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# calculating flow rate

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Member Posts: 41
square footage x 25 = total BTUs

take total BTUs and divide it by 10,000 (500(c) x 20(temp drop = 10,000)

I forget what c stands for. Something to do with the weight of water????? someone clarify please.

• Member Posts: 611
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Calculating flow rate

I have a 6 outlet manifold with 5/16' tubing: 159',145',151',136',130',32' each. Assuming no loss from distribution pipe and wide open (no adjustment) manifold. Is there a (simple) method for calculating the actual flow from a 15-58 circulator at it's different speeds.
• Member Posts: 1,790
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flow rates

The 500 comes from 8.33 lb/gallon for pure water multiplied by 60 minutes per hour and 1 btu/°F heat capacity of water. The "c" is a correction for the heat capacity difference of glycol solutions.

Most importantly, the temperature drop does not need to be 20°F. The only thing 20°F does is make the math easy to do in your head. The temp drop and flow rate (pump selection) are the numbers you get to play with. Increasing the temp drop might eliminate the need for a larger pump.

If you know your flow rate and temperature drop of an existing system, you can solve the equation for the heat output.

The original question relates to parallel piping loops of differing lengths. This is not an easy calculation by hand. The no-numbers answer is that the flow rates in the parallel loops naturally adjusts to make the head loss equal in each loop. I would suggest purchasing Siggy's Hydronic Design Studio. It does the matrix algebra for you, plus a whole lot more.
• Member Posts: 611
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flow rates

Thanks Anderw, I had a feeling this would be a hard thing to calculate. what If my loops were equal ?how would I set this problem up? this might help me to guesstimate the flow. Gpm=btu/500x delta T is the equation I'm used to. How does percent glycol plug in to this. I know it effects both carrying capacity and viscosity.
• Member Posts: 1,790
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Pump Selection

If the loops were equal, you could calculate the head loss as if you had one loop. Pick flow rates through one loop until it gives you a head loss close to what a 15-58 can provide; maybe 10ft or so. Then multiply the flow rate in that single loop by the number of loops.

For example, say you had 6 loops 140ft each. Calculate the head loss for 0.5gpm through 140ft of smooth tube. This is approx 13.4 ft of head loss (using Darcy-Weisbach). 0.5 gpm x 6 is 3 gpm, which is pretty close to the 15-58 on speed three at 12 ft. So you're probably somewhere in the 0.4 to 0.5 gpm range. The heat delivered to the space at a 20°F temperature drop at 3gpm is 30,000 btu/h. That gives you a really rough idea. With 5/16 tube the head loss increases very quickly with increasing flow rate, so a 15-58 is probably the practical low limit for pump selection.

That's what I love about hydronics, hydraulics, and heat transfer...It's all a balancing act among supply temp, temp drop, flow rate, pump curve, head loss, electrical consumption, boiler efficiency, ......
• Member Posts: 592
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C specific heat of water

I believe the C comes from the formulat Q= M C(T1-T2). This is the origin of the formula where Q is the quantity of heat energy, C is the specific heat (1), M is the mass(we use 8.33lbs = 1gallon), and T1-T2 is the delta T. The specific heat of a substance is the amount of heat per unit mass required to raise the temperature by one degree Celsius. For our purposes using water it is 1. I believe it changes minimally in value based on the initial starting temp of the water.

We modify it by assigning a time value such as gpm on the right side and muliplying by 60min/hr to get the left side to be btu's per hour.

Any engineers out there to confirm this??
• Member Posts: 6,928
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I could be wrong here--please correct me if so!!!

While the flow will vary among the individual loops, I believe that the pressure drop will stay constant among them.

In other words, calculate the calculate the flow and head loss (using the pump curve) for the longest (159') loop assuming that it the only loop.

The will be the head loss seen by each loop.

Then using the maximum head loss and loop length you can calculate the flow through each of the remaining loops. Add everything together and you have your total flow.

