2. Air flows through the tube shown in the diagram. Assume the air is an ideal f

Question

2. Air flows through the tube shown in the diagram. Assume the air is an ideal fluid with a density of 1.28 kg/m^3. Points 1 and 2 are at the same vertical height. a) What is the difference in pressure between point 1 and 2 (=p2 - p1)? HINT: The mercury (Hg) in the U-shaped tube is at rest. The difference in the pressure in the flowing air between points 1 and 2 is related to the difference in the height of the mercury in the two sides of the U-shaped tube. b) What are the air speeds at points 1 (v1) and point 2 (v2)? HINT: Start with Bernoulli's equation applied at points 1 and 2. c) What is the volume flow rate?

Explanation / Answer

P2 - P1 = rho_hg*g*h = 13600*9.8*0.1 = 13328 Pa

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continuity equation

A1*v1 = A2*v2

pi*r1^2*v1 = pi*r2^2*v2

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

d1^2*v1 = d2^2*v2

v1 = (d2/d1)^2*v2 = 25*v2

bernoullis pricilple

P2 + 0.5*rho*v2^2 = P1 + 0.5*rho*v1^2

P2 - P1 = 0.5*rho*(v1^2 - v2^2)

P2 - p1 = 0.5*rho*(25^2*v2^2 - v2^2)

P2 -P1 = 0.5*1.28*624*v2^2

13328 = 399.36*v2^2

v2 = 5.77 m/s

v1 = 5*v2 = 28.85 m/s

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r1 = 1mm = 1e-3 m

A1 = pi*r1^2
flow rate = A1*v1 = 3.14*1e-6*28.85 = 9.0589*10^-5 m^3/s

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