The cylindrical tube of a spray pump has radius $R$, one end of which has $n$ fine holes, each of radius $r$.

Question

The cylindrical tube of a spray pump has radius $R$, one end of which has $n$ fine holes, each of radius $r$. If the speed of the liquid in the tube is $v$, the speed of the ejection of the liquid through the holes is:

Accepted Answer

1. **Principle:** According to the **Equation of Continuity** for an incompressible fluid, the volume flow rate (flux) entering the tube must equal the total volume flow rate exiting through the holes ($A_1 v_1 = A_2 v_2$).
   
2. **Inlet Parameters:** 
   - Radius of the tube = $R$
   - Cross-sectional area of the tube ($A_1$) = $\pi R^2$
   - Speed of liquid in the tube ($v_1$) = $v$
   - Flow rate in = $A_1 v_1 = \pi R^2 v$
3. **Outlet Parameters:**
   - Number of holes = $n$
   - Radius of each hole = $r$
   - Total cross-sectional area of $n$ holes ($A_2$) = $n 	imes \pi r^2$
   - Speed of ejection ($v_2$) = ?
   - Flow rate out = $A_2 v_2 = n \pi r^2 v_2$
4. **Calculation:**
   Equating flow rates:
   $$ \pi R^2 v = n \pi r^2 v_2 $$
   $$ v_2 = \frac{\pi R^2 v}{n \pi r^2} $$
   $$ v_2 = \frac{v R^2}{n r^2} $$

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