TACO Hydronics Step by Step

In this program, we’re going to build a hydronic system step by step, and piece by piece from Heat Loss through start-up.

1

This may be an oversimplification – but any hydronic system needs to start with a heat loss. Do the MATH! This establishes a target for your heating system, as well as some documentation for your customer. There are tons of heat loss calculation methods out there – but all stem from IBR, ASHRAE or ACCA’s Manual J. These, plus all the heat loss calculators available in software form all use the same math formula. They may differ in what might be loosely termed as “safety” or “fudge” factor. Suffice it to say, however, that all neat loss formulas have plenty of safety built in and there’s no need on our part to add any extra! Heat loss comes from two separate elements – the first is air infiltration, or heat lost due to heated air leaking from the inside of the house to the outside and being replaced by cold air coming in to replace it. This air will need to be heated up to comfortable levels. Infiltration heat loss is – along with window loss – the largest single heat loss element. Heat loss calculations will also need to made for conductive heat transmission through walls, ceiling and floors, as well as through doors, windows and skylights. By any measure, however, heat loss is an educated estimation. You can go to great lengths to be as accurate as possible, but acceptable results can be found by simply rounding up your calculations. It’s also important to do a room by room heat loss analysis. This will enable you to adequately size the heat emitters – baseboard, radiators, radiant floor heat – for each room. A whole house heat loss is faster and simpler and will help you size the heating plant, but it doesn’t help you when it comes to making each room comfortable.

2

Let’s do a quick example. This exercise is meant to show you how a heat loss is done manually. Taco has training programs to help you do a more in depth heat loss analysis, as well as computer software that can design an entire system for you. It’s important, however, to know how to do a heat loss manually so that you’ll know what the numbers mean and how the software came up with them. For this example, let’s do a simple 10 by 15 room with 9 foot ceilings. The room has two outside walls with windows. When conducting a heat loss analysis, we need to determine the indoor design temperature – meaning the indoor temperature we wish to maintain during the heating season – in most cases the number to use is 70 degrees. The other number we need to determine is the outdoor design temperature, or the “coldest day of the year.” Our goals when designing a heating system is to keep the structure at 70 degrees when the weather reaches that outdoor design temperature. ASHRAE publishes recommended design temperatures for different parts of the country. For our example, let’s use an Outdoor Design Temperature of 0. Some people like to exaggerate this number to make sure their customers have “enough heat” if the temperature ever drops below the recommended temperature. This practice can often lead to needless oversizing of a boiler, which can add unnecessary cost to the job, as well as lower the overall efficiency of the system, since oversized boilers don’t run efficiently. In our example, we have an indoor design temperature of 70, with an outdoor design temperature of 0. We’ll use this information during the calculations in the form of DTD, or Design Temperature Difference. That’s the difference between what we want indoors when it’s at design conditions outdoors. We want 70 indoors when it’s 0 outdoors, so that’s a DTD of 70 degrees. Another factor needed for heat loss is the so-called infiltration factor. This is a number that helps us determine the rate of air leakage, and can vary based on how old the house is and how well sealed the structure may be. New construction tends to have rather low infiltration rates, while older homes may leak quite a bit. For our purposes, we’ll use the IBR factors for infiltration: 0.012 for rooms with one outside wall, 0.018 for rooms with two outside walls, and 0.027 for rooms with entries or with three outside walls. By way of explanation, 0.012 represents 2/3rds of an air change per hour, and the BTU’s needed to warm that air up again. Technically, 0.012 is the amount of heat needed to raise 2/3rds of a cubic foot of air one degree. Let’s do the math. To find the infiltration heat loss, we first need to find the volume of air in the room. We do that by multiplying the length times the width times the height. That tells us how many cubic feet of air we’re dealing with. We then multiply that times the DTD, and then multiply by the infiltration factor. In this example, we do that following calculation: L x W x H x DTD x Infiltration factor = infiltration loss. 10 x 15 x 9 x 70 x .018 = 1,701 BTUH

