Let's analyze the problem step-by-step. --- ## Given Data: - Viscosity, \( \mu = .020\, \text{Pa·s} \) - Gap between plates, \( h = 6\, \text{mm} = 6 \times 10^{-3}\, \text{m} \) - Bottom plate is fixed at \( y = \) - Top plate at \( y = h \), exposed to air - Shear stress, \( \tau_ = .12\, \text{Pa} \) - No pressure gradient in the x-direction --- ## Step 1: Understand the problem The fluid is between two large plates: - The bottom plate is stationary. - The top plate moves or exerts a shear stress due to air exposure. Since the top surface is exposed to air, the shear stress at the top surface is \( \tau_ \). --- ## Step 2: Find the velocity profile \( u(y) \) In a steady, laminar flow with no pressure gradient, the Navier-Stokes equations reduce to a simple form: \[ \frac{d^2 u}{dy^2} = \] which implies a linear velocity profile: \[ u(y) = A y + B \] ### Boundary conditions: - At \( y= \), bottom plate is fixed: \( u() = \) - At \( y=h \), shear stress \( \tau \) is related to velocity gradient: \[ \tau = \mu \frac{du}{dy} \] Given the shear stress at \( y=h \): \[ \left.\tau\right|_{y=h} = \tau_ = .12\, \text{Pa} \] --- ## Step 3: Find the velocity gradient Since \( \frac{du}{dy} \) is constant in a linear profile: \[ \frac{du}{dy} = \frac{\tau}{\mu} \] At \( y=h \): \[ \frac{du}{dy} = \frac{.12}{.020} = 6\, \text{m/s} \] This is the shear rate at the top surface. --- ## Step 4: Determine \( u(y) \) Because the flow is linear: \[ u(y) = \left( \frac{\tau}{\mu} \right) y + C \] Using boundary condition \( u() = \): \[ = + C \Rightarrow C= \] Therefore: \[ u(y) = 6 y \] At \( y=h=6\, \text{mm} = .006\, \text{m} \): \[ u(h) = 6 \times .006 = .036\, \text{m/s} \] --- ## **Final Results:** ### a) Velocity profile \( u(y) \): \[ u(y) = 6 y \quad \text{(m/s)} \] ### b) Maximum velocity \( u_{max} \): \[ u_{max} = u(h) = .036\, \text{m/s} \] ### c) Mean velocity \( \bar{u} \): \[ \bar{u} = \frac{1}{h} \int_^{h} u(y) dy = \frac{1}{h} \times \frac{1}{2} u(h) h = \frac{u(h)}{2} = .018\, \text{m/s} \] ### d) Shear stress per unit width \( q \): Since shear stress is uniform: \[ q = \tau_ \times h = .12\, \text{Pa} \times .006\, \text{m} = 7.2 \times 10^{-4}\, \text{N/m} \] --- ## **Summary:** - **Velocity profile:** \( u(y) = 6 y \, \text{m/s} \) - **Maximum velocity:** \( .036\, \text{m/s} \) - **Mean velocity:** \( .018\, \text{m/s} \) - **Shear stress per unit width:** \( 7.2 \times 10^{-4} \, \text{N/m} \) --- Let me know if you'd like further clarification!