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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.