3

Now let’s do the rest – which is all conductance heat loss through the walls, ceiling (if there’s a cold space above), floor (if there’s a cold space below), windows, doors and skylights. The formula for all these losses is the same: L x W X DTD x U. Length times width gives us the area and DTD is the design temperature difference. U represents the U-Value of the assembly. U-value represents the assembly’s ability to conduct heat from one side to the other. Obviously, when it comes to heat loss, we’d like as low of a U-value as possible. In fact, U-value is the inverse of R-value, which is an assembly’s ability to resist the flow of heat. If you know the R-value, simply divide 1 by the R-value and you’ll get the U-value. Let’s do the windows first. We look at the plan and see that we have two 3x5 windows on one wall and one 6x5 window on the other. Windows have U-values on their stickers when they’re installed, but there is plenty of data available – especially in heat loss computer programs – to help you determine the u-value of an existing window. For our purposes, we’re using standard wood framed, double-pane, low-E windows, which have an average U-value of .36. To find the window heat loss, we total up the square footage of all the windows, and count them as one big window – it’s easier that way. In this case, the room has 60 square feet of window area. We multiply 60 times the DTD of 70, and then multiply again by the U-value of .36, and we come up with a window loss of 1,512 BTUH. Walls are next – we want to use the NET area of the walls, and then multiply by the DTD and U-value. We have two outside walls here – one ten feet long and the other 15 feet long. To make it easier, we’ll simply add the lengths together and make it one big wall. (NOTE – with heating, north, south, east or west facing walls don’t matter because often we’ll be trying to heat the room when it’s dark out). In this case, we have 25 linear feet of wall times 9 feet high for a total area of 225 square feet. We then need to subtract the window area from that total – 225 minus 60 equals 165 square feet of net wall area. This is the number we’ll use for our calculations. We also need to find the U-value for the wall. Typically, we look at the level of insulation in the wall – in this case it’s 6 inches of fiberglass at an R-19. You can also look up the associated R-values for the exterior sheathing, house wrap and finish, as well as for the inside sheetrock, and that would be more accurate. But in the name of simplicity and safety margin, many folks will simply use the insulation R-value for the calculations. In this case, we divide 1 by 19 to find the wall U-value of .05. Let’s multiply it all out – 165 x 70 x .05 equals 578 BTUH lost through the walls. Outside doors, if there were any, would be calculated in the same way and their area would also be deducted from the Net Wall area.

4

Let’s finish up with the ceilings and the floors. We would only need to calculate either of these items if there were unheated spaces above or below the room. If the area below the room was heated, there would be no heat loss through the floor, nor would there be any loss through the ceiling if there was a heated space above the room. We use the same math formula. For the ceiling we multiply the length times the width times the DTD times the U-value. Again, for simplicity we’ll look at the insulation values. With R-38 in the ceiling, the math tells us the U-value would be 0.02. We multiply out 10 x 15 x 70 x .02 and we come up wth a ceiling loss of 210 BTUH. The floors have R-19 insulation, so we calculate a U-value of 0.05. We then multiply 10 x 15 x 70 x .05 for a floor loss of 525 BTUH. Finally, we add them all together for find the total heat loss of 4,526 BTUH for the room.

5

6

Let’s define some of the terms used in the chart. First is AGA Input for a gas boiler. This is the Gas Input needed per hour for the boiler to operated as rated. In this example, the AGA input is 120,000 BTUH. Since one therm of natural gas contains 100,000 BTUH, this means the input rate of the selected boiler is 1.2 therms per hour. Next is the DOE – or Department of Energy – Capacity. This is a federal rating – note on the boiler we’ve selected to DOE output is 101,000 BTUH. If we divide the DOE output by the AGA input, we come up with 84.1%, which happens to be the AFUE rating of the boiler. What is means is that for every 120,000 BTU’s worth of fuel you put into the boiler, the DOE says you’ll get 101,000 usable BTU’s out of it. Another item of note with the DOE capacity – it assumes the boiler and all the distribution piping is installed in a heated area and that any jacket losses and piping losses are usable heat that will help offset the heating load of the building. The I-B-R net output de-rates the DOE capacity by about 15%, under the assumption that 15% of the boiler’s output will be lost through the piping and jacket, or is needed to warm up the thermal mass of the boiler and the piping following an off-cycle. This is called the “pickup” allowance, and accounts for warming up cool cast iron boiler sections as well as cool steel or copper distribution piping. The IBR Net Output assumes the boiler and the distribution piping is in cold or unheated areas. Which one should you use? Well, it depends on where the boiler is to be installed. If the boiler and distribution piping are in a heated space – such as in a mechanical room in a finished, heated basement, then the DOE capacity would suffice. If the boiler and a good portion of the distribution piping is to be installed in an unheated basement, garage or crawlspace, then the IBR number is suitable.

