Chapter 12, Problem 15P

A natural-circulation pillow-block bearing with l/d = 1 has a journal diameter D of 2.500 in with a unilateral tolerance of ‒0.001 in. The bushing bore diameter B is 2.504 in with a unilateral tolerance of 0.004 in. The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 300 lbf in shaft-stirred air at 70°F with α = 1. The lateral area of the pillow-block housing is 60 in². Perform a design assessment using minimum radial clearance for a load of 600 lbf and 300 lbf. Use Trumpler’s criteria.

To determine

The design for load $600\text{lbf}$.

The design for load $300\text{lbf}$.

Answer to Problem 15P

The design has been successful for load $600\text{lbf}$.

The design has been successful for load $300\text{lbf}$.

Explanation of Solution

Write expression for minimum thickness.

    $c=\frac{b-d}{2}$                                                                                                       (I)

Here, bore diameter is $b$, and journal diameter is $d$

Write expression for journal radius.

    $r=\frac{d}{2}$                                                                                                            (II)

Write expression for radial clearance ratio.

    $c_r=\frac{r}{c}$                                                                                                         (III)

Write expression for nominal pressure.

    $P=\frac{W}{ld}$                                                                                                         (IV)

Here, length of bearing is $l$ , load on bearing is $W$.

Write expression for viscosity.

    ${\mu }^{\prime }={\mu }_o\exp \left[\frac{b}{T+95}\right]$                                                                                    (V)

Here, viscosity is ${\mu }^{\prime }$, viscosity at standard temperature is ${\mu }_o$,constant is $b$ , and temperature is $T$.

Write expression for Somerfield number.

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

Here Somerfield number is $S$, radial ratio is $\frac{r}{c}$, and rotational speed of shaft is $N$

Write expression for heat generated due to friction.

    $H_{gen}=\frac{2545}{1050}\left(W\times N\times c\right)\left(\frac{fr}{c}\right)$                                                                   (VII)

Here, Coefficient of friction is $f$,

Write expression for heat loss.

    $H_{loss}=\left(\frac{h_{CR}A}{\alpha +1}\right)\left(T_f-T_{\infty }\right)$                                                                         (VIII)

Here, overall coefficient of radiation and convection radiation is $h_{CR}$,$\alpha$ is constant dependent on lubrication scheme and bearing housing geometry, ambient temperature is $T_{\infty }$ ,and film temperature is $T_f$.

Write expression for heat loss by linear approximation.

    $H_{loss}=m{T}_f+b$                                                                                           (IX)

Here, constant is $m$ and constant is $b$.

Write expression for inlet temperature.

    $T_1=T_f-\left(\frac{\Delta T_f}{2}\right)$                                                                                           (X)

Here, film temperature is $T_f$, temperature rise of film $\Delta T_f$.

Write expression for maximum temperature of the lubricant.

    $T_{\max }=T_s+\Delta T_f$                                                                                           (XI)

Here, the sump temperature is $T_s$

Conclusion:

Substitute $2.500\text{in}$ for $d$,$2.504\text{in}$ for $b$ in Equation (I)

    $\begin{array}{c}c=\frac{2.504\text{in}-2.500\text{in}}{2}\\ =\frac{\mathrm{0.0.004}\text{in}}{2}\\ =0.002\text{in}\end{array}$

Substitute $2.500\text{in}$ for $d$ in Equation (II).

    $\begin{array}{c}r=\frac{2.500\text{in}}{2}\\ =1.250\text{in}\end{array}$

Substitute $1.250\text{in}$ for $r$, and $0.002\text{in}$ for $c$ in Equation (III).

    $\begin{array}{c}{c}_r=\frac{1.25\text{in}}{0.002\text{in}}\\ =625\end{array}$

Design for $600\text{lbf}$:

Substitute $600\text{lbf}$ for $W$ ,$2.5\text{in}$ for $l$ ,and $2.5\text{in}$ for $d$ in the Equation (IV).

    $\begin{array}{c}P=\frac{\left(600\text{lbf}\right)}{\left(2.5\text{in}\right)\left(2.5\text{in}\right)}\\ =\frac{\left(600\text{lbf}\right)}{\left(6.25{\text{in}}^2\right)}\\ =96\text{psi}\end{array}$

The nominal pressure $P$ is lower than given design pressure $300\text{psi}$, so The Trumpler’s design criteria is satisfied.

