Question

A journal bearing has shaft with a diameter of 75 mm. The bushing bore is 75.1 mm and the bushing is 37.5 mm long and supports a load of 2 kN. The journal speed is 720 RPM. Find the minimum film thickness, the heat loss rate (expressed in Watts), and the maximum lubricant pressure. Assume the lubricant is SAE 20 oil and the operating temperature is 60 C.

Answer 1

Explanation:

The given data is as follows.

   Length (l) = 37.5 mm,       Journal diameter (d) = 75 mm

   Diameter of bushing (D) = 75.1 mm,   Speed (N) = 720 rpm,

   Load (F) = 2 kN

Dynamic viscosity of SAE 20 at $60^{o}C$ is 12.2 cp

  Clearance = $(\frac{D}{2}) - (\frac{d}{2})$

           c = $(\frac{75.1}{2}) - (\frac{75}{2})$

              = 0.05 mm

Bearing pressure will be calculated as follows.

   Bearing pressure (p) = $\frac{F}{l \times d}$

                            = $\frac{2 \times 10^{3}}{37.5 \times 75}$

                            = $0.711 N/mm^{2}$

Now, sommerfield number (S) will be calculated as follows.

      S = $(\frac{r}{c})^{2} \times \frac{\mu \times N}{p}$

          = $(\frac{37.5}{0.05})^{2} \times \frac{12.2}{10^{9}} \times \frac{720}{60} \times \frac{1}{0.711}$

          = 0.006

As,   $\frac{l}{d} = \frac{37.5}{75} = \frac{1}{2}$

Now, using Raimondi and John Boyd data the values for other variables will be obtained.

Minimum film thickness variable = $(\frac{h_{o}}{c})$ = 0.03

Therefore, minimum film thickness $(h_{o}) = 0.03 \times 0.05$

                                           = 0.0015 mm

As, $(\frac{P}{P_{max}})$ = 0.126

Hence, maximum lubricant pressure will be calculated as follows.

       $P_{max} = \frac{0.711}{0.126}$

                      = $5.642 N/mm^{2}$

Due to friction, the heat loss rate will be as follows.

            $\frac{2 \pi NfFr}{10^{6}}$ kW

According to coefficient of friction variable = $\frac{r}{c} \times f = 0.61$

              f = $\frac{c}{r} \times 0.61$

                = $\frac{0.05}{37.5} \times 0.61$

                = 0.000813

Therefore, heat loss will be calculated as follows.

      Heat loss = $\frac{2 \pi \times \frac{720}{60} \times 0.000813 \times 2 \times 10^{3} \times 37.5}{10^{6}}$

                      = 4.6 Watt