In this screencasts we will work through a piping example to determine the losses associated
with the pipe network and flow characteristics so that we can determine the pressure drop
between two points.
For instance the difference between the pressure entering a house we will call point 1 and
that leaving the faucet we will call point 2.
So let's look at this problem set up in which we have water flowing from our city supply
network into our house and up to the second floor bathroom out of a faucet.
The goal in this problem is going to be determine the pressure drop between points 1 and 2.
Now the first I like to do is convert all of the values we are given to the same unit
system and collect any information we may need to help us along the way like properties
of the fluid.
So we have water coming in from our city line which if you do a quick google search, maybe
let's assume 60 degrees Fahrenheit for the system.
We will need the density and viscosity of the water at 60 degrees Fahrenheit.
And we are working in English units so you can look this up in a table, in a book, or
in an online reference and you should get the following properties.
I think we have everything we need and the units we need.
If we need to convert anything else we can do it at that time.
So now we need to determine the losses associated with this system and we know in determining
losses the velocity of the fluid plays a larger role.
Furthermore we need to determine if the flow is laminar or turbulent since that's going
to tell us how to determine the major frictional losses in the pipe system.
So let's start with determining the type of flow.
Now to determine the type of flow whether it is laminar or turbulent we can use the
dimensionless Reynolds number and the Reynolds number is represented by the density of the
fluid times the velocity of the fluid times the diameter over the viscosity.
We have the density, diameter and viscosity but we have not calculated the velocity and
we know we can convert the volumetric flow rate to a velocity using the cross sectional
area of the pipe.
So when we do that I get a velocity equal to 4.08 ft/s.
So now we can calculate our Reynolds number.
When we plug in all of these values you will see that the units do not directly cancel
out and that is because we have not written in our unit conversion for slug which 1lb
= 1slug*ft/s^2.
When we convert from slug to pound you will see that this is indeed dimensionless.
You will also see that our value is 2.8*10^4.
Which is clearly a turbulent flow.
Now when we do these types of problems we are going to approach them using an overall
energy balance between two points.
Recall from our problem set up that our two points are the inlet and the outlet.
So this is what the overall energy balance looks like.
Now we do not have any shaft work for our system so I have not written it in there.
But we do account for the pressure terms which are P_1 and P_2.
We account for the kinetic energy term at point 1 and point 2.
We account for the potential energy terms at point1 and point2.
Then we have to account for our losses.
I have written out our losses as two different terms.
Our major loss term which has to do with the frictional losses in the pipe and our minor
loss term which has to do with all the joints and connections between the pipes.
So this almost becomes a plug and chug type of problem at this point but we have to make
sure to use the units appropriately and we have to use the correct values.
Also it may not be familiar on how to calculate our frictional factor "f" or this "K_L" term
which is our minor loss coefficient for our minor losses.
So we are in the problem statement looking for P1 minus P2.
Alpha 1 and Alpha 2 we can assume is going to be unity since this is turbulent flow the
velocity profile are nearly uniform across the pipe and they are the same at the two
pipes so alpha is equal to one.
Now density we had as 1.94 slugs/ft^3.
Our velocity at point 1 we already calculated as 4.08ft/s.
So that takes care of this term.
Now for our second term we have rho*g*z1.
Since point 1 is the lowest point in our system we could use this as our starting point and
consider our elevation at this point to be equal to zero feet.
So this term just drops out.
So we are looking for P 1 minus P 2.
We have our next term set up to solve.
We said alpha 2 is equal to 1 but we do not know the velocity at point 2 so that is something
that we need to determine.
We use the same method we did before, the volumetric flow rate divided by the cross
sectional area at point 2.
This cross sectional area had a pipe diameter of 3/4 in.
Solving this gives an exit velocity of 7.26 ft/s.
Now we know everything to calculate that term.
For our next term rho*g*z2 we have our density.
For gravity we are going to use 32.174 ft/s^2.
Then we need to know what our z2 is.
The elevation at point 2 is just the height at point 2 relative to our elevation 0 at
point1.
You can see that we have 5 feet high here, 10 feet high there.
So a height of 15 feet.
Now we know everything we need to for that term.
We are finally left with our major and minor loss term.
Unfortunately this one is not as straight forward.
To determine the major loss we have to figure out what the frictional factor "f" is.
One way to do this is to use the moody diagram.
Now there is a screencast on using moody diagrams so you could check that out to get an idea
of how to use it appropriately.
I am going to pull it up here and just show you I am going to find it.
This is the moody diagram which could tell us what the frictional factor "f" is based
on the relative roughness of the pipe and the Reynolds number.
We have the Reynolds number which we calculated as 2.8*10^4 which is roughly about here.
Now the relative roughness is determined by the roughness of the pipe as well as the diameter
of the pipe.
So our roughness factor for our copper pipe can be looked up in a table and found to be
5*10^-6 feet.
Our diameter we already converted to feet as 0.0833 ft.
This give us a relative roughness factor of 6*10^-5 which falls right above this line
of 5*10^-5.
So we follow those up until they intersect.
We get a frictional factor of about 0.024.
You could also use the Col brook equation to analytically solve for the frictional factor.
So continuing to determine our values.
Our length is just the combined length of all of the pipes so let's go back to the diagram.
We have 10ft plus 5 plus 5 plus 10 plus 10 plus 5 which gives us 45 feet.
Lastly our diameter of our pipe which now we can solve for this term however it is not
clear what velocity we use here.
We want to use the velocity through the pipe so that is v1 that we calculate before.
So now we have everything needed to determine our major frictional losses in our piping
system.
We can move on to our last term, our minor losses.
For the minor loss term we need to determine the loss coefficients for each connection
and designated component like the contraction bin at the faucet.
So we look these up.
For each 90 degree bend which we have one here, one here, one there, one there, we don't
count the one at the faucet because we are given the loss coefficient for that faucet.
We have four time the loss coefficient of a 90 degree bend.
We can look this up in a table and we find that this loss coefficient for a 90 degree
bend that is for threaded components as it says here is equal to 1.5.
So we have 4 times 1.5 which gives us a value of 6.
This accounts for the losses through each of the bends.
Now we also have an open globe valve which we can look up in the table.
For an open globe valve we will find that the K_L is equal to 10.
Last thing we can see in this problem is that there is a K_L at the faucet.
This K_L is given to us as 2.5.
The sum of the K_L terms which gives us 18.5.
Again we have our density and our velocity through the piping system of 4.08.
Now we have everything to determine the minor loss term.
At this point it really is just plug and chug.
This is what the final equation should look like.
I have not included any of the units in any of this but I have double checked to make
sure that they all do cancel out and I have the right form.
When I add all this up correctly I get a Pressure drop equal to -1,480 which seems incredibly
high except do recall that because of our units we are working in lb/ft^2.
So this actually when converted is 10.3 psi or pounds per square inch which is a much
more reasonable value when talking about the pressure loss through a piping system such
as the one in the problem statement.
Now whether it's negative or positive doesn't really matter as long as we understand that
it is a pressure drop and that the pressure is higher at the inlet than it is at the outlet.
If we wanted to calculate what P 1 was we would just need to know P 2 which in this
case is just exiting at atmosphere so P 2 is just 0 which means that our inlet pressure
for our system is also 10.3 psi.
So hopefully this gives you an idea on how to calculate pressure drop given certain components
of our piping system and the flow characteristics.
In another screencasts we will use a very similar problem to determine the flow rate
if that was our unknown.