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HEATING VENTILATING AIR CONDITIONING GUIDE 1942
Table 3. Iron and Copper Elbow Equivalents
Fitting
Elbow, 90-dec..
Elbow, 45-dee.
Elbow, 90-dee lone turn...... Open return band.______ Open gate valve......... Open globe valve....... ....... Angle radiator valve.. . Kaaator._........ Boiler or heater. ..... .
Tee, per cent flowing through branch:
100...
5a. .......
25....................
................................
--------------- --- ------ --------- --
Iron Pipe
0.7 0.5 1.0 12.0 3.0
1.8 4.0
Copper
'Tubing
1.0 0.7 0.5 1.0 0.7 17.0 3.0 4.0 4.0
1.2 4.0 20.0
auu uiu> win icquire
pipe. Section PQ carries 10 Mbh and requires M in. pipe.
To size the return start from the boiler and proceed backwards. Section IR carries
40 Mbh and from Fig. 3 a 1-in. pipe is required. Section RS carries 30 Mbh which is only
slightly over the capacity of a 54-in. pipe, so use 54 in. Section ST carries 20 Mbh and
requires a 54-in. pipe. The radiator branches are determined in the same manner. It is
evident from the chart that it is impossible to maintain a constant friction loss per foot
and therefore as the delivery varies there will be a change in the desired friction loss per foot of pipe.
Table 4. Piping Check Chart
Load, Mbh
Supply Main
AB 98 BC 58 CD 38 DE 23 EF II FG 4
Return Main
HI 98 IJ 58 JK 54 KL 47 LM 35 MN 20
Radiator Circuits
CN 20 DM 15 EL 12 FK 7 GJ 4
Supply Return
Supply Return
Supply Return
Supply Return
Supply Return
Length I Elbows Ft
Pipe
SlZB In.
Unit Heai [ MlLlSCHB1
!
per Ft
Friction MilinchB1s
Total Loss Milinches
s37 1 24 16 1
90 12 0 16 1
240 90
155
220 240
50
9600
1080 2790 1980
2880 850
9.600 10,680 13,470 15.450
18,330 19.180
5
11 16
11 9 15
5
1 0 0 1
IK i1*
l 1
240 90
300
230
4320 1260 5400
2530
4.320
5.580
10.880 13.410
140 170
1260 2890
14,670 -. 17.560
3 4
3 4
14 15
3 4
98 1
13 2
19 17
20 20
19 17
5 17
316
%
a
g
g g
170 3910
170
1190
5,100
420 9250
96
2880
12.130
270 9180
270
9450
18.630
100 2200
100
2100 .
4.300
50
so
650 1300
1.950
CHAPTER 16. HOT WATER HEATING SYSTEMS AND PIPING
It is desirable to check the various circuits so that if the variation from the calculated resistance is too great, it may be compensated by adding additional resistance at the proper point. This may be accomplished by sizing the short circuits by the procedure previously outlined. Prepare a chart such as Table 4 to be used in calculating the resistance of each circuit.
Section AB carries 98 Mbh with a unit head of 240 milinches per foot. In section AB there are 37 ft of pipe and 154 in. elbow. At 240 milinches per foot this is equivalent to 9600 milinches total loss in this section. Section BC carries 58 Mbh with a length of 2 ft and 4 elbows. The unit loss in this section is 90 milinches per foot. Loss in this section is then 1080 milinches. Section CD carries 38 Mbh and has 16 ft of pipe and 1 elbow. The unit loss in 1-in. pipe is 155 milinches. The loss in this section is 2790 milinches. The balance of the supply main and the return main are handled in a similar manner.
The radiator circuits are then checked. The 20 Mbh radiator on this circuit has 3 ft of supply pipe and 13 elbow equivalents while the return is composed of 4 ft and 2 elbows. The unit loss in 54 'n- pipe at this delivery is 170 milinches per foot. The total loss in the supply is 3910 milinches. The loss in the return is 1190. Total loss in the radiator circuit is 5100 milinches. Check each radiator circuit in a similar manner.
The total calculated loss for the longest circuit was determined as 60,000 milinches. The maximum loss in the short circuit is 18,630 plus 13,410 plus 15,450 or a total of
Fig. 5. A Forced Circulation Direct Return System
47,490 milinches. This difference is caused by the variation in length of the two circuits and may be corrected by using a flow control in the return main to supply the additional resistance or by introducing resistance into each separate circuit to compensate for the difference. A10 per cent variation will cause no complication as the flow from the various pipes will not exactly follow the curves of Fig. 3 any closer than this value.
Example S. Design a two-pipe direct return forced circulation system with copper tubing and fittings for the piping layout as detailed in Fig. 5, based on a 20 F tempera ture drop.through the radiation.
The piping circuit from the boiler to the highest radiator on the farthest riser and back to the boiler is 250 ft of pipe. There are about 16 elbow equivalents having an equivalent pipe length of about 50 ft; so that the total equivalent pipe length is 300 ft.
Assume that a circulator is available which will provide a pressure head of 6 ft. Solution. Refer to Table 2, which indicates the total equivalent lengths for pressure heads from 2 to 12 ft. With a circulator having a 6 ft pressure head and a system with a total equivalent length of 300 ft, the piping system will be designed on a basis of 240 milinch. Checking the piping diagram it will be noted that sections AB and KA, both supply 117.6 Mbh. Referring to the 240 milinch column of Table 2, 1)4 in. is shown to be the necessary pipe size. Sections BC and JK carry 88.8 Mbh and require 1)4 in. tubing. Sections CD and IJ supply 67.2 Mbh and require 1)4 in. tubing. Sections DE and HI supply 43.2 Mbh, which requires 1 in. tubing. Sections EF and GH with a load of 14.4 Mbh require 54 in. tubing. The risers are pipe sized in a similar manner. To secure proper distribution of hot
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