×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Contact US

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Pump power for hoses feeding to nozzles

Pump power for hoses feeding to nozzles

Pump power for hoses feeding to nozzles

(OP)
I did a Civil Engineering course some years ago and from my textbook I had a question on a pump with hoses. I think I have the correct answer to part (a) as can be seen below, but I am struggling with part (b) and need some help.
This is my logic for part (b) as well, as some things that I am trying to understand :
To calculate power I believe that we need to find the pump term Hp in bernoulli equation

Z1 + P1 / (Rho * g) + V1^2 / 2g + Hp = Z2 + P2 / (Rho * g) + V2^2 / 2g + Hf (frictional Head losses)

Z1 = 0 m (Because this is the level of the pump) and Z2 = 3 m because that is where jet of water comes out nozzles

In part (a) the head at the nozzle was calculated while in the process of finding the diameter of hoses.
Is this head used in part (b) of this question.

In part (a) they say that jet velocity is 24 m/s when jet diameter is 37.5 mm and nozzle is at the same level as the pump.
What I want to understand is how does the jet velocity and diameter differ when the nozzle is 3 m above the pump ?

I assume that the pressures P1 and P2 = 0 because they are atmospheric.

V2 is jet velocity from nozzle and is V1 = 0 or is V1 = hose velocity ?

Here is the question below
A pump feeds two hoses, each of which is 45 m long and is fitted with a nozzle. Each nozzle has a coefficient of velocity of 0.97 and discharges a 37.5 mm diameter jet of water at 24 m/s when the nozzle is at the same level as the pump. If the power lost in overcoming friction in the hoses is not to exceed 20 per cent of the hydraulic power available at the inlet end of the hoses, calculate (a) the diameter of the hoses, taking f = 0.007, and (b) the power required to drive the pump if its efficiency is 70 percent and it draws its water supply from a level 3 m below the nozzle ?

These are the calculations I have so far

a) Head lost in friction Hf = 4 * f * L * v^2/ D * 2g, where L = Length of hose, D = Hose diameter, v = hose velocity and f = friction coefficient

If H = Head at inlet to pump and hoses

Head at nozzle Hp = H − 4 * f * L * v^2/ D * 2g or Hp = Head at inlet to hoses - frictional head loss in each hose.

If jet velocity = V and Cv = 0.97 Jet Velocity V=Cv √(2g ∗ Hp), where V = 24m/s

Hp = V^2/(2g ∗ Cv^2) = (24^2) / 2g * (0.97)^2, so Hp = 31.208 m

For continuity of flow, flow from hoses = flow from nozzle, so 1/4 Pi * D^2 * v = 1/4 * Pi * d^2 * V, where d = jet diameter

v = V * (d2/D2) = 24 *(0.0375)^2 / D^2 = v = 0.03375 / D2.

I have then put this in the frictional head loss formula 4 * f * L * V2/D * 2g = 4 * 0.007 * 45 * (0.03375 D^-2)^2 / (D * 19.62) = 1.43521875 ∗ 10−3 ∗ D−4/ D ∗ 19.62 = 1.43521875 ∗ 10−3/ (D5 ∗ 19.62).

Note 2g = 19.62

If power lost in overcoming friction in hoses is not to exceed 20 per cent of hydraulic power available at inlet end of hoses, then H = 5 * 4 * f * L * v^2/ D2g so Hp = 5 ∗ (4 * f * L * v2/ D * 2g) - (4 * f * L * v^2/ D2g) = 4 * (4 * f * L * v^2/ D * 2g) = 4 * 1.43521875 * 10^-3/ (D^5 * 19.62) Hp = 5.740875*10^-3/ (D^5*2g)

So 2g * Hp = 5.740875 * 10^-3/ D^5 = 19.62 * 31.2018 = 612.1798m

D^5 = 5.740875 * 10^-3 / 612.179, so D = (5.740875 * 10^-3/ 612.1798)^0.2 = 0.09872m

so Hose Diameter = 98.72 mm IN MY TEXT BOOK ANSWER WAS GIVEN AS 98.5mm, WOULD THIS BE DUE TO ROUNDING?

(b) Need to get an understanding on what I mentioned above

RE: Pump power for hoses feeding to nozzles

What is the question, is it about the rounding - - the difference between calculation and given is insignificant, I would have rounded it to 99mm - again fairly irrelevant as it will be governed by what hose sizes are available.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Pump power for hoses feeding to nozzles

Yes, it is not a practical problem as there is no method available to calculate pressure loss in hoses. Hose manufacturers supply use empirical data to supply fluid resistance in hoses.

I suspect that you have a rounding error. You should be using significant digits in your entire calculation.

