To calculate the discharge ($Q$) through a rectangular orifice, we use the formula:

$Q = C_d \cdot A \cdot \sqrt{2gH_{\text{avg}}}$

Where:

• $C_d$ is the coefficient of discharge ($C_d = 0.62$),
• $A$ is the area of the orifice ($A = \text{width} \cdot \text{depth}$),
• $g$ is the acceleration due to gravity ($g \approx 9.81 \, \text{m/s}^2$),
• $H_{\text{avg}}$ is the average head of water above the orifice, given by:

$H_{\text{avg}} = H_{\text{top}} + \frac{\text{depth}}{2}$

Where $H_{\text{top}}$ is the head above the top edge of the orifice.

Step-by-Step Calculation:

1. Given Data:

   • Width ($b$) = 3 m,
   • Depth ($d$) = 2 m,
   • $H_{\text{top}} = 4 \, \text{m}$,
   • $C_d = 0.62$,
   • $g = 9.81 \, \text{m/s}^2$.

2. Calculate $H_{\text{avg}}$:

$H_{\text{avg}} = 4 + \frac{2}{2} = 4 + 1 = 5 \, \text{m}.$

3. Calculate the area of the orifice ($A$):

$A = \text{width} \cdot \text{depth} = 3 \cdot 2 = 6 \, \text{m}^2.$

4. Substitute values into the discharge formula:

$Q = 0.62 \cdot 6 \cdot \sqrt{2 \cdot 9.81 \cdot 5}.$

5. Simplify the expression:

   • Calculate $\sqrt{2 \cdot 9.81 \cdot 5}$: $\sqrt{2 \cdot 9.81 \cdot 5} = \sqrt{98.1} \approx 9.9.$

   • Multiply all terms: $Q = 0.62 \cdot 6 \cdot 9.9 = 36.828 \, \text{m}^3/\text{s}.$

Final Answer:

The discharge through the rectangular orifice is approximately: $\boxed{36.83 \, \text{m}^3/\text{s}}$

Would you like a more detailed explanation or further assistance?

Here are 5 related questions for further exploration:

1. How would the discharge change if the width of the orifice was halved?
2. What would be the effect of changing the coefficient of discharge to 0.7?
3. How does the head ($H_{\text{avg}}$) influence the discharge rate?
4. Can the formula be used for circular orifices, and how would it differ?
5. How is the coefficient of discharge ($C_d$) determined in real-world scenarios?

Tip: Always ensure the water head is measured from the center of the orifice for accurate average head calculations.