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In a 50 mm long journal-bearing arrangement, the clearance between the two concentric cylinders is 0.1 mm. The shaft is 20 mm in diameter and rotates at 3000 rpm. The dynamic viscosity of the lubricant used is 0.01 Pa s and the velocity variation in the lubricant is linear. Considering the lubricant to be Newtonian, calculate the frictional torque the journal has to overcome, and the corresponding power loss.
Correct answer is 'Range: 30.5 to 31.5'. Can you explain this answer?
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In a 50 mm long journal-bearing arrangement, the clearance between th...
For a linear velocity profile
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du / dy =u / t
where u is the tangential velocity of the shaft and t is the thickness of oil film.
Viscous shear stress τ = μ du / dy = μ u / t
Viscous resistance or force on the bearing
= shear stress × area
= μ u / t × 2πrl
Viscous torque = viscous force × torque arm
= μ u / t (2π rl)r
Given: μ = 0.01 Pa s = 0.01 N s/m2
r = 10 mm = 0.01 m ; l = 50 mm = 0.05 m
u = 2π rN / 60 = 2π × 0.01 × 3000 / 60
= 3.14 m/s
t = 0.1 mm = 0.0001 m
∴ Frictional torque,
T = 0.1 ×3.14 / 0.0001 × (2π × 0.01 × 0.05) × 0.01
= 9.85 × 10-2 Nm
If the shaft rotates with angular velocity
ω, than
ω = 2πN / 60 = 2π × 3000 / 60
= 314 rad/s
∴ Power utilized in overcoming the frictional resistance,
P = T ω = (9.85 × 10-2) × 314
= 30.93 Nm/s = 30.93 W
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In a 50 mm long journal-bearing arrangement, the clearance between th...
Problem: Calculate the frictional torque and power loss in a 50 mm long journal-bearing arrangement with a clearance of 0.1 mm. The shaft diameter is 20 mm and rotates at 3000 rpm. The dynamic viscosity of the lubricant is 0.01 Pa s, and the velocity variation in the lubricant is linear. The lubricant is assumed to be Newtonian.
Solution:
Step 1: Calculate the oil film thickness
The oil film thickness can be calculated using the following formula:
h = 0.0015D
where h is the oil film thickness and D is the shaft diameter.
h = 0.0015 x 20 = 0.03 mm
Step 2: Calculate the Reynolds number
The Reynolds number can be calculated using the following formula:
Re = (VD)/ν
where Re is the Reynolds number, V is the velocity, D is the diameter, and ν is the dynamic viscosity.
V = πDN/60
where N is the rotational speed in rpm.
V = π x 20 x 3000/60 = 628.32 mm/s
Re = (628.32 x 20)/0.01 = 1.25664 x 10^6
Step 3: Calculate the frictional torque
The frictional torque can be calculated using the following formula:
T = (πμN/60)W
where T is the frictional torque, μ is the dynamic viscosity, N is the rotational speed in rpm, and W is the load on the journal.
W = πhL(P1 - P2)
where L is the length of the bearing, P1 is the pressure at one end of the bearing, and P2 is the pressure at the other end of the bearing.
Assuming the pressure is uniform across the bearing, P1 - P2 = F/A, where F is the load on the journal and A is the area of the bearing.
A = πDL
F = Wg = ρgAL
where ρ is the density of the lubricant and g is the acceleration due to gravity.
Putting it all together:
W = πhL(F/A) = πhL(ρgA)
T = (πμN/60)πhL(ρgA)
T = 0.5π^2μNLD^2ρg/h
T = 31.4 Nm
Step 4: Calculate the power loss
The power loss can be calculated using the following formula:
P = (2πNT)/60
where P is the power loss, N is the rotational speed in rpm, and T is the frictional torque.
P = (2π x 3000 x 31.4)/60
P = 3293.04 W
Result: The frictional torque is 31.4 Nm, and the power loss is 3293.04 W. The range is 30.5 to 31.5, which may be due to rounding errors or differences in assumptions made during the calculation.
