We're going to get very "old fashioned" and technical in this Newsletter. This one is for all you closet engineers out there who actually want to know how coils are selected, and the factors that go into picking a coil. If you are interested in how all this information is developed and the inter-relationship between the data, then this Newsletter is for you. Read on!
Example: Heat 7900 CFM from 50°F to 142°F using 190°F water, lvg. at 170°F
We are going to give you the actual "longhand" formulas and loads that people used to select this coil. By actually seeing the formulas you will get a much better idea of the selection process. You will also better understand how changing the input also changes the final selection. See if you can follow along. By the way, there used to be people who did this longhand on every job. Hard to believe, but true!
Step #1 - Calculate the load
CFM x 1.08 x (lvg. air temp - ent. air temp) = BTU/HR
7900 x 1.08 x (142-50) = 785,000 BTU/HR
Step #2 - Calculate G.P.M.
BTU/HR ÷ (500 x water temp. diff) = G.P.M.
785,000 ÷ (500 x 20) = 78.5 G.P.M.
Step #3 - Calculate face area
CFM ÷ air velocity = Face area
7900 CFM ÷ 500 F.P.M. = 15.8 sq. ft. required
Step #4 - Pick a coil size (any size you like)
27" F.H. x 84" F.L. = 15.75 sq.ft. (18 tubes/row)
Step #5 - Select a water velocity (charts available)
G.P.M. ÷ # tubes feed x factor (5/8" O.D.)
78.5 ÷ (18 x 1.17) = 5.1 ft./second
Step #6 - Select a "U" Factor
There used to be charts that showed each individual "U" factor based on ft./minute air velocity, ft./second water velocity, and fins/inch. This "U" factor is the actual heat transfer coefficient tied to a particular coil. You'll have to trust us on this one, but the "U" factor for 5.1 ft./second, 500 ft./minute and 8 fins/inch is 176
Step #7 - Figure the M.T.D. (mean temp difference)
Ent. Water - Lvg. Air (190 - 142) = 48 Relationship between the air temps and
Lvg. Water - Ent. Air (170 - 50) = 120 the water temps. Higher is better than lower.
(48-120) ÷ In(48/120)
M.T.D. (counterflow) = 78.6 M.T.D.
Step #8 - Actual # Rows Required
BTU/Hr. ÷ (M.T.D. x "U" Factor x Face area) = Theoretical # rows required
785,000 ÷ (78.6 x 176 x 15.8) = 3.6 Rows
Above is the actual formula required to calculate rows. Since it's impossible to build a coil exactly 3.6 rows, you must round up to 4 rows.
Believe it or not, when you plug all your numbers into a computer program, this is the set of calculations that takes place. Step #8 is exactly how coils are calculated. Everybody is always trying to build a coil with the fewest rows possible. Well, the way you do that is to make the M.T.D. larger, increase the "U" factor, or increase the face area. When you make any of those numbers larger, you decrease the number of rows.
How do you change the M.T.D.?
Make the water hotter, or the steam pressure higher
How can you make the "U" factor larger?
Make the water go faster, make the air go faster, or increase Fins/inch
There are other factors that increase "U". Coils operate at higher coefficients when wet rather than dry. "U" factors are different for 1/2" tube coils than 5/8".
How do you change the face area?
Make the coil bigger
If you do any of the above, you decrease the number of rows. These are all things that the computer does automatically for you when you change the data. Oftentimes, we are all guilty of just mechanically changing the data and never knowing how the actual calculation process takes place. Now you know! Knowledge is power, and the more knowledge you have regarding how coils work, the better you will be at the selection process. Please let us know if you have any questions or need any help with your coil selections. We're here when you need us!
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