Fluid Mechanics and Hydraulic Machinery Miscellaneous

Fluid Mechanics and Hydraulic Machinery Miscellaneous

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Fluid Mechanics and Hydraulic Machinery

Fluid Mechanics and Hydraulic Machinery Miscellaneous
  1. A model of a hydraulic turbine is tested at a head of 1/4th of that under which the full scale turbine works. The diameter of the model is half of that of the full scale turbine. If N is the RPM of the full scale turbine, then the RPM of the model will be








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    Given data:

    Model: Hm =
    Hp
    		;Dm =
    Dp
    49

    Nm = ?
    Prototype (Full scale turbine)
    Np = N
    CH =
    H
    (ND)2

    (CH)model = (CH)prototype
    Hm
    		 =
    Hp
    Nm²Dm²Np²Dp²

    Hp/4
    		 =
    Hp
    Nm² ×(Dp²/4)Np²Dp²

    Nm = N

    Correct Option: C

    Given data:

    Model: Hm =
    Hp
    		;Dm =
    Dp
    49

    Nm = ?
    Prototype (Full scale turbine)
    Np = N
    CH =
    H
    (ND)2

    (CH)model = (CH)prototype
    Hm
    		 =
    Hp
    Nm²Dm²Np²Dp²

    Hp/4
    		 =
    Hp
    Nm² ×(Dp²/4)Np²Dp²

    Nm = N

  1. The inlet angle of runner blades of a Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant through out the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is








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    Given data:
    Bi = 90°
    Vwo = 0

    Vfi = Vfo =
    Ui
    2


    Blade efficiency of the runner,
    ηH =
    Vwiui
    		 =
    U1²
    		 ∴  Vwi = ui
    gHgH

    By energy balance equation,
    ρQgH = ρQVwiUi +
    ρQV0²
    2

    gH = Ui² +
    Vfo²
    		 Ui² +
    Ui²
    28

    gH =
    9
    		Ui²
    8

    ∴  ηH =
    Ui²
    		 =
    8
    		= 0.8888
    (9/8)Ui²9

    = 88.88% ≈ 98%

    Correct Option: D

    Given data:
    Bi = 90°
    Vwo = 0

    Vfi = Vfo =
    Ui
    2


    Blade efficiency of the runner,
    ηH =
    Vwiui
    		 =
    U1²
    		 ∴  Vwi = ui
    gHgH

    By energy balance equation,
    ρQgH = ρQVwiUi +
    ρQV0²
    2

    gH = Ui² +
    Vfo²
    		 Ui² +
    Ui²
    28

    gH =
    9
    		Ui²
    8

    ∴  ηH =
    Ui²
    		 =
    8
    		= 0.8888
    (9/8)Ui²9

    = 88.88% ≈ 98%

  1. A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1: 4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW)will be








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    PM = Pp =
    Nm
    		3
    Dm
    		5
    NpDp

    = 300
    2000
    		3
    1
    		5 = 2.34 kW
    10005

    Correct Option: A

    PM = Pp =
    Nm
    		3
    Dm
    		5
    NpDp

    = 300
    2000
    		3
    1
    		5 = 2.34 kW
    10005

  1. The head loss for a laminar incompressible flow through a horizontal circular pipe is h1. Pipe length and fluid remaining the same, if the average flow velocity doubles and the pipe diameter reduces to half its previous value, the head loss is h2. The ratio h2 /h1. is








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    Head loss, h ∝
    uavg
    D2

    ∴ 
    h2
    		 =
    D1
    		2
    uavg,2
    		 = 22 = 2
    h1D2uavg,1

    h2
    		 = 8
    h1

    Correct Option: C

    Head loss, h ∝
    uavg
    D2

    ∴ 
    h2
    		 =
    D1
    		2
    uavg,2
    		 = 22 = 2
    h1D2uavg,1

    h2
    		 = 8
    h1

  1. Three parallel pipes connected at the two ends have flow-rates Q1, Q2 and Q3 respectively, and the corresponding frictional head losses are hL1, hL2 and hL3 respectively. The correct expressions for total flow rate (Q) and frictional head loss across the two ends (hL) are








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    Total flow rate Q = Q1 + Q2 + Q3
    Head loss h = hL1 = hL2 = hL3

    Correct Option: B

    Total flow rate Q = Q1 + Q2 + Q3
    Head loss h = hL1 = hL2 = hL3