Question 16.3: Design a plain surface journal bearing to carry a steady rad...

Step 1. Let’s select D = 2.50 in. Then R = 1.25 in.
Step 2. For p = 200 psi, L must be

L = F/pD=1500/(200)(2.50) = 3.00 in

For this value of L, L/D = 3.00/2.50 = 1.20. In order to use one of the standard design charts, let’s change L to 2.50 in so that L/D = 1.0. This is not essential, but it eliminates interpolation. The actual pressure is then

p = F/LD = 1500/(2.50)(2.50) = 240 psi

This is an acceptable pressure.

Steps. From Figure 16-3, C_d = 0.003 in is suitable for the diametral clearance, based on D = 2.50 in and n= 850 rpm, and C_r= C_d/2 = 0.0015 in. Also,

R/C_r = 1.25/0.0015 = 833

This value is used in later calculations.

Step 4. For the precision desired in this machine, use a surface finish of 16 to 32 pin, requiring a ground journal.
Step 5. Minimum film thickness (design value):

h_o= 0.00025D = 0.00025(2.50) = 0.0006 in (approx.)

Step 6. Film thickness variable:

h_o/C_r= 0.0006/0.0015 = 0.40

Step 7. From Figure 16-7, for h_o/C_r = 0.40  and L/D = 1, we can read S = 0.13.

Step 8. Rotational speed in revolutions per second:

n_s = n/60 = 850/60 = 14.2  rev/s

Step 9. Solve for the viscosity from the Sommerfeld number, S:

\mu =\frac{Sp}{n_s(R/C_r)^2} =\frac{(0.13)(240)}{(14.2)(833)^2}= 3.17 \times 10^{-6} reyn

Step 10. From the viscosity chart. Figure 16-6, SAE 30 oil is required to ensure a sufficient viscosity at 160°F. The actual expected viscosity of SAE 30 at 160°F is approximately 3.3\times 10^{-6} reyn.

Step 11. For the actual viscosity, the Sommerfeld number would be

S =\frac{\mu n_s(R/C_r)^2}{p} =\frac{(3.3\times 10^{-6})(14.2)(833)^2}{240}=0.135

Step 12. Coefficient of friction (from Figure 16-8):f(R/C_r) = 3.5 for S = 0.135 and L/D = 1. Now, because R/C_r = 833,

f = 3.5/833 = 0.0042

Step 13. Friction torque:

T_f = fFR = (0.0042)(1500)(l.25) = 7.88 lb.in

Step 14. Frictional power:

P_f= T_fn/63 000 = (7.88)(850)/63 000 = 0.106 hp

Comment A qualitative evaluation of the result would require more knowledge about the application. But note that a coefficient of friction of 0.0042 is quite low. It is likely that a machine requiring such a large shaft and with such high bearing forces also requires a large power to drive it. Then the 0.106-hp friction power would appear small.
Thermal considerations are also important to determine how much energy must be dissipated from the bearing. Converting the frictional power loss to thermal power gives

P_f= 0.106hp\frac{745.7 \ W}{hp} =79.0 \ W

Expressing this in U.S. Customary units gives

P_f= 79.0 \ W\frac{1 \ Btu/hr}{0.293 \ W} =270 \ Btu/hr