Chapter 12, Problem 16P

An eight-cylinder diesel engine has a front main bearing with a journal diameter of 3.500 in and a unilateral tolerance of ‒0.003 in. The bushing bore diameter is 3.505 in with a unilateral tolerance of +0.005 in. The bushing length is 2 in. The pressure-fed bearing has a central annular groove 0.250 in wide. The SAE 30 oil comes from a sump at 120°F using a supply pressure of 50 psig. The sump’s heat-dissipation capacity is 5000 Btu/h per bearing. For a minimum radial clearance, a speed of 2000 rev/min, and a radial load of 4600 lbf, find the average film temperature and apply Trumpler’s criteria in your design assessment.

To determine

The average film temperature.

The design of bearing by use of Trumpler’s criterion.

Answer to Problem 16P

The average film temperature $152.05°\text{F}$.

The design is not successful.

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 effective length

    $l^{\prime }=\frac{l-w}{2}$                                                                                                    (IV)

Here, length of bearing is $l$, and width of groove is $w$.

Write expression for slenderness ratio.

    $\lambda =\frac{l^{\prime }}{d}$                                                                                                           (V)

Write expression for nominal pressure.

    $P=\frac{W}{2l^{\prime }d}$                                                                                                      (VI)

Write expression for viscosity.

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

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

Write expression for Somerfield number.

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

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

Write expression for temperature rise in oil.

    $\Delta T=\frac{0.0123}{\left(1+1.5{\epsilon }^2\right)}\frac{\left(\frac{fr}{c}\right)S{W}^2}{p_s{r}^4}$                                                                        (IX)

Here, the supply pressure is $p_s$

Write expression for average temperature.

    $T_{avg}=T_s+\frac{\Delta T}{2}$                                                                                              (X)

Here, sump temperature is $T_s$.

Write expression for heat lost

    $H_{loss}=\rho \times C_p\times Q_s\times \Delta T$                                                                             (XI)

Here, the density of lubricant is $\rho$, heat capacity of lubricant.

Write expression for average film temperature

    ${\overline{T}}_f=\frac{T_f+T_{av}}{2}$                                                                                             (XII)

Conclusion:

Substitute $3.500\text{in}$ for $d$,$3.505\text{in}$ for $b$ in Equation (I).

    $\begin{array}{c}c=\frac{3.505\text{in}-3.500\text{in}}{2}\\ =\frac{0.0005\text{in}}{2}\\ =0.00025\text{in}\end{array}$

Substitute $3.500\text{in}$ for $d$ in Equation (II).

    $\begin{array}{c}r=\frac{3.500\text{in}}{2}\\ =1.75\text{in}\end{array}$

Substitute $1.75\text{in}$ for $r$, and $0.0025\text{in}$ for $c$ in Equation (III).

    $\begin{array}{c}{c}_r=\frac{1.75\text{in}}{0.0025\text{in}}\\ =700\end{array}$

Substitute $2.0\text{in}$ for $l$, and $0.25\text{in}$ for $c_a$ in Equation (IV).

    $\begin{array}{c}{l}^{\prime }=\frac{2.0\text{in}-0.25\text{in}}{2}\\ =\frac{1.75\text{in}}{2}\\ =0.875\text{in}\end{array}$

Substitute $0.875\text{in}$ for $l^{\prime }$, and $3.5\text{in}$ for $d$ in Equation (V).

    $\begin{array}{c}\lambda =\frac{0.875\text{in}}{3.5\text{in}}\\ =0.25\end{array}$

Substitute $4600\text{lbf}$ for $W$ ,$0.875\text{in}$ for $l$ ,and $3.5\text{in}$ for $d$ in the Equation (VI).

    $\begin{array}{c}P=\frac{\left(4600\text{lbf}\right)}{2\left(0.875\text{in}\right)\left(3.5\text{in}\right)}\\ =\frac{\left(4600\text{lbf}\right)}{2\left(3.0625{\text{in}}^2\right)}\\ =\frac{\left(4600\text{lbf}\right)}{\left(6.125{\text{in}}^2\right)}\\ =751.020\text{psi}\end{array}$

The, nominal pressure $P$ is greater than given design pressure, that is $300\text{psi}$.

