written 6.0 years ago by
teamques10
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modified 6.0 years ago
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Where f = friction factor
L= length of duct in meter
v = velocity of air in duct in m/sec
D= diameter of duct in meter
Discharge through duct is given by, Q= A.v …….(2)
Where A = cross sectional area of duct
Expression for the equivalent diameter of a circular duct corresponding to a rectangular duct when the quantity of air passing through both ducts is same
Substituting equation (2) in equation (1),
$\frac{h_f}{L}=\frac{(f(Q/A)^2)}{2gD}$
$D^{5}=\frac{64(a,b)^{2}}{2\pi^{2}(a+b)}$
D=1.265$\frac{a^{3}b^{3}}{a+b}^{1/5}$
ii.Expression for the equivalent diameter of a circular duct corresponding to a rectangular duct when the velocity of air flowing through both ducts is same
We can write velocity from equation (1) as
$\frac{a,b}{2(a+b)}$=$\frac{\frac{\pi}{4}D^{2}}{\pi.D}$
$D=\frac{2ab}{a+b}$