Calculating Coefficient of Discharge for an Orifice

This problem requires us to determine the coefficient of discharge for a small orifice given its diameter, the head of water above its center, and the actual discharge rate.

Given Information

• Diameter of the orifice, \( d = 20 \) mm
• Head of water over the center of the orifice, \( H = 1 \) m
• Actual discharge through the orifice, \( Q_{\text{actual}} = 0.85 \) litre/s

Goal

Find the coefficient of discharge, \( C_d \).

Required Formulas

The coefficient of discharge \( C_d \) is defined as the ratio of the actual discharge \( Q_{\text{actual}} \) to the theoretical discharge \( Q_{\text{theoretical}} \):

\( C_d = \frac{Q_{\text{actual}}}{Q_{\text{theoretical}}} \)

The theoretical discharge \( Q_{\text{theoretical}} \) is the product of the theoretical velocity \( v_{\text{theoretical}} \) and the area of the orifice \( A \):

\( Q_{\text{theoretical}} = A \times v_{\text{theoretical}} \)

The theoretical velocity \( v_{\text{theoretical}} \) through the orifice under a head \( H \) is given by Torricelli's theorem:

\( v_{\text{theoretical}} = \sqrt{2gH} \)

Where \( g \) is the acceleration due to gravity (approximately \( 9.81 \) m/s²).

The area of the circular orifice \( A \) is given by:

\( A = \frac{\pi}{4}d^2 \)

Step-by-Step Calculation

First, let's convert the given values to consistent units (SI units):

• Diameter \( d = 20 \) mm \( = 20 \times 10^{-3} \) m \( = 0.02 \) m
• Head \( H = 1 \) m (already in meters)
• Actual discharge \( Q_{\text{actual}} = 0.85 \) litre/s. Since 1 litre = \( 10^{-3} \) m³, \( Q_{\text{actual}} = 0.85 \times 10^{-3} \) m³/s.

Now, calculate the area of the orifice \( A \):

\( A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(0.02 \text{ m})^2 = \frac{\pi}{4}(0.0004 \text{ m}^2) = \pi \times 0.0001 \text{ m}^2 \approx 0.000314159 \text{ m}^2 \)

Next, calculate the theoretical velocity \( v_{\text{theoretical}} \) using \( g = 9.81 \) m/s²:

\( v_{\text{theoretical}} = \sqrt{2gH} = \sqrt{2 \times 9.81 \text{ m/s}^2 \times 1 \text{ m}} = \sqrt{19.62 \text{ m}^2/\text{s}^2} \approx 4.42945 \text{ m/s} \)

Now, calculate the theoretical discharge \( Q_{\text{theoretical}} \):

\( Q_{\text{theoretical}} = A \times v_{\text{theoretical}} \approx (0.000314159 \text{ m}^2) \times (4.42945 \text{ m/s}) \approx 0.0013913 \text{ m}^3/\text{s} \)

Finally, calculate the coefficient of discharge \( C_d \):

\( C_d = \frac{Q_{\text{actual}}}{Q_{\text{theoretical}}} = \frac{0.85 \times 10^{-3} \text{ m}^3/\text{s}}{0.0013913 \text{ m}^3/\text{s}} = \frac{0.00085}{0.0013913} \approx 0.611 \)

Rounding the result to two decimal places, the coefficient of discharge is approximately \( 0.61 \).

Summary of Results

The calculated coefficient of discharge is approximately 0.61.

Revision Table: Orifice Discharge Concepts

Additional Information: Orifice Coefficients

The coefficient of discharge \( C_d \) is a crucial parameter for real-world fluid flow calculations through orifices. It accounts for two main factors:

1. Contraction of the Jet (Vena Contracta): As fluid flows through the orifice, the streamlines converge, and the jet constricts to a minimum area slightly downstream of the orifice plane. This minimum area is called the vena contracta. The ratio of the area of the vena contracta to the area of the orifice is called the coefficient of contraction, \( C_c \). \( C_c = \text{Area of vena contracta} / \text{Area of orifice} \).
2. Losses in Velocity: Due to friction and other energy losses, the actual velocity at the vena contracta is slightly less than the theoretical velocity. The ratio of the actual velocity at the vena contracta to the theoretical velocity is called the coefficient of velocity, \( C_v \). \( C_v = \text{Actual velocity} / \text{Theoretical velocity} \).

The coefficient of discharge is related to the coefficient of contraction and the coefficient of velocity by the equation:

\( C_d = C_c \times C_v \)

For a sharp-edged orifice, typical values are \( C_c \approx 0.60 \) to \( 0.65 \), \( C_v \approx 0.95 \) to \( 0.99 \), resulting in \( C_d \approx 0.57 \) to \( 0.65 \). The calculated value of 0.61 in this problem falls within the typical range for \( C_d \).