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Design discharge of rural combined sewer

Short answer

Peak sewage is 0.0422 m3/s using 45 lpcd and peak factor 2.7. With tc=30 min, intensity is 19 mm/hr; storm runoff is 1.900 m3/s. Combined design discharge is 1.942 m3/s.

A population of 30000 is residing in a rural area of 60 hectares. If the average coefficient of runoff for this area is 0.60, time required to the entry port is 25 minutes and the time of travel from the entry port to the section of sewer under consideration for design is 5 minutes. What will be the design discharge for a combined sewer at the section to be considered if the average flow of sewage in rural is 45 lpcd and peak factor is 2.7?

Design discharge of a combined sewer = peak dry weather sewage flow + storm runoff.

Step 1: Peak sewage flow

Average flow = 45 lpcd (given as the sewage flow), peak factor 2.7.

Qsewage=30000×45×2.786400×1000=0.0422 m3/sQ_{sewage}=\frac{30000\times45\times2.7}{86400\times1000}=0.0422\ \text{m}^3/\text{s}

Step 2: Time of concentration and intensity

tc=te+tf=25+5=30 mint_c=t_e+t_f=25+5=30\ \text{min}

No rainfall formula is given, so I use the common empirical form i=760t+10i=\dfrac{760}{t+10} mm/hr (assumed):

i=76030+10=19.0 mm/hri=\frac{760}{30+10}=19.0\ \text{mm/hr}

Step 3: Storm water (Rational formula)

Qstorm=C i A360=0.60×19.0×60360=1.900 m3/sQ_{storm}=\frac{C\,i\,A}{360}=\frac{0.60\times19.0\times60}{360}=1.900\ \text{m}^3/\text{s}

Step 4: Design discharge

Q=0.0422+1.900=1.942 m3/sQ=0.0422+1.900=1.942\ \text{m}^3/\text{s}

Answer: Design discharge = 1.942 m³/s (the sewage is only about 2% of it).

✓ Answer checked by Claude Opus 5.5 (high effort)

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