Refer to figure 12.1 “Viscosity-temperature chart in SI units”, obtain the value of viscosity ${\mu }_o$ as $0.0136$, and constant $b$ as $1271.6°\text{F}$ for oil SAE 20 at film temperature $150°\text{F}$.

Substitute $0.0136$ for ${\mu }_o$, $1271.6°\text{F}$ for $b$, and $150°\text{F}$ for $T$ in Equation (VI).

    ${\mu }^{\prime }=\left(0.0136\right)\exp \left[\frac{\left(1271.6\right)}{\left(T_f+95°\text{F}\right)}\right]$

Substitute $1.25\text{in}$ for $r$, $0.002\text{in}$ for $c$, $33.33\text{rev}/\text{min}$ for $N$ and $96\text{psi}$ for $P$ in Equation (V).

    $\begin{array}{c}S={\left(\frac{1.250\text{in}}{0.002\text{in}}\right)}^2\frac{\left({\mu }^{\prime }\right)\left(1120\text{rev}/\text{min}\right)}{\left(96\text{psi}\right)}\\ ={\left(625\right)}^2\times \frac{\left({\mu }^{\prime }\right)\left(1120\text{rev}/\text{min}\right)}{\left(96\text{psi}\right)}\left|\frac{1\text{rev}/\text{s}}{60\text{rev}/\text{min}}\right|\\ ={\left(625\right)}^2\times \frac{\left({\mu }^{\prime }\right)\left(18.66\text{rev}/\text{s}\right)}{\left(96\text{psi}\right)}\\ =0.0760{\mu }^{\prime }\end{array}$

Substitute $600\text{lbf}$ for $W$, $1.25\text{in}$ for $r$, $0.002\text{in}$ for $c$, $18.67\text{rev}/\text{s}$ for $N$ and $96\text{psi}$ for $P$ in Equation (VII).

    $\begin{array}{c}{H}_{gen}=\left(242\right)\left[\left(600\text{lbf}\right)\times \left(18.67\text{rev}/\text{s}\right)\times \left(0.002\text{in}\right)\right]\left(\frac{fr}{c}\right)\\ =\left(242\right)\left[\left(600\text{lbf}\right)\times \left(18.67\text{rev}/\text{s}\right)\times \left(0.002\text{in}\right)\right]\left(\frac{fr}{c}\right)\left|\frac{12\text{ft}}{1\text{in}}\right|\\ =\left[\left(145200\text{lbf}\right)\times \left(18.67\text{rev}/\text{s}\right)\times \left(0.024\text{in}\right)\right]\left(\frac{fr}{c}\right)\left|\frac{12\text{ft}}{1\text{in}}\right|\\ =\left(54.3\right)\left(\frac{fr}{c}\right)\end{array}$

Substitute $2.7\text{Btu}/\left(\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)$ for $h_{CR}$, $60{\text{in}}^2$ for $A$, $1$ for $\alpha$ in Equation (VIII).

    $\begin{array}{c}{H}_{loss}=\left[\frac{\left(2.7\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)\left(60{\text{in}}^2\right)}{1+1}\right]\left(T_f-70°\text{F}\right)\\ =\left[\left(2.7\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)\left(30{\text{in}}^2\right)\right]\left(T_f-70°\text{F}\right)\left|\frac{12\text{ft}}{1\text{in}}\right|\left|\frac{144\text{ft}}{1{\text{in}}^2}\right|\\ =0.5625\left(T_f-70°\text{F}\right)\end{array}$

Calculate the value of $H_{gen}$,$\left(\frac{fr}{c}\right)$, $H_{loss}$, $S$,and $\mu$ at different film temperature $200°\text{F}$ and $240°\text{F}$ as in below table.

$T_f$	$\mu$	$S$	$\left(\frac{fr}{c}\right)$	$H_{gen}$	$H_{loss}$
$220$	$0.770$	$0.059$	$1.9$	$103.2$	$84.4$
$240$	$0.605$	$0.046$	$1.7$	$92.3$	$95.6$

Substitute $-0.545$ for $m$, $84.4$ for $H_{loss}$, and $223.1$ for $b$ in Equation (IX).