7

Let’s fast forward now to the completed heat loss, which is 75, 421 BTUH. We need to select a boiler that will handle that load. Here’s a chart from a gas boiler manufacturer showing their different size boilers. The numbers we want to look at for sizing are either the DOE capacity or the NET IBR RATING. We’ll explain those in a moment, but if we consider the heating load of 75,421 BTUH, is appears we could use either the S-90 model, which has a DOE capacity of 76,000, but only 66,000 NET IBR, or we could use the S-120, which has a DOE of 101,000 with a NET IBR of 88,000. both will cover the job without being oversized, but let’s look at it a little closer.

8

We’ve talked about pressure loss and pressure differential. In hydronics, pressure is typically referred to as head pressure. Technically, head is the total mechanical energy content of a fluid at a given point in a piping system. We use it to express pressure loss in the piping system. When water flows through the system, it will encounter friction loss due to the piping. The greater the flow through a given system, the greater the pressure loss. The circulator needs to produce the required amount of flow while overcoming that pressure loss. Pressure is generally measured in pounds per square inch, or PSI. In hydronics, we use head loss. Converting PSI to head loss is very simple. A column of water 2.31 feet, or 28 inches high, will have a gauge pressure at the bottom of 1 PSI. So simple math tells us that 1 PSI of pressure drop in a system equals 2.31 feet of head.

9

Why are residential hydronic systems typically filled to 12 psi? Well, if you have a plumbing license, you know that a column of water 2.31 feet (or 28 inches) high has a gauge pressure at the bottom of 1 psi. It doesn’t matter how big around the column is, the pressure is still 1 psi. If we multiply 2.31 feet by 12 pounds, we’ll come up with 27.72 feet. That means if the boiler’s in the basement, 12 psi is enough to pressurize the water in a closed loop system on the third floor of a structure, with a little extra for safety. Obviously, taller buildings, particularly commercial buildings, may need a higher static fill pressure. Those jobs also use boilers with higher pressure ratings as well as higher rated pressure relief valves.

10

It’s important to remember than when sizing a circulator, you do not need to take into account the height of the building. The physical height of the building does NOT equal the feet of head. Remember, this is a circulator, not a pump, and we’re dealing with a closed loop system, not a well or a sump pump system. The circulator does not need to lift the water to the top of the building due to the simple fact that what goes up must come down. The circulator doesn’t have to lift the water to the upper floors – the weight of the water coming back down the return side is a counterbalance. Think of the circulator as the motor on a ferris wheel. The motor doesn’t have to lift the weight of the people up – there are people on the other side of the wheel coming back down. All it has to do is overcome the friction loss of the bearing assemblies in the wheel. A circulator doesn’t have to lift the water – it only has to overcome the friction loss – or head loss – of the system.

11

Next on the to-do list is the pressure reducing valve ( and backflow preventer, if required by code), air eliminator and expansion tank assembly. The PRV, of course, reduces the water pressure going in to the system down from street pressure to whatever the static fill pressure is, usually 12 psi. The air eliminator controls and removes air bubbles from the system, and the expansion tank allows for pressurization of the system and gives the expanding water somewhere to go as it heats up. Why place the PRV here? Well, the way these things work is that there is in internal diaphragm and spring that exert pressure on the city water inlet side. When the system side reaches the desired pressure, the spring and diaphragm will close the city side inlet. The PRV will close and won’t allow any more water into the system as long as the system pressure is at least the desired, set amount – again, usually 12 psi. The only way the PRV will allow more water into the system is if the system pressure drops below 12 psi, but that won’t happen here. The point where the expansion tank connects to the heating system is called the point of no pressure change. The pressure at this point will always be at least the static fill pressure (unless there’s a leak in the system!). Because of this point of no pressure change, the PRV won’t be allowed to let more water into the system under normal working conditions. We’ll discuss the point of no pressure change in greater detail in a few moments. This is the Taco 3350 cartridge type pressure reducing valve. It works as a fast-fill type valve as well. When filling or purging the system, you simply press down on the top of the green cartridge – this manually opens the diaphragm and allows the system to be rapidly filled at street pressure. Release the cartridge and the system will stop filling. The pressure may be set by rotating the black dial on the side of the cartridge – they come preset to 12 psi. The 3350 can be installed in any positon and has a fully replaceable cartridge in case service is ever needed. In addition, the unique design allows you to replace the cartridge without having to drain the system, which can save lots of time and hassle. The cartridge also rotates, so no matter how it’s installed, you can always read the pressure indicator.

