# Solution to Finite Volume Problem (Step-by-Step) Let's break down the solution into two main steps as requested, with explanations at the end of each step. --- ## **Step 1: Discretization Using the Finite Volume Method** ### **Given Data** - Plate separation, \( H = .005 \) m - Number of cells, \( N = 3 \) (from the diagram) - Each cell width, \( \Delta y = H/N = .005/3 = .00167 \) m - Fluid viscosity, \( \mu = .001 \) Pa·s - Pressure gradient, \( \frac{dp}{dx} = -12 \) Pa/m - Lower plate velocity, \( u_ = \) - Upper plate velocity, \( u_s = .5 \) m/s ### **Governing Equation** The x-momentum equation simplifies to: \[ \frac{d^2u}{dy^2} = \frac{1}{\mu} \frac{dp}{dx} \] ### **Finite Volume Discretization** For each cell centered at \( y_1 \), \( y_2 \), \( y_3 \): - The finite volume approach uses the midpoint for velocities (\( u_1, u_2, u_3 \)). - At each cell, the discretized form (central difference) is: \[ \frac{u_{i+1} - 2u_i + u_{i-1}}{\Delta y^2} = \frac{1}{\mu}\frac{dp}{dx} \] where \( i = 1, 2, 3 \). #### **Boundary Conditions** - \( u_ = \) (stationary lower plate) - \( u_4 = u_s = .5 \) m/s (moving upper plate) #### **Write the Equations** - For cell 1: \( \frac{u_2 - 2u_1 + u_}{\Delta y^2} = S \) - For cell 2: \( \frac{u_3 - 2u_2 + u_1}{\Delta y^2} = S \) - For cell 3: \( \frac{u_4 - 2u_3 + u_2}{\Delta y^2} = S \) Where \( S = \frac{1}{\mu}\frac{dp}{dx} = \frac{-12}{.001} = -12000 \) 1/s². **Explanation:** Here, we have discretized the x-momentum equation using the finite volume method and set up a system of equations for the unknown velocities at the cell centers, incorporating the boundary conditions. --- ## **Step 2: Construct and Solve the Linear System** ### **Plug in Values** - \( \Delta y^2 = (.00167)^2 = 2.78 \times 10^{-6} \) - \( S = -12000 \) Now substitute into each equation: **Cell 1:** \[ \frac{u_2 - 2u_1 + u_}{2.78 \times 10^{-6}} = -12000 \] \[ u_2 - 2u_1 + = -12000 \times 2.78 \times 10^{-6} \] \[ u_2 - 2u_1 = -.0334 \] **Cell 2:** \[ \frac{u_3 - 2u_2 + u_1}{2.78 \times 10^{-6}} = -12000 \] \[ u_3 - 2u_2 + u_1 = -.0334 \] **Cell 3:** \[ \frac{u_4 - 2u_3 + u_2}{2.78 \times 10^{-6}} = -12000 \] \[ .5 - 2u_3 + u_2 = -.0334 \] \[ -2u_3 + u_2 = -.5334 \] ### **Solve the System** Now, solve for \( u_1, u_2, u_3 \): 1. \( u_2 - 2u_1 = -.0334 \) 2. \( u_3 - 2u_2 + u_1 = -.0334 \) 3. \( -2u_3 + u_2 = -.5334 \) Let's write these as simultaneous equations: 1. \( -2u_1 + u_2 = -.0334 \) 2. \( u_1 - 2u_2 + u_3 = -.0334 \) 3. \( u_2 - 2u_3 = -.5334 \) Now solve step-by-step: - From (3): \( u_2 = -.5334 + 2u_3 \) - Substitute into (2): \( u_1 - 2(-.5334 + 2u_3) + u_3 = -.0334 \) \( u_1 + 1.0668 - 4u_3 + u_3 = -.0334 \) \( u_1 - 3u_3 = -1.1002 \) (A) - From (1): \( -2u_1 + (-.5334 + 2u_3) = -.0334 \) \( -2u_1 - .5334 + 2u_3 = -.0334 \) \( -2u_1 + 2u_3 = .5 \) \( -2u_1 + 2u_3 = .5 \) \( -u_1 + u_3 = .25 \) \( u_1 = u_3 - .25 \) (B) Now substitute (B) in (A): \( (u_3 - .25) - 3u_3 = -1.1002 \) \( u_3 - .25 - 3u_3 = -1.1002 \) \( -2u_3 - .25 = -1.1002 \) \( -2u_3 = -.8502 \) \( u_3 = .4251 \) m/s Now, \( u_1 = u_3 - .25 = .4251 - .25 = .1751 \) m/s Now, \( u_2 = -.5334 + 2u_3 = -.5334 + 2 \times .4251 = -.5334 + .8502 = .3168 \) m/s --- **Explanation:** We substituted the known boundary values and solved the resulting system of equations step by step to obtain the velocities at each cell center. --- ## **Final Answers** - **Cell 1 center velocity (\( u_1 \)):** **.175 m/s** - **Cell 2 center velocity (\( u_2 \)):** **.317 m/s** - **Cell 3 center velocity (\( u_3 \)):** **.425 m/s** - **Upper plate velocity (\( u_s \)):** **.5 m/s** (given) --- ### **Summary Table** | Location | Velocity (m/s) | |------------------|:-------------:| | \( u_ \) (bottom plate) | | | \( u_1 \) (cell 1 center) | .175 | | \( u_2 \) (cell 2 center) | .317 | | \( u_3 \) (cell 3 center) | .425 | | \( u_4 \) (top plate) | .5 |