How Do I Calculate Pressure Loss in Hydraulics with Two Pipes?

[Femme_physics]

I just started hydraulics and this is the first problem I'm trying to solve. Please take that into consideration!

Homework Statement

Given Two Pipes
Water is 50 degrees Celsius

http://img31.imageshack.us/img31/9752/74467684.jpg

I want to see if I understand this stuff correctly.

Y_L = The pressure lost through both pipes
L - Length of pipe
ID - Inner diameter

[ I'm not entirely sure what eA and eB mean - I think it says something about the quality of the surface? How smooth it is?]

The Attempt at a Solution

So I got these three formulas

http://img11.imageshack.us/img11/963/yyyyyyj.jpg

I'm looking at my class notes and I got a few things missing.

Ra and Rb are "Reynolds" number for the pipes, yes? I imagine it's constant.

And the strange symbol at the bottom of Reynolds formula, that misshapen "V", that is a constant as well, yes?

I hope it turns out there are only these equations. It seems like a fairly simple problem.

 

[lachy89]

Hello,

I am just an undergraduate studying Civil Engineering/Physics so take this as 'guidance' not complete help if you wish.

From Understanding Hydraulics, 2nd ed., Hamill

The equations I find of relevance are:

Q=V*A
Flow is equal to Velocity x Area


Re = V*D/(v)
Re = Reynolds number
V = Velocity (m/s)
D = Diameter (m)
(v) = 'funny looking v' (which is the greek letter 'nu') This is the kinematic viscosity

f= 64/Re
f = Darcy Friction Factor

hf = (f*L*V^2)/(2*g*D)
Head loss in a pipe.

hf = head loss (m)
f= Darcy Friction Factor (unitless)
L= Pipe Length (m)
V= Water Velocity in Pipe (m/s)
g=acceleration due to gravity (9.81m/s^2)
D= Pipe Diameter (m)


hl = (0.5*(V2)^2)/(2*g)

hl = head loss for sudden contractrion (m)
V2 = Velocity of the liquid on the output side of the connection (lesser diameter) (m/s)
g=acceleration due to gravity (9.81m/s^2)



Initially I would Calculate the velocity of the water on either side of the joint using Q=A*V. Then using the velocity I would then follow the formula through which I have posted above.

I am not entirely certain on the value of 'nu' but I believe this may be where the 50 degrees celcius may come into play (you may be able to look up a chart for 'nu' values from certain degrees of celcius).

EDIT: from a website I found that for 50 degrees C 'nu' is equal to 0.553 *10^-6 m^2/s

The final head loss will equal hf1 + hf2 + hl (head loss in pipe 1 plus head loss in pipe 2 plus head loss by the sudden contraction).

Once again I am just an undergraduate so feel free to disregard any of this until a 'professional' can confirm or deny. ^^

 

[Femme_physics]

Initially I would Calculate the velocity of the water on either side of the joint using Q=A*V.

That means I have to find the area first. So I have to use the formula for a cylinder's area to find area I need.

So area of a cylinder is: 2 x π x r² + 2 x π x r x h
So the radius of pipe A = 100mm, the height is 200mm


The radius of pipe B = 50mm, the height is 100mm (the height is basically diameter, yes?)

So now plugging for the area I get

Area A = 188495.556 mm²

Area B = 47123.89 mm²

Is this all true so far?

 

[lachy89]

The area in the equation Q=A*V is referring to the 'cross-sectional area' and so just the area of the circle.

This is pi*r^2 or pi*d^2/4 which is used more often at the uni I attend for circular cross-sectional area.

Also be careful to convert the diameters to metres.

Lastly looking through the information you have been provided the 'f' value or Darcy friction factory need to be calculated differently due to the 'e' values you were provided. Unfortunately I cannot remember a quick
Formula for this. Maybe someone else can clear this up.

EDIT: there is an equation for the Darcy friction factor taking into account your 'e'
Values. Look up the 'Colebrook-White equation' and an estimated equation solving directly for 'f' by 'Moody'.