RE: Pump power for hoses feeding to nozzles

(OP)
I have given it some thought on how to get the answer of 31.1 kW and I am very close, I got 31.2 kW by doing it this way below.

Is this meant to be the way this is calculated, this is what I have been trying to figure out.



From part (a) the power head at the nozzle was calculated as 31.208 m, when the pump is at the same elevation or level as the nozzle. This is the kinetic energy from the water jet.

I think you know that Power Head = Total Head - Frictional Head Loss for transmission of power by pipeline or hoses in this case.

The question ignores small loss of head from nozzle (probably because it is not much in this case)



So now in part (b) we have to find pump head term Hp in order to find power.



Bernoulli equation for this is now :



Z1 + P1 / (Rho g) + V1^2 /2g + Hp = Z2 + P2 / (Rho g) + V2^2 /2g + Hf



For part (b) the 31.2018 m head available at nozzle = V2^2 / 2g term in bernoulli equation



Z1 = level of pump = 0 m

Z2 = 3 m (outlet of nozzle)



P1 = P2 = 0 (because inlet of pump and outlet of nozzle are at atmospheric pressure)

V1 = 0 (because it is assumed water is taken from where the velocity is 0 at pump inlet )



Note: V2 = Jet Velocity / Cv = 24 / 0.97 = 24.742 m/s and this is velocity leaving nozzle where it reaches atmosphere. This is where the water area is slightly smaller than 37.5 mm diameter at nozzle outlet and this velocity is slightly larger than when it initially leaves the nozzle at 24 m/s.

I don't think we need to know this for this question because I believe that we are using the 31.2018 m head at nozzle calculated in part (a)



and Hf = Frictional Head losses from hoses



Now power lost in overcoming friction in hoses is not to exceed 20 per cent of the hydraulic power available at the inlet end of the hoses.

So this means 80 % of head is available at nozzle and this is 4 times the 20 % lost in the hoses.



So Hf = 31.2018 / 4 = 7.8 m

or H = 31.2018 * (5/4) = 39.002 m Total Head



Rearranging Bernoulli Equation to find Hp

Hp = (Z2 - Z1) + (P2 - P1) / (Rho * g) + (V2^2 - V1^2) / 2g + Hf



with P1, P2 and V1 = 0



we have Hp = (Z2 - Z1) + V2^2 / 2g + Hf



Hp = 3 + 31.2018 + 7.8

Pump Head term Hp = 42.002 m



Power from one Hose = P = w *Q * Hp



Now we want to find the flow = Q for the power



flow Q is either



Flow through the pipe = ( 24 (0.0375^2) / (0.09872^2) ) * ( ( ( pi 0.09872^2)/4) = 0.026507 m^3/s

or Flow leaving nozzle = 24 * ( ( pi 0.0375^2) / 4 ) = 0.026507 m^3/s



so Flow = Q = 0.026507 m^3/s or 26.507 litres /sec



w = = Density per unit Weight = 9.81 * 1000 = 9,810 N/m^3



P = 9,810 * 0.026507 * 42.002 = 10,922.085 Watts



For 2 Hoses, P = 22,556.88 Watts (This is power output, which is less than input, due to efficiency)



As the pump has an efficiency of 70 per cent, then

Total Input Power = 21,844.17 / 0.70 = 31,205.96 Watts



so Power required to drive pump = 31.206 kW



this is just over 100 Watts more than the 31.1 kW answer

RE: Pump power for hoses feeding to nozzles

Hope you are happy with the result because until such times as you know the actual flow rate and head and power absorbed by the
pump that can only be established by running and measuring its performance, the theoretical calculations become meaningless.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Pump power for hoses feeding to nozzles

(OP)
@Artisi
The textbook answer is 31.1 kW and my answer is 31.2 kW, is the difference only due to rounding or have I missed a little bit of something that will make the difference ?

It is important that you confirm with me that my theory calculations are correct for this question .


I will be happy the most when that happens and a reply to what I have said will be much appreciated.


RE: Pump power for hoses feeding to nozzles

(OP)
@Artisi
I understand that there may be a bit of a difference between the theory result and actual real life result, but if the theory is reasonably good, then I don't think there should be too much of a difference.

RE: Pump power for hoses feeding to nozzles

Your theory might be ok but, if the pump is 69.5% efficient and not 70% or the power supply to the pump is 1 or 2 volts down, your power calculation is incorrect.
But please let us know the outcome once you run and confirm the results.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Pump power for hoses feeding to nozzles

(OP)
Hi @Artisi

Are you saying that the calculations I have for this theory question are correct for my answer of 31.2 kW and that the 0.1 kW difference is only due to rounding ?

