Thank you so much for making this available, ties everything together at a nice pace with just enough math to drive home the point, as opposed to math driving around until I'm lost.
Thanks, Jeffrey, for your comments. We are happy to post our Coffee with Caleffi webinar series to RUclips as a timeless educational reference. Happy learning!
Great job. I am actually an Aerospace Engineer, but focused on external aerodynamics and we rarely do actual math (its all computer simulations or physical testing). Now I am building my own house with radiant and found your explanation to be perfect for what I needed. Just the right balance of simplicity and detail. Got my math done and now I can move on with finding the right pump. ;) I'd give two thumbs up if they would let me.
Marine engineer turned Tech Ed teacher turned Steam Plant operator. Would have loved to have taught this to my students. Just a thought ... at 39:50 you discuss Darcy head loss as related to V^2. Many people might understand that KE = 1/2*mv^2. While energy is ft-lb ... and H = ft ... there is an anlaogus VOLUME-Velocity relationship to H loss as there is MASS-Velocity relationship to KE.
Having found John Siegenthaler's book and Caleffi's Idronics journals immensely helpful, I nevertheless would like to know: do you have an Equivalent Length table (44:01) for piping other than copper -- PP-AL-PP (polypropylene), for example? Also, are radiators considered to have equivalent lengths, and if so, how are they calculated?
From John himself: "In concept, any component in a hydronic system which has flow passing through it has an equivalent length for a given type and size of tubing. You can estimate that equivalent length by setting the head loss of the component at some known flow rate equal the head loss equation (Darcy Weisbach), and then solving for the length. If the tubing is smooth (i.e., copper, PEX, PEX-AL-PEX, etc.) the Darcy Weisbach equation reduces to: HL = (acL)(f)^1.75. Where HL = head loss (feet) a = fluid properties factor (figure 4-2 in idronics 16) c= pipe size factor (figure 4-3 in idronics 16) L = length of tube (feet) f = flow rate (gpm) So, let’s say the head loss of a radiator is 3 feet of head at a flow rate of 2 gpm, and you want equivalent length of it based on 1/2” copper tubing, and the water temperature in the system averages 130 ºF. a = 0.048 c = 0.33352 f = 2 HL = 3 So, 3 = (0.048 x 0.33352 x L)(2)^1.75 The only unknown is L. Solving for L . L = 3 / (0.048 x 0.33352 x[2^1.75]) L = 55.7 ft. So the equivalent length of the radiator is about 56 feet of 1/2” copper tube. Note: The equivalent length of the radiator would be different for other types and sizes of tubing. Hope this helps."
Thanks for the reply. But you have explained how to calculate the equivalent length of a radiator from the head loss across it. Actually my problem is the reverse: I need to calculate the head loss, using the formula you cite, FROM its equivalent length. I need to know the equivalent length of the radiators as well as those for the pipe fittings in order to get my system curve. Do radiator manufacturers typically give a radiator's equivalent length, or head loss, or pressure drop, in the specifications?
Hello, Nikos. Pipe size coefficients change based on pipe materials. You may want to search online for the coefficient of the pipe material you are working with. For example: When working with copper, consult the Copper Development Association website.
Thank you for your inquiry. The sizing of circulators is a science. We will refer you to the 16th edition of idronics (an education journal series): Circulation in Hydronic Systems. It will be a good reference for you! www.caleffi.com/sites/default/files/coll_attach_file/idronics_16_na_0.pdf
Thank you so much for making this available, ties everything together at a nice pace with just enough math to drive home the point, as opposed to math driving around until I'm lost.
Thanks, Jeffrey, for your comments. We are happy to post our Coffee with Caleffi webinar series to RUclips as a timeless educational reference. Happy learning!
Great job. I am actually an Aerospace Engineer, but focused on external aerodynamics and we rarely do actual math (its all computer simulations or physical testing). Now I am building my own house with radiant and found your explanation to be perfect for what I needed. Just the right balance of simplicity and detail. Got my math done and now I can move on with finding the right pump. ;) I'd give two thumbs up if they would let me.
We sure appreciate the accolades. Congratulations on completing your math homework! All the best on your new radiant home.
Wao. Really informative. Sucks that is imperial units.
amazing videos. Love your educational resources and products. Will keep getting you specified!
Thank you sooooo much!
Marine engineer turned Tech Ed teacher turned Steam Plant operator. Would have loved to have taught this to my students. Just a thought ... at 39:50 you discuss Darcy head loss as related to V^2. Many people might understand that KE = 1/2*mv^2. While energy is ft-lb ... and H = ft ... there is an anlaogus VOLUME-Velocity relationship to H loss as there is MASS-Velocity relationship to KE.
Thanks, very helpful video.
Glad it was helpful!
Having found John Siegenthaler's book and Caleffi's Idronics journals immensely helpful, I nevertheless would like to know: do you have an Equivalent Length table (44:01) for piping other than copper -- PP-AL-PP (polypropylene), for example? Also, are radiators considered to have equivalent lengths, and if so, how are they calculated?
From John himself:
"In concept, any component in a hydronic system which has flow passing through it has an equivalent length for a given type and size of tubing.
You can estimate that equivalent length by setting the head loss of the component at some known flow rate equal the head loss equation (Darcy Weisbach), and then solving for the length.
If the tubing is smooth (i.e., copper, PEX, PEX-AL-PEX, etc.) the Darcy Weisbach equation reduces to: HL = (acL)(f)^1.75.
Where HL = head loss (feet)
a = fluid properties factor (figure 4-2 in idronics 16)
c= pipe size factor (figure 4-3 in idronics 16)
L = length of tube (feet)
f = flow rate (gpm)
So, let’s say the head loss of a radiator is 3 feet of head at a flow rate of 2 gpm, and you want equivalent length of it based on 1/2” copper tubing, and the water temperature in the system averages 130 ºF.
a = 0.048
c = 0.33352
f = 2
HL = 3
So, 3 = (0.048 x 0.33352 x L)(2)^1.75
The only unknown is L. Solving for L .
L = 3 / (0.048 x 0.33352 x[2^1.75])
L = 55.7 ft.
So the equivalent length of the radiator is about 56 feet of 1/2” copper tube.
Note: The equivalent length of the radiator would be different for other types and sizes of tubing.
Hope this helps."
Thanks for the reply. But you have explained how to calculate the equivalent length of a radiator from the head loss across it. Actually my problem is the reverse: I need to calculate the head loss, using the formula you cite, FROM its equivalent length. I need to know the equivalent length of the radiators as well as those for the pipe fittings in order to get my system curve. Do radiator manufacturers typically give a radiator's equivalent length, or head loss, or pressure drop, in the specifications?
Thank you very much for all of this valuable information. Where someone can find the pipe size coefficient for more pipe types?
Hello, Nikos.
Pipe size coefficients change based on pipe materials. You may want to search online for the coefficient of the pipe material you are working with.
For example: When working with copper, consult the Copper Development Association website.
What about sizing circulator for radient floor heating
Thank you for your inquiry. The sizing of circulators is a science. We will refer you to the 16th edition of idronics (an education journal series): Circulation in Hydronic Systems. It will be a good reference for you! www.caleffi.com/sites/default/files/coll_attach_file/idronics_16_na_0.pdf
Love your videos