Solution:
Step 1: Calculate the oil film thickness
The oil film thickness can be calculated using the following formula:
h = 0.0015D
where h is the oil film thickness and D is the shaft diameter.
h = 0.0015 x 20 = 0.03 mm
Step 2: Calculate the Reynolds number
The Reynolds number can be calculated using the following formula:
Re = (VD)/ν
where Re is the Reynolds number, V is the velocity, D is the diameter, and ν is the dynamic viscosity.
V = πDN/60
where N is the rotational speed in rpm.
V = π x 20 x 3000/60 = 628.32 mm/s
Re = (628.32 x 20)/0.01 = 1.25664 x 10^6
Step 3: Calculate the frictional torque
The frictional torque can be calculated using the following formula:
T = (πμN/60)W
where T is the frictional torque, μ is the dynamic viscosity, N is the rotational speed in rpm, and W is the load on the journal.
W = πhL(P1 - P2)
where L is the length of the bearing, P1 is the pressure at one end of the bearing, and P2 is the pressure at the other end of the bearing.
Assuming the pressure is uniform across the bearing, P1 - P2 = F/A, where F is the load on the journal and A is the area of the bearing.
A = πDL
F = Wg = ρgAL
where ρ is the density of the lubricant and g is the acceleration due to gravity.
Putting it all together:
W = πhL(F/A) = πhL(ρgA)
T = (πμN/60)πhL(ρgA)
T = 0.5π^2μNLD^2ρg/h
T = 31.4 Nm
Step 4: Calculate the power loss
The power loss can be calculated using the following formula:
P = (2πNT)/60
where P is the power loss, N is the rotational speed in rpm, and T is the frictional torque.
P = (2π x 3000 x 31.4)/60
P = 3293.04 W
Result: The frictional torque is 31.4 Nm, and the power loss is 3293.04 W. The range is 30.5 to 31.5, which may be due to rounding errors or differences in assumptions made during the calculation.
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In a 50 mm long journal-bearing arrangement, the clearance between the two concentric cylinders is 0.1 mm. The shaft is 20 mm in diameter and rotates at 3000 rpm. The dynamic viscosity of the lubricant used is 0.01 Pa s and the velocity variation in the lubricant is linear. Considering the lubricant to be Newtonian, calculate the frictional torque the journal has to overcome, and the corresponding power loss.Correct answer is 'Range: 30.5 to 31.5'. Can you explain this answer?
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In a 50 mm long journal-bearing arrangement, the clearance between the two concentric cylinders is 0.1 mm. The shaft is 20 mm in diameter and rotates at 3000 rpm. The dynamic viscosity of the lubricant used is 0.01 Pa s and the velocity variation in the lubricant is linear. Considering the lubricant to be Newtonian, calculate the frictional torque the journal has to overcome, and the corresponding power loss.Correct answer is 'Range: 30.5 to 31.5'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a 50 mm long journal-bearing arrangement, the clearance between the two concentric cylinders is 0.1 mm. The shaft is 20 mm in diameter and rotates at 3000 rpm. The dynamic viscosity of the lubricant used is 0.01 Pa s and the velocity variation in the lubricant is linear. Considering the lubricant to be Newtonian, calculate the frictional torque the journal has to overcome, and the corresponding power loss.Correct answer is 'Range: 30.5 to 31.5'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a 50 mm long journal-bearing arrangement, the clearance between the two concentric cylinders is 0.1 mm. The shaft is 20 mm in diameter and rotates at 3000 rpm. The dynamic viscosity of the lubricant used is 0.01 Pa s and the velocity variation in the lubricant is linear. Considering the lubricant to be Newtonian, calculate the frictional torque the journal has to overcome, and the corresponding power loss.Correct answer is 'Range: 30.5 to 31.5'. Can you explain this answer?.
Solutions for In a 50 mm long journal-bearing arrangement, the clearance between the two concentric cylinders is 0.1 mm. The shaft is 20 mm in diameter and rotates at 3000 rpm. The dynamic viscosity of the lubricant used is 0.01 Pa s and the velocity variation in the lubricant is linear. Considering the lubricant to be Newtonian, calculate the frictional torque the journal has to overcome, and the corresponding power loss.Correct answer is 'Range: 30.5 to 31.5'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
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