In this screencast we are going to work through a pipe flow example problem, which the pipe
diameter is unknown.
Thus the solution will require an irritative numerical method type of approach.
We have a new water piping system that is being designed for an apartment complex.
Now the supply of water to this complex must provide 1200 gallons/min at a pressure of
25 psi, and this will come from the large diameter main city supply line which is maintained
at 65 psi.
We want to determine the appropriate pipe diameter to connect the city line to the apartment
complex line.
So we are going to use a plastic material for our pipe that is known to have an equivalent
roughness of 5 times the 10 to the -7 feet.
We also know that are going to use 100 ft of pipe, 4 threaded 90 elbows, 10 threaded
unions, and a swing check valve for forward flow.
The elevation difference between the main source and the apartment source is 10 ft.
We need to determine the pipe diameter to use.
So when we look at solving pipe flow problems we know to start with an energy balance that
looks like the following.
Now I have provided this in a certain form.
You might see it in different forms, but you should quickly realize if we need to solve
for the diameter D. Well this is not going to be a straight forward plug and chug form
problem.
There is a couple of different terms that relay in the diameter.
Such as the Velocity, Here is the diameter.
The friction factor that is a function of the diameter, and then we have those other
velocities terms.
So we are going to need to a computer solver or irritative approach to help us get our
answer.
So lets work on setting up the irritative approach, and at the end of the day I will
show you how to use the computer solver to make your life easier.
The first thing I like to doing is writing down what we know, and drawing our system.
So a quick drawing just to give us an idea.
I am going to call this point 1, which is our source.
Then we have a line.
and we know that we have 4 elbows.
So we have 1 turn, 2 turns, 3 turns, 4 turns, and this is going to go to our point 2, which
is our apartment duplex source.
So we know the pressures here, and we know the length of pipe.
We also know there is connections in between, and some kind of valve.
So I will just put a star there to represent our valve.
So we are given a flow rate of 1200 gal/min, I will go ahead and convert this to ft^3/sec,
and when I do this I get 2.674 ft^3/sec.
I like to change all my units as I start writing this out to one base set.
So we know pressure at P1, our city source to be 65 psi.
Again instantly converting this I get 9360 psf.
Our pressure at our apartment complex needs to be 25 psi, which converted gives me 3600
psf.
We know the equivalent roughness of our pipe.
It was told to use that it was 5x10^-7 ft.
We have a length of pipe that is 100 ft.
We also have for our minor losses 4 times a threaded 90 degree bend.
We have 10 threaded unions, and we have a swing check valve in the forward position.
So again I am using these to look up loss coefficient.
So I am going to go ahead and look up loss coefficients associated with these so I can
write them down.
For our threaded 90 degree bend one table I have uses a loss coefficient of 1.5.
For our threaded unions I get 0.08.
For our swing check valve I get 2.
This all comes from a fluid mechanics text book.
You could look these up probably in other hand books as well.
So again I can save myself the trouble know and just do the math, and I can see that the
sum of these loss coefficients is equal to 8.8.
Now based on the problem statement we are not given anything else, but looking at the
energy balance here.
We know we need a density of our fluid.
We are also going to need the viscosity of the fluid to get a Re number down the road.
So lets assume a temperature, and 60 degrees Fahrenheit is not a bad assumption for the
temperature of the city water.
So the density if we look this up in a table it is going to be 1.94 slugs/ft^3.
The viscosity at 60 degrees Fahrenheit is 2.34x10^-5 lb*s/ft^2 and lastly just to be
safe.
We will write gravity down at 32.174 ft/s^2.
Are we missing anything?
We are.
We are missing our elevation.
So we know that z2-z1 is equal to 10 ft.
Alright lets start with our energy balance.
So we can start simplifying it down.
So we know P1-P2.
Since we have P1 and P2.
z1, z2 we have reported as 10 feet.
We have our density and gravity.
We have our length, and we have the sum of our minor loss coefficients.
So what we are down to is looking for velocities, as well as diameter, and a friction factor.
Now what you might notice is that we don't have our KE parameter alpha in our equations.
Now we are going to assume turbulent flow in this problem since we have a flow rate
of 1200 gal/min.
So I have just used alpha equals 1 and didn't even add it to our equation.
