Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781337516877
Author: Das
Publisher: Cengage
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Chapter 11, Problem 11.18P
(a)
To determine
Calculate the time taken to reach
(b)
To determine
Calculate the consolidation settlement at
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- Date: Pag 396 How Astrong municipal ww with soluble BOD of 200 mill 2 20°C is treated in a bio-tower with an 5.4m depth of plastic packing. The hydraulic loading of the raw ww. is 0.86 m³/m² h, and the recirculation ratio is 3 The constants based on apilot-plant study arek = 0.35/hr and n=0.39. Calculate the effluent Soluble BOOD..arrow_forwardQ // A residential city with a growing population of 50,000 people is experiencing challenges related to wastewater management. The city council has proposed building a combined wastewater treatment plant to ensure that domestic and industrial wastewater is treated effectively, meeting regulatory standards for discharge. The community has identified the need for an engineering solution that considers environmental, technical, and operational aspects. Scenario Details: Each person in the city consumes an average of 200 liters of water per day, and the organic load generated per person is 40 grams daily. The city also houses light industries that generate 0.25 million liters per day (MLD) of wastewater with a BOD, concentration of 600 mg/L. Environmental regulations mandate that treated wastewater discharged from the plant must not exceed a BOD5 of 20 mg/L. The treatment plant must be designed to operate efficiently under seasonal temperature variations, with temperatures reaching 35°C…arrow_forward23) An HSS 10 × 6 × 5/16 with F, = 46 ksi is used as a column. The length is 16 feet. Both ends are pinned, and there is support against weak axis buckling at a point 6 feet from the top. Determine a. the design strength for LRFD. b.the allowable stress for ASD 6' 10' 16' y-axis x-axis (good luck) Falls 12Aarrow_forward
- + H.W ATF pland as illustrated in hig below. has. the following: a primary clarifier with a 16.8m diameter, 21m (SWD), and Single peripheral weir, an 26m diameter TE with a 2.1 deep stone filled bed; and a final settling Tank with a 15-2m diameter, 2.1m (SWD), and single peripheral weir, The normal operating recirculation ratio is NA 0.5 with return to the wet well from the bottom of the final during periods of low influent flow (fig② The daily wow flow is 5220 m³/d with an average BOD of 180-9/1 Calculate the loadings on all of the units, and the anticipated. effluent BOD 2 20°C and 16C * Assume @ efficiency of removal BOD load of 78% in 20 C Is of 68% 7416 68/74/6 Futurarrow_forwardW10 x 33 A992 steel [x= ry=9 Q2) Does the column shown in Figure below have enough available strength to support the given service loads? a. Use LRFD. b.Use ASD D= 180k L = 540* W14 x 90 13' A992 steel (Sto Sage)arrow_forwardStatics Beam Problem – Find Reactions, Shear, Moment, and Deflection Description: I need help solving this statics problem related to a beam subjected to multiple loads. The problem asks for: Determining the three reactions at the supports. Drawing the shear force diagram. Drawing the bending moment diagram. Drawing the deflected shape of the beam.arrow_forward
- subjectarrow_forwardIn one of the construction projects, a bulldozer is used for a weekly rent of $150, with a work rate of 10 hours per day, 50 minutes per hour, and 6 days per week. If you know that the dimensions of the bulldozer blade are 3 lengths and 1 m height, the slope of the pushed soil is (1:1), its speed when loaded is 45 m/min and when returning is 80 m/min, and the time for loading, unloading, and changing is 0.35 minutes, the transported soil is sandy, weighing 1650 kg/m3 by Bank volume and 1300 kg/m3 by loose volume, the transport distance is 300 m. If you know that using a tractor shovel costs $0.5 per m3, ma a decision about choosing the economical equipment to use in the project.arrow_forward/ An irrigation project requires digging and paving an irrigation canal with alength of (30 km) and a semi-circular section with a diameter of (8 m). The work mustmbe completed within one year, and in the event of delay, a fine delay of $200/day. You have two options for work: 1st Option: using standard equipment (excavators, graders and hand labor) had a cost of $10/m3, the job will be finish without delay. 2nd Option: Use a channel trimmer machine with the following specifications: Purchase price ($70,000), it can be sold after its economic life of (5 years) for ($30,000), its speed (10m/hr), engine diesel with (250 hp) and its work 48 minutes/hr, capacity of crankcase is (50 liters), the number of hours between changing the oil (100 hr), the maintenance cost is (80%) of its depreciation, the price of one liter of fuel ($0.5/liter) and of oil ($1/liter), the operators' wages ($15,000/year), annual operating hours (2000 hr), and the owner must pay taxes of 10% of the average value of…arrow_forward
- subjectarrow_forwardExample 5.2: Design Example of a Square Foundation for a Colum Figure A.3a shows a square column foundation with the followin conditions: Live load=L=675 kN Dead load D= 1125 kN Allowable gross soil-bearing capacity 145 kN/m² Column size-0.5 m x 0.5 m -20.68 MN/m² 1,413.7 MN/m² Let it be required to design the column foundation. 1.25m-D L-675 AN D-1125 AN (a) Column section- 05 m x 0.5marrow_forwardExample 5.3: Design Example of a Rectangular Foundation for Column This section describes the design of a rectangular foundation to support a column having dimensions of 0.4 m x 0.4 m in cross section. Other details are as follows: Dead load D= 290 kN Live load L= 110 kN Depth from the ground surface to the top of the foundation = 1.2 m Allowable gross soil-bearing capacity = 120 kN/m² Maximum width of foundation = B = 1.5 m f, = 413.7 MN/m² f=20.68 MN/m² Unit weight of soil = y = 17.27 kN/m³ Unit weight of concrete = y₁ = 22.97 kN/m³ General Considerations For this design, let us assume a foundation thickness of 0.45 m (Figure A.4a). The weight of foundation/m² = 0.45 y = (0.45) (22.97) = 10.34 kN/m², and the weight of soil above the foundation/m² = (1.2)y = (1.2) (17.27) = 20.72 kN/m² Hence, the net allowable soil- bearing capacity Iqata] =120-10.34-20.72 = 88.94 kN/m² The required area of the foundation = (D+L)/4(all) = (290 + 110)/88.94 = 4.5 m². Hence, the length of the foundation…arrow_forward
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