In this screencast we are going to look at a problem.
In which we are determining the power added by the pump to a fluid to move it from one
location to another.
Specifically this is a problem that is dealing with calculating a certain pressure or losses.
Using the Moody chart to determine frictional factors, and using this in our problem.
We have a ski resort that need to pump water at a certain temperature.
Though 1000 ft of 2 inch diameter steel pipe.
This is from one location to another, and we are given those elevations.
We are told that the volumetic flow rate nees to be a certain rate, and the pressure into
the snow blowing machine needs to have a certain pressure.
We are also given information to components of this piping system that are going to contribute
to the losses.
So try to set this up on your own and solve.
With any problem like this.
The best way to start is by drawing a diagram.
We are going to draw our hill, and our water reservoir down here.
Then we are going to draw our pipping system that comes from the water reservoir to some
kind of pump.
Comes out of our pump, and goes to our snow blowing machine.
We are also told there is a valve that the entrance of this system, and there is two
90 degree bends.
So at this point now that we have a diagram.
It is important to label points that we are interested in for this calculation.
At the entrance of our piping system, which we will call point 1, and at the exit of our
piping system into the machine.
I will designate this as point 2.
We know that this open to atmosphere and we know that the pressure in the snow blowing
machine has to be 20 psi, and that is probably safe to assume gauge pressure.
We are told that the volumetric flow rate Q has to be 0.25 ft^3/s.
We can say there is two 90 degree bends.
So make sure that we have those.
We also know that this is a steel system.
We also know that the entire length of this is 1000 ft, and that our elevation change
is 800 ft.
Which leads us the the second part of this problem solving.
It is to write down what we want.
So we want the horse power for our pump.
With any experience with problems like this we know that means we are going to need delta
p for the system.
We will also need energy losses.
Now we can look up power for a pump, and we can see that it is going to be equal to our
shaft head of the pump, the specific weight of the fluid we are moving, and the volumetric
flow rate.
So we can already see that we are given 2 pieces of information there.
We just need to find out what the shaft work.
The approach we are going to take with this problem is an energy balance.
So we can start by writing our general energy balance for a problem like this.
Which looks familiar to something you have seen before with the bernoulli balance, but
in this case we have added the two new terms.
One the shaft head, which is related to the power of either taking away or putting into
the system mechanically as well as the head loss, which are losses due to the piping system
or components.
As a refresher for the terms P is the pressure.
We have our specific weight in the numerator.
Our V is the velocity of the fluid at those points, g is the gravitational constant.
Z represents the height with respect to whatever reference we are using.
So we have our energy balance, and we know we need to figure out what the head loss.
So head loss is going to be written out as related to a frictional factor times the length
of the system over the diameter of the system multiplied by a Kinetic component.
So you can see here that we can calculate the velocity based on our volumetric flow
rate and our pipe diameter.
We are given the length, and the pipe diameter.
So all we would need is the frictional factor.
The frictional factor is going to be dependent on whether or not the flow is turbulent or
laminar.
We use the dimensionless term the Re number.
The Re number is a relationship between the inertial forces , which can be written as
rho, v,D. Over the viscous forces.
So we use viscosity or we can rewrite this as the velocity times the diameter over the
kinematic viscosity, and the kinematic viscosity for water at 10 degree Celsius is 2.026x10^-3
in^2/s.
The velocity of our fluid can be calculated as the volumetric flow rates over our cross-sectional
area.
When we do this calculation we get a velocity of 11.5 ft/s.
So know we have our velocity, we have our kinematic viscosity.
We can calculate our Re number by plugging in all the information that we have.
When I calculate this out I get a Re number of 1.4x10^5, which is greater then our standard
2100 for our transition from laminar to turbulent flow.
We can safely say that this flow is turbulent.
Where does that help us?
Well, we are going to take this information and go to what is called a moody chart to
help us relate a frictional factor to our flow in our pipe system.
Here is our moody chart.
It is a logarithmic plot.
Were we can see on the left side the y-axis we have our frictional factor.
On the bottom we have our Re number, and n the right we have what is called a relative
roughness, which is a relationship between the roughness of the pipe to the diameter
of the pipe.
This is important because as you recall we are looking for our head loss term, which
has our frictional factor within it.
So lets go to the plot.
We know we have a Re number of 1.4x10^5.
That puts us roughly right in this area.We will still need our relative roughness to
do this calculation.
So we have pulled out information up from a source.
What we are told is we have commercial steel.
Since we are working with feet as our units our roughness is 0.00015 since it is 10^-3.
We take that and divided it by our diameter, and we get a relative roughness of approximately
9x10^-4.
So we take that information and go up here to the axis to the right side, and we look
for where 9x10^-4 would be.
That squeezes us in right here on the right side.
So we would fall right in between these two lines, and we would go to where our Re number
is.
I am going to squeeze right inside both of them, and we will move up until we get right
about were we want too.
Lets say it is right here.Obviously there is error with reading a moody chart, but for
the most part you can see that on the left side the frictional factor wont be to dramatic.
So we are going to follow this line over, somewhere about there, and say that is about
0.021.
So know we have our frictional factor.
So lets go back to our main equation, and rewrite it into the form we are most interested
in.
We know if we use gauge pressure that p1 is equal to 0, since it is open atmosphere.
We also know that the velocity at point 1 is also going to be 0, assuming that the height
of the lake velocity is not relative compared to the velocity in the pipe.
So knowing that information we can rewrite our equation as the following.
We can do this on the right side, because we know our head loss term was the friction
factor, times the length, over the diameter, and our kinetic term.
So when we mix this together with our kinetic term we get the following.
So at this point it is an exercise of plug and chug.
So I will plug in our values into this equation.
So you can see it.You can see when I plug in all these number.
I end up getting about 1107 ft, as our pump head.
That we would need to add to the pump.
However we only calculated that of the major losses in the system, and we totally neglected
the minor losses.
Now is that an appropriate thing to do in this case.
Lets take a quick look at the minor losses and see how they stack up to the major losses.
So when we look at the minor losses of a system.
It is going to be the summation of all of the losses due to the components.
So you can look these values up.
They are reported differently, but at least in our case we are going to use a loss coefficient
and multiply it by a kinetic component.
So our loss coefficient for our valve assuming that it is fully open would be .05, Our two
90 degree bends each have a loss, somewhere between 0.2-1.5 depending on the type of elbow
that it is.
It can be flanged or threaded.
So lets just assume somewhere in the middle.
That is will have a kL ration of 1.
So that two of these will equal kL of 2.
So our minor losses due to the valve as well as the bends will be 2.05 times our kinetic
variable, and you can see that this comes out to a very small number with respect to
the major losses.
Now we could account for this just to be more accurate.
So our total shaft head that is necessary from the pump is going to be about 1111 ft,
and if we take this at put it into the equation of power.
It shows that we would need a shaft head multiplied by the specific weight of the fluid, and the
volumetric flow rate gives us a value of 1.7x10^4 lb*ft/s, but we can convert this to horsepower,
and we can see that our total pump power for our system is 31.5 Hp.
So we go back to our original problem statement, which asked us to determine the power added
by the pump, and we can safely say that we have calculated that, in this problem.