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QuestionWhat is key application of inverted siphons? What are major drawbacks of using inverted siphons?
AnswerInverted siphons are planned at locations in which sewer system is blocked by anti-establishment utilities, storm water drains. They are now and then known as depressed sewers as it is claimed that there is no real siphon action. They attach upstream and downstream sewers with U-shaped straight up alignment such that they forever running full. Drawbacks of inverted siphons are-(i) They induce extra head loss to sewer system which is disagreeable in hydraulic performance.(ii) U-shaped siphons construct sediment accumulation trouble and earlier experience showed that inverted siphons were simply blocked due to siltation; (iii) Protection of overturn siphons is complicated due to its isolation.
Parallel Pipe Systems or Multipath Pipeline Problems Two or more sections of piping (branches) of different diameters, lengths and/or pipe materials originating and terminating
Binomial Distribution - exact probability: For the binomial distribution having n = 10, y = 0.4, find p (X ≤ 4) using equations and verify that equation gives a better approxi
Question What is consideration in selecting an orientation of wing walls in designing of bridge abutments ? Answer There are three familiar arrangements of wing walls in
Transit Sheds and Warehouses: These are sheds constructed as steel frames or RCC frames with a steel truss, with galvanised iron or asbestos cement sheets. Transit sheds may b
Describe the two fundamental purpose of surveying.
pls explain about this field
Q. What is Critical steel ratio? Critical steel ratio - only consider 250mm of concrete from outer face The purpose of critical steel ratio is to control cracking pattern
Define Special considerations for underwater inspection? Once a diver enters the water, there environment changes completely. Visibility decreases and is often reduced to near
Tensile stresses It has been possible to introduce tensile stresses of upto 850 N/mm 2 in tendons with the use of this method. But the method is not suited in situations where
Determine the normal and shear stress elements: The state of stress at a point is given through the stress components σ x = 70 MPa, σ y = 10 MPa, and τ xy = - 40 MPa. Using
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