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A horizontal water jet of area 10 cm2 and velocity 5 m/s strikes on a vertical plate which is moving with velocity 2 m/s towards the nozzle.
The force acting on the plane is _____ (in N)
    Correct answer is '49'. Can you explain this answer?
    Verified Answer
    A horizontal water jet of area 10 cm2and velocity 5 m/s strikes on a v...
    Force on that vertical plate moving in the direction of jet:
    Relative velocity of jet with respect to plane
    • When plate is moving in direction of jet: vr = v – u
    • When plate is moving towards the jet: vr = v + u
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    Most Upvoted Answer
    A horizontal water jet of area 10 cm2and velocity 5 m/s strikes on a v...
    Calculation of force exerted by a water jet on a moving plate

    Given parameters:
    - Area of water jet, A = 10 cm2 = 0.001 m2
    - Velocity of water jet, Vj = 5 m/s
    - Velocity of the plate, Vp = 2 m/s

    1. Calculation of momentum flux of water jet
    - Momentum flux is the product of density, velocity and area of the fluid
    - Density of water, ρ = 1000 kg/m3
    - Momentum flux of water jet, J = ρ * Vj * A
    = 1000 * 5 * 0.001
    = 5 Ns/m2

    2. Calculation of change in momentum of water jet after striking the plate
    - Change in momentum is the product of mass flow rate and velocity difference
    - Mass flow rate, m = ρ * A * Vj
    = 1000 * 0.001 * 5
    = 5 kg/s
    - Velocity difference, ΔV = Vj - Vp
    = 5 - 2
    = 3 m/s
    - Change in momentum, ΔP = m * ΔV
    = 5 * 3
    = 15 Ns

    3. Calculation of force exerted by water jet on the plate
    - Force is the rate of change of momentum
    - Force, F = ΔP / t
    = (ΔP / Δt) * (Δt / t)
    = J * (Δt / t)
    - Δt is the time taken by water jet to strike the plate and rebound back
    - Let us assume that the rebound velocity is negligible, so Δt is the time taken by water jet to come to rest
    - Δt = Vj / g
    = 5 / 9.81
    = 0.509 s
    - Total time taken by water jet to strike and rebound is twice of Δt
    - t = 2 * Δt
    = 1.018 s
    - Force, F = 5 * (0.509 / 1.018)
    = 2.5 N

    4. Calculation of actual force exerted by water jet on the plate
    - The above calculation gives the force exerted by water jet on a stationary plate
    - In the given problem, the plate is moving towards the nozzle with a velocity of 2 m/s
    - Due to this, the actual force exerted on the plate will be higher than the calculated force
    - Let us assume that the water jet is perpendicular to the plate, so the angle between the plate and the jet is 90 degrees
    - Actual force, Factual = F * cosθ
    = 2.5 * cos90
    = 0 N (when plate is stationary)
    = 2.5 * cosθ (when plate is moving)
    - From the given problem, the plate is moving with a velocity of 2 m/s towards the nozzle
    - Let us assume that the plate is made up of a material having a coefficient of restitution of 0.8 (typical value for metals)
    - The water jet will strike the plate and rebound back with a velocity of 0.8 * 5 = 4 m/s (due to conservation of momentum)
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    A horizontal water jet of area 10 cm2and velocity 5 m/s strikes on a vertical plate which is moving with velocity 2 m/s towards the nozzle.The force acting on the plane is _____ (in N)Correct answer is '49'. Can you explain this answer?
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    A horizontal water jet of area 10 cm2and velocity 5 m/s strikes on a vertical plate which is moving with velocity 2 m/s towards the nozzle.The force acting on the plane is _____ (in N)Correct answer is '49'. 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 horizontal water jet of area 10 cm2and velocity 5 m/s strikes on a vertical plate which is moving with velocity 2 m/s towards the nozzle.The force acting on the plane is _____ (in N)Correct answer is '49'. 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 horizontal water jet of area 10 cm2and velocity 5 m/s strikes on a vertical plate which is moving with velocity 2 m/s towards the nozzle.The force acting on the plane is _____ (in N)Correct answer is '49'. Can you explain this answer?.
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