Pressure drop analysis on water pipe system w/pump

In summary, the conversation revolves around a project on fluids and determining the pressure drop in a complex pipe network. The speaker plans to use the Darcy-Weisbach equation and the continuity equation, but is unsure about the accuracy of their calculations and is seeking guidance. They also mention the importance of using correct units in calculations.
  • #1
The Real Nick
25
0
Hey everyone. Four days ago I didn't know anything about fluids but I started my internship and got assigned a project that is all about fluids! And I need some help, this topic can get really complicated but I'm trying to take it one step at a time.

Problem Description:
There is a pump attached to a tank (really it's two pumps and two tanks but let's keep it simple). The pump can produce 34.5 psi at about 4 gal/min flow rate, and I need to determine the pressure drop at the end of the system. The pipe network is complex, or at least I think so. It is probably 200 feet in total with plenty of bends, elevation changes, fittings(valves and whatnot), hubs where pipes branch off, and so on.

My Solution:
I was going to take the system one piece at a time, calculate the pressure drop after each piece, and then use that as my initial pressure for the next piece working my way around. I am going to do this in Excel so I can keep track of the pressure throughout the system. To get a rough estimate I was first going to look at all the bends, straight pieces, and elevation changes and then start adding more details. To do this I am using the Darcy-Weisbach equation but I'm not sure this is correct because my values are coming out wrong when I test it.

I'm using this currently:
P1 +rho/2(v1-v2)-[(rho*v2^2)/2]*(f*L/D+kb)-(z2-z1)*rho*g = P2

P1 - initial pressure
P2 - final pressure
Rho - density
v1 - initial velocity
v2 - final velocity
f - friction coefficient
D = inner pipe diameter
z1/z2 - heights

I think one of my problems may be with units, I have a ton of variables in Excel so I do need to straighten that out. But also one of my main concerns is that right now I am using 4 gal/min as my flow rate, and using that to calculate velocity, Reynold's number, and the friction coefficient. But shouldn't flow rate be a variable also? Otherwise the velocity will never change. I know in theory the flow rate and velocity are constant, but wouldn't they be changing in a system with friction?

If anyone can give me some guidance on this topic I would be really appreciative. I really want to make sure I do a good job with this internship and would love any help I can get.

Thanks in advance and let me know what questions you may have for me!
 
Engineering news on Phys.org
  • #2
The Real Nick said:
Hey everyone. Four days ago I didn't know anything about fluids but I started my internship and got assigned a project that is all about fluids! And I need some help, this topic can get really complicated but I'm trying to take it one step at a time.

Problem Description:
There is a pump attached to a tank (really it's two pumps and two tanks but let's keep it simple). The pump can produce 34.5 psi at about 4 gal/min flow rate, and I need to determine the pressure drop at the end of the system. The pipe network is complex, or at least I think so. It is probably 200 feet in total with plenty of bends, elevation changes, fittings(valves and whatnot), hubs where pipes branch off, and so on.

My Solution:
I was going to take the system one piece at a time, calculate the pressure drop after each piece, and then use that as my initial pressure for the next piece working my way around. I am going to do this in Excel so I can keep track of the pressure throughout the system. To get a rough estimate I was first going to look at all the bends, straight pieces, and elevation changes and then start adding more details. To do this I am using the Darcy-Weisbach equation but I'm not sure this is correct because my values are coming out wrong when I test it.

I'm using this currently:
P1 +rho/2(v1-v2)-[(rho*v2^2)/2]*(f*L/D+kb)-(z2-z1)*rho*g = P2

P1 - initial pressure
P2 - final pressure
Rho - density
v1 - initial velocity
v2 - final velocity
f - friction coefficient
D = inner pipe diameter
z1/z2 - heights

I think one of my problems may be with units, I have a ton of variables in Excel so I do need to straighten that out. But also one of my main concerns is that right now I am using 4 gal/min as my flow rate, and using that to calculate velocity, Reynold's number, and the friction coefficient. But shouldn't flow rate be a variable also? Otherwise the velocity will never change. I know in theory the flow rate and velocity are constant, but wouldn't they be changing in a system with friction?

If anyone can give me some guidance on this topic I would be really appreciative. I really want to make sure I do a good job with this internship and would love any help I can get.

Thanks in advance and let me know what questions you may have for me!

In addition to the Bernoulli equation, you should have come across the continuity equation in your study of fluids. Basically, the continuity equation says that the amount of fluid passing by one point in a pipe is equal to the amount of fluid passing by a point in the same pipe further down. In other words, between two points in the same pipe, no fluid is created or destroyed:

Q1 = Q2

Since Q = AV, then

A1V1 = A2V2

If your pump puts 4 gal. of water per minute into the piping system, then 4 gal. per minute is going to come out of the system somewhere. The velocity of the water will change depending on the pipe size, if more than one size is used in the system.

You didn't list the units of any of the quantities you are using to calculate the pressure in the system. You didn't explain why your calculations are not coming out the way you expect; the problem could be traced to a mismatch in the units.

If you care to post your calcs, we can see if the problem is readily apparent.
 
  • #3
SteamKing said:
In addition to the Bernoulli equation, you should have come across the continuity equation in your study of fluids. Basically, the continuity equation says that the amount of fluid passing by one point in a pipe is equal to the amount of fluid passing by a point in the same pipe further down. In other words, between two points in the same pipe, no fluid is created or destroyed:

Q1 = Q2

Since Q = AV, then

A1V1 = A2V2

If your pump puts 4 gal. of water per minute into the piping system, then 4 gal. per minute is going to come out of the system somewhere. The velocity of the water will change depending on the pipe size, if more than one size is used in the system.

