The water volume flow rate through the siphon is 0.005 m3/s, and the pipe diamet
ID: 2995501 • Letter: T
Question
The water volume flow rate through the siphon is 0.005 m3/s, and the pipe diameter is 25 mm. Assume that the flow is frictionless, water is incompressible with a density of 999 kg/m3, the atmosphere pressure is 101 kPa, and the gravity constant is g = 9.81 m/s2. Determine the velocity of the water in the pipe; If the height h = 4 m, determine the pressure at point A. Answers:(a)1.2m/s; 9.83kPa H6 - 3 7.12 from the textbook. Answer: PHI1 = F/muVD = constant To determine the drag on a balloon with a diameter of 1m in a wind speed of 5 m/s (the density of the air is 1.24 kg/m3 and the viscosity of the air is 1.8 Times 10 - 5 N.s/m2), a model with a diameter of 5cm is built for testing in water (the density of the water is 999 kg/m3 and the viscosity of the water is 1 Times 10 - 3 N.s/m2). What water speed is required to model the prototype? At this speed the model drag is measured to be 2000 N. What will be the corresponding drag on the prototype? Answers: (a) 6.90 m/s; (b) 522 N.Explanation / Answer
6-2
a)
V = Q/A
= 0.005 / (3.14 /4 * 0.025^2)
= 10.2 m/s
b)
Pa + 1/2*rho*V^2 = Patm - rho*g*h
Pa = 101*10^3 - 999*9.81*4 - 1/2*999*10.2^2
= 9831 Pa
= 9.83 kPa
6-4
Reynolds number of model = rho*v*d / u = 1.24*5*1 / (1.8*10^-5) = 3.44*10^5
Reynolds number of prototype = 999*v*0.05 / (1*10^-3) = 49950*v
Equating both, 3.44*10^5 = 49950*v
v = 6.9 m/s
Drag is proportional to rho* v^2 * d^2.
For air, rho * v^2 *d^2 = 1.24*5^2 * 1 = 31
For water, rho*v^2 *d^2 = 999* 6.9^2 * 0.05^2 = 118.9
Hence, D_proto / 2000 = 31 / 118.9
D_proto = 522 N
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