Tag: hydronics

Another follow-up article by Michael Housh… Enjoy!


In this article I thought I would show a pump curve and match it up with our system head-loss curve that we created in the last article, but before I do that I thought I would talk a little bit more about the system head-loss curve and why I said it was so valuable.

 

When we design a hydronic system we must match flow-rates and output according to our Manual-J and the specifications for the radiator/panel that is being utilized (sort of like Manual-S on the air-side of HVAC).  Here’s an example output specification for a baseboard radiator that I found on the web.

You’ll notice that it gives two different output rows (1 GPM or 4 GPM), you may also notice that the output is relatively close for both flow rates. It is safe to assume that we can move 1 GPM with less power consumed than moving 4 GPM.  Let’s check the difference using the Head-Loss Equation from the last article.


This system is so small, that most circulators are going to produce more than enough flow, but you could maybe imagine how this could greatly affect a much larger or more complicated piping arrangement.

Next let’s look at a pump curve, the following is for a Taco VT2218, this pump fits a ton of applications, is variable speed, and maintains a constant Delta-T across our system.  In today’s world, it doesn’t make sense in most applications to not utilize an advanced pump, in my opinion. Now using our 4 GPM and 5.13 ft. of head, we can see that we fall somewhere between speed 1 and 2 (or well within the shaded green area on the right), so this pump will be great for this application (and hopefully sized right for future exercises where make the piping arrangement more complex).

To end the article I thought I would follow up with more detail on the Flow Coefficient (Cv) mentioned in the last article for our air-separator.  It is not uncommon in design for items to have a Cv value instead of equivalent length. If you remember from the last article, Cv is the amount of flow (GPM) required at 60° water (important because of density) to create a 1 psi pressure drop across the device.  We can use the following equation to convert Cv to pressure drop for our design conditions using the following formula.

Given the above equation, we can solve for our pressure drop using the design criteria of 170° and a flow rate of 4 GPM.  The density of water at 170° is 60.81 (I have also created an interactive Water Density Chart online).

Using the equation from the determining pump flow article we can convert this pressure drop to feet of head.

Finally we can solve for the equivalent length for a given pipe size coefficient using a rearranged version of our system head-loss equation from the previous article.

I hope this has helped gain a deeper understanding of selecting a circulating pump and how to convert a Flow Coefficient (Cv) into an equivalent length.  In future articles, I hope to create a more complex piping arrangement to show the pump selection and equivalent lengths for those applications.

— Michael

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.  It should also be mentioned 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 a 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 you have 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 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 then 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, as opposed to 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 because of copyright I can’t publish them, 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 give the definition of 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, then you should know that a pressure difference can be converted to ft. of head, so there are some formulas that can be used to determine the equivalent length of a device for a certain pipe-size but 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 are going to 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 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 is something that is looked up in a chart as well, and for ¾” copper pipe is equal to 0.061957. I’m also going to pick a generic flow rate for the application of 4 GPM, so we can 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 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 some of the topics I have just skimmed the surface on.  While it may seem overwhelming, the main take away 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, but your ability to find the correct information/formulas/resources when they’re needed. Keep on learning!

— Michael

 

Here is another great explanation from Michael Housh from Housh Home Energy in Ohio.Thanks Michael!


I’m going to layout and compare the Sensible Heat Rate equations for both the air-side and water-side of HVAC, to help draw similarities and dive deeper into the science behind these equations.  This is the beginning of a series to try and help us all gain a deeper knowledge of where these equations come from. The more we learn about the two, the more similarities can be drawn between them. This will hopefully allow a technician to be more comfortable when faced with different systems in the field.  I should also note that while the equations can be complex, they are a great reference for those who would like to build them into spreadsheets (or other formats).

 

Sensible Heat Rate Equations:

 

Air

Water
Q = 1.08 * CFM * TQ = 500 * GPM * T
Where:Where:
Q = sensible heat transferred (Btu/hr)Q = sensible heat transferred (Btu/hr)
CFM = quantity of air (ft3/min)GPM = quantity of water (gallons/min)
T= dry bulb temperature difference (°F)T= dry bulb temperature difference (°F)

 

The only thing I will say about the Delta-T side is that it is the measurement of dry-bulb temperature, this is something I think all technicians know and have a decent grasp on.

 

Like most things in our industry these are “rules of thumb” equations, however, both derive from the same lower level equation, which is as follows:

 

Q = M * C * T

 

Where:

Q = sensible heat transferred (Btu/hr)

M = mass of the fluid (lb/ft3)

C = specific heat of the fluid (Btu/lb)

T= dry bulb temperature difference (°F)

 

I’ve often heard Bryan say that air-conditioning is about moving pounds of refrigerant.   We move pounds of refrigerant to create air-conditioning, and we move pounds of air to deliver that air-conditioning to the space.  As you may have gathered from the above equation the Sensible Heat Rate is derived from moving pounds of a substance (in our case air or water).

 

I’m not going to dive into the details of the above equation at this time, but wanted to share where both of these equations stem from. What I’d like to breakdown in this article is a deeper understanding of where the 1.08 (air) and 500 (water) constants come from.   Both CFM and GPM are actually what provides the “pounds” of the fluid (air is a fluid), based on density and specific heat.

 

Density is defined as its mass per unit of volume (or weight per unit of volume), and specific heat is the rate at which an object will give off or absorb thermal energy.  Both density and specific heat are moving targets, but in the “rule of thumb” below are the values that are used.

