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A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing is
- a)0.062
- b)0.125
- c)0.250
- d)0.785
Correct answer is option 'B'. Can you explain this answer?
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Micrometers and the length to diameter ratio is 1. The journal bearing operates at a temperature of 40°C. Determine the minimum film thickness and the power loss in the bearing.
To determine the minimum film thickness, we can use the Reynolds equation for journal bearings:
h = 0.5 * (c * U * L) / (μ * ω)
Where:
h = minimum film thickness
c = radial clearance
U = surface velocity
L = length of bearing
μ = viscosity
ω = angular velocity
First, let's calculate the surface velocity (U) using the formula:
U = (π * d * N) / 60
Where:
d = diameter of the bearing
N = revolutions per second
U = (π * 50 mm * 20) / 60 = 52.36 m/s
Now, we can calculate the minimum film thickness (h):
h = 0.5 * (50 μm * 52.36 m/s * 50 mm) / (20 mPas * 2π * 20 rev/s)
h = 0.5 * (0.05 mm * 52.36 m/s * 50 mm) / (20 mPas * 2π * 20 rev/s)
h ≈ 0.016 mm
Therefore, the minimum film thickness is approximately 0.016 mm.
To calculate the power loss in the bearing, we can use the following formula:
P = (F * U) / (2π * ω)
Where:
P = power loss
F = load
U = surface velocity
ω = angular velocity
First, let's convert the load to Newtons:
Load = 2.0 KN = 2000 N
Now, we can calculate the power loss (P):
P = (2000 N * 52.36 m/s) / (2π * 20 rev/s)
P = (2000 N * 52.36 m/s) / (2π * 20 rev/s)
P ≈ 829.4 W
Therefore, the power loss in the bearing is approximately 829.4 Watts.
To determine the minimum film thickness, we can use the Reynolds equation for journal bearings:
h = 0.5 * (c * U * L) / (μ * ω)
Where:
h = minimum film thickness
c = radial clearance
U = surface velocity
L = length of bearing
μ = viscosity
ω = angular velocity
First, let's calculate the surface velocity (U) using the formula:
U = (π * d * N) / 60
Where:
d = diameter of the bearing
N = revolutions per second
U = (π * 50 mm * 20) / 60 = 52.36 m/s
Now, we can calculate the minimum film thickness (h):
h = 0.5 * (50 μm * 52.36 m/s * 50 mm) / (20 mPas * 2π * 20 rev/s)
h = 0.5 * (0.05 mm * 52.36 m/s * 50 mm) / (20 mPas * 2π * 20 rev/s)
h ≈ 0.016 mm
Therefore, the minimum film thickness is approximately 0.016 mm.
To calculate the power loss in the bearing, we can use the following formula:
P = (F * U) / (2π * ω)
Where:
P = power loss
F = load
U = surface velocity
ω = angular velocity
First, let's convert the load to Newtons:
Load = 2.0 KN = 2000 N
Now, we can calculate the power loss (P):
P = (2000 N * 52.36 m/s) / (2π * 20 rev/s)
P = (2000 N * 52.36 m/s) / (2π * 20 rev/s)
P ≈ 829.4 W
Therefore, the power loss in the bearing is approximately 829.4 Watts.
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A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer?
Question Description
A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer?.
A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/second carries a load of 2.0 KN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm. The Sommerfeld number for the bearing isa)0.062 b)0.125 c)0.250 d)0.785 Correct answer is option 'B'. Can you explain this answer?.
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