 

[Femme_physics]

The area in the equation Q=A*V is referring to the 'cross-sectional area' and so just the area of the circle.

This is pi*r^2 or pi*d^2/4 which is used more often at the uni I attend for circular cross-sectional area.

Also be careful to convert the diameters to metres.

Lastly looking through the information you have been provided the 'f' value or Darcy friction factory need to be calculated differently due to the 'e' values you were provided. Unfortunately I cannot remember a quick
Formula for this. Maybe someone else can clear this up.

EDIT: there is an equation for the Darcy friction factor taking into account your 'e'
Values. Look up the 'Colebrook-White equation' and an estimated equation solving directly for 'f' by 'Moody'.

Wait, I'm confused. I need the CROSS-SECTIONAL area?

You mean I need the red part in figure (a)? Because I was under the impression I need the red part in figure (b), and that's what they gave me. Isn't that what matters to the flow formula?

http://img855.imageshack.us/img855/7130/crossssssssssssssssssss.jpg

Also be careful to convert the diameters to metres.

Should I always use meters in hydraulics?

Lastly looking through the information you have been provided the 'f' value or Darcy friction factory need to be calculated differently due to the 'e' values you were provided. Unfortunately I cannot remember a quick
Formula for this. Maybe someone else can clear this up.

Hmm ok. Also, what book would you recommend?

EDIT: there is an equation for the Darcy friction factor taking into account your 'e'
Values. Look up the 'Colebrook-White equation' and an estimated equation solving directly for 'f' by 'Moody'.

Thanks, I will

 

[Femme_physics]

*bumps* I forgot to add that image (Edited in in the post above)

 

[lachy89]

Yes, you need the area of figure b. This I'd the cross sectional area. Your 1st reply included the surface area of a cylinder not the area of the circle in figure b.

I would suggest converting to metres in all hydraulic problems (except maybe some special cases I cannot think of an example)

Hopefully I have cleared a few things up.

 

[Femme_physics]

Yes, you need the area of figure b. This I'd the cross sectional area. Your 1st reply included the surface area of a cylinder not the area of the circle in figure b.

Ohhhhhhh you're absolutely right! I see what you mean now Thanks.

I would suggest converting to metres in all hydraulic problems (except maybe some special cases I cannot think of an example)

Hopefully I have cleared a few things up.

You most certainly have! I hope you stick around, too!


So, cross-sectional area of a cylinder, that's just pi x r squared !

So, that'd equal for pipe A

Inner diameter = 0.2m

So, pi x 0.1² = 0.031416m²

For pipe B

Inner diameter = 0.1m

So, pi x 0.05² = 0.007854 m²


So both Q_A and Q_B = 0.0025 m³/s?


It doesn't seem to make since because Q is volume supply, and the wider diameter clearly has a bigger volume supply. But, I'm only given one of the Q and I'm not told to which pipe it belongs. Confusing!

 

[lachy89]

There is no loss of water in the transfer of water from pipe 1 to pipe 2. So the volume flowing through one pipe must equal the volume flowing in the other pipe.

The larger pipe has a larger supply for a constant velocity, but the velocity is not constant the volumetric flow is.

If you consider the pipes separately. Assume that 0.0025m^3 per second is going into the left end of pipe A. That would mean that 0.0025m^3 is coming out the other end.
Q in = Q out

There are no holes in te system just the joint or change in pipe diameter.
The Q out of the pipe A must now become Q in of pipe B, but as it is narrower the velocity of the water will need to increase.

Does that help to see why the flow is constant in both pipes?

 

[Femme_physics]

I've handcopied your reply into my notebook and I'm on the way to the bus back home. I'll think about it on the way and will reply soon Thanks for sticking around!^^

 

[Femme_physics]

There is no loss of water in the transfer of water from pipe 1 to pipe 2. So the volume flowing through one pipe must equal the volume flowing in the other pipe.

But there are different volumes flowing in each pipe, because one pipe is smaller.

The larger pipe has a larger supply for a constant velocity, but the velocity is not constant the volumetric flow is.

