Inclined smooth pipe with wall shear limit

Data. $D = 0.1$ m, $L = 60$ m, inclination 10°, $\rho = 900$ kg/m³ (R.D. 0.9), $\nu = 120$ mm²/s $= 1.2\times10^{-4}$ m²/s, outlet gauge pressure $p_2 = 120$ kPa, maximum mean wall shear $\tau_0 = 200$ Pa.

Which flow regime?

Mean wall shear in terms of $f$: $\tau_0 = \dfrac{f\rho V^2}{8}$.

• If laminar: $\tau_0 = \dfrac{8\mu V}{D}$, i.e. $V = \dfrac{\tau_0 D}{8\mu} = \dfrac{200(0.1)}{8(900)(1.2\times10^{-4})} = 23.15$ m/s, giving $Re = 19290$, which is not laminar. So the flow is turbulent.
• Turbulent: $f = 0.32\,Re^{-1/4}$ (valid since $Re < 10^5$).

(b) Flow rate

$$\tau_0 = \frac{\rho V^2}{8}(0.32)\left(\frac{\nu}{VD}\right)^{1/4} = 0.04\,\rho\left(\frac{\nu}{D}\right)^{1/4} V^{7/4}$$ $$200 = 0.04(900)\left(\frac{1.2\times10^{-4}}{0.1}\right)^{1/4} V^{1.75} = 6.700\,V^{1.75}$$ $$V = \left(\frac{200}{6.700}\right)^{1/1.75} = 6.963\ \text{m/s}$$

Check: $Re = \dfrac{6.963(0.1)}{1.2\times10^{-4}} = 5803 < 10^5$, so the formula applies; $f = 0.32Re^{-1/4} = 0.03666$ and $\dfrac{f\rho V^2}{8} = 200.0$ Pa.

$$Q = \frac{\pi}{4}D^2V = \frac{\pi}{4}(0.1)^2(6.963) = 0.05469\ \text{m}^3/\text{s} = 54.69\ \text{l/s}$$

(a) Maximum inlet pressure

Force balance on the liquid in the pipe along the incline (steady, uniform flow):

$$(p_1 - p_2)\frac{\pi D^2}{4} = \tau_0\,\pi D L + \rho g\,\frac{\pi D^2}{4}L\sin\theta$$ $$p_1 - p_2 = \frac{4\tau_0L}{D} + \rho g L\sin\theta = \frac{4(200)(60)}{0.1} + 900(9.81)(60)\sin10^\circ$$ $$p_1 - p_2 = 480000 + 91988 = 571988\ \text{Pa}$$ $$p_1 = 120\,000 + 480000 + 91988 = 691988\ \text{Pa} \approx 692\ \text{kPa (gauge)}$$

Answer: (a) $p_{1,max} \approx 692$ kPa gauge; (b) $Q = 54.7$ l/s ($V = 6.96$ m/s).