Therefore, Trumpler’s design criterion is not satisfied.

Trial (A):

Refer to figure 12.1 “Viscosity-temperature chart in SI units” to obtain the value of viscosity,${\mu }_o$ is $0.0141\times {10}^{-6}\text{reyn}$, and constant $b$ is $1360.0°\text{F}$ for oil SAE 30 at film temperature $150°\text{F}$.

Substitute $0.0141\times {10}^{-6}\text{reyn}$ for ${\mu }_o$, $1360.0°\text{F}$ for $b$, and $150°\text{F}$ for $T$ in Equation (VII).

    $\begin{array}{c}{\mu }^{\prime }=\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[\frac{\left(1360.0°\text{F}\right)}{\left(150°\text{F}+95°\text{F}\right)}\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[5.55\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\left(257.500\right)\\ =3.63075\times {10}^{-6}\text{reyn}\end{array}$

Refer to figure 12-13 “Viscosity-temperature chart in SI units” obtain the value of viscosity,$\mu$ for oil SAE 30 as $3.63\times {10}^{-6}\text{reyn}$ at temperature $150°\text{F}$.

Substitute $3.63\times {10}^{-6}\text{reyn}$ for $\mu$, $1.750\text{in}$ for $r$,$0.0025\text{in}$ for $c$, $33.33\text{rev}/\text{s}$ for $N$ and $751\text{psi}$ for $P$ in Equation (VIII).

    $\begin{array}{c}S={\left(\frac{1.750\text{in}}{0.0025\text{in}}\right)}^2\frac{\left(3.63\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ ={\left(700\right)}^2\times \frac{\left(3.63\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ =490000\times 0.1611\times {10}^{-6}\\ =0.078940\end{array}$

Refer to chart 12-18 “Coefficient of friction variable” to obtain the value of coefficient of friction variable $\left(\frac{fr}{c}\right)$ as $3.6$, at slenderness ratio $0.25$ ,and Somerfield number $0.078940$.

Refer to chart 12-16 “Minimum film thickness and eccentricity ratio” to obtain the value of eccentric ratio $\left(\epsilon \right)$ as $0.9$ at slenderness ratio of $0.25$ and Somerfield number of $0.078940$.

Substitute $0.915$ for $\epsilon$, $50\text{psi}$ for $p_s$, $3.6$ for $\left(\frac{fr}{c}\right)$,$0.0635$ for $S$ ,$4600\text{lbf}$ for $W$, and $1.75\text{in}$ for $r$ in Equation (IX).

    $\begin{array}{c}\Delta T=\frac{0.0123}{\left[1+1.5{\left(0.9\right)}^2\right]}\frac{\left(3.6\right)\left(0.0789\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\frac{\left(0.0123\right)}{\left(2.215\right)}\frac{\left(0.2840\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\left(5.553\times {10}^{-3}\right)\frac{\left(0.2840\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =71.2°\text{F}\end{array}$

Substitute $120°\text{F}$ for $T_s$, and $71.2°\text{F}$ for $\Delta T$ in Equation (X).

    $\begin{array}{c}{T}_{avg}=\left(120°\text{F}\right)+\frac{\left(71.2°\text{F}\right)}{2}\\ =\left(120°\text{F}\right)+\left(35.6°\text{F}\right)\\ =155.6°\text{F}\end{array}$

Trial (B):

Refer to figure 12.1 “Viscosity-temperature chart in SI units” to obtain the value of viscosity,${\mu }_o$ as $0.0141\times {10}^{-6}\text{reyn}$ and constant $b$ is $1360.0°\text{F}$ for oil SAE 30 at film temperature of $160°\text{F}$.

Substitute $0.0141\times {10}^{-6}\text{reyn}$ for ${\mu }_o$, $1360.0°\text{F}$ for $b$, and $150°\text{F}$ for $T$ in Equation (VII).

