A pump and its horizontal intake pipe are located 12 m beneath the surface of a reservoir . The speed of the water in the intake pipe causes the pressure there to decrease , in accord with Bernoulli's principle . What is the maximum speed with which water can flow through the intake pipe (assuming non-viscous flow ) ?

Options

A. 15 m/s
B. 18 m/s
C. 21 m/s
D. 24 m/s

Answer

To solve the problem of finding the maximum speed with which water can flow through the intake pipe, we will apply Bernoulli's principle. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a pump and an intake pipe located 12 meters below the surface of a reservoir. The pressure at the surface of the reservoir is atmospheric pressure, and we need to find the maximum speed of water in the intake pipe.

### Step 2: Apply Bernoulli's Equation
According to Bernoulli's principle, for two points along a streamline, the following equation holds:

\[ P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

Where:
- \( P \) = pressure
- \( \rho \) = density of the fluid (water, in this case)
- \( g \) = acceleration due to gravity
- \( h \) = height above a reference level
- \( v \) = velocity of the fluid

### Step 3: Define the Points
- Point 1: At the surface of the reservoir
- Point 2: At the intake pipe

At the surface (Point 1):
- Pressure \( P_1 = P_{\text{atm}} \) (atmospheric pressure)
- Height \( h_1 = 0 \) m (taking the surface as the reference level)
- Velocity \( v_1 \approx 0 \) m/s (the reservoir is large, so the velocity is negligible)

At the intake (Point 2):
- Pressure \( P_2 \) (unknown, but we will find it)
- Height \( h_2 = -12 \) m (12 m below the reference level)
- Velocity \( v_2 \) (this is what we want to find)

### Step 4: Substitute into Bernoulli's Equation
Substituting the known values into Bernoulli's equation gives:

\[ P_{\text{atm}} + \rho g (0) + \frac{1}{2} \rho (0)^2 = P_2 + \rho g (-12) + \frac{1}{2} \rho v_2^2 \]

This simplifies to:

\[ P_{\text{atm}} = P_2 - 12 \rho g + \frac{1}{2} \rho v_2^2 \]

### Step 5: Find Maximum Speed Condition
To find the maximum speed, we assume that the pressure \( P_2 \) at the intake pipe is at its minimum (which can be considered as zero gauge pressure in a vacuum scenario). Thus, we set \( P_2 = 0 \):

\[ P_{\text{atm}} = -12 \rho g + \frac{1}{2} \rho v_2^2 \]

Rearranging gives:

\[ \frac{1}{2} \rho v_2^2 = P_{\text{atm}} + 12 \rho g \]

### Step 6: Substitute Known Values
Using:
- \( P_{\text{atm}} = 1.01 \times 10^5 \, \text{Pa} \)
- \( \rho = 1000 \, \text{kg/m}^3 \)
- \( g = 9.81 \, \text{m/s}^2 \)

Substituting these values:

\[ \frac{1}{2} \cdot 1000 \cdot v_2^2 = 1.01 \times 10^5 + 12 \cdot 1000 \cdot 9.81 \]

Calculating the right-hand side:

\[ \frac{1}{2} \cdot 1000 \cdot v_2^2 = 1.01 \times 10^5 + 117720 \]
\[ \frac{1}{2} \cdot 1000 \cdot v_2^2 = 218720 \]

### Step 7: Solve for \( v_2 \)
Now, we can solve for \( v_2 \):

\[ 500 v_2^2 = 218720 \]
\[ v_2^2 = \frac{218720}{500} = 437.44 \]
\[ v_2 = \sqrt{437.44} \approx 20.92 \, \text{m/s} \]

### Conclusion
The maximum speed with which water can flow through the intake pipe is approximately **21 m/s**.

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