We have 3 unknown terms all involving velocity or diameter, and our KE terms has two different
velocity terms.
Lets simplify this to the easiest form to work with.
Which for me I am going to start plugging these values in and I will get it down to
the following.
So I have my energy balance on the left side of this purple line, and I have just broken
it up into each individual component.
I have put the values I know on the right side.
So one thing that we can easily get stuck on at this point.
Is trying to figure out what velocities we are looking for.
Now when we think of this system water is coming from a main water line into one that
we are building for the apartment complex.
Eventually out of some type of faucet in the building.
So the specified large diameter of the supply line.
Is there to give us the impression that the velocity in the main line is much smaller
then that in the line we are building.
So for the sack of the problem we assume that the velocity in the city line is V1 is negligible.
So V1 is much smaller then V2.
So now all the velocities in our problem is of V2 which is just the velocity through our
pipe networks.
Since the diameter is constant.
So now we have 1 unknown velocity V2, which shows up in 3 of the 5 terms.
We have 1 known diameter, and we have 1 unknown friction factor.
So the velocity is going to be solved by the relationship that volumetric flow, which is
represented by Q. Is divided by the cross-sectional area of the pipe, A, which gives us an average
fluid velocity V. So know we have our velocity as a function of D. So we can plug that into
our other equation.
Since we know Q. Now we are down to 2 unknowns f and D. Here we are going to plug our velocity
expression into this term, as well as this term and this term.
I have simplified a little bit pulling out D to the 4.
I have plugged in the known values we had before.
Make sure all our units match up, and when we do that we get the following expression.
So we have 2 unknowns.
The friction factor f and the diameter D. Now unfortunately the friction factors is
not so easy.
There are 2 ways to approach this.
One is the moody diagram, based on D we could solve for V, thus solve for the Re number,
calculate epsilon over D, and get f.
So this is a trail and error irritative approach, because it is all based on D. So we have to
guess a D, get f, and make sure this equation is satisfied.
The other approach we will talk about in a second.
After we do this guess and check system.
So lets write out our expression.
We have this one, equation 1, which is an equation of f and D.
We have the Re number, which we need for the moody diagram which is rho*v*D over mu.
Since we only want to work in terms of diameter.
We are going to plug in this expression up here, for our velocity, and when we do that
we get rho over mu times 4Q over pi times 1/D. Now we can simplify this since we know
rho, mu, and Q and get the following expression for the Re number.
Lastly we have to get epsilon over D. Epsilon on the problem statement was given to us.
Lets guess a D. A reasonable diameter for a pipe might be 3 inch line.
So our first irritation we are going to guess 0.25 feet.
So when we do that, right away one check we can make is to plug that in to our energy
equation up here.
So if we plug in a diameter of 0.25 feet you will find that the only solution for f is
that it is negative, and we are not going to get a negative f from the moody diagram.
So right now we know that 0.25 feet is not possible.
Lets try a diameter that is larger.
So when we use 0.4 feet I get a Re number of 7.06x10^5.
Epsilon over D 1.25x10^-6.
This gives me a friction factor of about 0.0125.
So now when we plug in our diameter into our energy balance the friction factor from our
energy balance gives us 0.007, which is not equal to what our moody chart gave us of 0.0125.
So we know that the diameter of 0.4 is not correct.
So we need to move down from the value in our moody chart to match up with the one from
our energy balance.
So I am going to guess a highers Re number.
So maybe a higher diameter.
So because we have this diameter to the 5th power very small changes are going to have
drastic affects.
So for my next trail I am just going to get 0.41 feet.
I will show you why.
When you do 0.41 ft I get a Re number that is equivalent to 6.88x10^5.
Now epsilon over the diameter changes to 1.22x10^6.
So as we are to expect the friction factor on trail 2 is equal to 0.0125.
Now I am using the exact same values because really this does not change how we read our
moody chart.
What does change is that when we us this in our energy balance we get a friction factor
of 0.0125 , which is offley close to what we are reading in our moody chart.
So all though I have only shown say 2 iterations you can get drastically get caught up on this.
If you are reading on the chart make small changes as you go along.
So a 5 inch diameter pipe would be most appropriate for our design of our case.
In another screencast I will show you a supper short cut method that will take all the data
and work we have already done, but use a computer solver to do all our work for us and give
us the value quickly.