Question 6.13: Fluid flows at an average velocity of 6 ft/s between horizon...

Part (a)

The viscosity µ = ρν = 3.8 × 10^{-5} slug/(ft · s). The spacing is 2h = 2.4 in = 0.2 ft, and D_h = 4h = 0.4 ft. The Reynolds number is

The flow is therefore turbulent. For reasonable accuracy, simply look on the Moody chart (Fig. 6.13) for smooth walls:

f ≈ 0.0173                h_f \approx f\frac{L}{D_h} \frac{V^2}{2g} = 0.0173\frac{100}{0.4}\frac{(6.0)^2}{2(32.2)} \approx 2.42  ft

Since there is no change in elevation,

This is the head loss and pressure drop per 100 ft of channel. For more accuracy, take D_{eff} = \frac{64}{96} D_h from laminar theory; then

and from the Moody chart read f ≈ 0.0189 for smooth walls. Thus a better estimate is

and          \Delta p = 1.9(32.2)(2.64) = 161  lbf/ft^2                    Better

The more accurate formula predicts friction about 9 percent higher.

Part (b)

Compute µ = ρν = 0.0038 slug/(ft · s). The Reynolds number is 6.0(0.4)/0.002 = 1200; therefore the flow is laminar, since Re is less than 2300.

You could use the laminar flow friction factor, Eq. (6.64)

f_{lam} = \frac{h_f}{(L/D_h)(V^2/2g)} = \frac{96 \mu}{\rho V(4h)} = \frac{96}{Re_{D_h}}                         (6.64)

from which                    h_f = 0.08 \frac{100}{0.4}\frac{(6.0)^2}{2(32.2)} = 11.2  ft

and                   \Delta p = 1.9(32.2)(11.2) = 684  lbf/ft^2

Alternately you can finesse the Reynolds number and go directly to the appropriate laminar flow formula, Eq. (6.63):

V = \frac{Q}{A} = \frac{h^2}{3\mu} \frac{\Delta p}{L} = \frac{2}{3} u_{max}                       (6.63)

or          \Delta p = \frac{3(6.0  ft/s) [0.0038  slug/(ft \cdot s)] (100  ft)}{(0.1  ft)^2} = 684  slugs/(ft \cdot s^2) = 684  lbf/ft^2

and            h_f = \frac{\Delta p}{\rho g} = \frac{684}{1.9(32.2)} = 11.2  ft

This is one of those (perhaps unexpected) problems where the laminar friction is greater than the turbulent friction.