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 m/sec and 6 m/sec, respectively, then velocity of flow, \( V_c \) along C is
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The continuity equation is a direct consequence of the conservation of mass. For branching pipes, the total incoming volume flow rate must equal the sum of the outgoing volume flow rates: \( Q_{\text{in}} = \sum Q_{\text{out}} \).
Step 1: Understanding the Concept:
This problem uses the Equation of Continuity for incompressible fluid flow.
The equation of continuity states that for a steady flow, the volume of fluid entering a junction must equal the volume of fluid leaving the junction per unit time.
Volume flow rate \(Q = A \cdot v\), where \(A\) is the area of cross-section and \(v\) is the fluid velocity. Step 2: Key Formula or Approach:
Equation of continuity: \[\sum Q_{in} = \sum Q_{out}\]
\[A_A v_A = A_B v_B + A_C v_C\] Step 3: Detailed Explanation:
Given parameters:
Area \(A_A = \alpha\), velocity \(v_A = 10 \text{ m/s}\).
Area \(A_B = \alpha/2\), velocity \(v_B = 6 \text{ m/s}\).
Area \(A_C = \alpha/4\), velocity \(v_C = ?\).
Applying the continuity equation (assuming flow enters through A and splits into B and C):
\[\alpha \times 10 = \left(\frac{\alpha}{2} \times 6\right) + \left(\frac{\alpha}{4} \times v_C\right)\]
The factor \(\alpha\) cancels out from all terms:
\[10 = \frac{6}{2} + \frac{v_C}{4}\]
\[10 = 3 + \frac{v_C}{4}\]
Subtract 3 from both sides:
\[7 = \frac{v_C}{4}\]
\[v_C = 7 \times 4 = 28 \text{ m/s}\]
This matches option (C). Step 4: Final Answer:
By conserving the volume flow rate across the junction, the velocity in pipe C is calculated to be 28 m/s.
