Water flows at mass flow rate lit through a 90 degree vertically oriented elbow
ID: 1850014 • Letter: W
Question
Water flows at mass flow rate lit through a 90 degree vertically oriented elbow of elbow radius R (to the centerline) and inner pipe diameter D as sketched. The outlet is exposed to the atmosphere. (Hint: This means that the pressure at the outlet is atmospheric pressure.) The pressure at the inlet must obviously be higher than atmospheric in order to push the water through the elbow. The irreversible head loss through the elbow is hL. Assume that the kinetic energy flux collection factor alpha is not unity, but is the same at the inlet and outlet of the elbow ( alpha 1 = alpha 2). Assume that the same tiling applies to the momentum flux collection factor beta (i.e..beta 1= beta 2). Using the head form of the energy equation, derive an expression for the gage pressure Pgage, 1 at the center of the inlet as a function of the other variables as needed. Neglecting the weight of the elbow itself and the weight of the water in the elbow, calculate the x and z components of the anchoring force required to hold the elbow in place. Your final answer for the anchoring force should be given as F = Fxi + Fz.k . Your answer for Fx should lie between -120 and -140 N, and your answer for F2 should lie between 80 and 90 N. Repeat Part (c) without neglecting the weight of the water in the elbow. Is it reasonable to neglect the weight of the water in this problem?Explanation / Answer
1/2 m v^2 = m g h so v^2 = 2gh v=sqrt(2gh) At a height of 3 m , v = sqrt(2 * 9.8 *3) = 7.7 m/s If you want the volume of water multiply the area * the speed. so for a hose of 12mm diameter. (0.012 m) area = pi() * r^2 = pi()* 0.006^2 And volume = pi() * 0.006^2 * 7.7 = 0.00087 m^3 /s = 0.87 litre per sec. which is 3100 litre per hour. In reality on a hose of this diameter the skin friction is greater than the amount of energy in kinetic energy even on a three metre length. This, according what you've described, is just a simple mechanics problem. It'll fall down like any other body from 10 feet .... tube or no tube. the radius of the tube is not relevant ..... Unless the tube has a variable radius. So, just treat it as an ordinary free fall problem. 1 foot = 30 cm approx. So, 10 feet => 300 cm => 0.3 m then, 2gs = v^2 -u^2 and u =0. so, 2 x 9.8 x 0.3 = v^2 v^2 = 5.88 => v=sqrt(5.88) = approx. 2.42 m/s
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