There are occasions when you might want to permanently change the amount of fluid you are pumping, or change the discharge head of a centrifugal pump. There are four ways you could do this:
- Regulate the discharge of the pump.
- Change the speed of the pump.
- Change the diameter of the impeller.
- Buy a new pump
Of the four methods the middle two are the only sensible ones. In the following paragraphs we will learn what happens when we change either the pump speed or impeller diameter and as you would guess other characteristics of the pump are going to change along with speed or diameter.
To determine what is going to happen you begin by taking the new speed or impeller diameter and divide it by the old speed or impeller diameter. Since changing either one will have approximately the same affect I will be referring to only the speed in this part of the discussion.
As an example:
The capacity, or amount of fluid you are pumping, varies directly with this number.
- Example: 100 Gallons per minute x 2 = 200 Gallons per minute
- Or in metric, 50 Cubic meters per hour x 0,5 = 25 Cubic meters per hour
The head varies by the square of the number.
- Example : a 50 foot head x 4 (22) = 200 foot head
- Or in metric, a 20 meter head x 0,25 ( 0,52) = 5 meter head
The horsepower required changes by the cube of the number.
- Example : a 9 Horsepower motor was required to drive the pump at 1750 rpm.. How much is required now that you are going to 3500 rpm?
- We would get: 9 x 8 (23) = 72 Horse power is now required.
- Likewise if a 12 kilowatt motor were required at 3000 rpm. and you decreased the speed to 1500 the new kilowatts required would be: 12 x 0,125 (0.53) = 1,5 kilowatts required for the lower rpm.
The following relationships are not exact, but they give you an idea of how speed and impeller diameter affects other pump functions.
The net positive suction head required by the pump manufacturer (npshr) varies by the square of the number.
- Example : A 3 meter NPSHR x 4 (22) = 12 meter N.P.S.H.R.
- Or: 10 foot NPSHR x 0.25 ( 0.52) = 2.5 foot N.P.S.H.R.
The amount of shaft run out ( deflection) varies by the square of the number
- As an example : If you put a dial indicator on the shaft and noticed that the total run out at 1750 rpm. was 0.005 inches then at 3500 rpm the run out would be 0.005″ x 4 (22), or 0.020 inches.
- Likewise if you had 0,07 mm. run out at 2900 rpm. and you slowed that shaft down to 1450 rpm the run out would decrease to 0,07 mm x 0,25 ( 0.52) or 0,018 mm.
The amount of friction loss in the piping varies by about 90% of the square of the number. Fittings and accessories varies by almost the square of the number.
- As an example : If the system head loss was calculated or measured at 65 meters at 1450 rpm., the loss at 2900 rpm. would be : 65 meters x 4 (22) = 260 x 0.9 = 234 Meters
- If you had a 195 foot loss at 3500 rpm the loss at 1750 rpm. would be : 195 x 0.25 (0.52) = 48.75 0.9 = 43.87 feet of head loss.
The wear rate of the components varies by the cube also
- Example : At 1750 rpm. the impeller material is wearing at the rate of 0.020 inches per month. At 3500 rpm the rate would increase to: 0.020 ” x 8 (23) or 0.160 inches per month. Likewise a decrease in speed would decrease the wear rate eight times as much.
I started this discussion by stating that a change in impeller speed or a change in impeller diameter has approximately the same affect. This is true only if you decrease the impeller diameter to a maximum of 10% . As you cut down the impeller diameter the housing is not coming down in size so the affinity laws do not remain accurate below this 10% maximum number.
The affinity laws remain accurate for speed changes and this is important to remember when we convert from jam packing to a balanced mechanical seal. We sometimes experience an increase in motor speed rather than a drop in amperage during these conversions and the affinity laws will help you to predict the final outcome of the change.