Question

1. If the air in a 12-inch diameter duct with an air flow of 1,500 ft³/min and a velocity of 14 ft/min flows into a 6-inch diameter duct, what are the new flow rate and velocity?
2. What velocity of air would be required to draw toxic vapors into a duct 24 inches away from a 12 inch round duct with an air flow rate of 1,500 ft³/min?
3. Compare the flow rates into a hood that needs to capture a contaminant at 12 inches vs 24 inches. Assume that the minimum capture velocity of 200 ft/min and an 8-inch round hood opening

Answer 1

(1). Given that, diameter = 12 inch , air flow rate Q1 = 1500 ft^3/min , V1 = 14 ft/min , Q2 ? and V2 ?

velocity in second duct =?

continuity equation A1V1 = A2V2

V2 = A1*V1/A2 = pi*(1^2)*14 / pi*(0.5)^2

V2 = 14/0.25 = 56 ft/min

volumetric flow rate Q2 = A*V

                 = pi*(0.5)^2*56 = 43.98 ft^3/min

(2).

. The velocity of air will be given by -

we know that, v = Q / A = Q / 4\pir2

where, Q = volumetric flow rate of air = 1500 ft3/s = 0.7079212 m3/s

r = distance away from a round duct = 24 inches = 0.6096 m

then, we get

v = (0.7079212 m3/s) / [4 (3.14) (0.6096 m)2]

v = [(0.7079212 m3/s) / (4.66 m2)]

(3).

The flow rate is given as,

Q = 1.4*P*x*V

where P = perimeter of tank

P = ?D = ?8 in = ?8/12 ft = 2.094 ft

x = distance of hood above tank

V = capture velocity = 200(ft/min)

For x = 12 in = 1ft

Q =1.4x 2.094 x1 x 200 = 586.32 ft3/min

For x = 24 in = 2ft

Q = 1.4 x 2.094 x2 x 200 = 1172.64 ft3/min

Hence, Q12 / Q24 = 0.5