You mention "if the pump is 69.5% efficient and not 70% or the power supply to the pump is 1 or 2 volts down, your power calculation is incorrect". What do you mean by this?

I don't have access to equipment to run and test or confirm the results, I just have this as a theory question to solve, so unfortunately I can't see the difference, I reckon it would be pretty interesting though to see what the difference between theory and actual is.
There may be a site on the internet that show difference between theory and actual.

I have this as a side interest as I studied Civil Engineering a couple of years ago and I still have my textbook, but these days I am doing web development.

RE: Pump power for hoses feeding to nozzles

Looking at your cals'gives me a headache, therefore there is no way I would even consider looking for any error 😉, my comment was *might be ok".
Having spent the majority of my working life in the pump industry, theory is ok to get somewhere near to operating flow, head, power, and overall operating eff. - but it's the actual operation that counts.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Pump power for hoses feeding to nozzles

(OP)
@artisi
OK, well lets look at this another way, we can discuss the approach steps I have taken to work out my answer for this particular question and then from that I can see if I have got this theory part correct.

We will not discuss the actual operation until we come to a conclusion as to whether my approach to this question is correct.

RE: Pump power for hoses feeding to nozzles

Assuming your theoretical approach is correct and there being no reason to think otherwise, it's now validation time, please let us know the outcome.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

RE: Pump power for hoses feeding to nozzles

Without checking the numbers, yes it looks correct.

RE: Pump power for hoses feeding to nozzles

Hi ,
In the equation the pressure must be expressed in absolute meaning P1 and P2 are equal to 1 atmosphere .
For head loss related to pipe , is f Moody factor or Fanning factor?
How is counted the head loss due to Nozzle in the equation of energy?

My 2 cents
Pierre

RE: Pump power for hoses feeding to nozzles

(OP)
Hi @Pierreick and @1503-44

For the head lost in friction formula 4fLv^2 /D 2g, f is the darcy-weisbach friction factor, so that can be the f Moody factor and it is not fanning factor.
I remember during my course years ago, I was told that America use 4f as just f, so then it would be fLv^2 /D 2g

With this question, I assumed that head loss at nozzle is negligible, so I just treated it as 0.
From this I have thought that the head at nozzle is from kinetic energy from jet only Hp = H - 4fLv^2/ D*2g

for part (a) I have calculated Head at nozzle Hp = V^2/(2g ∗ Cv^2) = (24^2) / 2g * (0.97)^2, so Hp = 31.208 m

I have used continuity of flow to arrive at an expression for pipe velocity in terms of jet velocity and knowing that head at nozzle is 4 times frictional head loss in pipe, I have calculated hose diameter.

Now this is where I need to clarify something.
I had been advised a while ago with this question, that the 31.208 m head at nozzle calculated in part (a) should be used in part (b) to work out the pump head term for power, is this correct and why should the same head at nozzle be used ?


So for part (b) instead of using Hp term for head at nozzle, I have used V2^2/ 2g for head at nozzle.
So Pump Head term (Hp) = (Z2 - Z1) + V2^2 / 2g + Frictional Head lost in a hose
From my calculations Pump Head term Hp = 42 m = 3 + 31.2 + 7.8

From there I have calculated power for one hose using Power = Density * Flow * Pump Head Term
Double it for 2 hoses and then divide by efficiency to get total power.

The most important thing here is to establish that using 31.208m head at nozzle calculated in part (a) should be used in part (b) and why ?

RE: Pump power for hoses feeding to nozzles

(OP)
@pierreick

Yes, I have it in my calculations that P1 = P2 = 0 (because inlet of pump and outlet of nozzle are at atmospheric pressure)
Also in part (b) for the bernoulli equation Z1 + P1 / (Rho g) + V1^2 /2g + Hp = Z2 + P2 / (Rho g) + V2^2 /2g + Hf
I have V1 = 0 because you assume the that water has come from a large tank or reservoir where the velocity is 0

Is this how the assumption should be made for this ?

RE: Pump power for hoses feeding to nozzles

Hi,
Please read my comments :
a) P is absolute pressure not gauge pressure , this means P1 and P2 are 1 atmosphere not 0 ! I understand that it will cancel out.
b) Now if F is Darcy friction factor , the head loss is f*L/D *V^2/2g (Darcy Weisbach equation).
c) My understanding, at the discharge of the pump there is a tee with 2 hoses in parallel (45 m length each).Somewhere this should appear in the system description .
d) you should clarify ( a sketch will hep) the location of the reference points (1&2)

My view .
Pierre


RE: Pump power for hoses feeding to nozzles

If there is a nozzle, you should include its head loss.

You can use either psiA or psiG. Same comment applies to absolute temperature. Just be consistant. It is only the differential value that is relevant. (And gas volumes are not involved here.)