12

Here’s a picture of what this part of the system should look like. You’ll note the 4900 air separator with the expansion tank and cold water fill connected to the ½” tapping at the bottom. As mentioned earlier, this is the perfect spot for the automatic fill/pressure reducing valve to connect to the system, because this is the one part of the system where the static pressure cannot change when the circulator operates. This spot, where the expansion tank connects to the piping system, is known as the point of no pressure change.

13

Why is this the point of no pressure change? Well, it’s the only area in the system where the circulator can’t change the pressure. Only three things can change the system pressure where the tank joins the system. First, the pressure will change if water is added to or removed from the system. Second, the system pressure will change if air is removed from or added to the expansion tank. And third, the pressure will change when the water is heated due to its expanding volume. The circulator can’t do any of these – it can’t add or remove water or air from the tank, and it can’t heat the water up to make it expand. The circulator has only one job, and that’s to create a pressure differential within the system in order to promote flow. And it can only create a pressure differential if it has a reference point – the circulator uses the expansion tank and the point of no pressure as that reference point.

14

So where should the circulator be in relation to the point of no pressure change? Well, let’s examine what will happen when you pump toward the expansion tank. Remember that the circulator must create a pressure differential – positive or negative – in order to move water. If the circulator is pumping into the expansion tank, it can’t raise the system pressure any higher than it already is, which is 12 PSI. So no matter how power the circulator is, the outlet pressure on the other side of the expansion tank will never be any higher than 12 PSI. If that’s the case, and if the circulator MUST create a pressure differential – what’s the only thing the circulator can do? Lower the suction side pressure and create the differential that way. That will create flow, but will also create a host of other issues.

15

A: when system is filled with cold water, the pre-charged pressure is equal to system fill pressure and diaphragm is flush against tank. B: As water temp increases, the expanded water is absorbed by the tank. C: As system water temp reaches max, the diaphragm flexes against the air cushion to allow for the increased water expansion

16

Here we show the circulator pumping into the expansion tank connection – the point of no pressure change. The circulator is off – the gauges read 12 psi at the tank, and at both the suction and discharge side of the circulator. The system is 18.5 feet high, so the static pressure drops roughly 8 psi – down to 4 psi – at the peak of the system. The weight of the water coming back down the return side is what brings the pressure back to 12 psi at the suction side. What happens when the circulator is turned on?

17

Now that the circulator is on, it’s creating a 4 psi pressure differential to create flow. Since the circulator can’t change the pressure at the tank – or point of no pressure change – the pressure there will remain 12 psi. The circulators differential will show itself as a pressure DROP on the suction side. A gauge on the outlet side will still read 12 psi, but the gauge on the suction side will drop to 8 psi. Since the system will lose 8 psi on it’s way to the peak, and add 8 psi back down to the suction side of the circulator, the pressure at the peak of the system will be zero, or close to it. This can create a system with air noise, or worse if there are manual air vents installed in the line – if the pressure in the system ever pulls negative, the air vents become air intake valves and can cause the system to get air bound.

18

Here we move the circulator so that it pumps away from the point of no pressure change. When the circulator is off, we have the same scenario as before, with the pressure reading 12 psi on either side of the circulator, and 4 psi at the highest point of the system.

19

When the circulator is turn on, since the system pressure on the suction side of the circulator can’t drop below 12 psi, the circulator will ADD its pressure differential to the system. The pressure reading on the outlet side of the system is now 16 psi. The system will lose 8 psi on its trip to the highest point – the pressure there will now be no lower than 8 psi – from that point back the weight of the water will increase the pressure back to the 12 psi at the expansion tank. This will keep air dissolved in the water solution and prevent the system from ever getting air bound or experiencing air related noise issues – such as a babbling brook sound. Keeping the air dissolved will also enhance system performance and circulator life.

