Question 11.12: (Refer to Example Problem 11.11.) Air at 68^◦ F flows in a p...

(a) In order to determine the pressure drop (and head loss) in flow of the water in the model, the major head loss equation, Equation 11.122 h_{f} = \frac{\tau _{w}L}{\gamma R_{h}} = \frac{\Delta p}{\gamma } = S_{f}L = C_{D}\rho v^{2} \frac{L}{\gamma R_{h}} = \frac{v^{2}L}{C^{2}R_{h}} = f \frac{L}{D} \frac{v^{2}}{2g} = \left(\frac{vn}{R_{h}^{2/3}} \right)^{2}L , is applied as follows:

where the friction factor, f is used to model the flow resistance. Because the Reynolds number, 2,000 < R < 4,000 (transitional pipe flow) is assumed, the friction  factor, f is a function of both ɛ/D and R, as illustrated by the Moody diagram in Figure 8.1. However the Colebrook equation, Equation 8.33 C = \sqrt{\frac{8g}{f} } , presents a mathematical representation of the Moody diagram in Figure 8.1. Furthermore, in order to determine the length, diameter, and the absolute pipe roughness of the model pipe, the model scale, λ (inverse of the length ratio) is applied. The fluid properties for water are given in Table A.2 in Appendix A.

D_{p}: = 3 ft                             L_{p}: = 1300 ft                             \varepsilon _{p} : = 0.03 ft                             \lambda : = 0.2

Guess value:                             D_{m}: = 0.1 ft                             L_{m}: = 1 ft                             \varepsilon _{m}: = 0.01 ft

Given

\lambda = \frac{D_{m}}{D_{p}}                             \lambda = \frac{L_{m}}{L_{p}}                             \lambda = \frac{\varepsilon _{m}}{\varepsilon _{p}}
\left ( \begin{matrix} D_{m} \\ L_{m} \\ \varepsilon _{m} \end{matrix} \right ) : = Find (D_{m}, L_{m}, \varepsilon _{m}) = \left ( \begin{matrix} 0.6 \\ 260 \\ 6 \times 10^{-3} \end{matrix} \right ) ft
slug: = 1 lb \frac{sec^{2}}{ft}                             \rho _{m} : = 1.936 \frac{slug}{ft^{3}}                             \mu _{m} : = 20.5 \times 10^{-6} lb \frac{sec}{ft^{2}}

g: = 32.174 \frac{ft}{sec^{2}}                             \gamma _{m}: = \rho _{m}. g = 62.289 \frac{lb}{ft^{3}}                             V_{m}: = 0.07 \frac{ft}{sec}

Guess value:                             h_{fm}: = 1 ft                             \Delta P_{m}: = 1 \frac{lb}{ft^{2}}                             f_{m}: = 0.01

Given

h_{fm} = f_{m} \frac{L_{m}}{D_{m}} \frac{V^{2}_{m}}{2g}                             \frac{1}{\sqrt{f_{m}} } = – 2 log \left(\frac{\frac{\varepsilon _{m}}{D_{m}} }{3.7} + \frac{2.51 }{R_{m}. \sqrt{f_{m}} } \right)
\Delta P_{m} = h_{fm} . \gamma _{m}
\left ( \begin{matrix} h_{fm} \\ \Delta P_{m} \\ f_{m} \end{matrix} \right ) : = Find (h_{fm, \Delta P_{m} , f_{m}})

h_{fm}= 1.622 \times 10^{-3} ft                             \Delta P_{m} = 0.101 \frac{lb}{ft^{2}}                             f_{m}= 0.049

(b) To determine the velocity flow of the air in the prototype pipe flow in order to achieve dynamic similarity between the model and the prototype for transitional pipe flow, the ɛ/D must remain a constant between the model and prototype as follows:

\left(\frac{\varepsilon}{D} \right)_{p} = \left(\frac{\varepsilon }{D } \right)_{m}
\frac{\varepsilon _{p}}{D_{p}} = 0.01                             \frac{\varepsilon _{m}}{D_{m}} = 0.01

Furthermore, the R must remain a constant between the model and prototype as follows:

The fluid properties for air are given in Table A.5 in Appendix A.

\rho _{p} : = 0.00231 \frac{slug}{ft^{3}}                             \mu _{p} : = 0.376 \times 10^{-6} lb \frac{sec}{ft^{2}}

Guess value:                             V_{p}: = 1 \frac{ft}{sec}                             R_{p}: = 3000

Given
R_{p} : = \frac{\rho _{p} . V_{p}. D_{p}}{\mu _{p}}
R_{p} = R_{m}
\left ( \begin{matrix} V_{p} \\ R_{p} \end{matrix} \right ) : = Find ( V_{p}, R_{p})

V_{p} = 0.215 \frac{ft}{s}                             R_{p} = 3.966 \times 10^{3}

(c) To determine the pressure drop (and head loss) in the flow of the air in the prototype in order to achieve dynamic similarity between the model and the prototype for transitional pipe flow, the friction factor, f must remain a constant between the model and the prototype (which is a direct result of maintaining a constant ɛ/D and a constant R between the model and the prototype) as follows:

\underbrace{\left[\frac{\frac{h_{f}}{v^{2}L} }{2gD} \right]_{p} }_{f_{p}} = \underbrace{\left[\frac{\frac{h_{f}}{v^{2}L} }{2gD} \right]_{m} }_{f_{m}}
\gamma _{p}: = \rho _{p}.g =0.074 \frac{lb}{ft^{3}}

Guess value:                             h_{fp}: = 1 ft                             \Delta p_{p}: = 1 \frac{lb}{ft^{2}}                             f_{p}: = 0.1

Given
f_{p} = \frac{h_{fp}}{\left(\frac{V^{2}_{p}. L_{p}}{2.g. D_{p}} \right) }                             \Delta p_{p} = h_{fp}. \gamma _{p}

f_{p} = f_{m}
\left ( \begin{matrix} h_{fp} \\ \Delta p_{p} \\ f_{p} \end{matrix} \right ) : = Find (h_{fp},\Delta p_{p}, f_{p})

h_{fp} = 0.015 ft                             \Delta p_{p} = 1.139 \times 10^{-3} \frac{lb}{ft^{2}}                             f_{p}: = 0.049

Therefore, although the similarity requirements regarding the independent π term, \varepsilon /D ((\varepsilon /D)_{p} = (\varepsilon /D)_{m} = 0.01) and the independent π term, R (“viscosity model”) ( R_{p} = R_{m} = 3.966 \times 10^{3} ) are theoretically satisfied, the dependent π term (i.e., the friction factor, f ) will actually/practically remain a constant between the model and its prototype ( f_{p} = f_{m} = 0.049 ) only if it is practical to maintain/attain the model velocity, pressure, fluid, scale, and cost.