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A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearly
- a)9 mm
- b)11 mm
- c)17 mm
- d)21 mm
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A thin cylinder contains fluid at a pressure of 500 MN/m2, internal d...
Σc = 9000
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=
t = 16.66 ≃ 17 mm
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A thin cylinder contains fluid at a pressure of 500 MN/m2, internal d...
Σc = 9000
=
t = 16.66 ≃ 17 mm
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A thin cylinder contains fluid at a pressure of 500 MN/m2, internal d...
Understanding the Problem
In this scenario, we need to determine the minimum thickness of a thin-walled cylinder subject to internal pressure. Given parameters include:
- Internal pressure (P) = 500 MN/m²
- Internal diameter (D) = 0.6 m
- Allowable tensile stress (σ) = 9000 N/m²
Thin Cylinder Formula
The formula for the thickness (t) of a thin-walled cylinder under internal pressure is:
t = (P * D) / (2 * σ)
Where:
- t = thickness of the cylinder
- P = internal pressure
- D = internal diameter
- σ = allowable tensile stress
Calculating the Thickness
1. Convert units:
- P = 500 MN/m² = 500,000,000 N/m²
- D = 0.6 m
2. Substituting values into the formula:
- t = (500,000,000 N/m² * 0.6 m) / (2 * 9000 N/m²)
3. Performing the calculations:
- t = (300,000,000 N/m) / 18,000 N/m²
- t = 16,666.67 mm
4. Convert to mm:
- t ≈ 16.67 mm
Conclusion
The minimum thickness of the cylinder must be approximately 16.67 mm. Since we are looking for a practical answer, we round this value up. The closest option is:
- c) 17 mm
Thus, the correct answer is option 'C'.
In this scenario, we need to determine the minimum thickness of a thin-walled cylinder subject to internal pressure. Given parameters include:
- Internal pressure (P) = 500 MN/m²
- Internal diameter (D) = 0.6 m
- Allowable tensile stress (σ) = 9000 N/m²
Thin Cylinder Formula
The formula for the thickness (t) of a thin-walled cylinder under internal pressure is:
t = (P * D) / (2 * σ)
Where:
- t = thickness of the cylinder
- P = internal pressure
- D = internal diameter
- σ = allowable tensile stress
Calculating the Thickness
1. Convert units:
- P = 500 MN/m² = 500,000,000 N/m²
- D = 0.6 m
2. Substituting values into the formula:
- t = (500,000,000 N/m² * 0.6 m) / (2 * 9000 N/m²)
3. Performing the calculations:
- t = (300,000,000 N/m) / 18,000 N/m²
- t = 16,666.67 mm
4. Convert to mm:
- t ≈ 16.67 mm
Conclusion
The minimum thickness of the cylinder must be approximately 16.67 mm. Since we are looking for a practical answer, we round this value up. The closest option is:
- c) 17 mm
Thus, the correct answer is option 'C'.
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Question Description
A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearlya) 9 mmb) 11 mmc) 17 mmd) 21 mmCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearlya) 9 mmb) 11 mmc) 17 mmd) 21 mmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearlya) 9 mmb) 11 mmc) 17 mmd) 21 mmCorrect answer is option 'C'. Can you explain this answer?.
A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearlya) 9 mmb) 11 mmc) 17 mmd) 21 mmCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearlya) 9 mmb) 11 mmc) 17 mmd) 21 mmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin cylinder contains fluid at a pressure of 500 MN/m2, internal diameter of the shell is 0.6 m and tensile stress in the materials is to be limited to 9000 N/m2. The shell must have a minimum thickness of nearlya) 9 mmb) 11 mmc) 17 mmd) 21 mmCorrect answer is option 'C'. Can you explain this answer?.
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