Question 16.EP.4: A large antenna mount weighing 12 000 lb is to be supported ...

We will choose the circular pad design for which the performance coefficients are available in Figure 16_12. The results of the design will specify the dimensions of the pads, the required oil pressure in the recess of each pad, the type of oil required and its temperature, the thickness of the film of oil when the bearings are supporting the load, the flow rate of oil required, and the pumping power required.

Step 1. From Figure 16_12, the minimum power required for a circular pad bearing would occur with the ratio R_{r }/R of approximately 0.50. For that ratio, the value of the load coefficient is a_{f}= 0.55. The pressure at the bearing recess will be somewhat below the maximum available of 500 psi because of the pressure drop in the restrictor placed between the supply manifold and the pad. Let’s design for a recess pressure of approximately 400 psi. Then, from Equation (16_6),F=a_{f} A_{p} p_{r}

A_{p}=\frac{F}{a_{f} p_{r}}=\frac{4000 \mathrm{lb}}{0.55\left(400 \mathrm{lb} / \mathrm{in}^{2}\right)}=18.2 \mathrm{in}^{2} But A_{p}=\pi D^{2} / 4. Then the required pad diameter is D=\sqrt{4 A_{p} / \pi}=\sqrt{4(18.2) / \pi}=4.81 \text { in } For convenience, let’s specify D = 5.00 in. Then the actual pad area will be A_{p}=\pi D^{2} / 4=(\pi)(5.00 \mathrm{in})^{2} / 4=19.6 \mathrm{in}^{2} The required recess pressure is then p_{r}=\frac{F}{a_{f} A_{p}}=\frac{4000 \mathrm{lb}}{0.55\left(19.6 \mathrm{in}^{2}\right)}=370 \mathrm{lb} / \mathrm{in}^{2} Also, R = D/2 = 5.00 in/2 = 2.50 in

Step 2. Specify the design value of the film thickness, h. It is recommended that h be between 0.001 and 0.010 in. Let’s use h = 0.005 in.
Step 3. Specify the lubricant and the operating temperature. Let’s select SAE 30 oil and assume that the maximum oil temperature in the oil film will be 120°F. A method of estimating the actual film temperature during operation may be consulted. (See Reference 6.) From the viscosity/temperature curves. Figure 16_6, the viscosity is approximately 8.3 × 10^{-6} reyn (Ib·s/in{2} ).

Step 4. Compute the oil flow through the bearing from Equation (16_7),Q=q_{f} \frac{F}{A_{p}} \frac{h^{3}}{\mu} The value of q_{f }= 1.4 can be found from Figure 16-12: Q=q_{f} \frac{F}{A_{p}} \frac{h^{3}}{\mu}=(1.4) \frac{4000 \mathrm{lb}}{19.6 \mathrm{in}^{2}} \frac{(0.005 \mathrm{in})^{3}}{8.3 \times 10^{-6} \mathrm{lb} \cdot \mathrm{s} / \mathrm{in}^{2}}

Step 5. Compute the pumping power required from Equation (16_8).P=p_{r} Q=H_{f}\left(\frac{F}{A_{p}}\right)^{2} \frac{h^{3}}{\mu} The value of H_{f} = 2.6 can be found from Figure 16_12: P=p_{r}Q=H_{f}\left(\frac{F}{A_{p}}\right)^{2} \frac{h^{3}}{\mu}=2.6\left(\frac{4000}{19.6}\right)^{2} \frac{(0.005 \mathrm{in})^{3}}{8.3 \times 10^{-6}}=1631 \mathrm{lb} \cdot \mathrm{in} / \mathrm{s} For convenience, we can convert this to horsepower: P=\frac{1631 \mathrm{lb} \cdot \text { in }}{\mathrm{s}} \frac{1.0 \mathrm{ft}}{12 \mathrm{in}} \frac{1.0 \mathrm{hp}}{550 \mathrm{lb} \cdot \mathrm{ft} / \mathrm{s}}=0.247 \mathrm{hp}