RE: Pump power for hoses feeding to nozzles

Somewhere you forgot about the 3m suction lift?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Pump power for hoses feeding to nozzles

Good catch, if so.
I thought if he listed it, he'd include it in the calcs.

RE: Pump power for hoses feeding to nozzles

(OP)
Hi LittleInch, 1503-44 and Pierreick

Actually LittleInch, a little while back, you replied to me and mentioned that I don't need velocity head at pump for part (b), is that because V1 velocity is 0 in a tank or reservoir that feeds the pump ?

You also mentioned use head of 31.2m from part (a) in the calculation for part (b), why do we do this ?
You got the same answer as my 31.2 kW.

Going by what you said the 31.2018 m head available at nozzle = V2^2 / 2g term in bernoulli equation

I have included the 3 m suction lift in part (b)

From my calculations above, using Bernoulli Equation and taking into account head from the pump

Z1 + P1 / (Rho g) + V1^2 /2g + H pump = Z2 + P2 / (Rho g) + V2^2 /2g + Frictional Head loss from hose

Z1 = level of pump = 0 m
Z2 = 3 m (outlet of nozzle)
and P1, P2, V1 = 0.
(Atmospheric pressure is usually 101.325 kPa, but we can put it as 0 in relation to other pressures and cancel out anyway)

Frictional head Loss is 1/4 of Head loss at nozzle

so H pump = 3 + 31.2 + 7.8 = 42 m (this includes 3 m suction lift)

Lastly I worked out power for 2 hoses running in parallel from pump and used efficiency to get 31.2 kW

Here is a diagram showing the Nozzle at the same level as the pump for part (a)


and Here is a diagram showing the Nozzle 3 m above the pump for part (b)


RE: Pump power for hoses feeding to nozzles

Hi ,
For those interested by fluid mechanic simulation , I encourage you to download from internet Pump system improvement modelling tool or PSIM2 , this software is free and you can build your own system using a set of example available .
Good luck
Pierre

RE: Pump power for hoses feeding to nozzles

(OP)
Hi @pierreick

I think you are referring to this https://www.aft.com/psim for the pump software
The one thing about this software is, does it just give the result from input with no calculation workings or does it give calculations as well.


The most important thing here though between you and I, is that you confirm with me that I have calculated this correctly and got the correct answer because I don't want to keep going back and forth forever and not come to a conclusion

I have scanned diagrams diagrams because you wanted clarification on the location of the reference points (1 and 2)

Your other comments
a) P is absolute pressure not gauge pressure , this means P1 and P2 are 1 atmosphere not 0 ! I understand that it will cancel out.
b) Now if F is Darcy friction factor , the head loss is f*L/D *V^2/2g (Darcy Weisbach equation).
c) My understanding, at the discharge of the pump there is a tee with 2 hoses in parallel (45 m length each).Somewhere this should appear in the system description .

Yes there is a tee with 2 hoses in parallel
The Darcy friction factor you are referring to is the American one where f = 4 f
My textbook uses Head Loss = 4 f*L/D * V^2/2g
and yes strictly speaking atmospheric pressure is 101.325 kPa, but a lot of the times with the questions in my book they say it is 0 because it is used as a reference point and if P1 = P2 it always cancels out.

RE: Pump power for hoses feeding to nozzles

(OP)
Hi @pierreick

I tried downloading that pump software from https://www.aft.com/psim , but it said that it was for Windows operating system.
I have a MacBook Air laptop, do you know if they have a Mac version I can download ?

Those calculations you have in excel are your own calculations aren't they.
They are pretty close to what I had in the past on this forum https://engineering.stackexchange.com/questions/31...

RE: Pump power for hoses feeding to nozzles

Hi,
Yes indeed , those are my calculations ! Unfortunately I cannot support you on software , I'm using windows operating system .
BTW you may be able to drop them (AFT) an email and ask for options.
Good luck
Pierre

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members! Already a Member? Login


Resources

Low-Volume Rapid Injection Molding With 3D Printed Molds
Learn methods and guidelines for using stereolithography (SLA) 3D printed molds in the injection molding process to lower costs and lead time. Discover how this hybrid manufacturing process enables on-demand mold fabrication to quickly produce small batches of thermoplastic parts. Download Now
Design for Additive Manufacturing (DfAM)
Examine how the principles of DfAM upend many of the long-standing rules around manufacturability - allowing engineers and designers to place a part’s function at the center of their design considerations. Download Now
Taking Control of Engineering Documents
This ebook covers tips for creating and managing workflows, security best practices and protection of intellectual property, Cloud vs. on-premise software solutions, CAD file management, compliance, and more. Download Now

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close