Chapter 2, Problem 129P

A $\text{10-cm}$ diameter cylindrical shaft rotates inside a $\text{50-cm-long}\text{10}\text{.3-cm}$ diameter bearing. The space between the shaft and the bearing is completely filled with oil whose viscosity at anticipated operating temperature is $\text{0}\text{.300}\text{N}\cdot {\text{s/m}}^{\text{2}}.$ Determine the power required to overcame friction when the shaft rotates at a speed of (a) $600\text{rpm}$ and (b) $\text{1200}\text{rpm}\text{.}$

To determine

(a)

The power required to overcome the friction when the shaft rotates at a speed of $600\text{rpm}$.

Answer to Problem 129P

The power required to overcome the friction when the shaft rotates at a speed of $600\text{rpm}$ is $247.55\text{W}$.

Explanation of Solution

Given information:

The diameter of cylindrical shaft is $10\text{cm}$, the length of shaft is $40\text{cm}$, the diameter of bearing is $10.3\text{cm}$, the viscosity of the oil filled between the shaft and the bearing is $0.3\text{N}\cdot \text{s}/{\text{m}}^2,$ and the speed of the shaft is $600\text{rpm}$.

Write the expression for the gap between shaft and the bearing.

  $l=\frac{D_b-D_s}{2}$...... (I)

Here, tthe gap between the shaft and the bearing is $l$, the diameter of the bearing is $D_b,$ and the diameter of the shaft is $D_s$.

Write the expression for the torque generated by the shaft.

  $T=\mu \frac{4{\pi }^2{R}^{3}nL}{l}$...... (II)

Here, the torque is $T$, the radius of the shaft is $R$, the speed of the shaft is $n$, the length of the shaft is $L,$ and the viscosity of the oil is $\mu$.

Write the expression for the radius of the shaft.

  $R=\frac{D_s}{2}$...... (III)

Write the expression for the angular velocity.

  $\omega =\frac{2\pi n}{60}$...... (IV)

Here, the angular velocity is $\omega$.

Write the expression for the power.

  $P=T\omega$...... (V)

Here, the power is $P$.

Calculation:

Substitute $10.3\text{cm}$ for $D_b$ and $10\text{cm}$ for $D_s$ in Equation (I).

  $\begin{array}{c}l=\frac{10.3\text{cm}-10\text{cm}}{2}\\ =0.15\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\\ =1.5\times {10}^{-3}\text{m}\end{array}$

Substitute $10\text{cm}$ for $D_s$ in Equation (III).

  $\begin{array}{c}R=\frac{10\text{cm}}{2}\\ =5\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.05\text{m}\end{array}$

Substitute $0.3\text{N}\cdot \text{s}/{\text{m}}^2$ for $\mu$, $0.05\text{m}$ for $R$, $600\text{rpm}$ for $n$, $40\text{cm}$ for $L$ and $1.5\times {10}^{-3}\text{m}$ for $l$ in Equation (II).

  $\begin{array}{c}T=\left(0.3\text{N}\cdot \text{s}/{\text{m}}^2\right)\frac{4{\pi }^2\times {\left(0.05\text{m}\right)}^3\times 600\text{rpm}\times 40\text{cm}}{1.5\times {10}^{-3}\text{m}}\\ =0.98696\text{N}\cdot \text{s}\times \left(600\text{rpm}\times \frac{1\text{rps}}{60\text{rpm}}\right)\times \left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =3.94\text{N}\cdot \text{m}\end{array}$

Substitute $600\text{rpm}$ for $n$ in Equation (IV).

  $\begin{array}{c}\omega =\frac{2\pi \times 600\text{rpm}}{60}\\ =\frac{3769.9}{60}\text{rad}/\text{s}\\ =62.83\text{rad}/\text{s}\end{array}$

Substitute $62.83\text{rad}/\text{s}$ for $\omega$ and $3.94\text{N}\cdot \text{m}$ for $T$ in Equation (V).

  $\begin{array}{c}P=3.94\text{N}\cdot \text{m}\times 62.83\text{rad}/\text{s}\\ =247.55\text{N}\cdot \text{m}/\text{s}\times \left(\frac{1\text{W}}{1\text{N}\cdot \text{m}/\text{s}}\right)\\ =247.55\text{W}\end{array}$

Conclusion:

The power required to overcome the friction when the shaft rotates at a speed of $600\text{rpm}$ is $247.55\text{W}$.

To determine

(b)

The power required to overcome the friction when the shaft rotates at a speed of $1200\text{rpm}$.

Answer to Problem 129P

The power required to overcome the friction when the shaft rotates at a speed of $1200\text{rpm}$ is $991.45\text{W}$.

Explanation of Solution

Given information:

The speed of the shaft is $1200\text{rpm}$.

Calculation:

Substitute $0.3\text{N}\cdot \text{s}/{\text{m}}^2$ for $\mu$, $0.05\text{m}$ for $R$, $1200\text{rpm}$ for $n$, $40\text{cm}$ for $L$ and $1.5\times {10}^{-3}\text{m}$ for $l$ in Equation (II).

  $\begin{array}{c}T=\left(0.3\text{N}\cdot \text{s}/{\text{m}}^2\right)\frac{4{\pi }^2\times {\left(0.05\text{m}\right)}^3\times 1200\text{rpm}\times 40\text{cm}}{1.5\times {10}^{-3}\text{m}}\\ =0.98696\text{N}\cdot \text{s}\times \left(1200\text{rpm}\times \frac{1\text{rps}}{60\text{rpm}}\right)\times \left(40\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =7.89\text{N}\cdot \text{m}\end{array}$

Substitute $1200\text{rpm}$ for $n$ in Equation (IV).

  $\begin{array}{c}\omega =\frac{2\pi \times 1200\text{rpm}}{60}\\ =\frac{7539.8}{60}\text{rad}/\text{s}\\ =125.66\text{rad}/\text{s}\end{array}$

Substitute $125.66\text{rad}/\text{s}$ for $\omega$ and $7.89\text{N}\cdot \text{m}$ for $T$ in Equation (V).

  $\begin{array}{c}P=7.89\text{N}\cdot \text{m}\times 125.66\text{rad}/\text{s}\\ =991.45\text{N}\cdot \text{m}/\text{s}\times \left(\frac{1\text{W}}{1\text{N}\cdot \text{m}/\text{s}}\right)\\ =991.45\text{W}\end{array}$

Conclusion:

The power required to overcome the friction when the shaft rotates at a speed of $1200\text{rpm}$ is $991.45\text{W}$.