A tank cotaining water has an orifice 10 m below the surface of water in the tank . If there is no wastage of energy ,find the speed of discharge.

Answer

To find the speed of discharge of water from an orifice located 10 meters below the surface of water in a tank, we can use the principle of conservation of energy, specifically Bernoulli's equation. Here’s a step-by-step solution:

### Step 1: Understand the setup
We have a tank filled with water, and there is an orifice (a small opening) located 10 meters below the surface of the water. We need to find the speed at which water discharges from this orifice.

### Step 2: Apply Bernoulli's equation
According to Bernoulli's principle, the total mechanical energy along a streamline is constant if there is no energy loss. The equation can be expressed as:

\[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]

Where:
- \( P \) = pressure at the point (at the surface of the water)
- \( \rho \) = density of the fluid (water)
- \( v \) = velocity of the fluid at that point
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height above a reference point

### Step 3: Set up the equation for the two points
1. At the surface of the water (Point 1):
- Pressure \( P_0 \) (atmospheric pressure)
- Velocity \( v_1 = 0 \) (the water is at rest)
- Height \( h_1 = 10 \, \text{m} \)

2. At the orifice (Point 2):
- Pressure \( P_2 = P_0 \) (atmospheric pressure)
- Velocity \( v_2 = v \) (the velocity we want to find)
- Height \( h_2 = 0 \, \text{m} \)

### Step 4: Write Bernoulli's equation for both points
At Point 1:
\[ P_0 + \frac{1}{2} \rho (0)^2 + \rho g (10) = P_0 + \rho g (10) \]

At Point 2:
\[ P_0 + \frac{1}{2} \rho v^2 + \rho g (0) = P_0 + \frac{1}{2} \rho v^2 \]

### Step 5: Equate the two points
Since there is no energy loss, we can equate the total energy at both points:

\[ P_0 + \rho g (10) = P_0 + \frac{1}{2} \rho v^2 \]

### Step 6: Simplify the equation
Cancel \( P_0 \) from both sides:

\[ \rho g (10) = \frac{1}{2} \rho v^2 \]

### Step 7: Cancel \( \rho \) from both sides
Assuming \( \rho \) is not zero, we can divide both sides by \( \rho \):

\[ g (10) = \frac{1}{2} v^2 \]

### Step 8: Solve for \( v^2 \)
Multiply both sides by 2:

\[ 20g = v^2 \]

### Step 9: Substitute the value of \( g \)
Using \( g = 9.81 \, \text{m/s}^2 \):

\[ v^2 = 20 \times 9.81 \]

### Step 10: Calculate \( v^2 \)
\[ v^2 = 196.2 \]

### Step 11: Take the square root to find \( v \)
\[ v = \sqrt{196.2} \approx 14.0 \, \text{m/s} \]

### Final Answer
The speed of discharge of water from the orifice is approximately **14.0 m/s**.

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