You didn't list the units of any of the quantities you are using to calculate the pressure in the system. You didn't explain why your calculations are not coming out the way you expect; the problem could be traced to a mismatch in the units.

If you care to post your calcs, we can see if the problem is readily apparent.

Hey Steamking, thanks for the reply! Your name is reassuring. Haha. I guess I was really tired when I made this post last night; sorry I didnt include units. Everything is in inches, and although I don't have the numbers or specific units infront of me currently, I will post them in a few hours when I can get at them again.

Maybe I could start by listing the units that I am using, in inches, with the associated conversation factor I used.

Also, I did use the Q=VA to calculate the velocity based on my flow rate. So, theoretically, the speed of the water will stay constant for a pipe of the same cross sectional area? I suppose where I was getting caught up was by looking at a coworkers old spreadsheet for an oxygen system. In his case the flow rate and velocity were changing constantly, probably due to many junctions and what not.

So, do you think it's fine to leave my flow rate as a constant? Maybe I should tackle one thing at a time, but I'm just thinking ahead to when the lines start to split off. I've read that you can treat the flow of water similarly to current in a circuit using Kirchoffs rules, is this correct?

Sorry that I have so many questions. I always need to get concepts Down fully before I can really understand and feel confident in what I'm doing, and I have a rushed background of studying on this topic.

Also, I am happy to post the calculations when I am able to look at them again.

Again, thanks for the help!
Nick
 
  • #4
Welcome to PF. It looks to me like you are approaching the task in the right way and have correctly identified the problem of flow (pump performance) being variable, not constant. So here's what I'd do:

Set up the spreadsheet such that flow is a variable with a single cell reference, so you can change it easily. Then adjust the flow until the flow and resulting pressure match a point on the pump curve.
 
  • #5
russ_watters said:
Welcome to PF. It looks to me like you are approaching the task in the right way and have correctly identified the problem of flow (pump performance) being variable, not constant. So here's what I'd do:

Set up the spreadsheet such that flow is a variable with a single cell reference, so you can change it easily. Then adjust the flow until the flow and resulting pressure match a point on the pump curve.

Hey Russ, thanks for your reply and vote of confidence! I am not quite sure what you mean exactly... wait I may have just Figured it out. Do you mean change the value of flow until the flow and pressure match a point on the flow rate vs pressure Graph associated with the particular pump I have? That sounds like a good idea. Would I do this assuming basically no bends or length in the darcy-weisbach equation? Or Should I just use Q=VA?

Thanks again!
 
  • #6
The Real Nick said:
Also, I did use the Q=VA to calculate the velocity based on my flow rate. So, theoretically, the speed of the water will stay constant for a pipe of the same cross sectional area? I suppose where I was getting caught up was by looking at a coworkers old spreadsheet for an oxygen system. In his case the flow rate and velocity were changing constantly, probably due to many junctions and what not.

You can't just pick up any old pressure drop calculation and assume that it applies to your situation.

Gases like oxygen don't always behave the same as liquids like water. Gases can become compressed as they flow, which is why the properties like velocity may be changing inside a pipe. Liquids like water can be treated as essentially incompressible, hence their velocity will be a constant as long as the flow rate and the pipe diameter are constant.

So, do you think it's fine to leave my flow rate as a constant? Maybe I should tackle one thing at a time, but I'm just thinking ahead to when the lines start to split off. I've read that you can treat the flow of water similarly to current in a circuit using Kirchoffs rules, is this correct?

Pipe networks can be analyzed in a fashion similar to analyzing an electrical network, but there is one important difference: unlike an electrical network where the voltage drop is proportional to the current and the resistance in a branch, in a fluid network, the pressure drop is proportional to the fluid velocity squared, so this complicates things a bit. The sum of the flows into a branch must equal the sum of the flows out, so this fact is equivalent to what happens in an electrical circuit.

Sorry that I have so many questions. I always need to get concepts Down fully before I can really understand and feel confident in what I'm doing, and I have a rushed background of studying on this topic.

Also, I am happy to post the calculations when I am able to look at them again.

Again, thanks for the help!
Nick

Before tackling a complex flow network, I recommend first that you get the bugs worked out of calculating flow thru a single pipe.
 
  • #7
The Real Nick said:
Hey Russ, thanks for your reply and vote of confidence! I am not quite sure what you mean exactly... wait I may have just Figured it out. Do you mean change the value of flow until the flow and pressure match a point on the flow rate vs pressure Graph associated with the particular pump I have? That sounds like a good idea. Would I do this assuming basically no bends or length in the darcy-weisbach equation? Or Should I just use Q=VA?

Thanks again!
Correct method, but instead of assuming no bends, you should be able to find tables for "equivalent length" of the fittings/elbows.
 
  • #8
Ask around at work for a copy of "the Crane book." Officially it is called Crane Tech Paper No. 410, Flow of Fluids through Valves, Fittings, and Pipe. It is an orange book less than 1/2 inch thick and it has nearly everything you need to know. If you spend your entire internship reading and understanding that one book you will have had a very worthwhile summer.

When I work away from the office I take my calculator and my Crane 410.
 
  • #9
SteamKing said:
You can't just pick up any old pressure drop calculation and assume that it applies to your situation.

