Air, with ρ = 0.00237 slug/ft^3 and ν = 0.000157 ft^2/s, is forced through a horizontal square 9-by 9-in duct 100 ft long at 25 ft^3/s. Find the pressure drop if ε = 0.0003 ft.

Question

Air, with ρ = 0.00237 slug/[latex]ft^3[/latex] and ν = 0.000157 [latex]ft^2[/latex]/s, is forced through a horizontal square 9-by 9-in duct 100 ft long at 25 [latex]ft^3[/latex]/s. Find the pressure drop if ε = 0.0003 ft.

Accepted Answer

Compute the mean velocity and hydraulic diameter:

[latex]V = \frac{25 ft^3/s}{(0.75 ft)^2} = 44.4 ft/s[/latex]

[latex]D_h = \frac{4A}{\mathscr{P}} = \frac{4(81 in^2)}{36 in} = 9 in = 0.75 ft[/latex]

From Table 6.4, for b/a = 1.0, the effective diameter is

[latex]D_{eff} = \frac{64}{56.91} D_h = 0.843 ft[/latex]

whence [latex]Re_{eff} = \frac{VD_{eff}}{ u} = \frac{44.4(0.843)}{0.000157} = 239,000[/latex]

[latex]\frac{\epsilon}{D_{eff}} = \frac{0.0003}{0.843} = 0.000356[/latex]

Table 6.4 Laminar Friction Constants f Re for Rectangular and Triangular Ducts
Rectangular		Isosceles triangle
b/a	[latex]f_{Re_{D_h}}[/latex]	θ, deg	[latex]f_{Re_{D_h}}[/latex]
0.0	96.00	0	48.0
0.05	89.91	10	51.6
0.1	84.68	20	52.9
0.125	82.34	30	53.3
0.167	78.81	40	52.9
0.25	72.93	50	52.0
0.4	65.47	60	51.1
0.5	62.19	70	49.5
0.75	57.89	80	48.3
1.0	56.91	90	48.0

From the Moody chart, read f = 0.0177. Then the pressure drop is

[latex]\Delta p = ho g h_f = ho g \left(f\frac{L}{D_h}\frac{V^2}{2g} ight) = 0.00237(32.2) \left[0.0177\frac{100}{0.75}\frac{44.4^2}{2(32.2)} ight][/latex]

or Δp = 5.5 lbf/[latex]ft^2[/latex]

Pressure drop in air ducts is usually small because of the low density.

Question 6.15: Air, with ρ = 0.00237 slug/ft^3 and ν = 0.000157 ft^2/s, is ...

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