Respuesta :
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 [tex]60^{o}C[/tex] is 12.2 cp
Clearance = [tex](\frac{D}{2}) - (\frac{d}{2})[/tex]
c = [tex](\frac{75.1}{2}) - (\frac{75}{2})[/tex]
= 0.05 mm
Bearing pressure will be calculated as follows.
Bearing pressure (p) = [tex]\frac{F}{l \times d}[/tex]
= [tex]\frac{2 \times 10^{3}}{37.5 \times 75}[/tex]
= [tex]0.711 N/mm^{2}[/tex]
Now, sommerfield number (S) will be calculated as follows.
S = [tex](\frac{r}{c})^{2} \times \frac{\mu \times N}{p}[/tex]
= [tex](\frac{37.5}{0.05})^{2} \times \frac{12.2}{10^{9}} \times \frac{720}{60} \times \frac{1}{0.711}[/tex]
= 0.006
As, [tex]\frac{l}{d} = \frac{37.5}{75} = \frac{1}{2}[/tex]
Now, using Raimondi and John Boyd data the values for other variables will be obtained.
Minimum film thickness variable = [tex](\frac{h_{o}}{c})[/tex] = 0.03
Therefore, minimum film thickness [tex](h_{o}) = 0.03 \times 0.05[/tex]
= 0.0015 mm
As, [tex](\frac{P}{P_{max}})[/tex] = 0.126
Hence, maximum lubricant pressure will be calculated as follows.
[tex]P_{max} = \frac{0.711}{0.126}[/tex]
= [tex]5.642 N/mm^{2}[/tex]
Due to friction, the heat loss rate will be as follows.
[tex]\frac{2 \pi NfFr}{10^{6}}[/tex] kW
According to coefficient of friction variable = [tex]\frac{r}{c} \times f = 0.61[/tex]
f = [tex]\frac{c}{r} \times 0.61[/tex]
= [tex]\frac{0.05}{37.5} \times 0.61[/tex]
= 0.000813
Therefore, heat loss will be calculated as follows.
Heat loss = [tex]\frac{2 \pi \times \frac{720}{60} \times 0.000813 \times 2 \times 10^{3} \times 37.5}{10^{6}}[/tex]
= 4.6 Watt