Gases like oxygen don't always behave the same as liquids like water. Gases can become compressed as they flow, which is why the properties like velocity may be changing inside a pipe. Liquids like water can be treated as essentially incompressible, hence their velocity will be a constant as long as the flow rate and the pipe diameter are constant.
Pipe networks can be analyzed in a fashion similar to analyzing an electrical network, but there is one important difference: unlike an electrical network where the voltage drop is proportional to the current and the resistance in a branch, in a fluid network, the pressure drop is proportional to the fluid velocity squared, so this complicates things a bit. The sum of the flows into a branch must equal the sum of the flows out, so this fact is equivalent to what happens in an electrical circuit.
Before tackling a complex flow network, I recommend first that you get the bugs worked out of calculating flow thru a single pipe.

Yeah, I came to the realization also that the oxygen system would behave Differently, I shouldn't have gotten so caught up on that aspect of the problem. I need to get my equation working in Excel and as long as there is nothing seriously wrong with the equation I posted I guess the issue may be coming down to units. I am in a conference all day today so I probably won't get back to the calculations until Tuesday. I will try and get the spreadsheet working with a simple system first and then move on. Right now My pressure drop is way too high, which is mainly driven by the velocity sqaured term. I will try and get that worked out soon. Thanks For your help!
 
  • #10
russ_watters said:
Correct method, but instead of assuming no bends, you should be able to find tables for "equivalent length" of the fittings/elbows.
Ive had some trouble tracking down kb values For bends, but ill give this a try, this seems like a really good way to simplify things.
 
  • #11
gmax137 said:
Ask around at work for a copy of "the Crane book." Officially it is called Crane Tech Paper No. 410, Flow of Fluids through Valves, Fittings, and Pipe. It is an orange book less than 1/2 inch thick and it has nearly everything you need to know. If you spend your entire internship reading and understanding that one book you will have had a very worthwhile summer.

When I work away from the office I take my calculator and my Crane 410.
Thank you! I will see if I can get my hands.on this Book. I've been looking for something like this.
 
  • #12
Ok let me run the units by you guys. Here's the equation:
P1+(ρ/2)(v1-v2)-(ρv22/2)(ƒL/D+Kb)-(z2-z1)ρg=P2
P1/P2 - lb/in2
ρ - lb/in3 (0.0361 for water @ °60 F)
v1/v2 - in/s
f - Unitless. Calculated via: 0.316/Re1/4
Re - Unitless. Calc. via: vD/(kinematic viscosity)
Kinematic Viscosity - in2/s (0.001515 @ 60 F)
L - in
D - in
z1/z2 - in
g - 32.2 ft/s (Well I guess I found an error!)
Except this doesn't change my calculation because I didn't have any height change. Fooey.
velocity I got from Q/A which is
Q - in3/s (15.437 in^3/s for 4 gal/min)
A - in2

How does this look to you guys? Anything else I should be looking out for? Thanks again!
 
  • #13
The Real Nick said:
Ok let me run the units by you guys. Here's the equation:
P1+(ρ/2)(v1-v2)-(ρv22/2)(ƒL/D+Kb)-(z2-z1)ρg=P2
P1/P2 - lb/in2
ρ - lb/in3 (0.0361 for water @ °60 F)
v1/v2 - in/s
f - Unitless. Calculated via: 0.316/Re1/4
Re - Unitless. Calc. via: vD/(kinematic viscosity)
Kinematic Viscosity - in2/s (0.001515 @ 60 F)
L - in
D - in
z1/z2 - in
g - 32.2 ft/s (Well I guess I found an error!)
Except this doesn't change my calculation because I didn't have any height change. Fooey.
velocity I got from Q/A which is
Q - in3/s (15.437 in^3/s for 4 gal/min)
A - in2

How does this look to you guys? Anything else I should be looking out for? Thanks again!

It's not clear what values of L and D you are using for the pipe.
 
  • #14
SteamKing said:
It's not clear what values of L and D you are using for the pipe.
Well the L values will be whatever the length of straight pipe I have. for the D term, its a 0.93 inch inner diameter pipe.
 
  • #15
BTW, g is the acceleration due to gravity, and its value is 32.2 ft/s2 (notice the units), or about 386.4 in/s2.

My copy of Crane TP-410 lists the ID of a 1" Schedule 80 steel pipe as 0.957 in.

For 1" clean carbon steel pipe, fT = 0.023, according to Crane, and is constant for fully turbulent flow, which for this size pipe is any Re > 700,000.
(If you are not familiar with the Moody diagram, this is where it comes in handy). The friction factor for pipe depends on the material, the roughness, and the size.

The kinematic viscosity of fresh water at 60 °F is 1.20828×10-5 ft2/s according to some tables I have for the properties of fresh and salt water.
This kinematic viscosity would also be equal to 0.00174 in2/s.
 
  • #16
SteamKing said:
BTW, g is the acceleration due to gravity, and its value is 32.2 ft/s2 (notice the units), or about 386.4 in/s2.

My copy of Crane TP-410 lists the ID of a 1" Schedule 80 steel pipe as 0.957 in.

For 1" clean carbon steel pipe, fT = 0.023, according to Crane, and is constant for fully turbulent flow, which for this size pipe is any Re > 700,000.
(If you are not familiar with the Moody diagram, this is where it comes in handy). The friction factor for pipe depends on the material, the roughness, and the size.