    $\begin{array}{c}{\mu }^{\prime }=\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[\frac{\left(1360.0°\text{F}\right)}{\left(160°\text{F}+95°\text{F}\right)}\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[5.33\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\left(207.12\right)\\ =2.92\times {10}^{-6}\text{reyn}\end{array}$

Refer to figure 12-13 “Viscosity-temperature chart in SI units” to obtain the value of viscosity for oil SAE 30 as $3.63\times {10}^{-6}\text{reyn}$ at temperature $150°\text{F}$.

Substitute $3.63\times {10}^{-6}\text{reyn}$ for ${\mu }^{\prime }$, $1.750\text{in}$ for $r$,$0.0025\text{in}$ for $c$, $33.33\text{rev}/\text{s}$ for $N$ and $751\text{psi}$ for $P$ in Equation (VIII).

    $\begin{array}{c}S={\left(\frac{1.750\text{in}}{0.0025\text{in}}\right)}^2\frac{\left(2.92\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ ={\left(700\right)}^2\times \frac{\left(2.92\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ =490000\times 0.129\times {10}^{-6}\\ =0.0635\end{array}$

Refer to chart 12-18 “Coefficient of friction variable” to obtain the value of coefficient of friction variable $\left(\frac{fr}{c}\right)$ as $3.6$ at slenderness ratio of $0.25$ and Somerfield number of $0.0635$.

Refer to chart 12-16 “Minimum film thickness and eccentricity ratio” to obtain the value of eccentric ratio $\left(\epsilon \right)$ is $0.915$ at slenderness ratio of $0.25$ and Somerfield number of $0.0635$.

Substitute $0.915$ for $\epsilon$, $50\text{psi}$ for $p_s$, $3$ for $\left(\frac{fr}{c}\right)$,$0.0635$ for $S$, $4600\text{lbf}$ for $W$, and $1.75\text{in}$ for $r$ in Equation (IX).

    $\begin{array}{c}\Delta T=\frac{0.0123}{\left[1+1.5{\left(0.915\right)}^2\right]}\frac{\left(3.6\right)\left(0.0635\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\frac{\left(0.0123\right)}{\left(2.255\right)}\frac{\left(0.2840\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\left(5.4525\times {10}^{-3}\right)\frac{\left(0.2840\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =46.9°\text{F}\end{array}$

Substitute $120°\text{F}$ for $T_s$, and $46.9°\text{F}$ for $\Delta T$ in Equation (X).

    $\begin{array}{c}{T}_{avg}=\left(120°\text{F}\right)+\frac{\left(46.9°\text{F}\right)}{2}\\ =\left(120°\text{F}\right)+\left(23.45°\text{F}\right)\\ =143.45°\text{F}\end{array}$

Trial (C):

Refer to figure 12.1 “Viscosity-temperature chart in SI units” obtain the value of viscosity as $0.0141\times {10}^{-6}\text{reyn}$, and constant $b$ as $1360.0°\text{F}$ for oil SAE 30.

Substitute $0.0141\times {10}^{-6}\text{reyn}$ for ${\mu }_o$, $1360.0°\text{F}$ for $b$, and $152.5°\text{F}$ for $T$ in Equation (VII).

    $\begin{array}{c}{\mu }^{\prime }=\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[\frac{\left(1360.0°\text{F}\right)}{\left(152.5°\text{F}+95°\text{F}\right)}\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[5.4949\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\left(243.45\right)\\ =3.43\times {10}^{-6}\text{reyn}\end{array}$

Substitute $3.43\times {10}^{-6}\text{reyn}$ for ${\mu }^{\prime }$, $1.750\text{in}$ for $r$,$0.0025\text{in}$ for $c$, $33.33\text{rev}/\text{s}$ for $N$ and $751\text{psi}$ for $P$ in Equation (VIII).

    $\begin{array}{c}S={\left(\frac{1.750\text{in}}{0.0025\text{in}}\right)}^2\frac{\left(3.43\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ ={\left(700\right)}^2\times \frac{\left(3.43\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ =490000\times 0.1611\times {10}^{-6}\\ =0.078940\end{array}$

Refer to chart 12-18 “Coefficient of friction variable” to obtain the value of coefficient of friction variable $\left(\frac{fr}{c}\right)$ as $3.4$ at slenderness ratio of $0.25$ and Somerfield number of $0.0746$.