    $\begin{array}{l}\left(84.4\right)=\left(-0.545\right)T_f+223.1\\ 84.4-223.1=\left(-0.545\right)T_f\\ -138.7=-0.545T_f\\ T_f=254.49°\text{F}\end{array}$

The value of $H_{gen}$,$\left(\frac{fr}{c}\right)$, $H_{loss}$, $S$, and $\mu$ at film temperature of $254°\text{F}$ is tabulated as below.

$T_f$	$\mu$	$S$	$\left(\frac{fr}{c}\right)$	$H_{gen}$	$H_{loss}$
$254$	$0.517$	$0.0393$	$1.63$	$92.3$	$93.2$

Write expression for minimum film thickens.

    $\frac{h_o}{c}=0.21$                                                                                                 (XII)

Solve Equation (XII) for $h_o$.

    $h_o=0.21c$                                                                                              (XIII)

Substitute $0.002\text{in}$ for $c$ in Equation (XIII).

    $\begin{array}{c}{h}_o=0.21\times 0.002\text{in}\\ =0.0042\text{in}\end{array}$

Refer to figure 12-24 “Temperature rise and Somerfield graph”

Write expression for temperature rise.

    $\Delta T_{°\text{F}}=\frac{P_{psi}}{9.70}\left[0.349109+6.00940S+0.047467S^2\right]$                                 (XIV)

Here, Temperature rise is $\Delta T_{°\text{F}}$ (in Fahrenheit), The pressure is $P_{psi}$(in psi)

Substitute $96\text{psi}$ for $P$, and $0.047$ for $S$ in Equation (XIV).

    $\begin{array}{c}\Delta T_{°\text{F}}=\frac{\left(96\text{psi}\right)}{9.70}\left[\begin{array}{l}0.349109+6.00940\times \left(0.047\right)\\ +0.047467\times {\left(0.047\right)}^2\end{array}\right]\\ =\left(9.89\text{psi}\right)\left[\begin{array}{l}0.349109+0.2824\\ +0.0001048\end{array}\right]\\ =\left(9.89\text{psi}\right)\left[0.6316\right]\\ =6.25°\text{F}\end{array}$

Substitute $237°\text{F}$ for $T_f$, and $6.31°\text{F}$ for $\Delta T_f$ in Equation (X).

    $\begin{array}{c}{T}_1=237°\text{F}-\left(\frac{6.31°\text{F}}{2}\right)\\ =233.85°\text{F}\end{array}$

Thus, the inlet temperature of the lubricant is $233.85°\text{F}$.

Substitute $6.31°\text{F}$ for $\Delta T_f$ and $233.8°\text{F}$ for $T_s$ in Equation (XI).

    $\begin{array}{c}{T}_{\max }=233.8°\text{F}+6.31°\text{F}\\ =240.1°\text{F}\end{array}$

Since the maximum temperature $T_{\max }$ is less than the given temperature $250°\text{F}$, Therefore Trumpler’s design criteria satisfied.

Write expression for minimum film thickness using Trumpler’s criteria.

    $h_o\ge \left(0.0002+0.00004d\right)$                                                                         (XV)

Substitute $2.5\text{in}$ for $d$ in Equation (XV)

    $\begin{array}{c}{h}_o\ge \left[\left(0.0002\text{in}\right)+0.00004\left(2.5\text{in}\right)\right]\\ \ge \left[\left(0.0002\text{in}\right)+\left(0.0010\text{in}\right)\right]\\ \ge 0.00030\text{in}\end{array}$

The minimum radial clearance $0.00042\text{in}$ is greater than $0.00030\text{in}$. So, the Trumpler’s design criteria satisfied.

Write the equation for $n_d$.

    $n_d\ge 2$                                                                                                       (XVI)

Substitute $2$ for $n_d$ in Equation (XIV).

    $2\ge 2$

So, the Trumpler’s criteria satisfied.

Thus, the design is successful for load $600\text{lbf}$.

Design for $300\text{lbf}$:

Substitute $300\text{lbf}$ for $W$ ,$2.5\text{in}$ for $l$ ,and $2.5\text{in}$ for $d$ in the Equation (IV).