The kinematic viscosity of fresh water at 60 °F is 1.20828×10-5 ft2/s according to some tables I have for the properties of fresh and salt water.
This kinematic viscosity would also be equal to 0.00174 in2/s.
It looks like the values we have for acceleration and kinematic viscosity are the same, I think I just listed the value of kinematic viscosity at 70 F instead of 60. My 60 F value is the same as yours. The f value that I calculated was 0.024 or so, so basically the same as yours. The only thing though is my Reynold's number is no where near 700,000... I think it was 35,000. Also, the OD is 1" and the wall thickness is 0.035" so the inner diameter of this piping is 0.93". I think they're called CRS pipes (corrosion resistant steel). Which is basically stainless steel. I need to get myself a copy of the ellusive Crane TP-410...
 
  • #17
Well I found a PDF of the TP-410 (I am going to buy one though) and after about 2 minutes of reading it I think I found my problem. It says the Darcy-Weisbach equation in it's standard form is used to calculate feet of fluid, which you can then convert to inches, and then divide by 27.8 to get psi. I forget what the 27.8 conversion factor is, it's like the amount of inches water can raise per 1 psi or something. So I was assuming that through that calculation I was getting PSI out as an answer, but really it was feet of fluid. I don't have the spreadsheet with me (has to stay at work) otherwise I would test it out. But just by thinking about it out, it still seems like my answer would be off.

I got a pressure drop of something like 30 psi, or what I thought was psi, so that would be 30 feet of fluid, or 360 inches, and then 13.33 psi or so. That's still way too much for a single 90 degree bend.
 
  • #18
The Real Nick said:
It looks like the values we have for acceleration and kinematic viscosity are the same, I think I just listed the value of kinematic viscosity at 70 F instead of 60. My 60 F value is the same as yours. The f value that I calculated was 0.024 or so, so basically the same as yours. The only thing though is my Reynold's number is no where near 700,000... I think it was 35,000. Also, the OD is 1" and the wall thickness is 0.035" so the inner diameter of this piping is 0.93". I think they're called CRS pipes (corrosion resistant steel). Which is basically stainless steel. I need to get myself a copy of the ellusive Crane TP-410...

To be clear, I didn't say the Reynolds number for your flow was 700,000, just that for fully turbulent flow inside a 1" pipe, the Reynolds number had to be at least 700,000.

Using a flow rate of 4 GPM and your velocity of 15.4 fps, the actual Reynolds No. for your flow is about 8200, which brings up another issue: the type of flow regime.

In fluid flow, there is laminar flow, fully turbulent flow, and a transition zone between the two. In each flow regime, there is a different formula for calculating the friction factor. The friction factor formulas are listed below:

Friction-Factor-Equations.jpg

The absolute roughness values for several different pipe materials are given in the table below:
8035245f7461e889d168d525c1443dcdee1eeb93_large.jpg

This article discusses the various friction factor formulae in more detail:

http://en.wikipedia.org/wiki/Darcy_friction_factor_formulae

The equation for friction factor calculation in the transition zone is known as the Colebrook equation. In its full glory, the friction factor f can be calculated only by using an iterative process. The article above gives several approximate versions of the Colebrook equation which avoid using iteration to calculate f.

If you are basing your pressure drop calculations on a friction factor which is valid only in the turbulent zone, but you have laminar flow or flow in the transition zone, then your pressure drop calculations will be in error.

The equation you have been using to calculate the friction factor is based on pipe materials which have no roughness whatsoever, i.e., a perfectly smooth material, which typically doesn't exist. The friction factors obtained from such an equation must therefore be considered approximations to the friction factor inside an actual pipe. In any event, this formula is valid only for flows with Re < 100,000.

For pipes which are not smooth, once fully turbulent flow develops, the friction factor becomes a constant, independent of the Reynolds number of the flow.
 
  • #19
The Real Nick said:
Well I found a PDF of the TP-410 (I am going to buy one though) and after about 2 minutes of reading it I think I found my problem. It says the Darcy-Weisbach equation in it's standard form is used to calculate feet of fluid, which you can then convert to inches, and then divide by 27.8 to get psi. I forget what the 27.8 conversion factor is, it's like the amount of inches water can raise per 1 psi or something. So I was assuming that through that calculation I was getting PSI out as an answer, but really it was feet of fluid. I don't have the spreadsheet with me (has to stay at work) otherwise I would test it out. But just by thinking about it out, it still seems like my answer would be off.

I got a pressure drop of something like 30 psi, or what I thought was psi, so that would be 30 feet of fluid, or 360 inches, and then 13.33 psi or so. That's still way too much for a single 90 degree bend.

IDK where this factor of 27.8 come from, but you can convert from feet of water to psi by a very simple calculation.

1 cubic foot of fresh water weighs 62.4 lbs., which implies that the pressure on the bottom of this 1-foot cube of water is 62.4 lbs. per square foot.

A head of 1 foot of water is also equivalent to a pressure of 62.4 lbs / ft2, which is equal to
(62.4 lbs. / ft2) / (144 in2 / ft2) = 0.433 psi.

To find the pressure in psi given so many feet of water head H, multiply the head value H by 0.433.
 
  • #20
SteamKing said:
To be clear, I didn't say the Reynolds number for your flow was 700,000, just that for fully turbulent flow inside a 1" pipe, the Reynolds number had to be at least 700,000.

Using a flow rate of 4 GPM and your velocity of 15.4 fps, the actual Reynolds No. for your flow is about 8200, which brings up another issue: the type of flow regime.

In fluid flow, there is laminar flow, fully turbulent flow, and a transition zone between the two. In each flow regime, there is a different formula for calculating the friction factor. The friction factor formulas are listed below:

Friction-Factor-Equations.jpg

The absolute roughness values for several different pipe materials are given in the table below:
This article discusses the various friction factor formulae in more detail:

http://en.wikipedia.org/wiki/Darcy_friction_factor_formulae

The equation for friction factor calculation in the transition zone is known as the Colebrook equation. In its full glory, the friction factor f can be calculated only by using an iterative process. The article above gives several approximate versions of the Colebrook equation which avoid using iteration to calculate f.

