Question 10.8: Figure 10-23 shows a closed, constant flow two-pipe water sy......

The first step is to select criteria for sizing of the pipe. Because the complete system is confined to an equipment room where noise is not critical, the velocity and head loss criteria may be relaxed somewhat. Let the maximum velocity be 5 ft/sec and the maximum head loss be about 7 ft per 100 ft in the main run. Somewhat higher values

may be used in the parallel circuits. The equivalent lengths for fittings, L_{f}, are assumed values for this example. Using Fig. 10-20 we select pipe sizes and create Table 10-3. The lost head for the three parallel circuits that begin at 3 and end at 8 may now be determined from the data in the table:

H_{c}  =  l_{34}  +  l_{45}  +  l_{c}  +  l_{67}  +  l_{78}  =  0.98  +  5.42  +  10.0  +  2.08  +  0.98  =  19.46  ft

H_{b}  =  l_{34}  +  l_{47}  +  l_{b}  +  l_{78}  =  0.98  +  5.96  +  10.0  +  0.98  =  17.65  ft

H_{a}  =  l_{38}  +  l_{a}  =  5.9  +  15.0  =  20.9  ft

At this point notice that the three parallel paths have different lost heads, with the specified flow rate for each coil. In order to balance out the required flow rates, paths b and c require some adjustment by a balancing valve to increase their lost head to that for path a, 20.9 ft. Each coil will then have the specified flow rate. Another approach to the balancing issue is to change the layout to reverse-return by moving the connection at point 3 to point 3′. Note that the path through all three coils would then be approximately the same length. Now the required pump head may be estimated by adding the parallel circuits to section 8-1, the chiller, and section 2-3:

H_{p}  =  l_{81}  +  l_{ch}  +  l_{23}  +  l_{38}  +  l_{a}

H_{p}  =  1.95  +  14.0  +  1.70  +  5.9  +  15.0  =  38.55  ft

The pump may then be specified to produce 60 gpm at about 39 ft of head.