Refer to chart 12-16 “Minimum film thickness and eccentricity ratio” to obtain the value of eccentric ratio $\left(\epsilon \right)$ as $0.905$ at slenderness ratio of $0.25$ and Somerfield number of $0.0746$.

Substitute $0.905$ for $\epsilon$, $50\text{psi}$ for $p_s$, $3.4$ for $\left(\frac{fr}{c}\right)$,$0.0746$ for $S$ ,$4600\text{lbf}$ for $W$, and $1.75\text{in}$ for $r$ in Equation (IX).

    $\begin{array}{c}\Delta T=\frac{0.0123}{\left[1+1.5{\left(0.905\right)}^2\right]}\frac{\left(3.4\right)\left(0.0746\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\frac{\left(0.0123\right)}{\left(2.2285\right)}\frac{\left(0.25364\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\left(5.519\times {10}^{-3}\right)\frac{\left(0.2367\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =63.2°\text{F}\end{array}$

Substitute $120°\text{F}$ for $T_s$, and $63.2°\text{F}$ for $\Delta T$ in Equation (X).

    $\begin{array}{c}{T}_{avg}=\left(120°\text{F}\right)+\frac{\left(63.2°\text{F}\right)}{2}\\ =\left(120°\text{F}\right)+\left(31.6°\text{F}\right)\\ =151.6°\text{F}\end{array}$

Trial (D):

Refer to figure 12.1 “Viscosity-temperature chart in SI units” to obtain the value of viscosity as $0.0141\times {10}^{-6}\text{reyn}$ and constant $b$ as $1360.0°\text{F}$ for oil SAE 30 at film temperature of $152°\text{F}$.

Substitute $0.0141\times {10}^{-6}\text{reyn}$ for ${\mu }_o$, $1360.0°\text{F}$ for $b$, and $152°\text{F}$ for $T$ in Equation (VII).

    $\begin{array}{c}{\mu }^{\prime }=\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[\frac{\left(1360.0°\text{F}\right)}{\left(152°\text{F}+95°\text{F}\right)}\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\exp \left[5.506\right]\\ =\left(0.0141\times {10}^{-6}\text{reyn}\right)\left(246.18\right)\\ =3.47\times {10}^{-6}\text{reyn}\end{array}$

Substitute $3.63\times {10}^{-6}\text{reyn}$ for $\mu$, $1.750\text{in}$ for $r$,$0.0025\text{in}$ for $c$, $33.33\text{rev}/\text{s}$ for $N$ and $751\text{psi}$ for $P$ in Equation (VIII).

    $\begin{array}{c}S={\left(\frac{1.750\text{in}}{0.0025\text{in}}\right)}^2\frac{\left(3.47\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ ={\left(700\right)}^2\times \frac{\left(3.47\times {10}^{-6}\text{reyn}\right)\left(33.33\text{rev}/\text{s}\right)}{\left(751\text{psi}\right)}\\ =490000\times 0.1611\times {10}^{-6}\\ =0.0754\end{array}$

Refer to chart 12-18 “Coefficient of friction variable” to obtain the value of coefficient of friction variable $\left(\frac{fr}{c}\right)$ as $3.4$ at slenderness ratio of $0.25$ and Somerfield number of $0.0754$.

Refer to chart 12-16 “Minimum film thickness and eccentricity ratio” to obtain the value of eccentric ratio $\left(\epsilon \right)$ as $0.902$ at slenderness ratio of $0.25$ and Somerfield number of $0.0754$.

Substitute $0.905$ for $\epsilon$, $50\text{psi}$ for $p_s$, $3.4$ for $\left(\frac{fr}{c}\right)$, $0.0746$ for $S$,$4600\text{lbf}$ for $W$, and $1.75\text{in}$ for $r$ in Equation (IX).