If you are basing your pressure drop calculations on a friction factor which is valid only in the turbulent zone, but you have laminar flow or flow in the transition zone, then your pressure drop calculations will be in error.

The equation you have been using to calculate the friction factor is based on pipe materials which have no roughness whatsoever, i.e., a perfectly smooth material, which typically doesn't exist. The friction factors obtained from such an equation must therefore be considered approximations to the friction factor inside an actual pipe. In any event, this formula is valid only for flows with Re < 100,000.

For pipes which are not smooth, once fully turbulent flow develops, the friction factor becomes a constant, independent of the Reynolds number of the flow.
Wow, thank you so much for posting all that great information. One thing though, the flow that I calculated in in3/s is 15.4, not the velocity. The velocity was something like 72 in/sec. But after just thinking about my math on that 72 in/sec seems like way too much... maybe that's where part of my error was coming from.

Anyways, so if you calculate the Re # with 15.4 in^3/sec being the flow rate not the velocity it should be well above the transition region, but also well below the fully turbulent region.That's why I just used the smooth pipe turbulent flow equation. I will look at the TP-410 document and the link you posted tomorrow, I'm super tired. I really do appreciate all the help you have given me on this topic though, I'm really glad I came to this forum for help.
 
  • #21
The Real Nick said:
Wow, thank you so much for posting all that great information. One thing though, the flow that I calculated in in3/s is 15.4, not the velocity. The velocity was something like 72 in/sec. But after just thinking about my math on that 72 in/sec seems like way too much... maybe that's where part of my error was coming from.

You are right. I made a mistake in calculating the flow velocity.

Q = (4 gal/min * 231 in3 / gal) / (60 sec / min.) = 15.4 in3 / sec.

ID = 0.93 in

A = π × ID2 / 4 = 0.679 in2

V = Q / A = (15.4 in3/s) / 0.679 in2 = 22.67 in / sec. = 1.889 fps [EDITED]

Re = V × ID / ν = 22.67 in / sec × 0.93 in / 0.00174 in2/sec

Re = 12,117 which is still in the transition zone for flow

The absolute roughness ε for commercial steel pipe is 0.00015 feet = 0.0018 in

The relative roughness ε / ID = 0.0018 / 0.93 = 0.0019

Using the Haaland equation to estimate f

1/√f = -1.8 log [(0.0019 / 3.7)1.11 + 6.9 / 12,117]

1/√f = 5.577

f = 0.0322

Plugging this value of f into the Colebrook equation to check:

f = 0.0324

which should be the friction factor to use to calculate pressure drop.

Calculating head loss for 100 feet of pipe:

hL = f (L/ID)×V2 / 2g

hL = 0.0324 (1200 / 0.93) 1.8892 / (2 ×32.2)

hL = 2.32 ft of water = 1.0 psi for 4 GPM flow rate thru 100 feet of pipe.
 
Last edited:
  • #22
SteamKing said:
You are right. I made a mistake in calculating the flow velocity.

Q = (4 gal/min * 231 in3 / gal) / (60 sec / min.) = 15.4 in3 / sec.

ID = 0.93 in

A = π × ID2 / 4 = 0.679 in2

V = Q / A = 15.4 in3 / 0.679 in2 = 22.67 in / sec. = 1.889 fps

Re = V × ID / ν = 22.67 in / sec × 0.93 in / 0.00174 in2/sec

Re = 12,117 which is still in the transition zone for flow

The absolute roughness ε for commercial steel pipe is 0.00015 feet = 0.0018 in

The relative roughness ε / ID = 0.0018 / 0.93 = 0.0019

Using the Haaland equation to estimate f

1/√f = -1.8 log [(0.0019 / 3.7)1.11 + 6.9 / 12,117]

1/√f = 5.577

f = 0.0322

Plugging this value of f into the Colebrook equation to check:

f = 0.0324

which should be the friction factor to use to calculate pressure drop.

Calculating head loss for 100 feet of pipe:

hL = f (L/ID)×V2 / 2g

hL = 0.0324 (1200 / 0.93) 1.8892 / (2 ×32.2)

hL = 2.32 ft of water = 1.0 psi for 4 GPM flow rate thru 100 feet of pipe.
Awesome! Thanks man. The only number that was different for me was velocity, but if I do a hand calc. I get the same value as you do, so I will check my velocity calculation. 1 psi for a 100 foot pipe seems very reasonable to me.

One thing that I still have a question about though is the Reynolds number and flow characterization. I read in a textbook that Re in and around the 2100 value are considered transition, but numbers above 2100 could be considered turbulent. I haven't really looked too hard at a Moody diagram and maybe it says otherwise there, but that is where I was getting my idea of the flow being turbulent and not transitioning. Either way though, it seems like the way you calculated everything is pretty straight forward so I may just stick with that. Thanks a ton!
 
  • #23
The Real Nick said:
One thing that I still have a question about though is the Reynolds number and flow characterization. I read in a textbook that Re in and around the 2100 value are considered transition, but numbers above 2100 could be considered turbulent. I haven't really looked too hard at a Moody diagram and maybe it says otherwise there, but that is where I was getting my idea of the flow being turbulent and not transitioning. Either way though, it seems like the way you calculated everything is pretty straight forward so I may just stick with that. Thanks a ton!

