A pump is designed as a horizontal cylinder with a piston of area `A` and and outlet orifice of area `a` arranged near the cylinder axis. Find the velocity of out flow of the liquid from the pump if the piston moves with a constant velocity under the action of a constant force F. The density of the liquid is `rho`.

Explanation

To find the velocity of outflow of the liquid from the pump, we can follow these steps:

### Step 1: Define the pressures at the two sections
- Let \( P_1 \) be the pressure just behind the piston (Section 1) and \( P_2 \) be the pressure at the outlet orifice (Section 2).
- The pressure at Section 1 can be expressed as:
\[ P_1 = P_a + \frac{F}{A} \]
where \( P_a \) is the atmospheric pressure, \( F \) is the constant force applied, and \( A \) is the area of the piston.

- The pressure at Section 2 is simply the atmospheric pressure:
\[ P_2 = P_a \]

### Step 2: Apply the continuity equation
- According to the continuity equation, the mass flow rate at Section 1 must equal the mass flow rate at Section 2:
\[ \dot{m}_1 = \dot{m}_2 \]
- This can be expressed as:
\[ \rho A v_1 = \rho a v_2 \]
where \( v_1 \) is the velocity of the piston and \( v_2 \) is the velocity of the outflow.
- We can simplify this to:
\[ A v_1 = a v_2 \]
- Rearranging gives:
\[ v_1 = \frac{a}{A} v_2 \]

### Step 3: Apply Bernoulli's equation
- According to Bernoulli's theorem, we can write:
\[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} \]
- Substituting for \( P_1 \) and \( P_2 \):
\[ \frac{P_a + \frac{F}{A}}{\rho g} + \frac{v_1^2}{2g} = \frac{P_a}{\rho g} + \frac{v_2^2}{2g} \]
- This simplifies to:
\[ \frac{F}{A \rho g} + \frac{v_1^2}{2g} = \frac{v_2^2}{2g} \]

### Step 4: Substitute \( v_1 \) into Bernoulli's equation
- Substitute \( v_1 = \frac{a}{A} v_2 \) into the equation:
\[ \frac{F}{A \rho g} + \frac{1}{2g} \left(\frac{a}{A} v_2\right)^2 = \frac{v_2^2}{2g} \]
- This leads to:
\[ \frac{F}{A \rho g} + \frac{a^2}{2A^2 g} v_2^2 = \frac{v_2^2}{2g} \]

### Step 5: Rearranging the equation
- Rearranging gives:
\[ \frac{F}{A \rho g} = \frac{v_2^2}{2g} - \frac{a^2}{2A^2 g} v_2^2 \]
- Factoring out \( v_2^2 \):
\[ \frac{F}{A \rho g} = \frac{v_2^2}{2g} \left(1 - \frac{a^2}{A^2}\right) \]

### Step 6: Solve for \( v_2^2 \)
- Multiply both sides by \( \frac{2g}{(1 - \frac{a^2}{A^2})} \):
\[ v_2^2 = \frac{2F}{\rho A} \cdot \frac{1}{(1 - \frac{a^2}{A^2})} \]

### Step 7: Take the square root to find \( v_2 \)
- Taking the square root gives:
\[ v_2 = \sqrt{\frac{2F}{\rho A} \cdot \frac{1}{(1 - \frac{a^2}{A^2})}} \]

### Step 8: Special case when \( a \) is negligible
- If \( a \) is very small compared to \( A \), we can approximate:
\[ v_2 \approx \sqrt{\frac{2F}{\rho A}} \]

### Final Answer
Thus, the velocity of outflow of the liquid from the pump is:
\[ v_2 = \sqrt{\frac{2F}{\rho A} \cdot \frac{1}{(1 - \frac{a^2}{A^2})}} \]
or approximately:
\[ v_2 \approx \sqrt{\frac{2F}{\rho A}} \quad \text{(if \( a \) is negligible)} \]