Question

3) A 6-foot diameter (T.D.) storm drain. (n=0.015) along a street was Constructed on a Uniform slope of 0.0032. During a recent storm the depths of flow mearned at two manholes A Cupstream) and B (downstreams 30off honizontal distance apart are 3. so ft & 4.30ft respectively if the flow rate is 85 % a determine the flow area, wetted permetes! and hydraulic radius in the storns drain at points A and a Estimate the head loss between the two manholes.

Please help solve this hydraulic problem correctly. Thank you.

Answer 1

0° = 64:32 Q=3 civen in Question circular drain & D = 6 foot, L=300ft (b/w Af B) = 0.015, 2=85 ft 3/86 Manhole A depth d = 3.50 fut (US) findy manhole B depth da = 4.3oft ② A, P, at AfB Head loss b/w two manholes Sole & At point A (manhole) ② Ast point B (manhole) Parte A = area of section-Area AAOC o=1.122 Radian - (Area of unfilled section) A = 18.91 ft 2 * 3ht = $4220 - 5(2rsing. soso P= 6.732 ft A8 (20-sin 20) R=A=2.80 ft Se t Auf = + (2x1.4033 –siml2x80-40)] Auf = 15.04 ft2 so O = cos(os) so now area of flow at 0= 80.40° area of flow = Falloon In Radian Total area- Auf 0=1.4033 Radian = 771 2 2² - Auf - 28.27-18. 14 = 17:1262 A so area is Az 13118 ft 2 Now → P=204 ffor 360°-21194 so for 10 = 2 so for 20 = 20$ P= 2x 14033X3 -8.41 (0.50 Ft A. Now » Hydraulic radius = R(1/2 = 2.038 ft. An 80 A = 17.12 fut? p=8.419 fut R=2.033 fut

1 Head loss due to only friction (neglecting minor head loss) 85= 18.015 0.015 85 = 18,015 ( 2.4165) the 2 = 0.6807 hf = 0.4633 so from energy equation head les=(4442) + (721 min.) - ho - ကားက head loss = 0.5448 oft JUUUUUUUUUUUUUUUU