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For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;
- a)2.564 m/s2
- b)7.91 m/s2
- c)15.753 m/s2
- d)5.182 m/s2
Correct answer is option 'B'. Can you explain this answer?
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For a fluid flow through a divergent pipe of length 2m having inlet a...
Given : R1 = 0.3 m R2 = 0.2 m
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Q = 0.5 m3/sec L = 2 m
Acceleration at exit;
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For a fluid flow through a divergent pipe of length 2m having inlet a...
Given data:
- Length of divergent pipe (L) = 2m
- Inlet diameter (D1) = 0.6m
- Outlet diameter (D2) = 0.4m
- Flow rate (Q) = 0.5 m3/sec
To find: Acceleration at the exit (a)
Formula used: Continuity equation
The continuity equation states that the mass flow rate (m) through a pipe is constant, and is given by:
m = ρ * Q
where,
ρ = density of fluid
Q = flow rate
Since the density of fluid is constant, we can write:
m1 = m2
where,
m1 = mass flow rate at inlet
m2 = mass flow rate at outlet
Using the formula for mass flow rate, we get:
m1 = ρ * Q1
m2 = ρ * Q2
where,
Q1 = flow rate at inlet
Q2 = flow rate at outlet
Since Q1 = Q2 = Q, we can write:
m1 = m2 = ρ * Q
Next, we can use the cross-sectional area of the pipe to relate the velocity (V) at the inlet and outlet, as follows:
m1 = ρ * Q1 = ρ * A1 * V1
m2 = ρ * Q2 = ρ * A2 * V2
where,
A1 = cross-sectional area at inlet
A2 = cross-sectional area at outlet
V1 = velocity at inlet
V2 = velocity at outlet
Since the pipe is assumed to be axial and uniform, we can assume that V1 = V2 = V. Therefore, we get:
A1 * V = A2 * V
or, A1 = A2 * (D1/D2)^2
Using the above equation, we can find A1 in terms of A2:
A1 = 1.44 * A2
Substituting this in the continuity equation, we get:
ρ * Q = ρ * A1 * V = ρ * 1.44 * A2 * V
or, Q = 1.44 * A2 * V
Solving for V, we get:
V = Q / (1.44 * A2)
Substituting the given values, we get:
V = 1.302 m/s
Now, to find the acceleration at the exit, we can use the Bernoulli's equation, which relates the pressure (P) and velocity (V) of a fluid:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
where,
P1 = pressure at inlet
P2 = pressure at outlet
Assuming that P1 = P2 (i.e., no significant pressure drop), we get:
(1/2) * ρ * V1^2 = (1/2) * ρ * V2^2
or, V2^2 = (V1^2 * A1^2) / A2^2
Substituting the given values, we get:
V2^2 = 3.063 m^2/s^2
Since V2 = Q / A2, we can write:
a = (
- Length of divergent pipe (L) = 2m
- Inlet diameter (D1) = 0.6m
- Outlet diameter (D2) = 0.4m
- Flow rate (Q) = 0.5 m3/sec
To find: Acceleration at the exit (a)
Formula used: Continuity equation
The continuity equation states that the mass flow rate (m) through a pipe is constant, and is given by:
m = ρ * Q
where,
ρ = density of fluid
Q = flow rate
Since the density of fluid is constant, we can write:
m1 = m2
where,
m1 = mass flow rate at inlet
m2 = mass flow rate at outlet
Using the formula for mass flow rate, we get:
m1 = ρ * Q1
m2 = ρ * Q2
where,
Q1 = flow rate at inlet
Q2 = flow rate at outlet
Since Q1 = Q2 = Q, we can write:
m1 = m2 = ρ * Q
Next, we can use the cross-sectional area of the pipe to relate the velocity (V) at the inlet and outlet, as follows:
m1 = ρ * Q1 = ρ * A1 * V1
m2 = ρ * Q2 = ρ * A2 * V2
where,
A1 = cross-sectional area at inlet
A2 = cross-sectional area at outlet
V1 = velocity at inlet
V2 = velocity at outlet
Since the pipe is assumed to be axial and uniform, we can assume that V1 = V2 = V. Therefore, we get:
A1 * V = A2 * V
or, A1 = A2 * (D1/D2)^2
Using the above equation, we can find A1 in terms of A2:
A1 = 1.44 * A2
Substituting this in the continuity equation, we get:
ρ * Q = ρ * A1 * V = ρ * 1.44 * A2 * V
or, Q = 1.44 * A2 * V
Solving for V, we get:
V = Q / (1.44 * A2)
Substituting the given values, we get:
V = 1.302 m/s
Now, to find the acceleration at the exit, we can use the Bernoulli's equation, which relates the pressure (P) and velocity (V) of a fluid:
P1 + (1/2) * ρ * V1^2 = P2 + (1/2) * ρ * V2^2
where,
P1 = pressure at inlet
P2 = pressure at outlet
Assuming that P1 = P2 (i.e., no significant pressure drop), we get:
(1/2) * ρ * V1^2 = (1/2) * ρ * V2^2
or, V2^2 = (V1^2 * A1^2) / A2^2
Substituting the given values, we get:
V2^2 = 3.063 m^2/s^2
Since V2 = Q / A2, we can write:
a = (
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For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. Can you explain this answer?
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For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. 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 For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. Can you explain this answer?.
For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. 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 For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. Can you explain this answer?.
Solutions for For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;a)2.564 m/s2b)7.91 m/s2c)15.753 m/s2d)5.182 m/s2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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