Merriam Webster defines affinity as: “a likeness based on relationship or causal connection.” The Affinity Laws are a heuristic governing pump sizing for one pump based on known constants for another of that same pump. Knowing of these laws helps, but really understanding to apply them in problem solving goes a long way. I’ve been holding off on writing this because the topic is one I had been struggling with. It’s important to know, so it’s worth the struggle.

I once had someone tell me, “I’m not paying you to learn all of this theory!” That’s fair enough, but know that the trade-off is between a culture of striving for a good enough versus one that’s striving for excellence. Which standard would you rather hold? If we’re honest, we can’t have it both ways. For this, I challenge you. Be honest. Hold your torches high!

- Pump performance involves relationships between performance [ie, head, shaft speed, volumetric flow rate] and power. If speed or impeller diameter are known with one pump, performance based on speed or impeller diameter change for a another of that same pump can be determined. Also, if these are known for just one pump with a VFD (variable frequency drive), which is a commonplace scenario, new H-Q (head-capacity relationship) and BHP (brake horsepower) curves with a different speed than published on a pump performance curve can be plotted.
- There are two sets of Affinity Laws, and both are based on the premise of a pump’s specific speed not being changed once its been calculated. One law holds impeller diameter constant. The other holds speed constant.
- What’s nice is that this can all be seen on a pump performance curve. These relationships are what we’re typically seeing most manufacturers including in their pump H-Q curves. The curve is designed to provide this information/knowledge as given information. What I’m writing about here is less about that knowledge, itself, and more about understanding relationships based on knowledge.
- Though you could just see it all on the curve and not bother with knowing the why and how of it, don’t cheat. Learn how it all ties together foundationally. This is like the difference between seeing a movie in just two dimensions versus in all three (seeing versus “seeing”). Wouldn’t it be more far more entertaining to have it all come to life in vivid detail – not once, but every single time?
- Here are the two sets of Affinity Laws presented in basic and practical terms:

**Affinity Laws Set 1.**

*Holding the impeller diameter, D, constant, let’s solve for speed:*

Q1/Q2 = N1/N2

- Q is capacity (flow in GPM) and N is speed (motor speed). 1 is the first baseline pump; 2 is a second of that same pump with a capacity change having affinity in predicting second pump speed based on capacity change.

*What that’s saying is this:*

Q1(capacity of baseline pump 1)/Q2(capacity change for pump 2) = [N1(speed of baseline pump 1)/N2(speed change for pump 2)].

Visualizing this, I see two of the same exact pumps sitting side-by-side. Each has the same unchanged impeller size inside the respective volutes. But each pump calls for a different capacity (flow in GPM). How will the flow difference in that second pump change the required motor speed, since capacity and speed are related? We’re about to find out!

Let’s solve an iterative trial and error problem with these known terms, referencing this pump curve:

Using an example pump curve, what speed is rquired with a full diameter impeller to make this rating: 3000 gpm @ 225′? Here, I’m referencing an ITT Goulds Model 3196 centrifugal pump performance curve, 6 x 8 – 15 at 1780 RPM. (This model has a 6″ discharge, an 8″ suction, and a 15″ full impeller diameter). Let’s try out 2200 RPM to see if it gets to the desired rating:

*Holding the impeller diameter, D, constant, let’s solve for speed:*

Q1/Q2 = N1/N2

- Q1 = 3000 GPM; H1 = 225 ft; N1 = 1780 RPM
- Q2 = Q1 * (N2/N1); Q2 = 3000 * (1780/2200) = 2427 gpm
- H2 = H1 * (N2/N1) squared; H2 = 225 (1780/2200) squared = 147′

This first test doesn’t work out. We’re trying for a motor speed to accommodate 3000 gpm given an untrimmed full 15″ impeller diameter. 2427 gpm isn’t a high enough flow for the full impeller to make sense on the example 1780 RPM pump curve. It’s below the 15″ full diameter curve line. Ditto for the 225 feet of head requirement. 147′ isn’t a high enough head for the full 15″ impeller size.

