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HEATING VENTILATING AIR CONDITIONING GUIDE 1941
Table 3. Iron and Copper Elbow Equivalents
Fitting
Tee, per cent flowing through branch: 100.............................................................................................................. ............. 50. _ ...................................................................... 25............................................................................
Iron Pipe
1.0 0.7 0.5 1.0 0.5 12.0 2.0 3.0 3.0
1.8 4.0 16.0
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
and this will require %-in. pipe. Section PQ carries 10 Mbh and requires Yz 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 %-in. pipe, so use % in. Section ST carries 20 Mbh and requires a %-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
Pipe Length
Ft
Elbows
Pipe Size In.
LossUnit Head
Milinches per Ft
Friction Milinches
Total Milinches
Supply Main
AB 98 BC 58 CD 38 DE 23 EF 11 FG 4
37 2 16
,9 12 16
1
i
240
9600
9.600
4 lX
90
1080
10.680
11
155
2790
13,470
0
X
220
1980
15.450
0
X
240
2880
18.330
1 54 50 850 19.180
Return Main
HI 98 IJ 58 JK 54 KL 47 LM 35 MN 20
5
5
IX
240
4320
4.320
11
1 IX
90
1260
5.580
16
11
300
5400
10.880
11
01
230
2530
13.410
9 01
140
1260
14.670
15
1
X
170
2890
17,560
Radiator Circuits
CN 20 DM 15 EL 12 FK 7 GJ 4
Supply Return
Supply Return
Supply Return
Supply Return
Supply Return
3 4
3 4
14 15
3 4
8 9
13 X 170 3910
2
X
170
1190
5,100
19 X 420 9250
17
X
96
2880
12.130
20 20
X X
270 9180
270
9450
18.630
19 17
X X
100 2200
100
2100
4,300
5' X 17 X
50 650 .
50
1300
1.950
CHAPTER 15. HOT WATER HEATINC 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 1Y 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 % in. 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. A 10 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 :8. 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 Yl. in. is shown to be the
necessary pipe size. Sections BC and JK carry 88.8 Mbh and require IY ini tubing.
Sections CD and IJ supply 67.2 Mbh and require \Y 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
M\>h require % in. tubing.
' ' "" '
`The risers are pipe sized in a similar manner. To secure proper distribution of hot
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