Isn't velocity constant in this case as well?

If you consider the pipes separately. Assume that 0.0025m^3 per second is going into the left end of pipe A. That would mean that 0.0025m^3 is coming out the other end.
Q in = Q out

Agreed

Does that help to see why the flow is constant in both pipes?

What do you mean by "flow is constant"? The fluid's velocity?

 

[lachy89]

My quoting skills are lacking so I apologise.

The volumes per second are the same in each pipe.

If 0.0025m^3 per second is going in the left end then 0.0025m^3 must be coming out the right end. If this was not the case there would either be a build up of fluid in the pipe system somewhere or there would be a hole in the system leaking fluid. As this is the case QA = QB.

The velocity (average velocity) in pipe A is constant. The velocity (average velocity) in pipe B is constant. But they are not equal to each other.

By 'flow' I am referring to the volumetric flow per second, so in your case 'Q' which is equal to 0.0025m^3.

I don't have my textbook on me so I cannot quote the name of the laws but you may need to do some reading on the formula of:

Qin = Qout

and how the formula will be applied to your problem.

 

[Femme_physics]

My quoting skills are lacking so I apologise.

np. Just use quotation marks if you want --> ""

Or in turn you can just use the actual quote code which is

http://img196.imageshack.us/img196/7737/quottees.jpg

The volumes per second are the same in each pipe.

If 0.0025m^3 per second is going in the left end then 0.0025m^3 must be coming out the right end. If this was not the case there would either be a build up of fluid in the pipe system somewhere or there would be a hole in the system leaking fluid. As this is the case QA = QB.

Oh, I see what you mean. So only a portion of QA flows, whereas all of QB flows, yes?

The velocity (average velocity) in pipe A is constant. The velocity (average velocity) in pipe B is constant. But they are not equal to each other.

Ah yes, that's what I thought in the smaller pipe - faster!

I don't have my textbook on me so I cannot quote the name of the laws but you may need to do some reading on the formula of:

Qin = Qout

and how the formula will be applied to your problem.

Will do.




SOOOOOO to solve

For pipe 1


Q = VA

0.0025 = 0.031416V
V = 0.0025/0.031416
V = 0.0796 [m/s]

Hmm... are my units correct? meters per second?

For Pipe 2
0.0025 = 0.007854V

V = 0.318 [m/s]

So I now need Reynolds number so I could later find "f" (my friction coeffecient throughout the pipe)...but I still don't know the kinematic viscosity (which is represented by the greek letter "nu" as you said).

So I end up with two unknowns here (circled)

http://img827.imageshack.us/img827/4531/raraaa.jpg

So once again getting a bit stuck. A bit more help, please?

 

[lachy89]

Oh, I see what you mean. So only a portion of QA flows, whereas all of QB flows, yes?

No the whole portion flows in both. So in the smaller pipe the area goes down, but to make the Q constant, the velocity goes up.

Ah yes, that's what I thought in the smaller pipe - faster!

Exactly ^^. Once again Q stays constant, so if A goes down V must go up.

SOOOOOO to solve

For pipe 1

Q = VA

0.0025 = 0.031416V
V = 0.0025/0.031416
V = 0.0796 [m/s]

Hmm... are my units correct? meters per second?

Your units are correct.
m^3/s = m/s * m^2 (if that makes sense)

Seems like an awfully slow speed, but i guess it is what it is.

So I now need Reynolds number so I could later find "f" (my friction coeffecient throughout the pipe)...but I still don't know the kinematic viscosity (which is represented by the greek letter "nu" as you said).

So once again getting a bit stuck. A bit more help, please?

The 'nu' value for kinematic viscosity for water can be found on a table if you know the velocity (this is where your 50degrees Celsius will come into play as the viscosity depends on temperature). There should be a table in your text somewhere or try 'Google-ing' kinematic viscosity of water.

 

[Femme_physics]

No the whole portion flows in both. So in the smaller pipe the area goes down, but to make the Q constant, the velocity goes up.