There are various Reynolds numbers quoted for the end of laminar flow and the beginning of turbulent flow. Typical Reynolds numbers in the range of 2100-2300 are common. Since turbulence must be studied experimentally, it is difficult to say when laminar flow ceases and fully turbulent flow is developed, since you can't easily do a calculation and say with certainty the flow is fully turbulent here but not there.

The modified Moody diagrams in Crane TP-410 on pages A-24 and A-25 show how the different zones of flow overlap
 
  • #24
SteamKing said:
There are various Reynolds numbers quoted for the end of laminar flow and the beginning of turbulent flow. Typical Reynolds numbers in the range of 2100-2300 are common. Since turbulence must be studied experimentally, it is difficult to say when laminar flow ceases and fully turbulent flow is developed, since you can't easily do a calculation and say with certainty the flow is fully turbulent here but not there.

The modified Moody diagrams in Crane TP-410 on pages A-24 and A-25 show how the different zones of flow overlap
Thanks for all the help SteamKing. I think I have the Excel spreadsheet working now after looking over everything you had helped me with. I did some test calculations like the one you posted of 100 feet of steel pipe and got the same result, along with a few other calculations I had done by hand.

I am having trouble figuring out what some of my Kb values should be though. Like if I have a tee fitting that's not threaded and comes back at an angle (so not a 90 degree tee). Things like that seem to be kind of 'custom' and left out of the tables for standard K values.

Also, if I am calculating the pressure drop from a pipe that goes straight up, do I calculate the pressure drop from the length of the pipe also? Or just from the height change? I want to make sure I'm not doubling up on the drop.

Oh and one last question (for now anyways) if I have a set of parallel pipes that come from one pipe, deviate into two paths, and then come back into one pipe, is it correct to say the pressure drop will be equal in both paths even though one path has bends and elevation increase/decrease and the other path is straight?

Thanks again!
 
Last edited:
  • #25
The Real Nick said:
Thanks for all the help SteamKing. I think I have the Excel spreadsheet working now after looking over everything you had helped me with. I did some test calculations like the one you posted of 100 feet of steel pipe and got the same result, along with a few other calculations I had done by hand.

I am having trouble figuring out what some of my Kb values should be though. Like if I have a tee fitting that's not threaded and comes back at an angle (so not a 90 degree tee). Things like that seem to be kind of 'custom' and left out of the tables for standard K values.

This sounds like a Y-strainer. The pressure drop will depend on what's inside. In large Y-strainers, there is a basket-type filter which collects debris to keep it from flowing downstream. The PD values for these units depend on the strainer used and are usually furnished by the manufacturer.

Here is a sample of some Y-strainers:

http://industrialstrainer.com/eaton-hayward-strainers/y-strainers/

Also, if I am calculating the pressure drop from a pipe that goes straight up, do I calculate the pressure drop from the length of the pipe also? Or just from the height change? I want to make sure I'm not doubling up on the drop.

The PD in this section will include the minor fitting losses (which also includes the friction due to the length of pipe) and the change in elevation.

Oh and one last question (for now anyways) if I have a set of parallel pipes that come from one pipe, deviate into two paths, and then come back into one pipe, is it correct to say the pressure drop will be equal in both paths even though one path has bends and elevation increase/decrease and the other path is straight?

In general, the flow going into each branch from the supply line will apportion itself such that the PD in each branch will be the same, so that when the branches rejoin into a single line, you won't see the flow from one branch trying to back up into the other, i.e., you won't see the branches acting as a loop for the flow. This is usually where after adding up the minor losses from the fittings and accounting for the lengths of straight pipe, plus any changes in elevation, you must iterate to find the flow in each branch which gives the same PD, making sure that the combined flow in equals the combined flow out.
 
  • #26
SteamKing said:
This sounds like a Y-strainer. The pressure drop will depend on what's inside. In large Y-strainers, there is a basket-type filter which collects debris to keep it from flowing downstream. The PD values for these units depend on the strainer used and are usually furnished by the manufacturer.

Here is a sample of some Y-strainers:

http://industrialstrainer.com/eaton-hayward-strainers/y-strainers/
The PD in this section will include the minor fitting losses (which also includes the friction due to the length of pipe) and the change in elevation.
In general, the flow going into each branch from the supply line will apportion itself such that the PD in each branch will be the same, so that when the branches rejoin into a single line, you won't see the flow from one branch trying to back up into the other, i.e., you won't see the branches acting as a loop for the flow. This is usually where after adding up the minor losses from the fittings and accounting for the lengths of straight pipe, plus any changes in elevation, you must iterate to find the flow in each branch which gives the same PD, making sure that the combined flow in equals the combined flow out.
Thanks your replying again. The fitting I am thinking of is called a weldment I believe. I guess I am really dealing with hose in some sections and tubes in other sections, so the fittings aren't typical threaded type pipe fittings but instead are more like hose connectors and couplings. I will have to see if the manufacturer of these fittings has some values for pressure drop and friction coefficients. The hose is a really nice flexible hose wrapped in braided stainless steel and what not, but I will probably have to try and get absolute roughness values for this material because it's not steel. Though, at a certain point in the system the hose transitions to steel tubes. There is a bunch of seemingly custom looking fittings that I'm sure I'll have to look up Friction values for.

My current challenge will be to try and work out how I am going to make my excel spreadsheet work with the pull-outs that pull flow for short periods of time. Another question on this topic: if a hose goes straight up to feed an appliance that is closed most of the time, do I need to calculate the pressure drop going up there when the valve is closed? Or only when it's open? Thanks again! I can't believe I knew nothing about fluids last week!
 