We tried a 2200 RPM motor speed for 3000 gpm @ 225′ and it didn’t work out. I’ll try again, this time @ 2000 RPM:

*Holding the impeller diameter, D, constant, let’s try again to solve for motor speed:*

Q1/Q2 = N1/N2

- Q1 = 3000 GPM; H1 = 225 ft; N2 = 1780 RPM
- Q2 = Q1 * (N2/N1); Q2 = 3000 * (1780/2000) = 2670 gpm
- H2 = H1 * (N2/N1) squared; H2 = 225 * (1780/2000) squared = 178′

Referencing the performance curve published for this pump, 2670 gpm @ 178′ does fall on the curve line for the 15″ full size impeller. (It wasn’t slow enough of a speed to be above the line.)* 2000 RPM speed is the test winner!* Awesome.

Q1/Q2 = N1/N2

- For this first set of Affinity Laws (that being the set with the impeller diameter held constant), it’s also true that:

H1/H2 = (N1/N2) squared

- H1(ft/hd requirement for baseline pump 1)/H2(ft/hd change for pump 2) = [N1(speed of baseline pump 1)/N2(speed change for pump 2) squared].

BHP1/BHP2 = (D1/D2) cubed

- BHP1(brake horsepower for baseline pump 1)/BHP2(brake horsepower change for pump 2) = [N1(speed of baseline pump 1)/N2(speed change for pump 2) cubed].

In summary, this first set of laws holds impeller diameter as constant. The second set of affinity laws is different. It holds speed constant in order to solve for impeller diameter trim.

**Affinity Laws Set 2:**

*The speed is held constant. Let’s solve for impeller trim.*

- Flow in GPM is the same as shaft speed (1780 RPM, for example).
- Head is shaft speed squared. (1780 * 1780 RPM).
- Power (BHP) is the cube of shaft speed (1780 * 1780 * 1780 RPM).

Q1/Q2 = D1/D2

- Q is capacity (flow in GPM) and D is impeller diameter (imp dia). 1 is the first baseline pump; 2 is a second of that same pump with an imp dia change having affinity in predicting second pump imp dia based on capacity change.

*What that’s saying is this:*

Q1(capacity of baseline pump 1)/Q2(capacity change for pump 2) = [D1 (imp dia of baseline pump 1)/D2(imp dia change for pump 2)].

Visualizing this scenario, I see two of the same exact pumps sitting side-by-side. Each runs on the same motor speed (1780 RPM, for example). But each pump calls for a different capacity (flow in GPM). How will the flow difference in that second pump change the impeller trim, since capacity and impeller diameter are related? *Let’s do this!*

In the second set of Affinity Laws (the set holding speed as the constant), it’s also true that:

- H1/H2 = (D1/D2) squared
- BHP1/BHP2 = (D1/D2) cubed

*To summarize*, the first set of Affinity Laws holds the impeller diameter as an unchanged constant in order to solve for motor speed required to accommodate a pump rating. In set one of the Affinity Laws, the problems can be solved by either changing capacity, feet of head, or brake horsepower to solve for speed. The second Affinity Laws set holds the speed constant in order to solve for impeller diameter trim. In set two, the problems can be solved by either changing capacity, feet of head, or brake horsepower to solve for impeller diameter trim.

*To conclude*, the Affinity Laws are a good rule of thumb, but *can have up to a 15%-20% margin of error* when solving for impeller trim. Slower motors tend to allow for greater impeller trim while following the Laws than higher specific speed motors. I personally find it interesting that capacity is equal to, while ft/head is squared and brake horsepower is cubed to solve for these variables. It’s so neat and tidy to have these variables line up for solving that way. I hope this information brings value to you. Please feel free to hold onto this for your next *learning “curve”* adventure!

With warm regards,

Jennifer Zadka

PS: Here’s my reference source: Pump Characteristics and Applications, 3rd Edition by Mike Volk.