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- An orifice meter was calibrated by the manufacturer for air at 25℃ and 1atm. It is installed in an air sampling system where it operates at 33℃, 0.9atm and reads a ΔP of 4.8″H2O. The calibration curve has the relationship Qi=41×(ΔP)1/2in slpm(liters per minute at 25℃, 1atm). What is the true gas flow rate in slpm?Can I get a thorough step by step for this question? Each month the speedy dry-cleaning company buys 1.75 barrel (0.28 m') of dry-cleaning fluid. The amount of fluid lost to the atmosphere as emission is 210 kg/month and the remaining remains as residue to be disposed of. The density of the dry-cleaning fluid is 1.5940 g/mL. Draw a mass-balance diagram and estimate the monthly % of mass residue to be dosposed of.Oil (SG=0.80) flows through aventuri meter as shown,a. Calculate the flowrate.b. Calculate thereading ‘‘h’’ shownby the differentialmanometer fittedto the pipelinewhich is filled with mercury of specificgravity 13.6.c. How will the flow rate change if the liquidflowing through the pipeline is replacedwith water.
- 1. THERE IS A LEAK IN A HORIZONTAL 300 MMDIAMETER PIPELINE. UPSTREAM FROM THE LEAK TWOGAGES 600 METER APART SHOWED A DIFFERENCE OF140 ???. DOWNSTREAM FROM THE LEAK TWO GAGES600 METER APART SHOWED A DIFFERENCE OF 126??? . ASSUMING ? = 0.025. HOW MUCH WATER ISBEING LOST FROM THE PIPE. show answer in litre per secA waste treatment lagoon having 3 cells, each 115,000 m2 in area, a minimum operating depth of 0.6 m, and a maximum operating depth of 1.5 m, receives 1,900 m3/d of wastewater having an average BOD5 of 122 mg/L. What is the BOD5 loading and detention time? The following data applies: the slope of the lagoon walls is disregarded in the calculations. a. 6.7 kg/ha-d, 54.5 daysb. 20.2 kg/ha-d, 163.4 daysc. 18.5 kg/ha-d, 177.6 daysd. 18.5 kg/ha-d, 54.5 dayse. None of theseAs the project engineer for your consulting firm, you are in charge of designing a flocculation tank to treat municipal drinking water. The design flow for the plant is 0.050 m3/s. The average water temperature is 10oC. The following design assumptions for the flocculation basin have been made:Number of tanks = 1Number of compartments in tank: 3Tapered G in 3 compartments: 60, 50, and 20 s-1Detention time = 30 min for entire flocculation basinDepth = 3 mTanks will have a square compartment plan (overhead) view1 Horizontal shaft paddle wheel per axis will be usedPaddle wheels will have 4 arms and 2 blades/armPaddle blades are 3 m long by 0.6 m wide. Assume blade spacing (diameter) of 1 and 2 m. Design the flocculation tank by providing the following for the first two compartments only:a) Tank and compartment dimensions in mb) Water power input in kWc) Rotational speed of the paddle wheel shaft in rpm
- A tank has 40L of sewage runoff. It initially consists of 60% water and 40% pollutants. Another sewage runoff with 70% water and 30% pollutant is being added to the tank at a rate of 3L/min. As the tank is being filled, the solution is also being drained at a rate of 4L/min. How much pollutant is in the tank after 7 mins?Show all problem solution's steps. Direct formula substitution solutions will have NO CREDIT POINTS. 1. A 75 mm diameter fire hose discharges water through a nozzle having a jet diameter of 25 mm. The lost head in the nozzle is 25% of the velocity head in the jet. If the pressure at the base of the nozzle is 625 KPa: compute the discharge in m3/min; the maximum horizontal range to which the stream can be thrown; the maximum vertical reach, and the diameter of the jet at a point two-third of the maximum vertical reach from the tip of the nozzle. Neglect air resistance.A raw sewage sample was brought in for total suspended solids and volatile solids analysis. A portion of the sample was poured into a tared evaporating dish and weighed. The sample was then heated at 105°C for 4 hours, weighed, heated at 550°C overnight, and weighed again. The data are as follows: • Tare wt. 42.9073g• Wt of sample 104.4680g• 105 °C wt. 45.4140g• 550 °C wt. 43.3236g What is the TSS in g/L? What percent of the solids are volatile?
- The water need of a district with a future population of 38800 and average daily water consumption qort = 189 lt / N / day will be met by leakage pipes fed double-sided (at the end of the collection box). Water layer thickness H = 5 m, k = 0.0018 m / s and n = 0.02 are given. Standard pipe diameters and costs are given as follows: 0.4 m (60 TL / m), 0.7 m (70 TL / m), 1.0 m (80 TL / m) and 1.3 m (85 TL / m). Size the leaking tube so that it is most economical (diameter, length and slope). Average flow velocity is between 0.2-0.4 m / s.not sure how to get this answer Answer: ???????? ???????? ???????? = ?? = 7570 ?%/???30 ? × 5 ? = 50.5 ???? = 2.1 ?/ℎ Total concentration of particles = 7240 /mL Total concentration of removed particles = 5199 /mL Overall particle removal efficiency = 5199/7240 = 71.8%Please do not give solution in image formate thanku. Q18. A groundwater basin of 144.7 km2 receives 126.7 cm of recharge on average each year. Groundwater loss from the basin in the form of evapotranspiration occurs at a mean rate of 62.9 cm/year (i.e. there is no groundwater discharge to the sea or rivers). Pumping draws groundwater from the basin at a rate of 0.65 m3/s. What is the rate of change in storage in GL/year? (note: 1 GL = 109 L)