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Finding force of water jet

  1. Sep 16, 2015 #1
    This is a simple question that I should've known, but I'll ask. Fluid jet with density $ flows with velocity V1 before it strikes a block, then goes in two perpendicular directions, like left, right with the same velocity V2 in opposite directions. Find the force on the block. My try: The decrease in kinetic energy density is converted into excess pressure energy, so it should be, but I'm a silly unsure person. Initial energy/vol = $V1*V1/2, Final energy= twice $ V2*V2/2 since there are two final flows. So the difference multiplied area of impact, force. $(V*V1/2 - V2*V2)*area = force. But the correct answer, they say, is $(V1*V1 -V2*V2)* area.
  2. jcsd
  3. Sep 16, 2015 #2


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    Gold Member

    The jet transports the mass flow m' with the velocity v1. Defining the direction of the jet in the first place in x-direction, after hitting the block, all the water flows in y-direction (perpendicular, so no component in x-direction). The force (in x-direction) is the difference of the mass flow's velocity in x-direction:

    Fx = m' ⋅ (vx2 - vx1)

    m' = ρ ⋅ vx1 ⋅ A → Fx = ρ ⋅ A ⋅ (vx22 - vx12)

    vx22 = 0 → Fx = ρ ⋅ A ⋅ (0 - vx12) = - ρ ⋅ A ⋅ vx12

    That's the force on the water jet, the same force on the block, but in opposite direction: FBl = ρ ⋅ A ⋅ vx12
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