1. (4 points) The inet of a two-dimensional duct, shown below, is of width b and

Question

1. (4 points) The inet of a two-dimensional duct, shown below, is of width b and contains a frictionless fluid of constant density . The velocity in one half of the duct is Ul and the other half is 0.5U1. The two streams do not mix. The outlet of the duct is of width 0.5b. The flow is parallel at the duct entrance and exit, so that the pressure is uniform across the two sections. By using the continuity equation and Bernoulli's equation, determine the pressure change pi - P2 along the duct contraction and the individual dimensions bi and b2 of the two streams at the exit section 0.5b 0.5U1 U2. 0.5b b, 0.5b 2b Station (iniet) Station 2 (outlet)

Explanation / Answer

1. given inlet of a 2D duct

frictionless fluid of constant density rho

inlet dimension = b

outlet dimension = 0.5b

from the ginve figure

from conservation of mass

rho*(0.5b)*0.5U1 = rho*b2*U2a

=> 0.25b*U1 = b2*U2a

and

0.5b*U1 = b1*U2b

from bernouillis theoerm

Pi + 0.5*rho*0.25U1^2 + rho*g*0.75b = Pf + 0.5*rho*U2a^2 + rho*g*(0.25b + b1 + b2/2)

Pf - Pi = 0.5*rho*(0.25U1^2 - U2a^2) + rho*g(0.5b - b1 - b2/2)

Pi + 0.5*rho*U1^2 + rho*g*0.25b = Pf + 0.5*rho*U2b^2 + rho*g(0.25b + b1/2)

Pf - Pi = 0.5*rho*(U1^2 - U2b^2) - rho*g(b1/2)

comparing

U1^2 - U2b^2 - gb1 = 0.25U1^2 - U2a^2 + gb - 2gb1 - gb2

U1^2 - U2b^2 + U2a^2 = gb(1 - U1/2*U2b - U1/4*U2a) + 0.25 U1^2

hence

Pf - Pi = rho*[((gb/2 - gb*U1/2*U2b - gb*U1/8*U2a + 0.25 U1^2 ]

and

b1 = 0.5bU1/U2b

b2 = 0.25*bU1/U2a

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