| Question: 14-05 | Code Section: 7 | Date: 24 February 2014 | OBC 2012 Reference: |
Question submitted: How many gang trapped washing machines discharging 6.5 gpm each, may discharge into a 4” trap? There are 2 ways of calculating this, but which one is the right answer? (note: FU= fixture unit)
Method #1:
Given, 1FU = 6.25 gpm and a 4” trap (from OBC table 7.4.10.2) has a hydraulic load of 6 FU.
A FU is equal to one cubic foot (6.25 gal.) of water drained in a 1 ¼” pipe over one minute.
6FU (4” trap) X 6.25 gpm = 6×6.25 = 37.5 gpm
37.5gpm ÷ 6.5gpm (wash machine output) = 5.77 wash machines
Method #2:
Given, from OBC Appendix B, conversion of gpm to l/s, divide gpm by 13.198
6.5 gpm (wash machine output) ÷ 13.198 = 0.49 litre/sec
From OBC table 7.4.10.3 , the maximum permitted flow for a 4” trap is 5.7 litres/sec.
5.7 litres/second ÷ 0.49 litres/second = 11.63 wash machines
Conversion factor for imperial to metric found on page 1 of appendix B in OBC 2012. All references from OBC 2012 Reg 332/12
Interpretation: Method #1 does not fit this scenario as Table 7.4.10.2 is used to calculate a hydraulic load for a fixture not listed in Table 7.4.9.3. A clothes washer for commercial use is listed in that table as a 2” minimum trap size and 2 FU’s. Method #2 is the appropriate table to use, as a clothes washer is a fixture that produces a semi-continuous flow. Table 7.4.9.3 provides us with the information for the trap size and load of a single fixture, this question deals with multiple washers connecting to a single trap. By following the math provided in the question, 11 machines would be considered acceptable to discharge through a single 4” P-trap.