in order not to reach the vapor pressure of the liquid inside the impeller, which will create vapor formation and cavitation as the action of the impeller suddenly increases the pressure again.
Pump manufacturer's may give you a required NPSH, we are calculating the static head needed in the pipe system in order to achieve an available NPSH of value 4.

Note that Ps is in your above equation the absolute atmospheric pressure above the liquid (in meters) plus the height of the column of liquid above center of impeller or static head.
That static head should be referred to the lower level of the liquid, which is the worse condition.


Bernouilli:
P1/q + V1^2/2 + gh1 = P2/q + V2^2/2 + gh2

Here we have a problem.
We have friction in the system, reason for which, total energy at surface of liquid = total energy at suction of impeller + energy wasted as friction (friction head).


So like you say lets convert the flow rate to m/s
200hl / h-1
= 3.33 hl/min
= 333 L/min
1L/min = 10^-2 m^3/min
333L/min = 0.333 m^3/min
= 0.00555m^3/sec Correct!

Cross Sec Area
= 3.14(0.075/2)^2
= 0.0044

Velocity = Flow/Cross Sec Area
= 1.26 m/s Correct!
 
  • #10
Lnewqban
Gold Member
1,256
710
NPSH is:

Ps/Unit Weight + Vs^2 / 2g - Pv / Unit Weight

101.3/1065 +1.26^2 / 2(9.81) - 99 / 1065
= 0.0831 which is > 0

You are missing the static height of liquid here.
Refer to previous post.
Unify your units, everything should be in meters.
 
  • #12
83
5
I followed Darcys Eq on this link:

Δpmajor_loss = λ (l / dh) (ρf v2 / 2)
(0.002) (8.5m)/(0.075m) (1065kgm-3)(1.26)^2 /2
=191.56
Can you give me the correct answer here if this is incorrect

The minimum vertical distance is shown in this formula: Book link

Hmax = (Pl - Pv)/qg - NPSH + (Vl^2-Vs^2)/2g - Friction

Hmax =
(101300Pa - 99,000Pa)/(1065kgm-3)(9.81m/s) - 4m + (1.26^2)/(2)(9.81m/s) - 191.45
= 0.2201 - 4m + 0.0809 - 191.56
= -3.699 - 191.56

I think im nearly there with that Hmax formula if I could get the friction by Darcys Eq right. Can you finish this off for me so I can see how to get the 3.94 metres please.
 
  • #13
Lnewqban
Gold Member
1,256
710
What are the units of that number (191.56) and how did you reach that result?
I am sorry, but I can't follow.

Please, refer to post #10 above.
Calculate each head in meters.
Remember that the expected way to solve this problem was the application of the Bernoulli equation.
 
  • #14
83
5
I used the linkk you gave me: 191.56 Pa

Δpmajor_loss = λ (l / dh) (ρf v2 / 2) (1)

where

Δpmajor_loss = major (friction) pressure loss in fluid flow (Pa (N/m2), psf (lb/ft2))

λ = Darcy-Weisbach friction coefficient - 0.002

l = length of duct or pipe (m, ft) - 8.5m

v = velocity of fluid (m/s, ft/s) 1.26 m/s

dh = hydraulic diameter (m, ft) 0.075 m

ρf = density of fluid (kg/m3, slugs/ft3) - 1065 kgm-3

I entered the results into the calc on the link and its giving 191 too.

I wish I had more time to spend on this but its been nearly a week since I posted this.
 
  • #15
Lnewqban
Gold Member
1,256
710
Calculating only friction head:

There is no schematic available, but in real life calculations, some equivalent length of tank-pipe transition (3.7 m), 90-degree elbow (1.2 m), gate isolation valve (2.6 m) and pipe reduction for pump entrance (1.5 m) should be added to the given length of pipe (8.5 m).
The value of the actual length plus the several equivalent lengths should be used in the Darcy equation.

##h_f={f_DV^2L}/{2gd}##
##h_f={(0.002)(1.257)^2(8.5+3.7+1.2+2.6+1.5)}/{(2)(9.81)(0.075)}##
##h_f=0.038~meters##

Please, see:
https://neutrium.net/fluid_flow/pressure-loss-in-pipe/

https://web.njit.edu/~barat/ChE396_Spring2011/EquivLengths.pdf

Calculating minimum height:

##Minimum~height=NPSH-(Atmospheric~pressure~absolute)+(Vapor~pressure~absolute)+(Friction~loss~head~in ~pipe)-(Velocity~head~at~pump~suction)##
##Minimum~height=NPSH-(P_{atm}/{\rho~g})+(P_v/{\rho~g})+({f_DV^2L}/{2gd})-(V^2/2g)##

:cool:
 
Last edited:
  • #16
83
5
Thanks,

So my velocity is correct - 1.26m/s
My Darcys Friction is correct - 1532.98Pa
However i need to covert to m 1532.98/(1065)(9.81) = 0.1467m

I need to convert Pa to m like you suggested. I didn't know how, i do now,

Ha = 101300/(1065)(9.81) = 9.6596m
Hp = 99000/(1965)(9.81) = 9.67581 m

Ha + Hs = NPSH + Hp + HDarcys
9.69596m + Hs = 4m + 9.67581m + 0.1467m
Hs = 3.93m

Cheers
 
Last edited:

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