9:33 PM eccphys1b.files.wordpress.com o AT&T; 151% 8 of 43 (2.1.20) Air is blowi

Question

9:33 PM eccphys1b.files.wordpress.com o AT&T; 151% 8 of 43 (2.1.20) Air is blowing through a horizontal tube which is connected to a U-shaped tube, as shown below. The left side of the U-tube is connected to a part of the horizontal tube with a diameter di 3.00 cm. The right side of the U-tube is connected to a part of the horizontal tube with a diameter d 2.00 cm. The liquid in the U-tube is water, except for a column of oill of height h 5.00 cm on the left side of the U-tube. The liquid levels on the right and left sides of the U-tube are equal. What is the speed of the air in the narrow part of the horizontal tube? Use the following densities: Pair = 1 29 kg/m, Prater 1000 kg/m3, and Aaa = 800 kg/m" V1 (2.1.30) A vertical cylinder filled with water has a 5.0-cm inner diameter. There is a 1.0 em hole near the bottom of the cylinder. The hole is a distance h beneath the water surface. (a) As a function of h, what is the speed at which water exits the hole? (b) As a function of h, what is the rate at which the water level in the cylinder drops (c) If the water level in the cylinder is initially 40.0 cm above the hole, how much time elapses before the water level is 30.0 cm above the hole?

Explanation / Answer

2.1.2

given data

diameter of left pipe, d1 = 3 cm

diameter of right pipe = d2 = 2 cm

velocity of flow in left pipe = v1

velocity of flow in right pipe = v2

density of oil, a = 800 kg/m^3

density of water, b = 1000 kg/m^3

density of air, c = 1.29 kg/m^3

now let pressure in left pipe be P1, in the right pipe be P2

then applying continuity equation

A1v1 = A2v2

pi(d1)^2/4 * v1 = pi (d2)^2 / 4 *v2

9*v1 = 4*v2

v1 = 4*v2/9

applying bernoullis equation

P1 + c*v1^2/2 = P2 + c*v2^2/2

P1 - P2 = c(v2^2 - v1^2)/2 = c*v2^2 (1 - 16/81)/2 = 65c*v2^2/162

Now, in the U tube if the liquids are level then

let height of water column in right side of U tube be H

then P1 + a*gh + b*g(H - h) = b*gH + P2

P1 - P2 = bgH - bgH + bgh - agh = gh(b - a)

h = 5 cm given

0.05*g(1000 - 800) = 65*1.29*v2^2/162

v = 13.767 m/s

so speed of air in the narrow part is 13.767 m/s

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