Chapter 12, Problem 1P

A full journal bearing has a journal diameter of 25 mm, with a unilateral tolerance of −0.03 mm. The bushing bore has a diameter of 25.03 mm and a unilateral tolerance of 0.04 mm. The l/d ratio is 1/2. The load is 1.2 kN and the journal runs at 1100 rev/min. If the average viscosity is 55 mPa · s, find the minimum film thickness, the power loss, and the side flow for the minimum clearance assembly.

To determine

The minimum film thickness for the bearing.

The power loss for the bearing.

The side flow for the minimum clearance assembly.

Answer to Problem 1P

The minimum film thickness for the bearing is $0.0045\text{mm}$.

The power loss for the bearing is $\text{11}\text{.2}\text{W}$.

The side flow for the minimum clearance assembly is $177.47{\text{mm}}^3/\mathrm{s}$.

Explanation of Solution

Write the expression for the minimum radial clearance.

$c_{\min }=\frac{b_{\min }-d_{\max }}{2}$ (I)

Here, the minimum radial clearance is $c_{\min }$, the bushing bore diameter of the bearing is $b_{\min }$ and the journal diameter of the bearing is $d_{\max }$.

Write the expression for the radius of the journal bearing.

$r=\frac{d_{\max }}{2}$ . (II)

Here, the radius of the journal bearing is $r$.

Write the expression for the radial clearance ratio of the bearing.

$r_c=\frac{r}{c}$ . (III)

Here, the radial clearance ratio of the bearing is $r_c$ and  the radial clearance is $c$.

Write the expression for the number of revolution per second.

$N=\frac{n}{60}$ (IV)

Here, the number of revolution per second is $N$ and the number of revolution per minute is $n$.

Write the expression for the characteristic pressure.

$P=\frac{W}{ld}$ . (V)

Here, the characteristic pressure of the bearing is $P$, the load on the bearing is $W$, the diameter of the bearing is $d$ and the length is $l$.

Write the expression for the bearing characteristic number.

$S={\left(\frac{r}{c}\right)}^2\frac{\mu N}{P}$ (VI)

Here, the bearing characteristic number is $S$ and the average viscosity of the oil is $\mu$.

Write the expression for the minimum thickness of the film.

$h_o=0.3\times c$ (VII)

Here, the minimum thickness of the film is $h_o$.

Write the expression for the coefficient of the friction.

$f=\frac{5.4}{r_c}$ (VIII)

Here, the coefficient of the friction is $f$.

Write the expression for the parasitic friction torque.

$T=fWr$ (IX)

Here, the parasitic friction torque is $T$.

Write the expression for the heat loss of the journal bearing.

$H_{\text{loss}}=2\pi NT$ (X)

Here, the heat loss of the journal bearing is $H_{\text{loss}}$.

Write the expression for the volumetric oil flow rate in to the bearing.

$Q=\left(5.1\right)rcNl$ (XI)

Here, the volumetric oil flow rate in to the bearing is $Q$.

Write the expression for the volumetric side flow leakage out of the bearing.

$Q_s=\left(0.81\right)Q$ (XII)

Here, the volumetric side flow leakage out of the bearing is $Q_s$.

Conclusion:

Substitute $25.03\text{mm}$ for $b_{\min }$ and $25\text{mm}$ for $d_{\max }$ in Equation (I).

$\begin{array}{c}{c}_{\min }=\frac{25.03\text{mm}-25\text{mm}}{2}\\ =\frac{0.03\text{mm}}{2}\\ =0.015\text{mm}\end{array}$

Substitute $25\text{mm}$ for $d_{\max }$ in Equation (II).

$\begin{array}{c}r=\frac{25\text{mm}}{2}\\ =12.5\text{mm}\end{array}$

Substitute $0.015\text{mm}$ for $c$ and $12.5\text{mm}$ for $r$ in Equation (III).

$\begin{array}{c}{r}_c=\frac{12.5\text{mm}}{0.015\text{mm}}\\ =833.3\end{array}$

Substitute $1100\text{rev}/\min$ for $n$ in Equation (IV).

$\begin{array}{c}N=\frac{1100\text{rev}/\min }{60}\\ =18.33\text{rev}/\mathrm{s}\end{array}$

Substitute $1.2\text{kN}$ for $W$, $25\text{mm}$ for $d$ and $12.5\text{mm}$ for $l$ in Equation (V).

