Chapter 10, Problem 32P

To determine

The total power required and the reduction in the required power input.

Explanation of Solution

Given:

Angular velocity $\left(\omega \right)$ is $200\text{rad}/\text{s}$.

Thickness of the oil film $\left(t\right)$ is 1.2 mm.

Dynamic viscosity of the SAE 10W oil at $20°\text{C}$ $\left({\mu }_{@20°\text{C}}\right)$ is $0.1\text{ Pa}\cdot \text{s}$.

Dynamic viscosity of the SAE 10W oil at $80°\text{C}$ $\left({\mu }_{@80°\text{C}}\right)$ is $0.0078\text{ Pa}\cdot \text{s}$.

Calculation:

Determine the expression of shear stress.

  $\begin{array}{c}{\tau }_w=\mu \frac{du}{dr}\\ =\mu \frac{V}{h}\\ =\mu \frac{\omega r}{h}\end{array}$

Determine the expression of shear force.

  $\begin{array}{c}dF={\tau }_{w}dA\\ =\mu \frac{\omega r}{h}dA\end{array}$

Determine the expression for torque.

  $\begin{array}{c}d\text{T}=rdF\\ =\mu \frac{\omega r^2}{h}dA\end{array}$

Differentiate the above equation.

  $\text{T}=\frac{\mu \omega }{h}{\displaystyle {\int }_A{r}^2}dA$

Determine the expression for power required.

  $\begin{array}{c}{\dot{W}}_{\text{sh}}=\omega \text{T}\\ =\frac{\mu {\omega }^2}{h}{\displaystyle {\int }_A{r}^2}dA\end{array}$

Determine the expression for power required from the top.

  $\begin{array}{c}{\dot{W}}_{\text{sh, top }}=\frac{\mu {\omega }^2}{h}{\displaystyle {\int }_{r=0}^{D/2}{r}^2}(2\pi r)dr\\ =\frac{2\pi \mu {\omega }^2}{h}{\displaystyle {\int }_{r=0}^{D/2}{r}^3}dr\\ ={\frac{2\pi \mu {\omega }^2}{h}\frac{r^4}{4}|}_{r=0}^{D/2}\\ =\frac{\pi \mu {\omega }^2{D}^4}{32h}\end{array}$

Determine the expression for power required from the bottom.

  ${\dot{W}}_{\text{sh, bottom }}=\frac{\pi \mu {\omega }^2{d}^4}{32h}$

Determine the expression for power required from the top.

  $\begin{array}{c}{\dot{W}}_{\text{sh, top }}=\frac{\mu {\omega }^2}{h}{\displaystyle {\int }_{r=0}^{D/2}{r}^2}\frac{4\pi L}{D-d}rdr\\ =\frac{4\pi \mu {\omega }^{2}L}{h(D-d)}{\displaystyle {\int }_{r=d/2}^{D/2}{r}^3}dr\\ ={\frac{4\pi \mu {\omega }^{2}L}{h(D-d)}\frac{r^4}{4}|}_{r=d/2}^{D/2}\\ =\frac{\pi \mu {\omega }^{2}L\left(D^2-d^2\right)}{16h(D-d)}\end{array}$

Determine the total power required.

  $\begin{array}{c}{\dot{W}}_{\text{sh, total }}={\dot{W}}_{\text{sh, top }}+{\dot{W}}_{\text{sh, bottom }}+{\dot{W}}_{\text{sh,side }}\\ =\frac{\pi \mu {\omega }^2{D}^4}{32h}\left(1+{\left(\frac{d}{D}\right)}^4+\frac{2L\left(1-{\left(\frac{d}{D}\right)}^4\right)}{D-d}\right)\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{sh total }}=\left\{\begin{array}{c}\frac{\pi \left(0.1\text{N}\cdot \text{s}/{\text{m}}^2\right){\left(200\text{rad}/\text{s}\right)}^2{\left(0.12\text{ m}\right)}^4}{32\left(0.0012\text{ m}\right)}\\ \left(1+{\left(\frac{1}{3}\right)}^4+\frac{2\left(0.12\text{ m}\right)\left(1-{\left(\frac{1}{3}\right)}^4\right)]}{\left(0.12\text{ m}-0.04\text{ m}\right)}\right)\left(\frac{1\text{ W}}{1\text{N}\cdot \text{m}/\text{s}}\right)\end{array}\right\}\\ =270\text{ W}\end{array}$

Thus, the total power required is $\underset{\_}{270\text{ W}}$.

Determine the power required at $80°\text{C}$.

  $\begin{array}{c}{\dot{W}}_{\text{sh, total },80°\text{C}}=\frac{{\mu }_{80°\text{C}}}{{\mu }_{20°\text{C}}}{\dot{W}}_{\text{sh, total }20°\text{C}}\\ =\frac{0.0078\text{N}\cdot \text{s}/{\text{m}}^2}{0.1\text{N}\cdot \text{s}/{\text{m}}^2}\left(270\text{ W}\right)\\ =21.1\text{ W}\end{array}$

Determine the reduction in the required power input.

  $\begin{array}{c}\text{Reduction}=W_{\text{sh, total,}20°\text{C}}-W_{\text{sh, total,}80°\text{C}}\\ =270\text{ W}-21.1\text{ W}\\ =248.9\text{ W}\end{array}$

Thus, the reduction in the required power input is $\underset{\_}{248.9\text{ W}}$.