Determine the size of a galvanized iron pipe needed to transmit water a distance of 180 m at 0.085 [latex]m ^{3}[/latex]/s with a loss of head of 9 m. (Take kinematic viscosity of water ν = 1.14 × [latex]10^{-6}[/latex] [latex]m ^{2}[/latex]/s, and the average surface roughness for galvanized iron = 0.15 mm).

From (11.6b), [latex]h_{f}=\frac{\Delta p^{*}}{\rho g}=f \frac{L}{D}\left(V^{2} / 2 g\right)[/latex] (11.6b) [latex]9=f \frac{180}{D}\left(\frac{0.085}{\pi D^{2} / 4}\right)^{2} \frac{1}{2 \times 9.81}[/latex] which gives [latex]D^{5}=0.012 f[/latex] (11.8) and [latex]\operatorname{Re}=\frac{0.085 D}{\left(\pi D^{2} / 4\right) \times 1.14 \times 10^{-6}}[/latex] [latex]=9.49 \times 10^{4} \frac{1}{D}[/latex] (11.9) First, a guess in f is made as 0.024. Then from Eq. (11.8) D = 0.196 m and from Eq. (11.9) [latex]\operatorname{Re}=4.84 \times 10^{5}[/latex] The relative roughness [latex]\varepsilon / D=\frac{0.15}{0.196} \times 10^{-3}=0.00076[/latex] With the values of Re and e/D, the updated value of f is found from Fig. 11.2 as 0.019. With this value of f as 0.019, a recalculation of D and Re from Eqs (11.8) and (11.9) gives D = 0.187 m, Re = 5.07 [latex]\times 10^{5}[/latex]. e/D becomes (0.15/0.187) [latex]\times 10^{-3}[/latex] = 0.0008. The new values of Re and ∈/D predict f = 0.0192 from Fig. 11.2. This value of f differs negligibly (by 1%) from the previous value of 0.019. Therefore the calculated diameter D = 0.187 m is accepted as the final value.

Question 11.3: Determine the size of a galvanized iron pipe needed to trans...

Introduction to Fluid Mechanics and Fluid Machines

Determine the size of a galvanized iron pipe needed to transmit water a distance of 180 m at 0.085 $m ^{3}$ /s with a loss of head of 9 m. (Take kinematic viscosity of water ν = 1.14 × $10^{-6}$ $m ^{2}$ /s, and the average surface roughness for galvanized iron = 0.15 mm).

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From (11.6b),

$h_{f}=\frac{\Delta p^{*}}{\rho g}=f \frac{L}{D}\left(V^{2} / 2 g\right)$ (11.6b)

9=f \frac{180}{D}\left(\frac{0.085}{\pi D^{2} / 4}\right)^{2} \frac{1}{2 \times 9.81}

which gives $D^{5}=0.012 f$ (11.8)

and $\operatorname{Re}=\frac{0.085 D}{\left(\pi D^{2} / 4\right) \times 1.14 \times 10^{-6}}$

$=9.49 \times 10^{4} \frac{1}{D}$ (11.9)

First, a guess in f is made as 0.024.

Then from Eq. (11.8) D = 0.196 m

and from Eq. (11.9) $\operatorname{Re}=4.84 \times 10^{5}$

The relative roughness $\varepsilon / D=\frac{0.15}{0.196} \times 10^{-3}=0.00076$

With the values of Re and e/D, the updated value of f is found from Fig. 11.2 as 0.019. With this value of f as 0.019, a recalculation of D and Re from Eqs (11.8) and (11.9) gives D = 0.187 m, Re = 5.07 $\times 10^{5}$ . e/D becomes (0.15/0.187) $\times 10^{-3}$ = 0.0008. The new values of Re and ∈/D predict f = 0.0192 from Fig. 11.2. This value of f differs negligibly (by 1%) from the previous value of 0.019. Therefore the calculated diameter D = 0.187 m is accepted as the final value.