Cavitation in Centrifugal Pump (Part-7) | How Cavitation occurs in Centrifugal Pump?

แชร์
ฝัง
  • เผยแพร่เมื่อ 3 พ.ย. 2024

ความคิดเห็น •

  • @Coreengineers
    @Coreengineers  2 ปีที่แล้ว +1

    Watch Centrifugal Pump all videos:👇 th-cam.com/play/PLrqNx2OqGF513HW1AN7M9sw0kTqnO1ESX.html

  • @MohsenRezaeimanesh
    @MohsenRezaeimanesh 10 หลายเดือนก่อน +1

    The type of impeller is open impeller

  • @arsalmash2812
    @arsalmash2812 ปีที่แล้ว +1

    Totally wrong how if you increase the area , pressure increases. Wrong concept

    • @Coreengineers
      @Coreengineers  ปีที่แล้ว

      Don't go by equations, but understand the intuition behind it.
      The pressure is not decided by the area in true sense, but by the force. Infact both are similar quantities. The reason pressure may be used instead of force is because in case of fluids, the force may not be localized but spread over an area 'due to fluidity of fluids'. If I keep an icecube on a plate it will exert a force, but as the ice melts to water, it spreads to occupy as much area of the plate. Obviously the height of cube has diminished causing lesser mass in a locality of the plate, so less force in that locality. But instead of just force, had we considered ratio of force to plate area, then melting process wouldn't change this quantity. Hence we define this ratio as pressure, and use it instead of force, where fluids are involved instead of rigid bodies.
      With above understanding, let us consider a cylindrical vessel of base area A, containing water up to height h. If hypothetically I could cut the vessel vertically in two parts, each having a semi-circular base of area A/2. Just using formula P=F/A, you may conclude the pressure must double since area is halved, but here the pressure remains the same. Why? Because the force too has gone half due to only half the water remaining in each vessel part.
      Coming to Bernoulli's principle -
      Pressure + Kinetic Energy + Potential Energy = constant, for flow of fluid through a pipe.
      Assuming pipe continues to remain at same height, no change of potential Energy, so -
      Consequently we may believe that reducing the area of tube, caused pressure to decrease. This seems to be a contradiction as per out original pressure formula P=F/A. But as was discussed earlier, Area independently cannot decide the pressure, so we need to consider force. What is the force in this tube scenario?
      The force is provided by the source of water motion, say a pump, or gravity (if water tank at top of building), which remains constant. This force is transmitted through water molecules to every point on the walls of the tube. With water jet ozzing out of tube faster(high energy), the force transmitted to tube wall is lesser. Hence the force is lesser causing the pressure to lessen. Thus when we consider force into picture, the apparent contradiction resolves. In summary, when talking of pressure, remember to take force into account