Six Lengths or Five: Take-off, Fall and Hanger Layout for a 100 mm Drain
A plumbing crew is laying out the horizontal 100 mm PVC building drain of a school canteen. The developed length from the upstream end of the run at the last wye to the base of the stack is 12.6 m, and the same 100 mm pipe continues up as a 3.0 m stack to the second-floor branch. The design slope is 1:60, and the invert at the base of the stack has been set at −0.55 m relative to the finished floor. Hangers are specified at maximum 1.2 m centres with a support within 300 mm of each hub. The pipe is supplied in 3.0 m lengths, the solvent-weld sockets take 55 mm of pipe per joint, and there are 9 joints in the whole 15.6 m of 100 mm pipe. The foreman wants to order 5 lengths, allow 5% waste, and start laying from the stack outward without setting a grade line.
1. Compute the upstream invert and state the fall over the 12.6 m run.
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The stack invert is the control, and the pipe rises as it runs upstream, so the upstream end is higher:
- Fall = length × slope = 12.6 m ÷ 60 = 0.210 m.
- Upstream invert = −0.55 m + 0.210 m = −0.34 m relative to finished floor.
Expressed as a check: 210 mm of fall over 12.6 m is 1:60, or about 1.67%, which is within the usual range for a 100 mm drain. Note that the 3.0 m stack is vertical and does not enter this calculation; only the horizontal run takes the slope. Two practical points follow: the upstream wye must be set at −0.34 m before the pipe is joined, because adding slope later means re-cutting the run; and the invert must be checked against the fixture it serves, so that the fixture drain still has room to connect at its own minimum slope and trap-arm length.
2. How many 3.0 m lengths should be ordered, and why is the foreman's figure of 5 wrong?
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Build the quantity from the actual pipe, then add waste and joint allowance:
- Pipe length = 12.6 + 3.0 = 15.6 m.
- Add 5% waste: 15.6 × 1.05 = 16.38 m.
- Add joint allowance: 9 joints × 0.055 m = 0.495 m.
- Total = 16.38 + 0.495 = 16.875 m.
- 16.875 ÷ 3.0 = 5.63, so round up to 6 lengths (18.0 m).
The foreman's 5 lengths give 15.0 m, which is 1.9 m short of the requirement before any cutting error. The joint allowance matters because each socket swallows 55 mm that cannot also be part of the measured run; ignoring it is how a crew ends up short at the last fitting. Order 6, then prepare a cut list so offcuts from the 18.0 m are used on the shorter pieces rather than thrown away, and keep the spare for the stack and future repairs.
3. How many hangers are needed on the 12.6 m horizontal run, where do they go, and what is wrong with laying from the stack outward with no grade line?
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Support count: 12.6 m ÷ 1.2 m = 10.5 spaces, so use 11 spaces of not more than 1.2 m, which gives 12 supports including one at each end - and each end support must land within 300 mm of the hub, so the intermediate spacing is adjusted slightly rather than the end ones being omitted. Check the governing code for the exact spacing and for the requirement at changes of direction.
Layout: set a grade line or laser from the stack invert at −0.55 m to the upstream target at −0.34 m, mark it, and hang to the line. Hangers hold the pipe at the grade; they are not a way to pull a mis-set run into position.
Laying from the stack outward is acceptable only because the downstream invert is the fixed control; without a grade line, each 3.0 m length is laid 'about right' and the error accumulates over 12.6 m, so the upstream end can finish above or below the fitting it must meet, and the pipe can end up backfalling in the middle. Set the line, check the invert at every joint, and do not close the trench until the run has been water-tested and the grade verified.