A two-dimensional diverging duct is being designed to diffuse the high speed air

Question

A two-dimensional diverging duct is being designed to diffuse the high speed air

exiting a wind tunnel. The x-axis is the centreline of the duct (it is symmetric about the x-axis),

and the top and bottom walls are to be curved in such a way that the axial wind

speed u decreases approximately linearly from u1= 300 m/s at section 1 to u2= 100 m/s at

section 2. Meanwhile, the air density ? is to increase approximately linearly from rho1=0.85

kg/m3 at section 1 to rho2=1.2 kg/m3 at section 2. The diverging duct is 2.0 m long and is

1.60 m high at section 1 (only the upper half is sketched in the Figure below; the half height

at section 1 is 0.80 m).

Find: (a) the y-component of velocity, v(x, y) in the duct. (b) Plot the approximate shape

of the duct, ignoring friction on the walls (the shape in the figure below is just a sketch,

you need to plot the streamlines based on u and v) (c) What should be the half height of

the duct at section 2.

Explanation / Answer

a)

At any section at distance x from left end,

u(x) = 300 - (300-100)*x/2 = 300 - 100*x

Density = rho(x) = 0.85 + (1.2 - 0.85)*x/2 = 0.85 + 0.175*x

Area = A(x)

Continuity eqn:

[del (rho*u) / (del x)] + [del (rho*v) / (del y)] = 0

[del (255 - 32.5x - 17.5*x^2) / (del x)] + [del (rho*v) / (del y)] = 0

( -32.5 - 35*x) + [del (rho*v) / (del y)] = 0

del (rho*v) / (del y) = 35*x + 32.5

Partially integrating, rho*v = (35*x + 32.5)*y + f(x)

At x,y = 0 we have v = 0.

Hence, f(x) = 0

Hence, rho*v = (35*x + 32.5)*y

or v = (35*x + 32.5)*y / (0.85 + 0.175*x)

b)

Continuity eqn:

rho(x)*u(x)*A(x) = constant

(300 - 100*x)*(0.85 + 0.175*x)*A(x) = constant

A(x) = constant / [(300 - 100*x)*(0.85 + 0.175*x)]

At section 1, A(x) = 2*0.8 = 1.6 m, rho(x) = 0.85 kg/m^3, u(x) = 300 m/s

Hece, constant = 1.6*0.85*300 = 408

Hence, A(x) = 408 / [(300 - 100*x)*(0.85 + 0.175*x)]

This can be plotted to get the shape.

c)

A(x) = 408 / [(300 - 100*x)*(0.85 + 0.175*x)]

At x = 2 m, we get A(2) = 408 / [(300 - 100*2)*(0.85 + 0.175*2)]

A(2) = 3.4 m

Half height = 3.4/2 = 1.7 m

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