Question 11.14: Air at 68^◦ F flows in a prototype circular pipe with a diam...

(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 11.148 h_{f,min} = \frac{\Delta p}{\gamma } = k \frac{v^{2}}{2g} , is applied as follows:

where the minor loss coefficient, k is used to model the flow resistance. Because the Reynolds number, R> 4000 (turbulent pipe flow with a pipe component) is assumed, the minor loss coefficient, k is only a function of ɛ/D and the geometry of the bend, and is independent of R, as illustrated in Figure 8.9. Furthermore, in order to determine the diameter and the absolute pipe roughness of the model pipe and the round radius of the 90^{\circ} model bend, 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}: = 4 ft                               r_{p}: = 8 ft                               \varepsilon _{p} : = 0.008 ft                               \lambda : = 0.25

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

Given

\lambda = \frac{D_{m}}{D_{p}}                               \lambda = \frac{r_{m}}{r_{p}}                               \lambda = \frac{\varepsilon _{m}}{\varepsilon _{p}}
\left ( \begin{matrix} D_{m} \\ r_{m} \\ \varepsilon _{m} \end{matrix} \right ) : = Find (D_{m}, r_{m}, \varepsilon _{m}) = \left ( \begin{matrix} 1 \\ 2 \\ 2 \times 10^{-3} \end{matrix} \right ) ft

Thus, the minor loss coefficient, k for the model is determined from Figure 8.9.

\frac{r_{m}}{D_{m}} = 2                               \frac{\varepsilon _{m}}{D_{m}} = 2 \times 10^{-3}                              k_{m}: = 0.35

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}: = 70 \frac{ft}{sec}

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

Given

h_{fm} = k_{m} \frac{v^{2}_{m}}{2g}
\Delta P_{m} = h_{fm} . \gamma _{m}
\left ( \begin{matrix} h_{fm} \\ \Delta P_{m} \end{matrix} \right ) : = Find (h_{fm}, \Delta P_{m})

h_{fm}= 26.652 ft                               \Delta P_{m} = 1.66 \times 10^{3} \frac{lb}{ft^{2}}

(b)–(c) 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 turbulent pipe flow with a pipe component flow, and 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 turbulent pipe flow with a pipe component, the geometry, L_{i}/L must remain a constant between the model and prototype as follows:

\left(\frac{L_{i}}{L} \right)_{p} = \left(\frac{L_{i} }{L } \right)_{m}
\frac{r_{p}}{D_{p}} = 2                              \frac{r_{m}}{D_{m}} = 2

And, 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}} = 2 \times 10^{-3}                              \frac{\varepsilon _{m}}{D_{m}} =2 \times 10^{-3}

However, because the minor loss coefficient, k is independent of R for turbulent flow with a pipe component, R does not need to remain a constant between the model and the prototype.
(b)–(c) 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 turbulent pipe flow with a pipe component, and 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 turbulent pipe flow with a pipe component, 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 L_{i}/L between the model and the prototype and applying the “pressure model” similitude scale ratio; specifically the velocity ratio, v_{r} given in Table 11.1) as follows:

\underbrace{\left[\frac{\frac{h_{f}}{v^{2}} }{2g} \right]_{p} }_{c_{D_{p}} = k_{p}} = \underbrace{\left[\frac{\frac{h_{f}}{v^{2}} }{2g} \right]_{m} }_{c_{D_{m}} = k_{m}}
v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\sqrt{\frac{\Delta p}{\rho } }\right) _{p} }{\left(\sqrt{\frac{\Delta p}{\rho } }\right) _{m}} = \Delta p_{r}^{\frac{1}{2} } \rho _{r}^{\frac{-1}{2} }

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}}                               \gamma _{p} : = \rho _{p}. g = 0.074 \frac{lb}{ft^{3}}

Guess value:                              V_{p}: = 1 \frac{ft}{sec}                               h_{fp}: = 1 ft                             \Delta p_{p}: = 1 \frac{lb}{ft^{2}}                               k_{p}: = 0.01

Given

k_{p} = \frac{h_{fp}}{\left(\frac{V^{2}_{p}}{2.g} \right) }                               \frac{V_{P}}{\Delta p_{p}^{\frac{1}{2} }. \rho _{p}^{\frac{- 1}{2}} } = \frac{V_{m}}{\Delta p_{m}^{\frac{1}{2} }. \rho _{m}^{\frac{- 1}{2}} }

k_{p} = k_{m}                               \Delta p_{p} = h_{fp}. \gamma _{p}
\left ( \begin{matrix} V_{p} \\ h_{fp} \\ \Delta p_{p} \\ k_{p} \end{matrix} \right ) : = Find ( V_{p}, h_{fp},\Delta p_{p}, k_{p})

V_{p} = 89.476 \frac{ft}{s}                               h_{fp} = 43.546 ft                               \Delta p_{p} = 3.236 \frac{lb}{ft^{2}}                               k_{p} = 0.35

Furthermore, the Euler number, E remains a constant between the model and the prototype as follows:

E_{m}: = \frac{\rho _{m}. V^{2}_{m}}{\Delta p_{m}} = 5.714                               E_{p}: = \frac{\rho _{p}. V^{2}_{p}}{\Delta p_{p}} = 5.714

Therefore, although the similarity requirements regarding the independent π term, \varepsilon /D ((\varepsilon /D)_{p} = (\varepsilon /D)_{m} = 0.002 ; the independent π term, L_{i}/L (r_{p} /D_{p} = r_{m} /D_{m} = 2); and the dependent π term, E (“pressure model”) (E_{p} = E_{m} = 5.714) are theoretically satisfied, the dependent π term (i.e., the minor loss coefficient, k) will actually/practically remain a constant between the model and its prototype (k_{p} = k_{m} = 0.35) only if it is practical to maintain/attain the model velocity, pressure, fluid, scale, and cost. Furthermore, because the minor loss coefficient, k is independent of R for turbulent flow with a pipe component, R does not need to remain a constant between the model and the prototype as follows:

R_{m} = 6.611 \times 10^{6}                               R_{p} : = \frac{\rho _{p} . V_{p}. D_{p}}{\mu _{p}} =2.199 \times 10^{6}