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A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.
Correct answer is between '4711,4713'. Can you explain this answer?
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A jet of water of diameter 50 mm moving with a velocity of 40 m/s, str...
Concept:
Force exerted by the jet in the direction of jet,
Fx = Mass per sec × [V1x - V2x]
Fx = ρaV[V - (-V cos θ)] = ρaV[V + V cos θ]
= ρaV2[1 + cos θ]
Fy = Mass per sec × [V1y - V2y]
Fy = ρaV[0 - V sin θ] = -ρaV2 sin θ
Calculation:
Velocity of jet, V = 40 m/s
Angle of deflection = 120° = 180° - θ or θ = 60°
Force exerted by the jet on the curved plate in the direction of the jet:
Fx = ρaV2 [1 + cos θ]
= 1000 × .001963 × 402 × [1 + cos 60°] = 4711.15 N
Most Upvoted Answer
A jet of water of diameter 50 mm moving with a velocity of 40 m/s, str...
°.
We can use the principle of conservation of momentum to find the force exerted by the jet on the plate. The momentum of the jet before it strikes the plate is given by:
p1 = m*v1
where m is the mass of water flowing per second and v1 is the velocity of the jet before it strikes the plate. The mass of water flowing per second can be calculated using the formula:
m = A*v
where A is the cross-sectional area of the jet and v is the velocity of the jet. Therefore:
m = π*(d/2)^2*v
where d is the diameter of the jet. Substituting the given values, we get:
m = π*(50/2)^2*40 = 157079.63 kg/s
The momentum of the jet after it strikes the plate is given by:
p2 = m*v2
where v2 is the velocity of the jet after it is deflected through an angle of 120°. We can use the law of reflection to find v2. The angle of incidence and the angle of reflection are equal, so the angle between the jet and the plate after reflection is 120°. Therefore, the angle of incidence is (180-120)/2 = 30°. The velocity of the jet can be resolved into two components - one perpendicular to the plate and one parallel to the plate. The component perpendicular to the plate is reflected back with the same velocity, while the component parallel to the plate is not affected. Therefore, the velocity of the jet after reflection is given by:
v2 = 2*v1*cos(30°) = v1*sqrt(3)
Substituting the given values, we get:
v2 = 40*sqrt(3) m/s
Therefore, the momentum of the jet after it strikes the plate is:
p2 = m*v2 = 157079.63*40*sqrt(3) = 27291916.4 kg m/s
The change in momentum of the jet is:
Δp = p2 - p1 = 27291916.4 - 628318.53 = 26663597.87 kg m/s
The force exerted by the jet on the plate in the direction of the jet is given by:
F = Δp/t
where t is the time taken by the jet to change its momentum. This time can be calculated using the formula:
t = l/v1
where l is the distance travelled by the jet after it strikes the plate. The distance travelled can be calculated using the formula:
l = r*θ
where r is the radius of curvature of the plate and θ is the angle through which the jet is deflected. Substituting the given values, we get:
r = d/2 = 25 mm = 0.025 m
θ = 120° = 2/3 π
l = r*θ = 0.025*2/3*π = 0.0524 m
Therefore, the time taken by the jet to change its momentum is:
t = l/v1 = 0.0524/40 = 0.00131 s
Substituting the values of Δp and t, we get:
F = Δp/t = 26663597.87/0.00131 = 2.04*10^10 N
We can use the principle of conservation of momentum to find the force exerted by the jet on the plate. The momentum of the jet before it strikes the plate is given by:
p1 = m*v1
where m is the mass of water flowing per second and v1 is the velocity of the jet before it strikes the plate. The mass of water flowing per second can be calculated using the formula:
m = A*v
where A is the cross-sectional area of the jet and v is the velocity of the jet. Therefore:
m = π*(d/2)^2*v
where d is the diameter of the jet. Substituting the given values, we get:
m = π*(50/2)^2*40 = 157079.63 kg/s
The momentum of the jet after it strikes the plate is given by:
p2 = m*v2
where v2 is the velocity of the jet after it is deflected through an angle of 120°. We can use the law of reflection to find v2. The angle of incidence and the angle of reflection are equal, so the angle between the jet and the plate after reflection is 120°. Therefore, the angle of incidence is (180-120)/2 = 30°. The velocity of the jet can be resolved into two components - one perpendicular to the plate and one parallel to the plate. The component perpendicular to the plate is reflected back with the same velocity, while the component parallel to the plate is not affected. Therefore, the velocity of the jet after reflection is given by:
v2 = 2*v1*cos(30°) = v1*sqrt(3)
Substituting the given values, we get:
v2 = 40*sqrt(3) m/s
Therefore, the momentum of the jet after it strikes the plate is:
p2 = m*v2 = 157079.63*40*sqrt(3) = 27291916.4 kg m/s
The change in momentum of the jet is:
Δp = p2 - p1 = 27291916.4 - 628318.53 = 26663597.87 kg m/s
The force exerted by the jet on the plate in the direction of the jet is given by:
F = Δp/t
where t is the time taken by the jet to change its momentum. This time can be calculated using the formula:
t = l/v1
where l is the distance travelled by the jet after it strikes the plate. The distance travelled can be calculated using the formula:
l = r*θ
where r is the radius of curvature of the plate and θ is the angle through which the jet is deflected. Substituting the given values, we get:
r = d/2 = 25 mm = 0.025 m
θ = 120° = 2/3 π
l = r*θ = 0.025*2/3*π = 0.0524 m
Therefore, the time taken by the jet to change its momentum is:
t = l/v1 = 0.0524/40 = 0.00131 s
Substituting the values of Δp and t, we get:
F = Δp/t = 26663597.87/0.00131 = 2.04*10^10 N
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A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer?
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A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer?.
A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer?.
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A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer?, a detailed solution for A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer? has been provided alongside types of A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate.Correct answer is between '4711,4713'. Can you explain this answer? theory, EduRev gives you an
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