A pipe $A$ is connected with other pipes $B$ and $C$ as shown in the figure. The areas of cross-section of $A, B$ and $C$ are respectively $\alpha, \frac{\alpha}{2}$ and $\frac{\alpha}{4}$. If the velocities of flow of water through $A$ and $B$ are $10\text{ m/sec}$ and $6\text{ m/sec}$, respectively, then velocity of flow, $V_{c}$ along $C$ is
$28\text{ m/sec}$
To find the velocity of flow along pipe \( C \), we can use the principle of conservation of mass or continuity equation for incompressible flow. According to this principle, the mass flow rate must remain constant throughout the connected pipes.
The continuity equation is given by:
\(A_{1}V_{1} = A_{2}V_{2} + A_{3}V_{3}\)
Where:
- \(A_{1}, A_{2}, A_{3}\) are the cross-sectional areas of pipes \( A, B, C \) respectively.
- \(V_{1}, V_{2}, V_{3}\) are the velocities of flow through pipes \( A, B, C \) respectively.
Given:
- \(A_{1} = \alpha\)
- \(A_{2} = \frac{\alpha}{2}\)
- \(A_{3} = \frac{\alpha}{4}\)
- \(V_{1} = 10\, \text{m/sec}\)
- \(V_{2} = 6\, \text{m/sec}\)
Substitute these values into the continuity equation:
\(\alpha \times 10 = \frac{\alpha}{2} \times 6 + \frac{\alpha}{4} \times V_{3}\)
Solving for \(V_{3}\):
\(10\alpha = 3\alpha + \frac{\alpha}{4}V_{3}\)
\(7\alpha = \frac{\alpha}{4}V_{3}\)
Dividing both sides by \(\frac{\alpha}{4}\):
\(V_{3} = \frac{7\alpha \times 4}{\alpha}\)
\(V_{3} = 28\, \text{m/sec}\)
Thus, the velocity of flow along pipe \( C \) is 28 m/sec.
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