$\begin{array}{c}P=\frac{1.2\text{kN}}{12.5\text{mm}\left(25\text{mm}\right)}\\ =\left(\frac{1.2\text{kN}}{312.5{\text{mm}}^2}\right)\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\\ =\left(3.84\text{N}/{\text{mm}}^2\right)\left(\frac{1\text{MPa}}{1\text{N}/{\text{mm}}^2}\right)\\ =3.84\text{MPa}\end{array}$

Substitute $833.3$ for $r_c$, $18.33\text{rev}/\mathrm{s}$ for $N$, $3.84\text{MPa}$ for $P$ and $\text{55}\text{mPa}\cdot \text{s}$ for $\mu$ in Equation (VI).

$\begin{array}{c}S={\left(833.3\right)}^2\left[\frac{\text{55}\text{mPa}\cdot \text{s}\left(18.33\text{rev}/\mathrm{s}\right)}{3.84\text{MPa}}\right]\\ ={\left(833.3\right)}^2\left[\frac{1008.15\text{mPa}\cdot \text{rev}}{3.84\text{MPa}}\right]\left(\frac{{\text{10}}^{-\text{3}}\text{Pa}}{\text{1}\text{mPa}}\right)\left(\frac{1\text{MPa}}{{\text{10}}^6\text{Pa}}\right)\\ =0.182\end{array}$

Substitute $0.015\text{mm}$ for $c$ in Equation (VII).

$\begin{array}{c}{h}_o=0.3\left(0.015\text{mm}\right)\\ =0.0045\text{mm}\end{array}$

Thus, the minimum film thickness for the bearing is $0.0045\text{mm}$.

Substitute $833.3$ for $r_c$ in Equation (VIII).

$\begin{array}{c}f=\frac{5.4}{833.3}\\ =0.00648\end{array}$

Substitute $0.00648$ for $f$, $\text{12}\text{.5}\text{mm}$ for $r$ and $\text{1}\text{.2}\text{kN}$ for $W$ in Equation (IX).

$\begin{array}{c}T=\left(0.00648\right)\left(\text{1}\text{.2}\text{kN}\right)\left(\text{12}\text{.5}\text{mm}\right)\\ =\left(0.0972\text{kN}\cdot \text{mm}\right)\left(\frac{{10}^3\text{N}}{\text{kN}}\right)\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right)\\ =0.0972\text{N}\cdot \text{m}\end{array}$

Substitute $0.0972\text{N}\cdot \text{m}$ for $T$ and $18.33\text{rev}/\mathrm{s}$ for $N$ in Equation (X).

$\begin{array}{c}{H}_{\text{loss}}=2\pi \left(0.0972\text{N}\cdot \text{m}\right)\left(18.33\text{rev}/\mathrm{s}\right)\\ =2\pi \left(1.7816\text{N}\cdot \text{m}\cdot \text{rev}/\mathrm{s}\right)\left(\frac{1\text{W}}{\text{1}\text{N}\cdot \text{m}/\mathrm{s}}\right)\\ =11.194\text{W}\\ \approx \text{11}\text{.2}\text{W}\end{array}$

Thus, the power loss for the bearing is $\text{11}\text{.2}\text{W}$.

Substitute $\text{12}\text{.5}\text{mm}$ for $r$, $\text{12}\text{.5}\text{mm}$ for $l$, $0.015\text{mm}$ for $c$ and $18.33\text{rev}/\mathrm{s}$ for $N$ in Equation (XI).

$\begin{array}{c}Q=5.1\left( \text{12}\text{.5}\text{mm}\right)\left(0.015\text{mm}\right)\left(18.33\text{rev}/\mathrm{s}\right)\left(\text{12}\text{.5}\text{mm}\right)\\ =\left(11.9531{\text{mm}}^3\right)\left(18.33\text{rev}/\mathrm{s}\right)\\ =219.10{\text{mm}}^3/\mathrm{s}\end{array}$

Substitute $219.10{\text{mm}}^3/\mathrm{s}$ for $Q$ in Equation (XII).

$\begin{array}{c}{Q}_s=0.81\left(219.10{\text{mm}}^3/\mathrm{s}\right)\\ =177.47{\text{mm}}^3/\mathrm{s}\end{array}$

Thus, the side flow for the minimum clearance assembly is $177.47{\text{mm}}^3/\mathrm{s}$.