Simple Pump Selection Exercise

Here is another excellent article from Michael Housh, owner of Housh Home Energy and a regular contributor to HVAC School. Thanks, Michael!


I thought I would go through a simple example of sizing a hydronic circulator for an application. This is a made-up scenario, but I sketched out a 20’ x 20’ square home with baseboard radiators that encompass mostly all of the exterior walls. I should also mention that this article does not go into detail on choosing a radiator/panel; this is a hypothetical situation to show the steps involved in determining the total head loss (resistance) through a piping arrangement.

I equate the following to what Manual D would be for duct design (the air side of HVAC). Before I move on, there are a few points that I need to make clear. Because hydronic systems are closed-loop, it’s important to realize that the lift (or height) from the pump to the highest point of the system has no effect on sizing the circulator; the only thing that matters is the drag/resistance of the pipe, valves, and fittings in the system. So, whether a system is like the one in the drawing or a loop that travels 1000’ in the air, the process to size the circulator is the same.

Another thing I will just gloss over here is that for any of the following formulas, it’s assumed that the piping system has turbulent flow (Reynold’s number between 2,300-200,000). Turbulent flow means the water molecules are not traveling in a straight line; they gravitate from the center to the wall of the pipe and back again; this adds increased head/resistance, but it allows heat to be transferred through the wall of the pipe/radiator more easily.

While this is an imaginary project, and I’m going with ¾” piping for the entire loop, it is important when designing and selecting the size of the loop that we keep our velocity inside of the pipe between 2-4 ft/sec. The 2 ft/sec is the minimum velocity required to be able to get air to move downwards throughout the piping system. If we get above 4 ft/sec, we can have noise issues or damage the pipes.

The primary difference between this exercise and doing a Manual D is that we must consider every pipe, valve, and fitting in the entire circuit instead of only using the longest total equivalent length (supply and return) for design.

For some, it may be easier to think of this like an electrical circuit (Ohm’s Law); we add the resistance through the entire circuit to come up with the total resistance (ft. head) our system will have to pump against.

 

The next step is totaling up all the equivalent lengths of everything using charts available (most all of the charts that I use are in Modern Hydronic Heating by John Siegenthaler, and I can’t publish them because of copyright, but engineeringtoolbox.com is a good place to look for these types of resources). I created a spreadsheet from the data in my drawing and came up with the following data.


You may notice that the air-separator has a Cv (flow coefficient). I’m not going to go into details in this article on how I determined it was equivalent to 11’ of ¾” copper, but I will define Cv. Cv is the flow rate of 60° water to make a 1 psi pressure drop across the device. If you read my last article on setting up a pump, you should know that a pressure difference can be converted to ft. of head. There are some formulas that you can use to determine the equivalent length of a device for certain pipe sizes, but that is a bit complex for this article. Below, I combined the important totals to give us our total equivalent length of 162.8’ of ¾” copper.

While this article is getting long, the above information is gold!  It allows us to create a fingerprint for our system and determine the performance over a range of flow rates. With this information, we can create the system head-loss curve, which is defined by the following equation:

The reason I say this is gold is that we are able to create our system fingerprint over an array of flow rates, which will give us the head-loss we must deal with, as in the following graph:

In our hypothetical system, we will design around a 180° output temperature from the boiler and a 20° delta-T, giving us an average temperature of 170°. Since density and viscosity are dynamic, this is the number we will use to determine the fluid properties factor for our application. We can solve for this value using the following formula:

Next, all we need is to determine our pipe size coefficient, which is just a multiplier based on the internal dimensions and roughness of our tubing. This can be looked up in a chart as well; ¾” copper pipe is equal to 0.061957. I’m also going to pick a generic flow rate for the application of 4 GPM to wrap things up. So, now we can substitute all of our values into our head-loss equation, solving for our design flow rate of 4 GPM.

So, currently, we would need to select a pump that would produce 4 GPM at 5.13 ft. of head. We would do so by matching up the manufacturer’s data/pump curve to match our desired output. This is a deep subject, some of which carries over into the air side of HVAC, but I have just skimmed the surface on some of the topics. While it may seem overwhelming, the main takeaway from this article is going to be to relate the piping system to an electrical diagram.

In future articles, this will allow us to break down complex piping systems into more manageable pieces. As with most things in our industry, it’s not always about the information you have in your head; it's about your ability to find the correct information/formulas/resources when they’re needed. Keep on learning!

—Michael

 

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