Water is maintained at a height of 10 min a tank. Calculate the diameter of orifice needed at the base of the tank to discharge water at the rate of `26.4 m^(3)` per minute.

Answer

To solve the problem of calculating the diameter of the orifice needed at the base of the tank to discharge water at a rate of 26.4 m³ per minute, we can follow these steps:

### Step 1: Convert the discharge rate from cubic meters per minute to cubic meters per second.
Given:
- Discharge rate \( Q = 26.4 \, \text{m}^3/\text{min} \)

To convert this to cubic meters per second:
\[ Q = \frac{26.4 \, \text{m}^3}{60 \, \text{s}} = 0.44 \, \text{m}^3/\text{s} \]

### Step 2: Use the equation of continuity.
The equation of continuity states that the discharge \( Q \) can be expressed as:
\[ Q = A \cdot v \]
where \( A \) is the area of the orifice and \( v \) is the velocity of water exiting the orifice.

### Step 3: Calculate the velocity of water exiting the orifice using Torricelli's theorem.
According to Torricelli's theorem, the velocity \( v \) of water flowing out of an orifice is given by:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)) and \( h \) is the height of the water column above the orifice (10 meters in this case).

Substituting the values:
\[ v = \sqrt{2 \cdot 9.8 \, \text{m/s}^2 \cdot 10 \, \text{m}} = \sqrt{196} \approx 14 \, \text{m/s} \]

### Step 4: Substitute \( v \) back into the equation of continuity to find the area \( A \).
From the equation of continuity:
\[ A = \frac{Q}{v} \]
Substituting the values we have:
\[ A = \frac{0.44 \, \text{m}^3/\text{s}}{14 \, \text{m/s}} \approx 0.0314 \, \text{m}^2 \]

### Step 5: Calculate the radius \( r \) of the orifice.
The area \( A \) of the orifice can also be expressed in terms of the radius \( r \):
\[ A = \pi r^2 \]
Setting the two expressions for \( A \) equal gives:
\[ \pi r^2 = 0.0314 \]
Solving for \( r \):
\[ r^2 = \frac{0.0314}{\pi} \approx 0.01 \quad \Rightarrow \quad r \approx 0.1 \, \text{m} \]

### Step 6: Calculate the diameter \( d \) of the orifice.
The diameter \( d \) is twice the radius:
\[ d = 2r = 2 \times 0.1 \, \text{m} = 0.2 \, \text{m} \]

### Final Answer:
The diameter of the orifice needed at the base of the tank is:
\[ \boxed{0.2 \, \text{m}} \]