    $\begin{array}{c}P=\frac{\left(300\text{lbf}\right)}{\left(2.5\text{in}\right)\left(2.5\text{in}\right)}\\ =\frac{\left(300\text{lbf}\right)}{\left(6.25{\text{in}}^2\right)}\\ =48\text{psi}\end{array}$

The nominal pressure $P$ is lower than given design pressure $300\text{psi}$, so the Trumpler’s design criteria is satisfied.

Refer to figure 12.1 “Viscosity-temperature chart in SI units”, obtain the value of viscosity ${\mu }_o$ as $0.0136$, and constant $b$ as $1271.6°\text{F}$ for oil SAE 20 at film temperature $150°\text{F}$.

Substitute $0.0136$ for ${\mu }_o$, $1271.6°\text{F}$ for $b$, and $150°\text{F}$ for $T$ in Equation (V).

    ${\mu }^{\prime }=\left(0.0136\right)\exp \left[\frac{\left(1271.6\right)}{\left(T_f+95°\text{F}\right)}\right]$

Substitute $1.25\text{in}$ for $r$, $0.002\text{in}$ for $c$, $33.33\text{rev}/\text{min}$ for $N$ and $96\text{psi}$ for $P$ in Equation (VI).

    $\begin{array}{c}S={\left(\frac{1.250\text{in}}{0.002\text{in}}\right)}^2\frac{\left(\mu \right)\left(1120\text{rev}/\text{min}\right)}{\left(48\text{psi}\right)}\\ ={\left(625\right)}^2\times \frac{\left(\mu \right)\left(1120\text{rev}/\text{min}\right)}{\left(48\text{psi}\right)}\left|\frac{1\text{rev}/\text{s}}{60\text{rev}/\text{min}}\right|\\ ={\left(625\right)}^2\times \frac{\left(\mu \right)\left(18.66\text{rev}/\text{s}\right)}{\left(48\text{psi}\right)}\\ =0.152\mu \end{array}$

Substitute $300\text{lbf}$ for $W$, $1.25\text{in}$ for $r$, $0.002\text{in}$ for $c$, $18.67\text{rev}/\text{s}$ for $N$ and $96\text{psi}$ for $P$ in Equation (VII).

    $\begin{array}{c}{H}_{gen}=\left(242\right)\left[\left(300\text{lbf}\right)\times \left(18.67\text{rev}/\text{s}\right)\times \left(0.002\text{in}\right)\right]\left(\frac{fr}{c}\right)\\ =\left(242\right)\left[\left(300\text{lbf}\right)\times \left(18.67\text{rev}/\text{s}\right)\times \left(0.002\text{in}\right)\right]\left(\frac{fr}{c}\right)\left|\frac{12\text{ft}}{1\text{in}}\right|\\ =\left[\left(145200\text{lbf}\right)\times \left(18.67\text{rev}/\text{s}\right)\times \left(0.024\text{in}\right)\right]\left(\frac{fr}{c}\right)\left|\frac{12\text{ft}}{1\text{in}}\right|\\ =\left(27.15\right)\left(\frac{fr}{c}\right)\end{array}$

Substitute $2.7\text{Btu}/\left(\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)$ for $h_{CR}$, $60{\text{in}}^2$ for $A$,$1$ for $\alpha$ in Equation (VIII).

    $\begin{array}{c}{H}_{loss}=\left[\frac{\left(2.7\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)\left(60{\text{in}}^2\right)}{1+1}\right]\left(T_f-70°\text{F}\right)\\ =\left[\left(2.7\text{Btu}/\text{h}\cdot {\text{ft}}^2\cdot °\text{F}\right)\left(30{\text{in}}^2\right)\right]\left(T_f-70°\text{F}\right)\left|\frac{12\text{ft}}{1\text{in}}\right|\left|\frac{144\text{ft}}{1{\text{in}}^2}\right|\\ =0.5625\left(T_f-70°\text{F}\right)\end{array}$

The value of $H_{gen}$,$\left(\frac{fr}{c}\right)$, $H_{loss}$, $S$, and $\mu$ at different film temperatures are tabulated as below.

