Flow Rate Calculation for a Small Hole

This problem involves calculating the volume of water flowing per unit time through a small opening in a tank. We can solve this using principles from fluid dynamics, specifically applying Torricelli's Law and accounting for real-world flow conditions with a coefficient of discharge.

Discharge Calculation Using Given Parameters

To find the rate of flow (also known as discharge, $Q$), we need to consider the following given information and formulas:

• Height of Water ($h$): The water level in the tank is $h = 3.2 \text{ m}$ above the hole.
• Area of the Hole ($A$): The cross-sectional area of the hole is given as $A = 2 \text{ mm}^2$. For calculations, we must convert this to square meters ($\text{m}^2$) since other units are in meters and seconds.

  $1 \text{ mm} = 10^{-3} \text{ m}$

  $1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$

  So, $A = 2 \times 10^{-6} \text{ m}^2$.

• Coefficient of Discharge ($C_d$): This value, $C_d = 0.75$, adjusts the theoretical flow rate to account for factors like friction and the narrowing of the water stream as it exits (vena contracta).
• Acceleration due to Gravity ($g$): We are given $g = 10 \text{ m/s}^2$.
• Formula for Discharge ($Q$): The rate of flow is calculated using the formula:

  $Q = C_d \times A \times v_{theoretical}$

  where $v_{theoretical}$ is the theoretical velocity of the water exiting the hole.
• Torricelli's Law: This law gives the theoretical velocity ($v_{theoretical}$) of fluid flowing out of an orifice under gravity.

  $v_{theoretical} = \sqrt{2gh}$

Step-by-Step Calculation of Water Discharge

Let's perform the calculation step-by-step:

1. Calculate the theoretical velocity ($v_{theoretical}$):

   Substitute the values of $g$ and $h$ into Torricelli's Law:

   $v_{theoretical} = \sqrt{2 \times (10 \text{ m/s}^2) \times (3.2 \text{ m})}$

   $v_{theoretical} = \sqrt{64 \text{ m}^2/\text{s}^2}$

   $v_{theoretical} = 8 \text{ m/s}$

2. Convert Area to Standard Units:

   As determined earlier, the area $A$ in square meters is:

   $A = 2 \times 10^{-6} \text{ m}^2$

3. Calculate the Actual Rate of Flow ($Q$):

   Now, use the discharge formula with the calculated velocity, converted area, and the given coefficient of discharge:

   $Q = C_d \times A \times v_{theoretical}$

   $Q = 0.75 \times (2 \times 10^{-6} \text{ m}^2) \times (8 \text{ m/s})$

   Multiply the numerical values:

   $Q = (0.75 \times 2 \times 8) \times 10^{-6} \text{ m}^3/\text{s}$

   $Q = (1.5 \times 8) \times 10^{-6} \text{ m}^3/\text{s}$

   $Q = 12 \times 10^{-6} \text{ m}^3/\text{s}$

Therefore, the rate of flow of water through the hole is approximately $12.0 \times 10^{-6} \text{ m}^3/\text{s}$.