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An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)
Correct answer is between '5.15,5.19'. Can you explain this answer?
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An orifice of 20 mm diameter provided in a bottom most side of tank of...
Volume of liquid leaving the tank = Volume of liquid flowing through orifice in time dt
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An orifice of 20 mm diameter provided in a bottom most side of tank of...
Given Information:
- Diameter of the orifice = 20 mm
- Cross-sectional area of the tank = 0.25 m^2
- Initial liquid level in the tank = 1.5 m
- Final liquid level in the tank = 0.5 m
- Coefficient of discharge = 0.6
To find:
Time taken for the liquid surface in the tank to fall from 1.5 m to 0.5 m.
Assumptions:
- The orifice is a small orifice, so the velocity of the liquid flow through the orifice can be considered constant.
- The head loss due to friction in the orifice is neglected.
- The flow through the orifice is steady and incompressible.
Solution:
Step 1: Calculate the velocity of the liquid flow through the orifice.
- The cross-sectional area of the orifice can be calculated using the formula:
A = πr^2, where r is the radius of the orifice.
- Given that the diameter of the orifice is 20 mm, the radius can be calculated as:
r = 20 mm / 2 = 10 mm = 0.01 m
- Substituting the value of r in the formula, we get:
A = π(0.01)^2 = 0.000314 m^2
- The velocity of the liquid flow through the orifice can be calculated using the formula:
Q = Av, where Q is the flow rate, A is the cross-sectional area, and v is the velocity.
- Rearranging the formula, we get:
v = Q / A
- The flow rate Q can be calculated using the formula:
Q = CdA√(2gh), where Cd is the coefficient of discharge, A is the cross-sectional area, g is the acceleration due to gravity, and h is the head.
- Given that Cd = 0.6, A = 0.000314 m^2, g = 9.81 m/s^2, and h = 1.5 m, the flow rate Q can be calculated as:
Q = 0.6 * 0.000314 * √(2 * 9.81 * 1.5) = 0.00682 m^3/s
- Substituting the values of Q and A in the formula for velocity, we get:
v = 0.00682 / 0.000314 = 21.74 m/s
Step 2: Calculate the time taken for the liquid surface to fall from 1.5 m to 0.5 m.
- The distance traveled by the liquid surface can be calculated as the difference between the initial and final liquid levels:
Distance = 1.5 m - 0.5 m = 1 m
- The time taken can be calculated using the formula:
Time = Distance / Velocity
- Substituting the values of Distance and Velocity, we get:
Time = 1 m / 21.74 m/s = 0.0459 s
- Converting the time from seconds to minutes, we get:
Time = 0.0459 s * 60 = 2.754 min
- Diameter of the orifice = 20 mm
- Cross-sectional area of the tank = 0.25 m^2
- Initial liquid level in the tank = 1.5 m
- Final liquid level in the tank = 0.5 m
- Coefficient of discharge = 0.6
To find:
Time taken for the liquid surface in the tank to fall from 1.5 m to 0.5 m.
Assumptions:
- The orifice is a small orifice, so the velocity of the liquid flow through the orifice can be considered constant.
- The head loss due to friction in the orifice is neglected.
- The flow through the orifice is steady and incompressible.
Solution:
Step 1: Calculate the velocity of the liquid flow through the orifice.
- The cross-sectional area of the orifice can be calculated using the formula:
A = πr^2, where r is the radius of the orifice.
- Given that the diameter of the orifice is 20 mm, the radius can be calculated as:
r = 20 mm / 2 = 10 mm = 0.01 m
- Substituting the value of r in the formula, we get:
A = π(0.01)^2 = 0.000314 m^2
- The velocity of the liquid flow through the orifice can be calculated using the formula:
Q = Av, where Q is the flow rate, A is the cross-sectional area, and v is the velocity.
- Rearranging the formula, we get:
v = Q / A
- The flow rate Q can be calculated using the formula:
Q = CdA√(2gh), where Cd is the coefficient of discharge, A is the cross-sectional area, g is the acceleration due to gravity, and h is the head.
- Given that Cd = 0.6, A = 0.000314 m^2, g = 9.81 m/s^2, and h = 1.5 m, the flow rate Q can be calculated as:
Q = 0.6 * 0.000314 * √(2 * 9.81 * 1.5) = 0.00682 m^3/s
- Substituting the values of Q and A in the formula for velocity, we get:
v = 0.00682 / 0.000314 = 21.74 m/s
Step 2: Calculate the time taken for the liquid surface to fall from 1.5 m to 0.5 m.
- The distance traveled by the liquid surface can be calculated as the difference between the initial and final liquid levels:
Distance = 1.5 m - 0.5 m = 1 m
- The time taken can be calculated using the formula:
Time = Distance / Velocity
- Substituting the values of Distance and Velocity, we get:
Time = 1 m / 21.74 m/s = 0.0459 s
- Converting the time from seconds to minutes, we get:
Time = 0.0459 s * 60 = 2.754 min
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An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer?
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An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer?.
An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An orifice of 20 mm diameter provided in a bottom most side of tank of constant cross sectional area of 0.25 m2. The time taken (in min) to fall of liquid surface in tank from 1.5 m to 0.5 m is (coefficient of discharge = 0.6)Correct answer is between '5.15,5.19'. Can you explain this answer?.
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