Extremely easy with the "Rad Pad", but unfortunately it does not list 5/16" tube. It lists 3/8; 1/2; 5/8; and 3/4 (based on "typical" pex tube).

You should be able to use any number of on-line head loss calculators--the simplest of which assume water as the medium and merely ask for length, inside diameter and "type" of tube. To work "backwards" with a known head loss, guess at the flow rate/velocity until the desired head loss is calculated.
• Member Posts: 592
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excel sheet

I use this little spreadsheet sometimes for some quickie calculations. I think its correct.
• Member Posts: 592
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I think the pressure drop is different

I think it goes something like this -
The longer the loop the higher the pressure drop. What stays constant is the total gpm flow rate through all the loops.
• Member Posts: 1,790
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Pipe Network

You have to solve simultaneous equations to find the flow rate in each loop. It's simple with a computer, but quickly gets tedious to solve by hand. I'm a little rusty, but I think with 6 loops you have 12 equations and 12 unknowns. Mathcad makes short work of this sort of thing, but then so does Siggy's program, and then some.
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Parallel and Series

Pipes in series have the same flow rate and different pressure drops.

Pipes in parallel have the same pressure drop and different flow rates.
• Member Posts: 6,928
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Given that two manifolds (supply and return) put the individual loops into parallel wouldn't my suggestion work?
• Member Posts: 1,790
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Simultaneous Equations

The pressure drop is the same in all the loops, but you don't know what that is. And without that information, you can't calculate the flow rates. That's six unknown flow rates and six unknown pressure drops. You do know the desired overall flow rate to the manifold and that the sum of the flow rates in the parallel loops equals that overall flow rate. You also know that the pressure drop is equal in all of the loops.

The head loss is the same in all the loops. That gives you the first five equations. HL1=HL2, HL2=HL3, HL3=HL4, HL4=HL5, HL5=HL6

Darcy-Weisbach applied to each of the 6 loops gives you six more equations. HL1=f*L/D*(V1^2)/(2g) + K*(V1^2)/(2g), for each loop. f is the Darcy Friction Factor. K is the minor loss coefficient.

The sum of the flow rates in the parallel loops equated to the overall flow rate is the final equation. q1+q2+q3+q4+q5+q6=Q

Simultaneously solve those twelve equations and all of the pressure drops and flow rates fall out.
• Member Posts: 6,928
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\"I see!\", Said the Blind Man

Started a calculation based on 3/8" tube and saw the problem! As soon as I add the flow of another loop (but keep the pressure drop the same) the result is no longer on the head loss curve!!!

So yes, for six loops of different length so we're certainly talking about multiple equations with multiple variables.

There [might] however be a fairly simple and reasonably accurate way to "weight" loops of different lengths such that you can initially treat the loops as if they're all the same length and then apply a correction.
• Member Posts: 6,928
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Rough estimate:

Between about 3 and 3 1/2 gpm @ about 7' - 8' of head with the "Super Brute" 15-58 on low speed.

Given your heat loss for those loops I'd suspect that delta-t is very low.
• Member Posts: 100
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Flow meters

5/16 tube sounds like you used Uponor manifolds. The flow meters are readable with a guage and will give you gpm and let you balance the loops. I have noticed most pictures that are posted do not show flow meters on manifolds. How do most people balance these? Is there an other way?
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Balancing

Ideally the loops are balanced in the tube layout, and the flowmeters become an unnecessary expense. However, that is not always possible, and the flowmeters are handy if you have widely varying loop lengths.
• Member Posts: 244
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Calculating Flow From 15-58 pump - use pump curve

If it is an installed and operating system you can compare the actual pump differential pressure in feet of head to the pump curve to get the operating point.

If you are designing a new system you have to calculate the pressure drop (head loss) due to the parallel flow through the tubes.

Pump curve is attached. The red line curving upward is the system head loss. The operating point is where the system head loss crosses the pump curves for the different speeds. Gives total pump flow rate.

Check Grundfos web-site for pump selection pump curves.

Is this an operating system or a system you are designing ?
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