Please show work, will award 5 stars to an answer with work! A two-dimensional d
ID: 1862667 • Letter: P
Question
Please show work, will award 5 stars to an answer with work!
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 rho is to increase approximately linearly from rho 1=0.85 kg/m3 at section 1 to rho 2=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). ? (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