Question

A liquid flows under steady and incompressible flow conditions from station 1 to station 4 through pipe sections P, Q, R, and S as shown in figure. Consider, $d, V$, and $h$ represent the diameter, velocity, and head loss, respectively, in each pipe section with subscripts ‘P’, ‘Q’, ‘R’, and ‘S’. $\Delta h$ represents the head difference between the inlet (station 1) and outlet (station 4). All the pipe sections are placed on the same horizontal plane for which the figure shows the top view.
Which one of the following options is correct for the given flow loop?

• A. $\Delta h = h_P + h_Q + h_R + h_S$ and $V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 = V_S d_S^2$
• B. $\Delta h = h_P + h_Q + h_R$ and $V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 = V_S d_S^2$
• C. $\Delta h = h_P + h_Q + h_R$ and $V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 + V_S d_S^2$
• D. $\Delta h = h_P + h_Q + h_R + h_S$ and $V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 + V_S d_S^2$

Hint:

For pipe networks: - Series pipes → head losses add. - Parallel pipes → head losses are equal, and flow rates add. Always apply continuity and energy equations systematically.

Answer 1

The Correct Option is C

Step 1: Identify flow arrangement.
From the figure: - Flow enters at station 1 → section P → section Q. - Then, the flow splits into two parallel paths: R (upper path) and S (lower path). - Finally, both paths recombine at station 4 and exit. Thus, sections P and Q are in \emph{series}, while R and S are in \emph{parallel}.

Step 2: Head loss distribution.
- For series pipes: head losses add. So, \[ \Delta h = h_P + h_Q + h_{RS}. \] - For parallel pipes: the head loss through each branch is equal (since they connect the same junctions). Hence, \[ h_R = h_S. \] So the total head loss between inlet and outlet is: \[ \Delta h = h_P + h_Q + h_R. \] (Note: we could equivalently write with $h_S$ instead of $h_R$).

Step 3: Continuity equation (discharge).
- For incompressible flow: discharge is conserved. - Inlet discharge $Q = V_P A_P = V_P \frac{\pi d_P^2}{4}$. - Similarly, $V_Q d_Q^2 = V_P d_P^2$. When the flow splits: \[ V_Q d_Q^2 = V_R d_R^2 + V_S d_S^2. \] Thus, continuity condition: \[ V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 + V_S d_S^2. \]

Step 4: Match with given options.
- Head loss: $\Delta h = h_P + h_Q + h_R$. - Continuity: $V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 + V_S d_S^2$. This exactly matches **Option (C)**. Final Answer:
\[ \boxed{\Delta h = h_P + h_Q + h_R \text{and} V_P d_P^2 = V_Q d_Q^2 = V_R d_R^2 + V_S d_S^2} \]