--- title: "Prandtl mixing length and sudden expansion loss" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:prandtl-mixing-length-log-profile" source: https://www.bctengineer.com/q/semester_4/hydraulics/prandtl-mixing-length-log-profile site: BCT Engineering --- # Prandtl mixing length and sudden expansion loss Hydraulics · Chapter 1: Pipe flow - Asked in: 2081 Chaitra, 2+3+3 marks; 2071 Magh, 2+3+3 marks (asked 2 times) - Video: [Prandtl's Velocity Distribution In Pipes - Turbulent Flow - Fluid Mechanics 2](https://www.youtube.com/watch?v=Teqo-xCW2z8) — Ekeeda - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/prandtl-mixing-length-log-profile - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question Explain Prandtl's mixing length theory. Show that the velocity distribution in pipe for turbulent flow is logarithmic. Derive an expression for head loss due to sudden expansion of pipe. ## Short answer Prandtl's mixing length theory gives turbulent shear stress tau_t = rho l^2 (du/dy)^2 with l = kappa y. Integration yields logarithmic velocity distribution u = (u*/kappa) ln y + C. For sudden expansion, head loss is h_e = (V1 - V2)^2/(2g). ## Answer #### Prandtl's mixing length theory In turbulent flow, fluid lumps move across the layers and carry momentum with them. Prandtl assumed that a lump keeps its identity over a distance $l$ (the **mixing length**) and then mixes with the new layer, like the mean free path in kinetic theory. - Velocity difference between layers a distance $l$ apart: $u' \approx l\,\dfrac{du}{dy}$ - The transverse fluctuation $v'$ is of the same order as $u'$, so $v' \approx l\,\dfrac{du}{dy}$ - Turbulent (Reynolds) shear stress: $\tau_t = -\rho\,\overline{u'v'} = \rho\, l^2 \left(\dfrac{du}{dy}\right)^2$ Near a wall the eddy size grows with distance from the wall, so Prandtl took $l = \kappa y$, where $\kappa \approx 0.4$ is von Karman's constant. Total shear stress is $\tau = \mu\dfrac{du}{dy} + \rho l^2\left(\dfrac{du}{dy}\right)^2$; away from the wall the viscous part is negligible. #### Logarithmic velocity distribution in a pipe Near the wall the shear stress is nearly constant and equal to the wall shear $\tau_0$. Let $u_* = \sqrt{\tau_0/\rho}$ be the shear velocity and $y$ the distance from the wall. $$ \tau_0 = \rho\,\kappa^2 y^2\left(\frac{du}{dy}\right)^2 \;\Rightarrow\; \frac{du}{dy} = \frac{u_*}{\kappa\, y} $$ Integrating: $$ u = \frac{u_*}{\kappa}\ln y + C $$ so velocity varies with the logarithm of the distance from the wall. The constant $C$ is fixed by the wall condition: - Smooth pipe: $\dfrac{u}{u_*} = 5.75\log_{10}\dfrac{u_* y}{\nu} + 5.5$ - Rough pipe: $\dfrac{u}{u_*} = 5.75\log_{10}\dfrac{y}{k} + 8.5$ Here $\dfrac{1}{\kappa}\ln y = \dfrac{2.303}{0.4}\log_{10} y = 5.75\log_{10} y$. Measured velocities in pipes follow this law over almost the whole cross-section, and for $y = R$ at the axis, $\dfrac{u_{max}-u}{u_*} = 5.75\log_{10}\dfrac{R}{y}$. #### Head loss due to sudden expansion ```text 1 2 Section 1: A1, V1, p1 (small pipe) ____ _______ Section 2: A2, V2, p2 (large pipe) | |_____| | V1 ~~~~ eddies V2 Control volume between 1 and 2 |____ _______ (eddy zone pressure = p1 on the annulus) |_____| ``` Take a control volume from section 1 (just before the expansion) to section 2 (where the jet has spread over the full area). Eddies in the corner zone keep the pressure on the annular area $(A_2 - A_1)$ equal to $p_1$. Friction on the short length is neglected. Momentum equation (flow direction): $$ p_1 A_2 - p_2 A_2 = \rho Q (V_2 - V_1), \qquad Q = A_2 V_2 $$ $$ \frac{p_1 - p_2}{\gamma} = \frac{V_2(V_2 - V_1)}{g} $$ Energy equation between 1 and 2 (same level), with head loss $h_e$: $$ \frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_e \;\Rightarrow\; h_e = \frac{p_1-p_2}{\gamma} + \frac{V_1^2 - V_2^2}{2g} $$ Substituting the momentum result: $$ h_e = \frac{V_2^2 - V_1V_2}{g} + \frac{V_1^2 - V_2^2}{2g} = \frac{2V_2^2 - 2V_1V_2 + V_1^2 - V_2^2}{2g} $$ $$ \boxed{h_e = \frac{(V_1 - V_2)^2}{2g}} $$ Using continuity $V_2 = V_1A_1/A_2$, this is also $h_e = \left(1 - \dfrac{A_1}{A_2}\right)^2\dfrac{V_1^2}{2g}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In Prandtl's mixing length theory, what does the mixing length l represent? A. The pipe length over which friction acts B. The distance a fluid lump travels keeping its identity before mixing with the new layer C. The thickness of the laminar sub-layer D. The distance between the pipe wall and the axis Answer: B. Like a mean free path, l is the distance a lump of fluid moves across layers before it mixes and gives up its momentum. (Key point: Mixing length idea) 2. With u' ~ v' ~ l du/dy, what is the turbulent shear stress in Prandtl's theory? A. rho l (du/dy) B. mu l^2 (du/dy)^2 C. rho l^2 (du/dy)^2 D. rho l (du/dy)^2 Answer: C. tau_t = -rho u'v' with both fluctuations proportional to l du/dy gives rho l^2 (du/dy)^2. (Key point: Turbulent shear stress) 3. Which assumptions lead to du/dy = u*/(kappa y) and hence the logarithmic profile? A. l = kappa y and shear stress near the wall roughly constant at tau_0 B. l constant everywhere and shear stress zero C. Viscous stress dominates and l = 0 D. Shear stress increases linearly with y from the wall Answer: A. Putting l = kappa y and tau = tau_0 in tau = rho l^2 (du/dy)^2 gives du/dy = u*/(kappa y), which integrates to a log law. (Key point: Log law derivation) 4. Which two equations are combined to get the head loss h_e = (V1 - V2)^2/(2g) for a sudden expansion? A. Continuity and Hagen-Poiseuille B. Momentum and energy equations, with pressure p1 on the annular shoulder C. Bernoulli's equation only, with no loss term D. Darcy-Weisbach and continuity Answer: B. The momentum equation gives p1 - p2, which is substituted in the energy equation to find the loss. 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