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A sharp-edged orifice in the side of a large tank discharges water under a constant pressure head of 4 m. The orifice diameter is 50 mm and the jet diameter is 40 mm. The water jet moves through a horizontal distance of 2.25 m. The jet falls a vertical distance of 0.33 m. Calculate:

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Given:

  • Head, H=4 mH = 4 \text{ m}
  • Orifice diameter, Do=50 mm=0.05 mD_o = 50 \text{ mm} = 0.05 \text{ m}
  • Jet diameter, Dj=40 mm=0.04 mD_j = 40 \text{ mm} = 0.04 \text{ m}
  • Horizontal distance of jet, x=2.25 mx = 2.25 \text{ m}
  • Vertical distance of jet fall, y=0.33 my = 0.33 \text{ m}
  • Acceleration due to gravity, g=9.81 m/s2g = 9.81 \text{ m/s}^2

Part 3.1· 1 step

The area of the orifice.

  1. Calculate the area of the orifice (AoA_o) using its diameter.

    Ao=π4Do2A_o = \frac{\pi}{4} D_o^2 Ao=π4(0.05 m)2A_o = \frac{\pi}{4} (0.05 \text{ m})^2 Ao=π4(0.0025 m2)A_o = \frac{\pi}{4} (0.0025 \text{ m}^2) Ao≈0.0019635 m2A_o \approx 0.0019635 \text{ m}^2 Ao≈0.00196 m2A_o \approx \boxed{\text{0.00196 m}^2}

Part 3.2· 1 step

The area of the jet.

  1. Calculate the area of the jet (AjA_j) using its diameter.

    Aj=π4Dj2A_j = \frac{\pi}{4} D_j^2 Aj=π4(0.04 m)2A_j = \frac{\pi}{4} (0.04 \text{ m})^2 Aj=π4(0.0016 m2)A_j = \frac{\pi}{4} (0.0016 \text{ m}^2) Aj≈0.0012566 m2A_j \approx 0.0012566 \text{ m}^2 Aj≈0.00126 m2A_j \approx \boxed{\text{0.00126 m}^2}

Part 3.3· 1 step

The theoretical velocity of the jet.

  1. Calculate the theoretical velocity (VthV_{th}) using Torricelli's law.

    Vth=2gHV_{th} = \sqrt{2gH} Vth=2×9.81 m/s2×4 mV_{th} = \sqrt{2 \times 9.81 \text{ m/s}^2 \times 4 \text{ m}} Vth=78.48 m2/s2V_{th} = \sqrt{78.48 \text{ m}^2/\text{s}^2} Vth≈8.8589 m/sV_{th} \approx 8.8589 \text{ m/s} Vth≈8.86 m/sV_{th} \approx \boxed{\text{8.86 m/s}}

Part 3.4· 2 steps

The actual velocity of the jet.

  1. Calculate the time (tt) it takes for the jet to fall vertically.

    The vertical motion is given by y=12gt2y = \frac{1}{2}gt^2.

    t=2ygt = \sqrt{\frac{2y}{g}} t=2×0.33 m9.81 m/s2t = \sqrt{\frac{2 \times 0.33 \text{ m}}{9.81 \text{ m/s}^2}} t=0.669.81 s2t = \sqrt{\frac{0.66}{9.81} \text{ s}^2} t=0.067278 s2≈0.25938 st = \sqrt{0.067278 \text{ s}^2} \approx 0.25938 \text{ s}

  2. Calculate the actual velocity (VaV_a) using the horizontal distance and time.

    The horizontal motion is given by x=Vatx = V_a t.

    Va=xtV_a = \frac{x}{t} Va=2.25 m0.25938 sV_a = \frac{2.25 \text{ m}}{0.25938 \text{ s}} Va≈8.6748 m/sV_a \approx 8.6748 \text{ m/s} Va≈8.67 m/sV_a \approx \boxed{\text{8.67 m/s}}

Part 3.5· 1 step

The theoretical discharge.

  1. Calculate the theoretical discharge (QthQ_{th}) using the theoretical velocity and orifice area.

    Qth=AoVthQ_{th} = A_o V_{th} Qth=(0.0019635 m2)×(8.8589 m/s)Q_{th} = (0.0019635 \text{ m}^2) \times (8.8589 \text{ m/s}) Qth≈0.017389 m3/sQ_{th} \approx 0.017389 \text{ m}^3/\text{s} Qth≈0.0174 m3/sQ_{th} \approx \boxed{\text{0.0174 m}^3/\text{s}}

Part 3.6· 1 step

The actual discharge.

  1. Calculate the actual discharge (QaQ_a) using the actual velocity and jet area.

    Qa=AjVaQ_a = A_j V_a Qa=(0.0012566 m2)×(8.6748 m/s)Q_a = (0.0012566 \text{ m}^2) \times (8.6748 \text{ m/s}) Qa≈0.010902 m3/sQ_a \approx 0.010902 \text{ m}^3/\text{s} Qa≈0.0109 m3/sQ_a \approx \boxed{\text{0.0109 m}^3/\text{s}}

Part 3.7· 1 step

The coefficient of delivery (coefficient of discharge, CdC_d).

  1. Calculate the coefficient of discharge (CdC_d) using the actual and theoretical discharges.

    Cd=QaQthC_d = \frac{Q_a}{Q_{th}} Cd=0.010902 m3/s0.017389 m3/sC_d = \frac{0.010902 \text{ m}^3/\text{s}}{0.017389 \text{ m}^3/\text{s}} Cd≈0.6269C_d \approx 0.6269 Cd≈0.627C_d \approx \boxed{\text{0.627}}

Part 3.8· 1 step

The coefficient of velocity (CvC_v).

  1. Calculate the coefficient of velocity (CvC_v) using the actual and theoretical velocities.

    Cv=VaVthC_v = \frac{V_a}{V_{th}} Cv=8.6748 m/s8.8589 m/sC_v = \frac{8.6748 \text{ m/s}}{8.8589 \text{ m/s}} Cv≈0.9792C_v \approx 0.9792 Cv≈0.979C_v \approx \boxed{\text{0.979}}

Part 3.9· 1 step

The coefficient of contraction (CcC_c).

  1. Calculate the coefficient of contraction (CcC_c) using the areas of the jet and orifice.

    Cc=AjAoC_c = \frac{A_j}{A_o} Cc=0.0012566 m20.0019635 m2C_c = \frac{0.0012566 \text{ m}^2}{0.0019635 \text{ m}^2} Cc≈0.6400C_c \approx 0.6400 Cc≈0.640C_c \approx \boxed{\text{0.640}}
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