20

21

What we call a “pump” in hydronics is a bit of a misnomer. It’s not really a pump – it’s a centrifugal pump, or a circulator. It doesn’t lift water the way a well pump lifts water – it circulates water in a closed loop system by creating a pressure differential, meaning it takes the fluid that comes into it at a lower pressure and sends it back out at a higher pressure. When the circulator creates a pressure differential, mother nature takes over because high pressure will always go to low pressure, thus creating flow through the system. The trick for the circulator is to create enough of a pressure differential to produce adequate flow through the system. How does the circulator create a pressure differential? Simple – by using centrifugal force.

22

Here’s a line drawing of the guts of a Taco 007 circulator. You see flow entering at the bottom of the circulator. There’s a cast iron or bronze casing for the circulator – known as the volute. You also see the impeller – which is part of the circulator cartridge which houses all the moving parts of the 007. The cartridge fits into the steel motor housing and the motor itself, which makes the impeller go round and round.

23

Here’s a closer look at the cartridge. In the picture, you see the open faced impeller --With the 00 series, it’s a water lubricated cartridge, meaning there’s always water in the cartridge. You’ll also hear this type of circulator referred to as a wet-rotor circulator. As the system is filled and pressurized during initial start-up, system water is forced into the empty cartridge through the hollow ceramic shaft. In turn, the air in the cartridge is forced out, also through the ceramic shaft. Once the system is filled and pressurized, small amounts of air may remain in the cartridge. This remaining air will be purged from the cartridge as soon as the pump is turned on. The spinning of the shaft and rotor assembly creates a centrifugal suction that replaces the air with water. Once the cartridge is full of water the hollow ceramic shaft acts as a mini expansion tank. It provides the exact amount of space so the water in the cartridge can expand and contract, as the water heats and cools, without bringing new system water into the cartridge.

24

Here’s a closer look at the impeller itself. When the circulator is in operation, water flows into the suction side of the volute –which is always wider and larger than the discharge side -- and then into the eye of the impeller. As the impeller rotates, the vanes “slap” the water from the inside of the impeller to the outside of the impeller – adding velocity to the fluid. The fluid then moves to the discharge side of the volute – which is smaller and narrower. The collection chamber on the discharge side turns the kinetic energy of the fluid – energy due to velocity, into pressure. So the water comes into the circulator at a low pressure, and due to centrifugal force and the design of the circulator, leaves at a higher pressure. Some things to know about impellers – the thickness, diameter and construction all play a role in the performance of the circulator – just as much as motor horsepower. First, different circulators will have thicker or thinner impellers – the thickness of the impeller determines its flow capacity – the thicker the impeller, the more flow the circulator will be able to produce. The diameter of the impeller is also important. The larger the diameter –meaning the bigger around the impeller is, the more velocity the circulator can impart of the fluid. The more velocity, the more pressure the circulator can produce. Horsepower also plays a role with any type of impeller – the faster the impeller spins, the more flow AND pressure the circulator can produce. Also not the drawing in the center represents a closed vane impeller. The picture on the side shows an open vane impeller. Open vane impellers are used in circulators designed to provide high flow and relatively low head. These types of circulators are called flat curve circulators – as you will see when we look at performance curves. The Taco 007 and 0010 are considered flat curve circulators and are used in radiator and baseboard applications which require higher flow rates but have lower overall head losses. Close vane impellers are used for higher head and medium to high flow circulators, such as the Taco 008 and 00R. These are designed for lower flow systems that produce higher head losses, which are typically found in most radiant applications.

25

26

27

28

29

30

31

32

To size a circulator properly, you gotta do the math. By that we mean a heat loss analysis of the structure so you know what the job really needs. The flow rate needed for a particular job or a particular zone is based solely on the heat loss. Any heat loss analysis should be conducted on a room by room basis, since each room will have its own unique heating requirements and will need baseboard or a radiant sized for its unique needs. When you have each room figured out, then you can group rooms together into zones, and determine what the zone heat loss would be. This is important when zoning by circulator, since you will select each circulator based on the zone flow requirement. When zoning with zone valves, it will be important to know the flow rate for the entire job so the circulator will be big enough. Once we know the BTUH load of the zone or of the entire job, depending on whether we’re zoning with zone valves or circulators, we then need to calculate the actual flow rate using the Universal Hydronics formula: GPM (or gallons per minute) = BTUH divided by Delta T times 500. The BTUH is, of course, the heating load. To determine the flow rate, we’ll divide that load by the Delta T ( or temperature drop) times 500. The Delta T is the temperature drop across the piping circuit – typically with baseboard or radiators the Delta T is 20 (meaning the water goes out to the zone at 180 and returns to the boiler at 160). 500 is a constant, and represents the weight of one gallon of water (8.33 lbs) times the number of minutes in an hour (60) times the specific heat characteristic of the fluid used – in this case water (it takes one BTU to raise the temperature of one gallon of water one degree in one hour).

