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CivilHydraulicsmediumFE ExamCoursework

Discharge Through a Sharp-Edged Orifice

Water discharges through a sharp-edged circular orifice of diameter d=50 mmd = 50\ \text{mm} in the side of a tank.

The head above the orifice centreline is H=3.0 mH = 3.0\ \text{m}, and the discharge coefficient is Cd=0.62C_d = 0.62.

Take g=9.81 m/s2g = 9.81\ \text{m/s}^2. Find the discharge.

Given
d50 mmOrifice diameter
H3 mHead above orifice
Cd0.62 -Discharge coefficient
Hint 1

Torricelli gives the ideal jet velocity from the head; C_d corrects it for contraction and losses.

Hint 2

Q = C_d·A·√(2gH).

Hint 3

Convert the diameter to metres before computing the area.

Worked solution — try the problem first

Q=CdA2gHQ = C_d A\sqrt{2gH}

Orifice area: A=πd24=π(0.050)24=π(2.5×10−3)4=1.963×10−3 m2A = \frac{\pi d^2}{4} = \frac{\pi(0.050)^2}{4} = \frac{\pi(2.5\times10^{-3})}{4} = 1.963\times10^{-3}\ \text{m}^2

Theoretical (Torricelli) velocity: V=2gH=2(9.81)(3.0)=58.86=7.672 m/sV = \sqrt{2gH} = \sqrt{2(9.81)(3.0)} = \sqrt{58.86} = 7.672\ \text{m/s}

Discharge: Q=0.62×1.963×10−3×7.672=9.34×10−3 m3/sQ = 0.62 \times 1.963\times10^{-3} \times 7.672 = 9.34\times10^{-3}\ \text{m}^3/\text{s}

Q=9.34×10−3 m3/s=9.34 L/s\boxed{Q = 9.34\times10^{-3}\ \text{m}^3/\text{s} = 9.34\ \text{L/s}}

What CdC_d absorbs. Two separate effects: the jet contracts to a vena contracta smaller than the orifice (Cc≈0.64C_c \approx 0.64), and viscous losses slightly reduce velocity (Cv≈0.97C_v \approx 0.97). Their product Cd=CcCv≈0.62C_d = C_c C_v \approx 0.62 is the value used here.

Concepts:Orifice flowTorricelli's theoremDischarge coefficient

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