$T_f$	$\mu$	$S$	$\left(\frac{fr}{c}\right)$	$H_{gen}$	$H_{loss}$
$210$	$0.879$	$0.136$	$3.15$	$85.525$	$78.75$
$230$	$0.68$	$0.103$	$2.9$	$78.135$	$90$

Substitute $-0.369$ for $m$, $163$ for $b$ and $90$ for $H_{loss}$ in Equation (IX).

    $\begin{array}{l}\left(90\right)=\left(-0.369\right)T_f+223.1\\ 90-163=\left(-0.369\right)T_f\\ -73=-0.369T_f\\ T_f=197.8°\text{F}\end{array}$

The value of $H_{gen}$,$\left(\frac{fr}{c}\right)$, $H_{loss}$, $S$ and $\mu$ at film temperature $197.8°\text{F}$ are tabulated as in below.

$T_f$	$\mu$	$S$	$\left(\frac{fr}{c}\right)$	$H_{gen}$	$H_{loss}$
$197.8$	$0.652$	$0.112$	$2.51$	$82.23$	$82.32$

Write expression for minimum film thickens.

    $\frac{h_o}{c}=0.42$                                                                                                (XVII)

Solve Equation (XVII) for $h_o$.

    $h_o=0.42c$                                                                                              (XVIII)

Substitute $0.002\text{in}$ for $c$ in Equation (XVIII).

    $\begin{array}{c}{h}_o=0.42\times 0.002\text{in}\\ =8.4\times {10}^{-5}\text{in}\end{array}$

Thus, the minimum radial clearance is $8.4\times {10}^{-5}\text{in}$.

Write expression for temperature rise.

    $\Delta T=\frac{P}{9.70}\left[0.349109+6.00940S+0.047467S^2\right]$                                   (IXX)

Substitute $48\text{psi}$ for $P$, and $0.048$ for $S$ in Equation (IXX).

    $\begin{array}{c}\Delta T=\frac{\left(48\text{psi}\right)}{9.70}\left[\begin{array}{l}0.349109+6.00940\times \left(0.047\right)\\ +0.047467\times {\left(0.047\right)}^2\end{array}\right]\\ =\left(4.90\text{psi}\right)\left[\begin{array}{l}0.349109+0.2824\\ +0.0001048\end{array}\right]\\ =\left(4.90\text{psi}\right)\left[0.6316\right]\\ =5.23°\text{F}\end{array}$

Substitute $237°\text{F}$ for $T_f$, and $5.23°\text{F}$ for $\Delta T_f$ in Equation (X).

    $\begin{array}{c}{T}_1=237°\text{F}-\left(\frac{5.23°\text{F}}{2}\right)\\ =234.385°\text{F}\end{array}$

Thus, the inlet temperature of the lubricant is $234.385°\text{F}$.

Substitute $5.23°\text{F}$ for $\Delta T_f$ and $234°\text{F}$ for $T_s$ in Equation (XIV).

    $\begin{array}{c}{T}_{\max }=234°\text{F}+5.23°\text{F}\\ =239.23°\text{F}\end{array}$

Since the maximum temperature $T_{\max }$ is less than the given temperature $250°\text{F}$, So the Trumpler’s design criteria is satisfied.

Write expression for minimum film thickness using Trumpler’s criteria.

    $h_o\ge \left(0.002+0.00004d\right)$                                                                           (XX)

Substitute $2.5\text{in}$ for $d$ in Equation (XX)

    $\begin{array}{c}{h}_o\ge \left[\left(0.002\text{in}\right)+0.00004\left(2.5\text{in}\right)\right]\\ \ge \left[\left(0.002\text{in}\right)+\left(0.0010\text{in}\right)\right]\\ \ge 0.0030\text{in}\end{array}$

Since the minimum radial clearance $0.0030\text{in}$ is greater than $8.4\times {10}^{-5}\text{in}$, so the Trumpler’s design criteria is satisfied.

Write equation for $n_d$.

    $n_d\ge 2$                                                                                                       (XXI)

Substitute $2$ for $n_d$ in Equation (XXI)

    $2\ge 2$

So, the Trumpler’s criteria is satisfied.

Thus, the design is successful for load $300\text{lbf}$.