--- title: "Couette flow with a pressure gradient" subject: "Fluid Mechanics" chapter: 6 key: "semester_4/fluid_mechanics_mechanical:couette-flow-moving-plate-pressure-gradient" source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/couette-flow-moving-plate-pressure-gradient site: BCT Engineering --- # Couette flow with a pressure gradient Fluid Mechanics · Chapter 6: Viscous Effects - Asked in: year not given, 4+4 marks - Video: [Couette Flow](https://www.youtube.com/watch?v=epp_mgfTmOQ) — LearnChemE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/couette-flow-moving-plate-pressure-gradient - Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-6 ## Question (a) Derive the velocity distribution for steady laminar flow between two parallel plates when the upper plate moves with velocity $U$ and the lower plate is fixed, with a constant pressure gradient $dp/dx$ along the flow. Obtain the discharge per unit width. (b) Discuss the effect of the pressure gradient on the velocity profile. For $U = 3\ \text{m/s}$, gap $h = 4\ \text{mm}$ and oil of viscosity 0.1 Pa s, find the shear stress on the moving plate when $dp/dx = 0$ and when $dp/dx = -10\ \text{kPa/m}$. ## Short answer Velocity u = (Uy/h) - (1/(2μ))(dp/dx)(hy - y^2); q = Uh/2 - (h^3/(12μ))(dp/dx). Pressure gradient alters profile: zero gives linear, favorable bulges outward, adverse may reverse flow. Shear stress on moving plate τ_h = μU/h + (h/2)(dp/dx): 75 Pa for dp/dx=0, 55 Pa for dp/dx=-10 kPa/m. ## Answer #### (a) Velocity distribution For steady, fully developed laminar flow, the x-momentum (Navier-Stokes) equation reduces to $$ \mu\frac{d^2u}{dy^2} = \frac{dp}{dx} $$ Integrating twice: $u = \dfrac{1}{2\mu}\dfrac{dp}{dx}y^2 + C_1y + C_2$. Boundary conditions: $u = 0$ at $y = 0$, and $u = U$ at $y = h$. Then $C_2 = 0$ and $C_1 = \dfrac{U}{h} - \dfrac{h}{2\mu}\dfrac{dp}{dx}$. $$ u = \frac{U y}{h} - \frac{1}{2\mu}\frac{dp}{dx}\left(hy - y^2\right) $$ The first term is the **Couette** (linear) part due to the moving plate, and the second is the **Poiseuille** (parabolic) part due to the pressure gradient. **Discharge per unit width:** $$ q = \int_0^h u\,dy = \frac{Uh}{2} - \frac{h^3}{12\mu}\frac{dp}{dx} $$ #### (b) Effect of pressure gradient Let $P = -\dfrac{h^2}{2\mu U}\dfrac{dp}{dx}$. | Case | $dp/dx$ | Profile | |---|---|---| | Simple Couette flow | $0$ | Straight line from 0 to $U$ | | Favourable gradient | $< 0$ ($P > 0$) | Profile bulges outward, velocity above the linear value | | Adverse gradient | $> 0$ ($P < 0$) | Profile bends inward; if $dp/dx > 2\mu U/h^2$ the velocity near the fixed plate reverses (back flow) | ```text dp/dx = 0 dp/dx < 0 dp/dx > 0 (large) /| _.| | / | .' | _| / | / | <-'/ /___| /_____| /___| ``` Shear stress: $\tau = \mu\dfrac{du}{dy} = \mu\dfrac{U}{h} - \left(\dfrac{h}{2} - y\right)\dfrac{dp}{dx}$ At the moving plate ($y = h$): $\tau_h = \dfrac{\mu U}{h} + \dfrac{h}{2}\dfrac{dp}{dx}$ #### Numerical $\mu U/h = 0.1 \times 3/0.004 = 75\ \text{Pa}$. - For $dp/dx = 0$: $\tau_h = 75\ \text{Pa}$. - For $dp/dx = -10\,000\ \text{Pa/m}$: $$ \tau_h = 75 + \frac{0.004}{2}(-10\,000) = 75 - 20 = 55\ \text{Pa} $$ (The shear stress on the fixed plate would be $75 + 20 = 95\ \text{Pa}$.) The discharge per unit width rises from $Uh/2 = 0.006\ \text{m}^2/\text{s}$ to $0.006 + \dfrac{(0.004)^3 \times 10\,000}{12 \times 0.1} = 0.00653\ \text{m}^2/\text{s}$. **Answer: $\tau_h = 75\ \text{Pa}$ for $dp/dx = 0$, and $55\ \text{Pa}$ for $dp/dx = -10\ \text{kPa/m}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a moving upper plate and constant dp/dx, what are the two parts of the velocity profile? A. Two parabolic parts B. A linear Couette part U y/h and a parabolic Poiseuille part from the pressure gradient C. A constant part and a linear part D. A logarithmic and a linear part Answer: B. u = U y/h - (1/2mu)(dp/dx)(hy - y^2): the first term is Couette, the second Poiseuille. (Key point: Velocity distribution) 2. What is q? A. q = U h - h^3/(12 mu) dp/dx B. q = U h/2 + h^3/(12 mu) dp/dx C. q = U h/2 - h^3/(12 mu) dp/dx D. q = U h/3 - h^3/(8 mu) dp/dx Answer: C. Integrating u from 0 to h gives Uh/2 - (h^3/12mu)(dp/dx). (Key point: Discharge per unit width) 3. When does flow reverse near the fixed plate? A. For any favourable gradient B. When dp/dx = 0 C. When U = 0 only D. For an adverse gradient with dp/dx > 2 mu U/h^2 Answer: D. A sufficiently large positive (adverse) pressure gradient bends the profile so velocity near the fixed plate is negative. (Key point: Back flow) 4. U = 3 m/s, h = 4 mm, mu = 0.1 Pa s, dp/dx = -10 kPa/m. What is the shear stress on the moving plate? A. 95 Pa B. 75 Pa C. 20 Pa D. 55 Pa Answer: D. tau_h = mu U/h + (h/2) dp/dx = 75 + 0.002 x (-10000) = 55 Pa. 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