EBK FUNDAMENTALS OF GEOTECHNICAL ENGINE
5th Edition
ISBN: 8220101425829
Author: SIVAKUGAN
Publisher: CENGAGE L
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Chapter 5, Problem 5.3P
To determine
Find the zero-air void unit weight for a soil.
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Please explain all your steps. Thank you.
10000
Quizl:Design a grit chamber for a horizontal velocity 250 mm/s, minimum flow
5,000 m³/day, maximum 20,000 m³/day, average flow is 12,500m³/day, S.O.R=
0.017 m/s, For a parabolic channels A =2/3 WD, detention time = 1 min
Q
Ac == HxW
vs for inorganic particles = SOR
As = LxW
As
V=Qxt=LxWxH
Z=CxW2
Q (m³/day) |
Ac (m²)
W (m)
| Z (m)
Can you please explain how to draw the shear and moment diagrams. Thank you.
Chapter 5 Solutions
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- Q2: Design an activated sludge process to treat a waste flow of 15000 m³/day with a BODs of 180 mg/L following primary treatment. The effluent BOD, and SS are to be 20 m mg/L. Assume Xr = 15000 mg/L, X = 2500 mg/L, 0c = 10 day, Y = 0.60, kd=0.05. Determine 1- the reactor volume, 2- the Sludge production rate, 3- the circulation rate, 4- the hydraulic retention time, and 5- the oxygen required? C₁-Ce C 1 XV = 1 +0.532 QxC₁₂ VXF F= - 1+r (1+0.1r)² = Qr= dt Өс Qx Xr-X V R = [= Q YQ(So-S) Oc dx XV 1+Kd0c O₂demand =1.47 (So-S)Q-[1.14Xr(Qw)] Qw total sludge production/X, S = BOD, effluent - 0.63 * SS Warrow_forwardمارت حولة ملانول 60 Design Deceleration tank and screen for W.W.T.P of flow-144000 m² day Design Data: D.T=1-3 min 1-3b, d=1-1.5 m 60 Design Data: velocity= 0.6: 1.5 m/sec . A-Qdfv=b.d, b=2d 233 Design Data: bars used circular of p= 1.5:3 cm. rec. of (1:2 cm) x (2:6 cm) spacing between bars 2.5:5.0 cm fine screen, 2.5 7.5 cm coarse screen Sloping angle on hz (6)=45:60 = depth of water depth in approaching channel=d = ⚫nbars nspacings +1 ⚫b=nbars xo+nspacings I spacing net inclined area = 2 Aapproach channel=nspacing. spacing. L. 1=d/ sin Check: = 1.13 C 2.56 1134(6.05 P. 45.30 *velocity just befor the screen=v₁ = Qa(m³/sec) screen-b.d ≥ 0.6 m/sec. * velocity through the screen=v2 16mm use 2 Qa(m³/sec) nscreens (spacings Spacing).d ≤1.5 m/sec IsP. 22-65 12²² head losses = Ah = 1.4 x ≤0.1 m 2g bav usp = Ubar Q 23.65 k 148.72arrow_forwardCan you please explain the steps of this problem especially the variables for Q when solving for the shear force? Thanks.arrow_forward
- Please show all your steps. Thank you.arrow_forwardThe solution to this problem is 10.5kip. Please explain all the steps.arrow_forwardQ1: Design of trickling filter system single-stage with 3m depth, effluent BOD, of 20 mg/l, influent BODs =250mg/l, flow = 2.63 m³/min and hydraulic flow rate or hydraulic loading rate 25 m/day, r=3arrow_forward
- 04 Q4 A waste effluent of 1.25 m/s with BOD, 183 mg/L, DO=0 mg/L and T = 20 °C is to be discharged into a river of 8 m³/s flow, BOD, = 2mg/L, DO -9.14 mg/L and T= 15 °C. At 20 °C, (K₁) is 0.3/day and (K2) is 0.9/day. The average velocity of the river is 0.8 m/s. 1) Is DO min within the environmental limitations? 2) At what distance is the maximum deficit located. 3) Draw the oxygen sag curve? Given the saturation concentration at 15 °C and at mixed temperature = 10.15 mg/Larrow_forwardCalculate 1- the effluent BOD, of a two-stage trickling filter with the following flows, BOD, and dimensions? Q-5000 m³/day, influent BOD,-280 mg/L, volume of first filter-1000 m³, volume of second filter-800 m³, filter depth-2 m, r₁=1, z=1.25. Also, 2- calculate organic loading rate (BOD, kg/day) 3- Hydraulic loading (m³/m²/d), 4- efficiency of each stage 5- overall removal efficiency.arrow_forwardQ3: Determine the force in each member of the shown truss, and state whether they are tension or compression. 40 kN 3 m 10 kN A 25-1.25 m m -3.5 m- 3 m Barrow_forward
- Q3 Design a secondary clarifier for an activated sludge process with a recycle rate of 25 percent, a MLSS conc. 2500 mg/L, peak flow 9,000 m²/day, depth of tank 3 m and solid loading rate = 4 kg/m²/hrarrow_forwardA- Design grit removal chamber for a W.W.P with hourly flow equal 5000 m'h 410 markarrow_forward05 An average operating data for conventional activated sludge treatment plant is as follows: (1) Wastewater flow, Q-50000m/d (2) Volume of aeration tank, V-12000m (3) Influent BOD, Y.- 300 mg/1 (4) Effluent BOD, YE - 25 mg/1 (5) Mixed liquor suspended solids (MLSS), X₁ = 2500mg/1 (6) Effluent suspended solids, Xg30mg/1 (7) Waste sludge suspended solids, X-9700mg/1 (8) Quantity of waste sludge, Q., -220m³/d Based on the information above data, determine: (a) Aeration period (hrs.) (b) Food to microorganism ratio (F/M) (kg BOD per day/kg MLSS) (c) Percentage efficiency of BOD removal (d) Sludge age (days)arrow_forward
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