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Q10.16:
The cylindrical tube of a spray pump has a cross-section of $8.0 cm^2$ one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is $1.5 m min^{-1}$, what is the speed of ejection of the liquid through the holes ?

Solution :

Step 1: Establishing Given Parameters and Unit Conversion

First, we define the known physical quantities and convert them into the International System of Units (SI) for consistency:

  • Cross-sectional area of the cylindrical tube ($A_1$): $8.0 \text{ cm}^2 = 8.0 \times 10^{-4} \text{ m}^2$ [Conversion factor: $1 \text{ cm}^2 = 10^{-4} \text{ m}^2$].
  • Velocity of liquid inside the tube ($v_1$): $1.5 \text{ m min}^{-1} = \frac{1.5}{60} \text{ m s}^{-1} = 0.025 \text{ m s}^{-1}$.
  • Number of holes ($n$): $40$.
  • Diameter of each hole ($d$): $1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}$.
  • Radius of each hole ($r$): $d/2 = 0.5 \times 10^{-3} \text{ m} = 5.0 \times 10^{-4} \text{ m}$.

Step 2: Determining Area of Ejection

The total area of the 40 holes ($A_2$) through which the liquid is ejected is the sum of the areas of the individual circular orifices:

$A_2 = n \times (\pi r^2)$

$A_2 = 40 \times \pi \times (5.0 \times 10^{-4} \text{ m})^2$

$A_2 = 40 \times 3.14159 \times 25 \times 10^{-8} \text{ m}^2$

$A_2 = 3.14159 \times 10^{-5} \text{ m}^2 \approx 3.14 \times 10^{-5} \text{ m}^2$

Step 3: Application of the Equation of Continuity

Per the Equation of Continuity for an incompressible, non-viscous fluid in steady flow, the product of cross-sectional area and fluid velocity remains constant [Per the Principle of Conservation of Mass]:

$A_1 v_1 = A_2 v_2$

Where $v_2$ is the speed of ejection of the liquid through the holes. Solving for $v_2$:

$v_2 = \frac{A_1 v_1}{A_2}$

Step 4: Algebraic Calculation

Substitute the values calculated in the previous steps:

$v_2 = \frac{(8.0 \times 10^{-4} \text{ m}^2) \times (0.025 \text{ m s}^{-1})}{3.14159 \times 10^{-5} \text{ m}^2}$

$v_2 = \frac{2.0 \times 10^{-5}}{3.14159 \times 10^{-5}} \text{ m s}^{-1}$

$v_2 \approx 0.6366 \text{ m s}^{-1}$

Visual Representation of Fluid Flow

Area A1, Velocity v1 n holes (Area A2), Velocity v2

Final Solution: The speed of ejection of the liquid through the holes is $0.64 \text{ m s}^{-1}$.


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