  • #27
The Real Nick said:
Thanks your replying again. The fitting I am thinking of is called a weldment I believe. I guess I am really dealing with hose in some sections and tubes in other sections, so the fittings aren't typical threaded type pipe fittings but instead are more like hose connectors and couplings. I will have to see if the manufacturer of these fittings has some values for pressure drop and friction coefficients. The hose is a really nice flexible hose wrapped in braided stainless steel and what not, but I will probably have to try and get absolute roughness values for this material because it's not steel. Though, at a certain point in the system the hose transitions to steel tubes. There is a bunch of seemingly custom looking fittings that I'm sure I'll have to look up Friction values for.
You know more about the construction details of your system than I do.

For metal piping systems, there are three methods of assembling all the pieces: screwing, welding, or by bolting with flanges. In my experience, a "weldment" refers to anything, be it plates, shapes, or pipe and fittings, which is assembled from from smaller pieces joined together by welding.

If your system is composed of other pieces than metal pipe and fittings, then you'll probably have to get PD data from manufacturers. There may be some data floating around the web, but you'll need to search for it.

My current challenge will be to try and work out how I am going to make my excel spreadsheet work with the pull-outs that pull flow for short periods of time. Another question on this topic: if a hose goes straight up to feed an appliance that is closed most of the time, do I need to calculate the pressure drop going up there when the valve is closed? Or only when it's open? Thanks again! I can't believe I knew nothing about fluids last week!

It sounds like you are planning a more ambitious modeling project than calculating a few pressure drops. When the model gets complicated like this, designing and constructing it with spreadsheets may get unwieldy. It may be time to consider using software designed for such things. There are a couple of candidates to consider:

http://www.cadreanalytic.com/cadreflo.htm

http://eng-software.com/products/pipe-flo/

http://www.abzinc.com/

There may be other solutions, but these are a little familiar to me. You'll have to contact each for pricing their respective packages.

In general, these packages allow the user to construct a piping system interactively, and the system may contain active elements, like valves, which can be opened or closed, to allow the user to recalculate how the flow in the system responds to changes in its configuration. You can also input pump data and performance curves to see if the pump(s) will provide the flow performance desired for the system as a whole.
 
  • #28
SteamKing said:
You know more about the construction details of your system than I do.

For metal piping systems, there are three methods of assembling all the pieces: screwing, welding, or by bolting with flanges. In my experience, a "weldment" refers to anything, be it plates, shapes, or pipe and fittings, which is assembled from from smaller pieces joined together by welding.

If your system is composed of other pieces than metal pipe and fittings, then you'll probably have to get PD data from manufacturers. There may be some data floating around the web, but you'll need to search for it.
It sounds like you are planning a more ambitious modeling project than calculating a few pressure drops. When the model gets complicated like this, designing and constructing it with spreadsheets may get unwieldy. It may be time to consider using software designed for such things. There are a couple of candidates to consider:

http://www.cadreanalytic.com/cadreflo.htm

http://eng-software.com/products/pipe-flo/

http://www.abzinc.com/

There may be other solutions, but these are a little familiar to me. You'll have to contact each for pricing their respective packages.

In general, these packages allow the user to construct a piping system interactively, and the system may contain active elements, like valves, which can be opened or closed, to allow the user to recalculate how the flow in the system responds to changes in its configuration. You can also input pump data and performance curves to see if the pump(s) will provide the flow performance desired for the system as a whole.

Yeah, this is definitely a complicated project. As an intern they mainly want me to get a pretty decent, "rough" estimate. But I'm trying to make it as accurate as possible and also learn something along the way. I am working for a rather large company and do have access to some pipe-system analysis software such as Easy 5. My team lead basically told me that learning to use those programs may take longer than just doing it with excel and using the cad models for all my measurements. It could probably go either way, and in truth I may end up doing it both ways, or if the excel option doesn't seem to be panning out I may have to switch. It seems to be working pretty well so far as I am slowly but surely ironing out some of the details.

I am definitely going to have to look up the specs. for all the fittings, adapters, etc, that are in the system. One thing I am trying to figure out currently is the Kb value for bends that aren't 90 degrees, mainly for 45 degree bends. I've found some "equivalent lengths" for 45 degree bends that I am using temporarily but I noticed that when I convert these to a Kb value it ends up being slightly higher than most 90 degree bends, depending on the r/D ratio. Something like 0.28 per 45 degree bend.

One thing I am trying to work through conceptually is when one section of the system branches off and feeds an appliance, such as a toilet, but is only flowing water to that toilet very temporarily, do I still calculate the pressure loss up to the valve even when it's closed? Or only when the valve is open/toilet is being flushed? My hunch was to say only when it was open, so currently I have an if/then statement that only fills in the values for Length, Height Change, Bends, and Fittings when I enter a flow rate through that branch.

Let me know what you think! Thanks again.
 
  • #29
If there is no flow in a branch, then it's easy: the velocity = 0, and the PD = 0.

What often happens in cases of modeling a complex system is that the worst-case flow scenario, the one with the highest overall PD, is chosen for analysis.

In the case of fitting K value data, these will only be applicable for fully-turbulent flows. If you have a flow in the transition zone, or even the laminar zone, the K-values shown in handbooks like Crane will not give accurate PD results when used for Reynold's No. flows which aren't in the fully turbulent zone. The people who wrote the Crane TP-410 thought they could get away by assuming that fitting K values could be used regardless of the flow regime in which a piping system operated, but later experiments showed that using turbulent flow K values in non-turbulent situations gave inaccurate results.