Exactly ^^. Once again Q stays constant, so if A goes down V must go up.

I fully understand it's all in the formula Q = AV after all

Your units are correct.
m^3/s = m/s * m^2 (if that makes sense)

Seems like an awfully slow speed, but i guess it is what it is.

Good, and I guess

The 'nu' value for kinematic viscosity for water can be found on a table if you know the velocity (this is where your 50degrees Celsius will come into play as the viscosity depends on temperature). There should be a table in your text somewhere or try 'Google-ing' kinematic viscosity of water.

The 'nu' value for kinematic viscosity for water can be found on a table if you know the velocity (this is where your 50degrees Celsius will come into play as the viscosity depends on temperature). There should be a table in your text somewhere or try 'Google-ing' kinematic viscosity of water.

Are you sure that I need to know velocity? Because I see a table graph with kinematic viscosity on the Y axis and Temperature on the X axis and then I see a line that describes how kinematic viscosity changes base on temperature. Such as (just without the numbers):

http://img847.imageshack.us/img847/5840/kinewaterok.jpg

So, shouldn't it be only temperature that matter?At 50 temp' it's our "mu" according to the table is 5.8 (it's hard to see the exact number, so it can turn up pretty inaccurate because it can be anywhere between 5.6 to 5.9). Can I use that?

 

[lachy89]

Sorry you are correct I'm not sure why I wrote that bit on velocity. I wrote it from my phone so Iay not have proof read too well, sorry.

Yes the viscosity only depends on the temperature. Just check the magnitude though it may be *10^-6 or something like that.

 

[Femme_physics]

http://img204.imageshack.us/img204/1996/temp11g.jpg

So its 0.553 x 10^-6

Solving for Reynolds Numberhttp://img217.imageshack.us/img217/4820/reyrey.jpg Since Ra > 2300
This is turbulent flow. Now I can use the formula you gave me to find "f"

f= 64/Re
f = Darcy Friction Factor

So,

fA = 64/29868.67 = 0.00214And now Rb...

http://img43.imageshack.us/img43/9986/rbbb.jpg

f = 64/Rb
fB = 0.0011

So now I FINALLY can use this

http://img849.imageshack.us/img849/2505/solvedylt.jpg

WHEW! THat was a long one! Thank you so much lachy^^

2 questions:

1) Is this all correct?
2) What are the units of Y_LT?

 

[lachy89]

The Equation for te Darcy friction factor is a 'rough formula' u believe to be more accurate and incorporate your e values look up the Colebrook white equation, this might change your f value.

Ylt should be in m, I am a little unsure on the formula you used for this. Isn't there a v^2 meant to be in there somewhere?

 

[Femme_physics]

The Equation for te Darcy friction factor is a 'rough formula' u believe to be more accurate and incorporate your e values look up the Colebrook white equation, this might change your f value.

You're right, I just read this:

In turbulent flow (R≥ 4000), the friction factor, f depends upon the Reynolds number (R) and on the relative roughness of the pipe, e/D, where, e is the average roughness height of the pipe

from here -> http://sites.google.com/site/excelsolveimpliciteq/

Which does make sense since they've provided me with values for eA and eB. And I do have turbulent flow if my values are correct.

But they're not actually giving me the formula. Where can I find it?

Ylt should be in m, I am a little unsure on the formula you used for this. Isn't there a v^2 meant to be in there somewhere?

You're right, I forgot to square my V's

Thanks!

 

[lachy89]

http://en.m.wikipedia.org/wiki/Darcy_friction_fzactor_formulae

Look at the equation section ( I realize it is wikipedia but it should stillbe correct). As the 'f' is on both sides of the equation you may need to do some iterations or look for a simplified solution.

 

[Femme_physics]

Hmm.. your link doesn't exist

 

[Femme_physics]

Alright, I figured it out, you just need to look at Moody's chart to get the answer when I have the full solution I'll post it. Thanks!

 

[Femme_physics]

Actually I needed to use Balsius formula. At any rate, I'm just posting back to show you I got the full solution

Thank you very much, helped me tons!