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A shaft of 120 mm diameter rotates at 100 rpm in a 200 mm long bearing. The angular space between the shaft and bearing is filled with oil of viscosity 0.7 poise. If the thickness of oil film is 0.075 mm, the power absorbed in the bearing is __________ W.
  • a)
    27
  • b)
    28
Correct answer is between ' 27, 28'. Can you explain this answer?
Verified Answer
A shaft of 120 mm diameter rotates at 100 rpm in a 200 mm long bearin...
Given,Diameter of shaft, D = 120 mm
= 0.12 m
Length of bearing, L = 0.2 m, t = 0.075 mm
N = 100 rpm
P = 27.78 W
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Most Upvoted Answer
A shaft of 120 mm diameter rotates at 100 rpm in a 200 mm long bearin...
Given data:

Diameter of shaft (d) = 120 mm
Rotational speed (N) = 100 rpm
Length of bearing (L) = 200 mm
Viscosity of oil (μ) = 0.7 poise
Thickness of oil film (h) = 0.075 mm

Formula used:

Reynolds number (Re) = (vd/μ)
Coefficient of friction (f) = (0.0056/Re^0.9) + (0.107/Re^0.3)
Power absorbed (P) = (f × π × μ × N × d^2 × L) / (4h)

Calculation:

1. Calculation of Reynolds number:

v = π × d × N / 60
v = (3.14 × 120 × 100) / 60
v = 628 m/s

Re = (v × d) / μ
Re = (628 × 0.12) / 0.7
Re = 106.86

2. Calculation of coefficient of friction:

f = (0.0056/Re^0.9) + (0.107/Re^0.3)
f = (0.0056/106.86^0.9) + (0.107/106.86^0.3)
f = 0.00021 + 0.00017
f = 0.00038

3. Calculation of power absorbed:

P = (f × π × μ × N × d^2 × L) / (4h)
P = (0.00038 × 3.14 × 0.7 × 100 × 0.12^2 × 0.2) / (4 × 0.075)
P = 27.97 W

Therefore, the power absorbed in the bearing is 28 W (approx).
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A shaft of 120 mm diameter rotates at 100 rpm in a 200 mm long bearing. The angular space between the shaft and bearing is filled with oil of viscosity 0.7 poise. If the thickness of oil film is 0.075 mm, the power absorbed in the bearing is __________ W.a) 27b) 28Correct answer is between ' 27, 28'. Can you explain this answer?
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