It's important to know how to estimate the net positive suction head available in the
system so that you can prevent your pump from cavitating.
Reducing the risk of cavitation will ensure the life of your pump which is often a considerable
investment.
We start off with a reservoir, that supplies water to our pump.
The water exits the reservoir through a rounded entrance,
passes through a 90 degree bend and then heads into our pump.
The pump after doing its job discharges the water into another tank higher up in elevation.
The reservoir on the suction side of the pump is at an elevation of 1150 feet the pump inlet
is at an elevation of 1140 feet and the reservoir on the discharge side of the pump is at an
elevation of 1380 feet.
The length of the discharge pipe is a 6200 foot, 30 inch diameter pipe, with a friction
factor of 0.020.
For the suction pipe the length is 40 feet the diameter is 12 inches and the friction
factor is 0.025.
Our target flow rate for the system is 4000 gallons per minute and the fluid temperature
is at 70 degrees Fahrenheit.
The first mistake students typically make on this problem is considering the entire
system shown on this page.
The net positive suction head available is computed at the suction flange of the pump
and only considers the piping system and reservoir on the suction side of the pump –
this area right here.
You can ignore the discharge pipe length, diameter, friction factor and the elevation
of the discharge reservoir.
They're not relevant to this calculation.
To make later computations easier, I like to calculate the area of the pipe first and
then the velocity of the water in the pipe.
For the pipe area, a 12 inch diameter pipe is 0.785 square feet.
the velocity is found by converting the 4000 GPM into cubic feet per second, so I divide
by 448.8 then divide that by the area of the pipe and I get 11.4 feet per second.
The formula to compute net positive suction head available is the atmospheric pressure
plus the elevation difference between the reservoir and the pump minus the friction
head, minus the minor losses and minus the vapor pressure head.
Each of these terms are a quick lookup or need to be calculated, so let’s tackle each
from left to right.
There are many resources that provide information about the atmospheric pressure at different
elevations.
This table shown in this video come from Pumping Station Design, 3rd edition.
The appropriate atmospheric pressure for our example is measured at the suction flange
of the pump.
In this case we want to know the atmospheric pressure at 1140 feet above sea level.
This means we need to interpolate between the values of 14.2 psi and 13.7 psi because
our elevation is between 1000 and 2000 feet.
We start with 14.2 minus 0.14 because we are 140 feet or 14% of the distance between 1000
and 2000 feet.
Then we times that by 14.2 - 13.7.
The result is 14.1 pounds per inch squared.
This result is in pressure, but I want it to be in pressure head, or units of feet to
make the final step easier.
Therefore I need to convert this value into feet.
To do that, I multiply this number by a 144 divided by
the unit weight of the water, gamma, at 70 degrees Fahrenheit.
This means I have to go look up that value from my water properties table.
And this is another key mistake students often make is that they simply use 62.4 pounds per
cubic foot for the unit weight of water, rather than accounting for the temperature of the
fluid.
The unit weight of water at 70 degrees Fahrenheit is 62.3 pounds per cubic foot.
So the atmospheric pressure head is computed as 32.6 feet.
The next step is to calculate the static head of the fluid acting on the suction flange
of the pump.
This is just a fancy way of saying find the difference between the reservoir elevation
and the pump elevation, which is 10 feet.
The next calculation is to find the friction losses in the pipeline leading from the suction
reservoir to the suction flange of the pump.
In the problem statement I gave you the friction factor, so you know to use the Darcy-Weisbach
equation.
This equation is the friction factor times the pipe length times the velocity squared
divided by the diameter of the pipeline times 2g.
Plugging in our values we get 0.025 * 40 feet times 11.4 ft/s squared divided by 1.5 feet
times 64.4.
We hit the enter button on our Calculator and get a result of 2.0 feet.
The next step is to calculate the minor losses which is also written with a summation sign
in the front.
The minor losses are computed by summing up all of the minor loss coefficients, or K values,
and multiplying these values by the velocity head, V squared over 2g.
For our simplified system we have an entrance loss coefficient of 0.25 plus
a 90 degree bend coefficient of 0.25 times 11.4 squared divided by 64.4 . This gives
us 1.5 feet.
Now we look up the vapor pressure head which I find as 0.36 pounds per square inch.
Vapor pressure is the amount of pressure that the layer of vapor exerts on the surface of
water.
All standing water has a certain amount of vapor pressure acting on it.
So, for this computation we need to pull out the impact from the vapor.
Once again, we multiply by 144 and divide by the unit
weight at 70 degrees or 62.3 to get a Vapor pressure head of
0.84 feet.
For the final step we calculate the net positive suction head available by just adding and
subtracting numbers we picked up along the way.
32.59 feet plus 10 feet minus 2.0 feet minus 1.5 feet minus 0.84 feet.
Which is 37.3 feet.
Given that we typically use a 5 foot buffer between the net positive suction head available
and the net positive suction head required, this result tells us that we cannot select
a pump who's net positive suction head required is greater than 32.3 feet – at least for
potable water applications.
About 40% of my students try to keep pressure in pounds per square foot then try to convert
in the final step at the end.
They almost always make a mistake because of the large amount of computations and then
make a mess of their final answer.
Just make sure each of the terms are in units of feet before going into the final calculation.
It will make your life so much easier.
Unless you end up as a water resources engineer you may never do this calculation again.
However, for all engineers you need to understand what drives the problem of cavitation.
There are two general causes: the first is not enough net positive suction head available
and the second is too much net positive suction head required by the pump.
If you know each of the terms in the net positive suction head available calculation then you
should easily be able to understand the impact of changing any one of those values while
keeping the others constant.
For example, if atmospheric pressure increases what is the impact on net positive suction
head available?
It increases.
If there is a design change to the system that increases the minor losses then the net
positive suction head available will decrease.
Every engineer no matter what your favorite kind of project is to work on should understand
this.