Chapter 2, Problem 129P

To determine

(a)

The Relationship for the torque transmitted by an MR clutch for N plates attached to the input shaft rotating at an angular speed of $\omega$.

Answer to Problem 129P

The Relationship for the torque transmitted by an MR clutch for N plates attached to the input shaft rotating at an angular speed of $\omega$ is $T=4\pi N\left[{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right]$.

Explanation of Solution

Given:

The shear stress used for the Herschel-Bulkley constitute model is $\tau ={\tau }_{\nu }+K{\left(\frac{du}{dy}\right)}^m$, Yield stress is $900\text{kPa}$, consistency index is $58\text{Pa}\cdot {\text{s}}^m$, Power index is $0.82$, the number of clutch plates attached to input shaft are $N$, angular shaft is stationary, and angular speed of the input shaft is $\omega$.

Write the expression for the velocity gradient.

  $\frac{V}{h}=\frac{du}{dy}$   ...... (I)

Here, velocity is $V$, fluid thickness is $h$ and velocity gradient is $du/dy$.

Write the expression for shear stress.

  $\tau ={\tau }_y+K{\left(\frac{du}{dy}\right)}^m$   ...... (II)

Here, yield stress is is ${\tau }_y$, shear stress is $\tau$ and consistency index is $K$.

Power index is $m$.

Substitute Equation in Equation (II).

  $\tau ={\tau }_y+K{\left(\frac{V}{h}\right)}^{0.82}$   ...... (III)

Write the expression for the velocity of shaft.

  $V=r\omega$   ...... (IV)

Here, angular velocity is $\omega$ and radius of the shaft is $r$.

Substitute Equation (IV) in Equation (III).

  $\tau ={\tau }_y+K{\left(\frac{r\omega }{h}\right)}^{0.82}$   ...... (V)

Write the expression for the force.

  $dF=\tau dA$   ...... (VI)

Here, area is $dA$.

Substitute ${\tau }_y+K{\left(\frac{r\omega }{h}\right)}^{0.82}$ for $\tau$ in Equation (VI).

  $dF=\left[{\tau }_y+K{\left(\frac{r\omega }{h}\right)}^{0.82}\right]\left(2\pi rdr\right)$   ...... (VII)

Write the expression for the torque generated.

  $dT=rdF$   ...... (VIII)

Substitute $\left[{\tau }_y+K{\left(\frac{r\omega }{h}\right)}^{0.82}\right]\left(2\pi rdr\right)$ for $dF$ in Equation (VIII).

  $dT=r\left[{\tau }_y+K{\left(\frac{r\omega }{h}\right)}^{0.82}\right]\left(2\pi rdr\right)$   ...... (IX)

Calculation:

Integrate both sides of the Equation (IX).

  $\begin{array}{c}{\displaystyle \int dT}={\displaystyle \underset{R_1}{\overset{R_2}{\int }}r\left[{\tau }_y+K{\left(\frac{r\omega }{h}\right)}^{0.82}\right]\left(2\pi rdr\right)}\\ =2\pi \left[{\tau }_y{\displaystyle \underset{R_1}{\overset{R_2}{\int }}{r}^{2}dr+K{\left(\frac{\omega }{h}\right)}^{0.82}{\displaystyle \underset{R_1}{\overset{R_2}{\int }}{r}^{2.82}dr}}\right]\\ T=2\pi \left[{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right]\end{array}$

Write the expression for the torque generated by the both surfaces of the plate.

  $T=2\left[2\pi \left\{{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right\}\right]$

Write the expression for the torque generated by the $N$ number of clutch plate.

  $\begin{array}{c}T=N\left[4\pi \left\{{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right\}\right]\\ =4\pi N\left[{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right]\end{array}$

Conclusion:

The torque generated by the MR clutch is $4\pi N\left[{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right]$.

To determine

(b)

Torque transmitted by the clutch.

Answer to Problem 129P

Torque transmitted by the clutch is $81517.71\text{N}\cdot \text{m}$.

Explanation of Solution

Given information:

The number of plate clutch is $11$, the inner radius of clutch is $50\text{mm}$, the outer radius of clutch is $200\text{mm}$, the speed of the input shaft is $3000\text{rpm}$ and fluid thickness is $1.5\text{mm}$.

Write the expression for the angular speed of the shaft.

  $\omega =\frac{2\pi \stackrel{·}{n}}{60}$   ...... (X)

Here, speed of the shaft is $\stackrel{·}{n}$.

Write the expression for the torque generated.

  $T=4\pi N\left[{\tau }_y\left(\frac{1}{3}\right)\left(R_2^3-R_1^3\right)+K{\left(\frac{\omega }{h}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left(R_2^{3.82}-R_1^{3.82}\right)\right]$   ...... (XI)

Calculation:

Substitute $2400\text{rpm}$ for $\stackrel{·}{n}$ in Equation (X).

  $\begin{array}{c}\omega =\frac{2\pi \left(2400\text{rpm}\right)}{60}\\ =\frac{15079.644}{60}\\ =251.31\text{rad}/\text{s}\end{array}$

Substitute $11$ for $N$, $50\text{mm}$ for $R_1$, $200\text{mm}$ for $R_2$, $251.31\text{rad}/\text{s}$ FOR $\omega$, $1.5\text{mm}$ for $h$, $58$ for $K$, $900\text{Pa}$ for ${\tau }_y$ In Equation (XI)

  $\begin{array}{c}T=4\pi \left(11\right)\left[\begin{array}{l}\left(900\text{Pa}\right)\left(\frac{1}{3}\right)\left({\left(200\text{mm}\right)}^3-{\left(50\text{mm}\right)}^3\right)+\\ 58{\left(\frac{251.32\text{rad}/\text{s}}{1.5\text{mm}}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left({\left(200\text{mm}\right)}^{3.82}-{\left(50\text{mm}\right)}^{3.82}\right)\end{array}\right]\\ =4\pi \left(11\right)\left[\begin{array}{l}\left(900\text{Pa}\right)\left(\frac{1}{3}\right)\left({\left(200\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^3-{\left(50\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^3\right)+\\ 58{\left(\frac{251.32\text{rad}/\text{s}}{1.5\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left({\left(200\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^{3.82}-{\left(50\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^{3.82}\right)\end{array}\right]\end{array}$$\begin{array}{c}T=4\pi \left(11\right)\left[\begin{array}{l}\left(900\text{Pa}\right)\left(\frac{1}{3}\right)\left({0.2}^3{\text{m}}^{\text{3}}-{0.05}^3{\text{m}}^{\text{3}}\right)+\\ 58{\left(\frac{\left(251.31\text{rad}/\text{s}\right)}{0.0015\text{m}}\right)}^{0.82}\left(\frac{1}{3.82}\right)\left({0.2}^{3.82}{\text{m}}^{\text{3}}-{0.05}^{3.82}{\text{m}}^{\text{3}}\right)\end{array}\right]\\ =44\pi \left[\left(300\text{Pa}\times 0.0078\right){\text{m}}^{\text{3}}\text{Pa}+\left(291837.97\times 0.02169\right){\text{m}}^{\text{3}}\text{Pa}\right]\\ =44\pi \left[2.36{\text{m}}^{\text{3}}\text{Pa}+587.36{\text{m}}^{\text{3}}\text{Pa}\right]\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)\\ =81517.71\text{N}\cdot \text{m}\end{array}$

Conclusion:

The torque generated is $81517.71\text{N}\cdot \text{m}$.