 

Density (lb/ft3)Specific Heat (Btu/lb)
Air @ 70°F & at sea-level (14.7 psia)0.0750.24
Water @ 60°F62.371.0

 

I’m going to solve for the water-side first.  We have to take into account that our measurement for water is Gallons Per Minute, so for anyone who doesn’t know, there are approximately 7.48 gallons in 1 cubic foot.  Using the density from the table above we can solve for the weight of one gallon of water 62.37 / 7.48 = 8.34 lb @ 60°F.  Since our end result of the Sensible Heat Rate equation is Btu/hour we have to convert our GPM -> GPH (gallons per hour).  So, our 500 is a simplification of the following values:

 

Constant = 8.34 (lbs/gal) * 60 (min) * 1 (specific heat) = 500

 

When we provide the GPM in the Sensible Heat Rate equation for water, we have already accounted for its density (mass), specific heat, and converted to gallons per hour.

 

Next, let’s look at the air-side.  Here we have to account (just like in the water-side), that our measurement is in Cubic Feet per Minute, and since we are solving for Btu/hour we will have to convert CFM – > CFH (cubic feet per hour), we also have to use the density to account for the mass of air that we are moving, and the specific heat.  So, our 1.08 is a simplification of the following values:

 

Constant = .075 (density [ lbs/ft3]) * 60 (min) * 0.24 (specific heat) = 1.08

 

So, just like the water side, when we provide CFM to the Sensible Heat Rate equation, we have already accounted for its density (mass), specific heat, and converted to cubic feet per hour.

 

I hope I haven’t utterly confused you on such a technical topic, but stay tuned for more in the series to help bring the Sensible Heat Rate equation (and the air / water side) closer together.

— Michael Housh

Michael Housh from Housh Home Energy in Ohio wrote this tip to help techs determine the air side charge on a pressure tank. Thanks, Michael!


Determining the air-side charge of an expansion tank in a hydronic heating system is a relatively easy task.  A properly sized and charged tank is designed to keep the system pressure about 5.0 psi lower than the pressure relief while the system is at maximum operating temperature.

 

The proper air-side charge is equal to the static pressure of the fluid at the inlet of the tank plus an additional 5.0 psi allowance for the pressure in the top of the system.  The air-side of the tank should be checked and adjusted before adding water to the system, if the tank is already installed and the system has pressure in it, the pressure should be drained at the tank to 0 psi before testing the pressure on the tank.

 

The formula for calculating the air-side pressure is relatively easy and directly related to the highest point in the system from the inlet of the expansion tank.

 

Pa = H * (Dc / 144) + 5

Where:

Pa = air-side pressure in the expansion tank (psi)

H = height from the inlet of the tank to the highest point in the system (ft)

Dc = density of water at its coldest state / typically filling (lb/ft3)

 

The above graph shows us the relationship between density of water and temperature between 50°F – 250°F.

 

A lot of the “rule of thumb” equations for hydronic systems are based on the density of water @ 60°F is 62.37, so we could simplify the above equation into a rule of thumb equation by first solving for the density (Dc).

 

Dc = 62.37 / 144 =0.433

 

Substituting ‘Dc’ into the original equation would give us a slightly less complicated equation that can be used as a rule of thumb to solve for the air-side pressure.

 

Pa = H * 0.433 + 5

 

Below is a graph that shows us this rule of thumb equation and the required air-side pressure based on the height of the system piping.

— Michel H.

This tech tip was written by a friend of HVAC School, Brian Mahoney HVAC instructor at Western Suffolk BOCES/Wilson Tech. Thanks Brian!


The podcast on delta T for A/C the other day got me to thinking about the formula I learned in school about calculating the GPM of a hydronic system using a handy formula. We will be using the following values:

Td – temp difference of your supply vs return

Net boiler output(btu) use the boiler plate rating or get fancy and do an efficiency test and multiply your rated input multiplied by your efficiency rating. On an oil system, the unit could be down-fired.

It may be rated for 1 gallon per hour (140,000 BTU per hour input, but it may be firing with a .85 gallon per hour nozzle. So you have to do the math:
1 gallon of #2 fuel oil contains about 140,000 BTUs. Multiply that by .85 (your nozzle size) and you get 119,000 btu/hr input. Input would be 119,000 x .80 efficiency = 95,200.

500 – a constant which stands for a pound of water times 60 minutes – 8.33 x 60 = 499.8 (we fudge a bit.)

This is the weight of water at 60 degrees. You could look up the weight at the temp you are working with and multiply by sixty but it wouldn’t be far off.

To find a system’s gallon per hour:
BTU/ (500 x TD)
100,000/(500 x 20)
100,000 / 10,000= 10 GPH

Nice, but is there anything else you can do with this? How about a room that’s not warm enough. Is your baseboard supplying enough heat? You could look up the specs for that product, maybe. But what if it has dirty fins or mud in the pipe that is affecting temperature transfer. How would you know?

By using your Testo temp clamps on either end of the baseboard you find your temperature difference and using the data from the last calculation you solve for net BTU output of the baseboard

Btu = GPH x 500 x td
10 x 500 x 2 = 10,000 btu/hr

Now you know what you are getting. So you can check the specs of that baseboard and see if it’s giving you its rated output. If it is you don’t have enough baseboard or you have a problem with the room; thermal bypass for instance.

If it’s not performing as rated and the fins are clean you have an internal problem such as mud in the pipe insulating it.

Just something for the wet-heads.

— Brian M.

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