Chemical engineers, who design piping systems for plants which use a variety of fluids other than water or petroleum, needed a more accurate method for calculating pressure drops when the flow regime wasn't fully turbulent. Based on experiments, it seems they have decided to return to using equivalent-lengths to calculate PDs for fittings and straight pipe.

Also, other people have taken to using L/D values for fitting PD calculations even in the turbulent flow region. The Hydraulic Institute (HI) publishes a book similar to the Crane TP-410 which contains losses for various types of fittings. The HI values are given not as a single number of equivalent lengths for a given fitting, but instead are presented in chart form. As these charts are plotted in log-log form, the K value for a particular fitting, especially elbows and other bends, is actually obtained from a curve of K-value versus fitting size (diameter).

The link below is to the Engineering Data Book published by the HI:

http://estore.pumps.org/Guidebooks/EngData.aspx

Handbooks like Ludwig's "Applied Process Design for Chemical and Petrochemical Plants" contain extensive chapters on calculating PD for liquid and gas flows, complete with listing of PD data for various fittings. Well worth your time searching out this information of the web (Google "Ludwig's Applied Process Design"). The latest version of this book is, I believe, the fourth edition, but the earlier editions, though re-arranged some, still contain valuable information about analyzing PD in a variety of piping systems.

http://www.aiche.org/sites/default/files/cep/20071249.pdf
 
  • #30
SteamKing said:
If there is no flow in a branch, then it's easy: the velocity = 0, and the PD = 0.

What often happens in cases of modeling a complex system is that the worst-case flow scenario, the one with the highest overall PD, is chosen for analysis.

In the case of fitting K value data, these will only be applicable for fully-turbulent flows. If you have a flow in the transition zone, or even the laminar zone, the K-values shown in handbooks like Crane will not give accurate PD results when used for Reynold's No. flows which aren't in the fully turbulent zone. The people who wrote the Crane TP-410 thought they could get away by assuming that fitting K values could be used regardless of the flow regime in which a piping system operated, but later experiments showed that using turbulent flow K values in non-turbulent situations gave inaccurate results.

Chemical engineers, who design piping systems for plants which use a variety of fluids other than water or petroleum, needed a more accurate method for calculating pressure drops when the flow regime wasn't fully turbulent. Based on experiments, it seems they have decided to return to using equivalent-lengths to calculate PDs for fittings and straight pipe.

Also, other people have taken to using L/D values for fitting PD calculations even in the turbulent flow region. The Hydraulic Institute (HI) publishes a book similar to the Crane TP-410 which contains losses for various types of fittings. The HI values are given not as a single number of equivalent lengths for a given fitting, but instead are presented in chart form. As these charts are plotted in log-log form, the K value for a particular fitting, especially elbows and other bends, is actually obtained from a curve of K-value versus fitting size (diameter).

The link below is to the Engineering Data Book published by the HI:

http://estore.pumps.org/Guidebooks/EngData.aspx

Handbooks like Ludwig's "Applied Process Design for Chemical and Petrochemical Plants" contain extensive chapters on calculating PD for liquid and gas flows, complete with listing of PD data for various fittings. Well worth your time searching out this information of the web (Google "Ludwig's Applied Process Design"). The latest version of this book is, I believe, the fourth edition, but the earlier editions, though re-arranged some, still contain valuable information about analyzing PD in a variety of piping systems.

http://www.aiche.org/sites/default/files/cep/20071249.pdf
Thanks for all your help! I did get the project finished by the way. I figured I should jump back on here and give you an update. I ended up doing it all in excel, and it may be a little dumbed down but everything behaves as it should when I draw flow from different areas in the line. Thanks again!
 

1. What is pressure drop analysis on water pipe system with pump?

Pressure drop analysis on water pipe system with pump is a method used to determine the amount of pressure loss that occurs in a water pipe system due to friction and other factors. It involves calculating the difference in pressure between two points in the system, typically the inlet and outlet of a pump, and analyzing the causes of pressure loss.

2. Why is pressure drop analysis important in water pipe systems?

Pressure drop analysis is important in water pipe systems because it helps to ensure that the system is functioning efficiently and effectively. By identifying areas of high pressure loss, adjustments can be made to improve the overall performance of the system and reduce energy costs.

3. What factors can contribute to pressure drop in a water pipe system?

There are several factors that can contribute to pressure drop in a water pipe system, including pipe diameter, pipe length, flow rate, fluid viscosity, and fittings such as elbows and valves. These factors can cause friction, which results in a decrease in pressure as the fluid flows through the system.

4. How is pressure drop calculated in a water pipe system?

Pressure drop is calculated using the Darcy-Weisbach equation, which takes into account the factors mentioned above, such as pipe diameter, length, and flow rate. This equation can be solved using various methods, including hand calculations or computer software.

5. How can pressure drop be reduced in a water pipe system?

There are several ways to reduce pressure drop in a water pipe system, such as increasing the pipe diameter, reducing the length of the pipe, or using smoother pipes with less friction. Additionally, minimizing the number of fittings and maintaining a consistent flow rate can also help to reduce pressure drop.

Similar threads

  • General Engineering
Replies
11
Views
2K
Replies
4
Views
3K
  • General Engineering
Replies
4
Views
1K
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
691
Replies
1
Views
2K
  • General Engineering
Replies
1
Views
1K
  • General Engineering
Replies
10
Views
3K
Replies
5
Views
909
Replies
14
Views
2K
Back
Top