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hot water circulating manifolds
Glenn Sossin_2
Member Posts: 592
There are some key factors that you didn't mention - the size/rating of the boiler & the heating load of the loops. The manifolds are collection points to distribute and collect the system water flows. The key factor in determining it's size is the # of gpm that have to pass through it. You could do this using computer software, but since your questioning the old timer, lets look at it the conventional way using charts/graphs/tables. I will try to outline the concepts involved in a simple manner.
The rule of thumb for residential heating is each 10,000 btu's of heat is equivalent to a 1gpm flow with a 20deg delta T (difference between the supply & return temps). If you had a 100,000 btu input boiler, you could size the header so it could carry 10gpm of water. Actually it would be somewhat less because some of that heat energy went up the chimney. For the purpose of this example, assume 10gpm is the flow rate we need to extract the full capacity of the boiler based on a 20degree delta T.
Now lets look at the heat load. Supposing the 3 zones had 33ft of baseboard each and were 100ft long circuits. The 99 ft of baseboard would yield/require approximately 60,000 btu's - approximately 20,000 btu's for each circuit. Each circuit would therefore would require/carry 2 gpm. The header has to carry all the water from all the baseboard circuits to and from the boiler. How do we decide what size it should be?
It has little to do with the boiler tapping. In fact, the front and rear sections of a typical residential boiler are the same regardless if it's 3 sections or 9 sections.
So lets see if the old timer was safe using the above example constraints. Assume the header was made out of copper. Take a look at the attachment labeled friction loss 1 114. Using the top half of the page that represents 1" tubing, go down the left hand GPM column to 6 gpm. Now follow that row across to the middle section labeled Type L Tubing. You should come up with a value of 2.75ft . This represents the friction/head loss when you try to pump 6 gpm though 100ft of 1" L.
Wait a second,.... our headers are maybe 10ft long in total. That means the head loss through a 10ft section of 1" L, at 6 gpm is only .275 ft (2.75 x .10). You will learn, this is a very low number.
Now lets look at the head loss through the 3/4 copper loop that contains the 33ft of baseboard. Look at the attachment labeled 38 12 34 . Find the section labeled 3/4. Go down the right hand column to 2 gpm (remember thats approximately how many gallons of water we needed to handle the 33ft of baseboard in that loop). Follow across to the right to the section labeled Type L Tubing. You should come up with a number of 1.44 . This is the friction loss for pumping 2 gpm through 100ft of 3/4 L copper. For the purposes of this example, assume the losses through the boiler, valves and various fittings to be 4ft of head. Now we can add all these together to get the total head of the system. We had .275 ft for the header, 1.44 ft for the 100ft baseboard circuit, and another 4ft for boiler piping and fittings. When calculating system head loss, we take the highest value amongst all the heating circuits. So we have .275 ft (header) + 1.44 ft for the 100ft baseboard circuit and 4ft for the boiler piping and fittings. Thats a total of just under 6ft of head. Now we can select a pump based on handling 6gpm at 6ft of head if we use a 1" copper header.
Now look at the attachment titled grundfoss pump curve. The horizontal x axis is gpm and the vertical y axis is ft of head (friction loss). Plot out the point for 6 ft of head and 6 gpm. Now look at that point in relation to the red curve marked 7. At 6 gpm, this pump will produce around 14ft of head. We calculated we needed approximately 6 ft of head. This means that the curve for pump 7 has plenty of capacity to spare.
Now look at pump curve # 8. At 6 gpm, this pump will barely make 3 ft of head pressure. We determined we needed 6 ft so this pump is out. Pump 6 or 7, both very popular models will work just fine with the 1" header.
But what about a 3/4" header? Same procedure. Look at the friction loss for pumping 6 gpm though 10ft of 3/4" Type L. You should come up with 9.80 ft of head. Remember, this value is for 100ft and we have 10ft, so we multiply by .10 and get a head value of .98 ft . Now we add them all up, the header, the baseboard circuit, and boiler/fitting/valve losses. We would then have, .98 + 1.44 + 4. This comes out be approx 7ft. Again, back to the pump curve. If you plot out 6gpm and 7ft of head, pump 6 or 7 will have no problem with this scenario.
If you could follow all of that, think about this, what would happen if he tried to have just 1 circulator servicing all 100ft of baseboard in a series loop configuration making the loop 300ft long. What happens now. Look at the 3/4 table, get the value at 6 gpm and multiply it by 3 to equal 300ft worth of tubing. Then add the number up. We would need a pump that would provide 6gpm at approximately 35ft of head. None of the pumps on this chart could do it. You could find one but it would would be costly - 6 - 10 times the cost of a typical residential circulator.
So in the end, whether the old timer actually calculated all of this, it looks like his 3/4" header would work. Sometimes the old timers are right just based on experience.