33

So let’s do one. Let’s say we’re zoning with circulators, and we have a zone that has 27,000 BTU’s worth of baseboard, or roughly 45 feet. It’s a good sized zone. Since it’s baseboard, we’re using a 20 degree Delta T, and we’re also using 100% water – no glycol. To determine the flow rate, we simply divide the load – 27,000 – by the Delta T of 20 times 500. If we math it out, we will divide 27,000 by 10,000 (20 x 500) to determine an actual flow rate for the zone of 2.7 gallons per minute. What size pipe should we use for this zone? Well, the guidelines for pipe sizing are as follows: 2 to 4 gallons per minute of flow, use ¾” M copper; 4-9 GPM, use 1 inch; 8 to 14 GPM, use inch and a quarter; 14 to 22 GPM, use inch and a half. These all fall within hydronics guidelines for pipe sizing and keeping flow velocities at no less than 2 feet per second and no more than 4 feet per second. At velocities greater than 4 feet per second, the system will produce velocity noise and customer complaints. At velocities lower than 2 feet per second, dissolved oxygen will tend to come out of solution and cause air problems within the system.

34

To determine the head loss of a zone, start by measuring the total length of the zone, including the element. Once that is done, multiply the total by 1.5 to allow for fittings, valves, etc. Fittings and valves produce pressure drop in a system that are the equivalent of a few feet of pipe each, so multiplying by 1.5 accounts for most basic fittings and valves. We may need to add later for items such as flowchecks, 3 way valves (if used) and other high head loss items, but the .5 multiplier will take care of fittings and isolation valves. That multiplication will give you the total developed length of the circuit. Next, take that number and multiply by .04 – this number represents 4 feet of head per 100 feet of copper pipe. That head number applies as long as the pipe has been sized according to the velocity guidelines shown in the previous slide. The end product is the head loss for the zone. Let’s do an example. We measure out zone to be a total of 80, including all the element and the supply and return piping to the boiler. If we multiply 80 times 1.5, we get a total developed length of 120 feet. Let’s multiply that times .04, and we get a head loss for the zone of 4.8’ of head. To size the circulator, we’ll need one that can provide a total of 2.7 gallons per minute and overcome 4.8 feet of head.

35

If we take a look at the Taco “00” series performance curve chart – we can determine which circulator we should use for this zone. First, on the bottom axis, we find the flow rate – in this example, it’s 2.7 gallons per minute. On the vertical axis we have head loss – in this example it’s just under 5 feet of head. We follow the two lines until they intersect to find our operating point of 2.7 GPM at 4.8 feet of head. Next, we look at the performance curves to find out which circulator would make the best selection. In this example, a 005, 005 or a 007 would be good choices – with the most likely choice being the 007, since it’s the most common and most readily available.

36

We spoke earlier about open vaned impellers and closed vaned impellers. Circulators such as the 007 and the 0010 have open vaned impellers – producing high flow rates but a relatively flat performance curve. Pumps such as this can handle a wide range of flow rates in systems with relatively low head losses, such as baseboard and radiator systems. Flat curve circulators are also preferable in zone valve jobs. The amount of flow needed will depend on how many zone valves may be calling at a particular time. As you can see with the flat curve circulators – big changes in flow will result in very small changes in head produced which will help keep the system quiet. Steeper curve circulators, such as the 008 and some of the European models, will require pressure differential bypass valves to prevent velocity noise in zone valve applications. Another option, of course, would be the Taco 00-VS variable speed setpoint circulator set to a Delta T operating mode. Steeper curve circulators, such as the 008, it’s close cousin the 00R and the 0011 and 0013, are better for higher head applications, including radiant floor heating applications, as well as high head loss fan coils, air handlers and heat exchangers.

