1. Assumptions and symbols
- Flow model: steady, incompressible, Newtonian fluid, fully developed, axisymmetric, circular pipe of diameter \(D=2R\).
- Symbols: \(\mu\) (dynamic viscosity), \(\rho\) (density), \(g\) (gravity), \(x\) (axial), \(r\) (radial), \(U\) (area-mean velocity), \(u(r)\) (axial velocity), \(Q\) (discharge), \(\Delta p\) (pressure drop), \(L\) (pipe length).
- Reynolds number: \[ \mathrm{Re} = \frac{\rho U D}{\mu}\,, \] laminar if \(\mathrm{Re} \lesssim 2100\) in straight smooth pipes.
2. Velocity profile
From the axial momentum balance with no-slip at \(r=R\):
\[ u(r) \;=\; \frac{-1}{4\mu}\left(\frac{dp}{dx}\right)\,\left(R^2 - r^2\right)\,. \]
- Centerline velocity: \[ u_{\max} \;=\; u(0) \;=\; \frac{-1}{4\mu}\left(\frac{dp}{dx}\right)R^2. \]
- Parabolic shape: \(u(r) = u_{\max}\!\left(1-\frac{r^2}{R^2}\right)\).
- Mean vs. max: \(U = \dfrac{1}{\pi R^2}\int_0^R 2\pi r\,u(r)\,dr = \dfrac{u_{\max}}{2}\), hence \(u_{\max}=2U\).
3. Discharge and pressure drop
- Discharge: \[ Q \;=\; \int_0^R 2\pi r\,u(r)\,dr \;=\; \frac{\pi R^4}{8\mu}\!\left(-\frac{dp}{dx}\right). \]
- Hagen–Poiseuille law: over length \(L\), \[ \Delta p \;=\; \frac{8\mu L Q}{\pi R^4} \;=\; \frac{32\,\mu L\,U}{R^2} \;=\; \frac{128\,\mu L\,U}{D^2}. \]
4. Piezometric head gradient
- Head loss: \[ h_f \;=\; \frac{\Delta p}{\rho g}. \]
- Gradient: \[ i \;\equiv\; \frac{dh}{dx} \;=\; \frac{1}{\rho g}\left(-\frac{dp}{dx}\right) \;=\; \frac{32\,\mu\,U}{\rho g\,D^2}. \]
- Darcy–Weisbach form: \[ h_f \;=\; f_D\,\frac{L}{D}\,\frac{U^2}{2g},\qquad f_D \;=\; \frac{64}{\mathrm{Re}}\ \ (\text{laminar}). \]
5. Shear stress distribution
- Radial variation: \[ \tau_{rx}(r) \;=\; -\,\frac{r}{2}\,\frac{dp}{dx}. \]
- Wall shear: \[ \tau_w \;=\; \tau_{rx}(R) \;=\; -\,\frac{R}{2}\,\frac{dp}{dx} \;=\; \frac{4\mu U}{R} \;=\; \frac{8\mu U}{D}. \]
- Friction factors: \[ \tau_w \;=\; f_F\,\rho U^2 \;=\; \frac{f_D}{8}\,\rho U^2,\qquad f_F \;=\; \frac{16}{\mathrm{Re}},\ \ f_D \;=\; \frac{64}{\mathrm{Re}}. \]
6. Power transmission through a pipeline
With a total available head \(H\) at inlet and only frictional loss in a pipe (length \(L\), diameter \(D\)), the delivered power is
\[ P \;=\; \rho g Q\,(H - h_f),\qquad h_f \;=\; f_D\,\frac{L}{D}\,\frac{U^2}{2g},\quad U=\frac{4Q}{\pi D^2}. \]
- Maximum power condition: \(\dfrac{dP}{dQ}=0 \Rightarrow h_f = \dfrac{H}{3}\).
- Efficiency at max power: \(\eta_{\max} = \dfrac{P_{\text{out}}}{\rho g Q H} = 1 - \dfrac{h_f}{H} = \dfrac{2}{3}\).
- Corresponding discharge: \[ Q_{\max P} \;=\; \frac{\pi D^2}{4}\,U_{\max P},\quad U_{\max P} \;=\; \sqrt{\frac{2g}{f_D}\,\frac{D}{L}\,\frac{H}{3}}. \] For laminar, substitute \(f_D=64/\mathrm{Re}\) with \(\mathrm{Re}=\rho U D/\mu\) to solve implicitly for \(U_{\max P}\) if the regime remains laminar.
7. Losses in pipe flow
- Major (friction) loss: \[ h_f \;=\; f_D\,\frac{L}{D}\,\frac{U^2}{2g},\qquad f_D \;=\; \frac{64}{\mathrm{Re}}\ \ \text{(laminar)}. \]
- Minor losses: entrances, exits, bends, valves, contractions/expansions: \[ h_m \;=\; K\,\frac{U^2}{2g},\quad \text{with geometry-dependent } K. \]
- Total head loss: \[ h_{\text{tot}} \;=\; h_f + \sum h_{m,i}. \]
8. Equivalent length for minor losses
Convert a minor loss with coefficient \(K\) into an “equivalent” pipe length \(L_{\text{eq}}\) that would cause the same loss as friction:
\[ K\,\frac{U^2}{2g} \;=\; f_D\,\frac{L_{\text{eq}}}{D}\,\frac{U^2}{2g} \;\;\Rightarrow\;\; L_{\text{eq}} \;=\; \frac{K\,D}{f_D}. \]
- Laminar regime: since \(f_D=64/\mathrm{Re}\propto 1/U\), \(L_{\text{eq}}\) depends on \(U\) (and properties). Use with care for low \(\mathrm{Re}\).
- Total effective length: \[ L_{\text{eff}} \;=\; L + \sum_i L_{\text{eq},i}, \] then apply Darcy–Weisbach with \(L_{\text{eff}}\) to estimate total head loss.
9. Quick worked example
Given: Water at 20°C, \(\mu=1.0\times10^{-3}\ \text{Pa·s}\), \(\rho=998\ \text{kg/m}^3\); \(D=0.02\ \text{m}\), \(L=5\ \text{m}\), \(U=0.1\ \text{m/s}\). Check laminar, find \(\Delta p\), \(h_f\), \(\tau_w\), \(u_{\max}\).
- Reynolds: \[ \mathrm{Re}=\frac{\rho U D}{\mu}=\frac{998\times 0.1\times 0.02}{1.0\times10^{-3}}\approx 1996\ (\text{laminar}). \]
- Pressure drop: \[ \Delta p=\frac{128\,\mu L U}{D^2}=\frac{128(10^{-3})(5)(0.1)}{0.02^2}\approx 160\ \text{Pa}. \]
- Head loss: \[ h_f=\frac{\Delta p}{\rho g}\approx \frac{160}{998\times 9.81}\approx 0.0163\ \text{m}. \]
- Wall shear: \[ \tau_w=\frac{8\mu U}{D}=\frac{8(10^{-3})(0.1)}{0.02}=0.04\ \text{Pa}. \]
- Velocity peak: \[ u_{\max}=2U=0.2\ \text{m/s}. \]
10. Notes and cautions
- Entrance effects: fully developed assumption holds beyond entrance length \(L_e/D \sim 0.05\,\mathrm{Re}\) (laminar, smooth).
- Units and consistency: keep \(D\) in metres, \(\mu\) in Pa·s, \(\rho\) in kg/m\(^3\), \(g\) in m/s\(^2\).
- Property dependence: viscosity is temperature-sensitive; check \(T\).