The rule of thumb for residential heating is each 10,000 btu's of heat is equivalent to a 1gpm flow with a 20deg delta T (difference between the supply & return temps). If you had a 100,000 btu input boiler, you could size the header so it could carry 10gpm of water. Actually it would be somewhat less because some of that heat energy went up the chimney. For the purpose of this example, assume 10gpm is the flow rate we need to extract the full capacity of the boiler based on a 20degree delta T.
Now lets look at the heat load. Supposing the 3 zones had 33ft of baseboard each and were 100ft long circuits. The 99 ft of baseboard would yield/require approximately 60,000 btu's - approximately 20,000 btu's for each circuit. Each circuit would therefore would require/carry 2 gpm. The header has to carry all the water from all the baseboard circuits to and from the boiler. How do we decide what size it should be?
It has little to do with the boiler tapping. In fact, the front and rear sections of a typical residential boiler are the same regardless if it's 3 sections or 9 sections.
So lets see if the old timer was safe using the above example constraints. Assume the header was made out of copper. Take a look at the attachment labeled friction loss 1 114. Using the top half of the page that represents 1" tubing, go down the left hand GPM column to 6 gpm. Now follow that row across to the middle section labeled Type L Tubing. You should come up with a value of 2.75ft . This represents the friction/head loss when you try to pump 6 gpm though 100ft of 1" L.
Wait a second,.... our headers are maybe 10ft long in total. That means the head loss through a 10ft section of 1" L, at 6 gpm is only .275 ft (2.75 x .10). You will learn, this is a very low number.
Now lets look at the head loss through the 3/4 copper loop that contains the 33ft of baseboard. Look at the attachment labeled 38 12 34 . Find the section labeled 3/4. Go down the right hand column to 2 gpm (remember thats approximately how many gallons of water we needed to handle the 33ft of baseboard in that loop). Follow across to the right to the section labeled Type L Tubing. You should come up with a number of 1.44 . This is the friction loss for pumping 2 gpm through 100ft of 3/4 L copper. For the purposes of this example, assume the losses through the boiler, valves and various fittings to be 4ft of head. Now we can add all these together to get the total head of the system. We had .275 ft for the header, 1.44 ft for the 100ft baseboard circuit, and another 4ft for boiler piping and fittings. When calculating system head loss, we take the highest value amongst all the heating circuits. So we have .275 ft (header) + 1.44 ft for the 100ft baseboard circuit and 4ft for the boiler piping and fittings. Thats a total of just under 6ft of head. Now we can select a pump based on handling 6gpm at 6ft of head if we use a 1" copper header.
Now look at the attachment titled grundfoss pump curve. The horizontal x axis is gpm and the vertical y axis is ft of head (friction loss). Plot out the point for 6 ft of head and 6 gpm. Now look at that point in relation to the red curve marked 7. At 6 gpm, this pump will produce around 14ft of head. We calculated we needed approximately 6 ft of head. This means that the curve for pump 7 has plenty of capacity to spare.
Now look at pump curve # 8. At 6 gpm, this pump will barely make 3 ft of head pressure. We determined we needed 6 ft so this pump is out. Pump 6 or 7, both very popular models will work just fine with the 1" header.
But what about a 3/4" header? Same procedure. Look at the friction loss for pumping 6 gpm though 10ft of 3/4" Type L. You should come up with 9.80 ft of head. Remember, this value is for 100ft and we have 10ft, so we multiply by .10 and get a head value of .98 ft . Now we add them all up, the header, the baseboard circuit, and boiler/fitting/valve losses. We would then have, .98 + 1.44 + 4. This comes out be approx 7ft. Again, back to the pump curve. If you plot out 6gpm and 7ft of head, pump 6 or 7 will have no problem with this scenario.
If you could follow all of that, think about this, what would happen if he tried to have just 1 circulator servicing all 100ft of baseboard in a series loop configuration making the loop 300ft long. What happens now. Look at the 3/4 table, get the value at 6 gpm and multiply it by 3 to equal 300ft worth of tubing. Then add the number up. We would need a pump that would provide 6gpm at approximately 35ft of head. None of the pumps on this chart could do it. You could find one but it would would be costly - 6 - 10 times the cost of a typical residential circulator.
So in the end, whether the old timer actually calculated all of this, it looks like his 3/4" header would work. Sometimes the old timers are right just based on experience.
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Comments
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sizing a feed and return hot water heating manifold
hi guys, just went to a job in my neighborhood in queens,New York and found a botched up heating manifold job. first let me tell you that there is a weil mclain hot water boiler with an 1 1/4 feed and return.comin out of the boiler they (the old contractor) reduced the 1 1/4 to 3/4 and ran 3 zones off the 3/4. when i asked the contractor about this. he said he was doing it for many years and noone ever complained. i told him it was wrong and said the outlets sizes are there for a reason. am i missing something?0 -
hot water heating manifolds
i understand the the math on the heating and all. but looking at this job its a diaster the guy choked down an 1 1/4 monoflow lop down to 3/4 and im finding more problems as im going thru the job. its going to be a long summer and fall0
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