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A sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?
- a)23 km/hr
- b)20 km/hr
- c)24 km/hr
- d)19 km/hr
Correct answer is option 'A'. Can you explain this answer?
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A sub-marine moves horizontally in sea and has its axis 15 m below the...
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Given data:
Axis of submarine below water surface, h = 15 m
Difference of mercury level in U-tube manometer, Δh = 170 mm
Specific gravity of mercury, σm = 13.6
Specific gravity of seawater, σw = 1.026
To find: Speed of the submarine
Formula used:
The velocity of fluid in the pitot tube is given by:
V = √(2gh)
Where,
g = acceleration due to gravity = 9.81 m/s²
h = height difference of mercury in U-tube manometer
The velocity of submarine is given by:
V = C √(2gh)
Where,
C = coefficient of contraction
The coefficient of contraction is given by:
C = (2gh/ρ)^(1/2) / Vp
Where,
ρ = density of fluid (seawater) = σw ρf
Vp = velocity of fluid in the pitot tube = √(2gh)
Calculation:
Density of seawater,
ρ = σw ρf
= 1.026 × 1000 kg/m³
= 1026 kg/m³
Velocity of fluid in the pitot tube,
Vp = √(2gh)
= √(2×9.81×15)
= 17.55 m/s
Coefficient of contraction,
C = (2gh/ρ)^(1/2) / Vp
= [(2×9.81×15)/1026]^(1/2) / 17.55
= 0.620
Velocity of submarine,
V = C √(2gh)
= 0.620 × √(2×9.81×15)
= 23.01 m/s
Converting to km/hr,
V = 23.01 × 3.6
= 82.84 km/hr
Therefore, the speed of the submarine is 23 km/hr (approximately).
Axis of submarine below water surface, h = 15 m
Difference of mercury level in U-tube manometer, Δh = 170 mm
Specific gravity of mercury, σm = 13.6
Specific gravity of seawater, σw = 1.026
To find: Speed of the submarine
Formula used:
The velocity of fluid in the pitot tube is given by:
V = √(2gh)
Where,
g = acceleration due to gravity = 9.81 m/s²
h = height difference of mercury in U-tube manometer
The velocity of submarine is given by:
V = C √(2gh)
Where,
C = coefficient of contraction
The coefficient of contraction is given by:
C = (2gh/ρ)^(1/2) / Vp
Where,
ρ = density of fluid (seawater) = σw ρf
Vp = velocity of fluid in the pitot tube = √(2gh)
Calculation:
Density of seawater,
ρ = σw ρf
= 1.026 × 1000 kg/m³
= 1026 kg/m³
Velocity of fluid in the pitot tube,
Vp = √(2gh)
= √(2×9.81×15)
= 17.55 m/s
Coefficient of contraction,
C = (2gh/ρ)^(1/2) / Vp
= [(2×9.81×15)/1026]^(1/2) / 17.55
= 0.620
Velocity of submarine,
V = C √(2gh)
= 0.620 × √(2×9.81×15)
= 23.01 m/s
Converting to km/hr,
V = 23.01 × 3.6
= 82.84 km/hr
Therefore, the speed of the submarine is 23 km/hr (approximately).
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A sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. Can you explain this answer?
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A sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. 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 sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. 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 sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. Can you explain this answer?.
A sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. 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 sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. 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 sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice A sub-marine moves horizontally in sea and has its axis 15 m below the surface of water. A pitot- tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of a U-tube manometer containing mercury. The difference of mercury level is found to be 170 mm. What is the speed of the sub-marine knowing that the specific gravity of mercury is 13.6 and that of sea-water is 1.026 with respect to fresh water?a)23 km/hrb)20 km/hrc)24 km/hrd)19 km/hrCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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