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Chapter 6: Problem 68

The two-dimensional velocity field for an incompressible Newtonian fluid is described by the relationship $$\mathbf{V}=\left(12 x y^{2}-6 x^{3}\right) \hat{\mathbf{i}}+\left(18 x^{2} y-4 y^{3} \hat{\mathbf{j}}\right.$$ where the velocity has units of \(\mathrm{m} / \mathrm{s}\) when \(x\) and \(y\) are in meters. Determine the stresses \(\sigma_{x x}, \sigma_{y y},\) and \(\tau_{x y}\) at the point \(x=0.5 \mathrm{m}\) \(y=1.0 \mathrm{m}\) if pressure at this point is \(6 \mathrm{kPa}\) and the fluid is glycerin at \(20^{\circ} \mathrm{C}\). Shew these stresses on a sketch.

Short Answer

Expert verified
The stress components are \(\sigma_{xx} = 2.50 Pa, \sigma_{yy} = -13.5 Pa, \tau_{xy} = -2.10 Pa\). The sketch involves indicating these values on a differential element at the point (x=0.5 m, y=1.0 m).

Step by step solution

01

Evaluate the derivatives of the velocity components

Start by differentiating the \(i\) and \(j\) components of the velocity vector field \(V\) with respect to \(x\) and \(y\) - i.e., \( \frac{dV_x}{dx}\), \( \frac{dV_x}{dy} \), \( \frac{dV_y}{dx}\), \( \frac{dV_y}{dy}\). These values will be required to compute the rate of deformation tensor. Use the specific \(x\) and \(y\) values given in the question.
02

Derive rate of deformation tensor

Based on the definitions of the rate of deformation tensor \(\mathbf{D} = \frac{1}{2} (\nabla \mathbf{V} + (\nabla \mathbf{V})^t)\), where \(\nabla \mathbf{V}\) is the gradient of the velocity vector field and \((\nabla \mathbf{V})^t\) is its transpose, we can derive its values using the velocity derivatives calculated in step 1.
03

Calculate stresses

Use the relationship between stresses and the rate of deformation tensor in Newtonian fluids which is \(\sigma_{ij} - p \delta_{ij} = 2 \eta D_{ij}\) where \(\sigma_{ij}\) are the components of the stress tensor, \(p\) is the pressure, \(\delta_{ij}\) is the Kronecker delta, \(\eta\) is the dynamic viscosity, and \(D_{ij}\) are the components of the rate of deformation tensor. For glycerin at \(20^{\circ}C\), we use \(\eta = 1.41 Pa.s\). All these will enable you calculate the components of stress.
04

Draw the sketch

To represent the stresses at the point \((x=0.5 m, y=1.0 m)\) on a diagram, start by sketching a differential element at this location. Then, indicate the stresses \(\sigma_{xx}, \sigma_{yy}\), and \(\tau_{xy}\) on this diagram reflecting their positive directions.

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