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A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)
- a)6.60
- b)6.86
Correct answer is between ' 6.60, 6.86'. Can you explain this answer?
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Verified Answer
A journal bearing has journal diameter 50 mm and a bush length of 50 ...
Tangential velocity of shaft,
Y = 0.02
Shear force on shaft = T x A = 1309πDL
= 1309 x π x (50)2 x 10-6 = 10.28 N
= 0.257 Nm
= 6.73 W
Most Upvoted Answer
A journal bearing has journal diameter 50 mm and a bush length of 50 ...
To calculate the power loss in a journal bearing, we need to consider the frictional losses due to the viscosity of the lubricant and the rotation of the shaft.
Given:
Journal diameter = 50 mm
Bush length = 50 mm
Shaft rotation speed = 250 rpm
Viscosity of lubricant = 0.4 poise
Clearance = 0.02 mm
1. Calculate the shaft speed in rad/s:
Shaft speed = 2π × rotation speed in rpm
Shaft speed = 2π × 250/60 rad/s
Shaft speed = 26.18 rad/s
2. Calculate the oil film thickness:
Oil film thickness = (journal diameter - bush length) / 2
Oil film thickness = (50 - 50) / 2
Oil film thickness = 0 mm
Since the oil film thickness is 0 mm, there is direct contact between the journal and the bush. In this case, the power loss can be calculated using the following equation:
Power loss = (0.2 × viscosity × shaft speed × clearance) / oil film thickness
3. Calculate the power loss:
Power loss = (0.2 × 0.4 × 26.18 × 0.02) / 0
Power loss = Infinity
Since the oil film thickness is 0 mm, the power loss becomes infinite.
Therefore, it is not possible to calculate the power loss for a journal bearing with a clearance of 0.02 mm and no oil film thickness.
Hence, the answer cannot be determined and the given options of 6.60 and 6.86 are incorrect.
Given:
Journal diameter = 50 mm
Bush length = 50 mm
Shaft rotation speed = 250 rpm
Viscosity of lubricant = 0.4 poise
Clearance = 0.02 mm
1. Calculate the shaft speed in rad/s:
Shaft speed = 2π × rotation speed in rpm
Shaft speed = 2π × 250/60 rad/s
Shaft speed = 26.18 rad/s
2. Calculate the oil film thickness:
Oil film thickness = (journal diameter - bush length) / 2
Oil film thickness = (50 - 50) / 2
Oil film thickness = 0 mm
Since the oil film thickness is 0 mm, there is direct contact between the journal and the bush. In this case, the power loss can be calculated using the following equation:
Power loss = (0.2 × viscosity × shaft speed × clearance) / oil film thickness
3. Calculate the power loss:
Power loss = (0.2 × 0.4 × 26.18 × 0.02) / 0
Power loss = Infinity
Since the oil film thickness is 0 mm, the power loss becomes infinite.
Therefore, it is not possible to calculate the power loss for a journal bearing with a clearance of 0.02 mm and no oil film thickness.
Hence, the answer cannot be determined and the given options of 6.60 and 6.86 are incorrect.
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A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. Can you explain this answer?
Question Description
A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. 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 journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. 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 journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. Can you explain this answer?.
A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. 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 journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. 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 journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. Can you explain this answer?.
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A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. Can you explain this answer?, a detailed solution for A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. Can you explain this answer? has been provided alongside types of A journal bearing has journal diameter 50 mm and a bush length of 50 mm. The shaft is rotating at 250 rpm and viscosity of lubricant is 0.4 poise. The power loss (in W) will be __________. (Take clearance = 0.02 mm)a) 6.60b) 6.86Correct answer is between ' 6.60, 6.86'. Can you explain this answer? theory, EduRev gives you an
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