    $\begin{array}{c}\Delta T=\frac{0.0123}{\left[1+1.5{\left(0.902\right)}^2\right]}\frac{\left(3.4\right)\left(0.0746\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\frac{\left(0.0123\right)}{\left(2.22040\right)}\frac{\left(0.25364\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =\left(5.5395\times {10}^{-3}\right)\frac{\left(0.2367\right){\left(4600\text{lbf}\right)}^2}{\left(50\text{psi}\right){\left(1.750\text{in}\right)}^4}\\ =64.1°\text{F}\end{array}$

Substitute $120°\text{F}$ for $T_s$ and $64.1°\text{F}$ for $\Delta T$ in Equation (X).

    $\begin{array}{c}{T}_{avg}=\left(120°\text{F}\right)+\frac{\left(64.1°\text{F}\right)}{2}\\ =\left(120°\text{F}\right)+\left(32.05°\text{F}\right)\\ =152.05°\text{F}\end{array}$

Write average film temperature.

    $T_{avgfilm}=\frac{T_f+T_{avg}}{2}$.                                                                     (XI)

Substitute $152.5°\text{F}$ for $T_{avg}$, and $151.6°\text{F}$ for $T_f$ in Equation (X).

    $\begin{array}{c}{T}_{avgfilm}=\frac{151.6°\text{F}+152.5°\text{F}}{2}\\ =\frac{304.1°\text{F}}{2}\\ =152.05°\text{F}\end{array}$

Thus, the average film temperature is $152.05°\text{F}$.

Substitute the $64.1°\text{F}$ for $\Delta T$ and $0.0311\text{lb}/{\text{in}}^3$ for $\rho$ and $0.42$ for $C_p$, $1.047{\text{in}}^3/\text{s}$ for $Q_s$ in Equation (XI).

    $\begin{array}{c}{H}_{loss}=\left(0.0311\text{lb}/{\text{in}}^3\right)\times \left(0.42\text{Btu}/\text{lb}\cdot °\text{F}\right)\times \left(1.047{\text{in}}^3/\text{s}\right)\times \left(64.1°\text{F}\right)\\ =\left(0.0311\text{lb}/{\text{in}}^3\right)\times \left(0.42\text{Btu}/\text{lb}\cdot °\text{F}\right)\times \left(1.047{\text{in}}^3/\text{s}\right)\\ =\left(0.03256\text{lb}/\text{s}\right)\left(26.922\text{Btu}/\text{lb}\right)\left|\frac{60\text{h}}{1\text{s}}\right|\left|\frac{60\text{h}}{1\text{s}}\right|\\ =3160\text{Btu}/\text{h}\end{array}$

Thus, the heat loss is $3160\text{Btu}/\text{h}$.

Refer to chart of “Minimum film thickness variable and eccentricity ratio” obtain the value of film thickness variable $\left(\frac{h_o}{c}\right)$ as $0.098$ at Somerfield number $0.0457$

Write expression for minimum film thickens.

    $\frac{h_o}{c}=0.098$                                                                                                 (XII)

Solve Equation (XII) for $h_o$.

    $h_o=0.098c$                                                                                               (XIII)

Substitute $0.0025\text{in}$ for $c$ in Equation (XIII).

    $\begin{array}{c}{h}_o=0.098\times 0.0025\text{in}\\ =2.45\times {10}^{-4}\text{in}\end{array}$

Thus, the minimum radial clearance is $2.45\times {10}^{-4}\text{in}$.

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

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

Substitute $3.5\text{in}$ for $d$ in Equation (XIV)

    $\begin{array}{c}{h}_o\ge \left[\left(0.002\text{in}\right)+0.00004\left(3.5\text{in}\right)\right]\\ \ge \left[\left(0.002\text{in}\right)+\left(0.00014\text{in}\right)\right]\\ \ge 0.00214\text{in}\end{array}$

Write expression for maximum temperature of the lubricant.

    $T_{\max }=T_s+\Delta T$                                                                                          (XIV)

Here, the sump temperature is $T_s$

Substitute $64.1°\text{F}$ for $\Delta T$ and $120°\text{F}$ for $T_s$ in Equation (XIV).

    $\begin{array}{c}{T}_{\max }=120°\text{F}+64.1°\text{F}\\ =184.1°\text{F}\end{array}$

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

Thus, the design is not successful.