37

38

39

Sizing expansion tanks involves calculating water volume within a system as well as the temperature increase, etc. This simple rule can help you move tank sizing along. Tank manufacturers offer what they call 15 gallon, 30 gallon and 60 gallon tanks, based on how much water is in the system. This can also relate to BTUH requirement. For the most part, if the system requirement/output is less than or equal to 50,000 BTUH, a 15 gallon tank will suffice. From 50 to 150,000 BTUH, use a 30 gallon tank, and 150 up to 200,000 BTUH, use a 60 gallon tank. This will handle most residential applications.

40

The next step is to size the primary boiler loop. In hydronics, all pipe sizing is done based on flow, or Gallons per minute. To find the flow, we use the universal hydronics formula – GPM = BTUH divided by Delta T times 500. BTUH is the total heating load. Delta T is the temperature drop across the piping circuit – in this case we’ll use a 20 degree temperature drop across the primary, and 500 is a constant representing the weight of 1 gallon of water (8.33 pounds) times 60 (the number of minutes in an hour) times the specific gravity of the fluid used (water has a specific gravity of 1) times the specific heat of the fluid (water has a specific heat of 1). 8.33 times 60 times 1 times 1 equals 499 and change, we’ll call it 500. If we calculate out our flow rate, we’ll divide 75,421 BTUH by 10,000 (20x500) to determine a primary flow rate of 7.54 gallons per minute. The pipe is always sized to handle the required flow. Standard hydronics guidelines tell us that flow between 2 and 4 gallons per minute will require ¾” copper pipe. 4 to 9 GPM requires 1”. 8 to 14 is 1¼” and 14 to 22 GPM is 1½” pipe. This pipe sizing keeps systems within the velocity guidelines of hydronics which require a velocity of at least 2 feet per second and no more than 4 feet per second when using copper pipe. At velocities less than 2 feet per second the fluid won’t be able to keep air in solution and will sound like a babbling brook. At velocities greater than 4 feet per second the water will create velocity noise and sound like a freight train in the mechanical room. With a calculated flow rate of 7.54, rounded up to 7.6 GPM, the guidelines tell us we’ll need to use 1” pipe.

41

Here’s your basic cast iron Taco air scoop. It’s been around forever – it’s a one piece air separator with no moving parts. There’s an 1/8” tapping on top for a Hyvent and a ½” tapping at the bottom for the expansion tank and cold water fill connections.

42

How does an air scoop work? When water enters the scoop, it enters a larger area, so it’s velocity slows down and the pressure decreases slightly. As the velocity slows down and the pressure decreases, the air bubbles will separate from the water. There are turning vanes inside the scoop that will scoop the bubbles into the upper chamber, where they will vent out through the hy-vent at the top of the scoop. It’s a simple process with no moving parts. The scoop will only work, however, if there is a minimum of 18 inches of straight pipe on the inlet side. The air bubbles need to be on the top of the flow as the water enters the scoop, otherwise the vanes won’t catch the bubbles and the scoop won’t remove the air. The 18 inches is needed to create laminar flow - meaning the flow will smooth out and air will be allowed to float to the top. If the scoop is installed right after an elbow, the flow will be turbulent and the air will be mixed with the water and the scoop will be unable to remove the air.

43

Taco offers three levels of air eliminators that fall under the good-better-best categories. First the traditional Taco air scoop with a Taco Hy-Vent. A better alternative – one capable of removing lots of air rapidly from the system– is the Taco VorTech. And the best option of all – one that removes more air and smaller microbubbles than any other eliminator on the market, is the Taco 4900. We’ll discuss each of these indivudually.

44

The next step up in air removal is the Taco 4900 air separator. The 4900 is a high performance air separator, removing 3x times more air than the Spirovent according to independent studies. In addition, unlike the Sprirovent, the 4900 can be used with glycol in the system. No minimum pipe run is required with the 4900. The 4900 uses Pall rings to remove the air from the system. Like the air scoop, the 4900 is a large chamber. With will slow the velocity down and lower the pressure slightly. This allows disolved air to come out of solution – these microbubbles cling to the pall rings. There are dozens of pall rings in the 4900, representing a huge amount of area for the bubbles to cling to. Bubbles will continue to grab onto the pall rings and, over time, the bubbles will grow to a point where they will rise through the 4900 – the float mechanism will go up, opening the air vent to let the air out.

45