Question

For a fully turbulent flow ($\text{Re} > 10^5$) in a pipe of diameter $d$, with a constant pressure gradient, the volumetric flow rate ($Q$) of an incompressible fluid varies with the diameter as:

• A. $d$
• B. $d^2$
• C. $d^{2.5}$
• D. $d^4$

Hint:

In laminar flow, the flow rate scales as $Q \propto d^4$ according to the Hagen-Poiseuille equation. However, in fully turbulent flow, the scaling changes to $Q \propto d^{2.5}$ due to turbulent momentum transport.

Answer 1

The Correct Option is C

Concept: To find how the volumetric flow rate $Q$ scales with pipe diameter $d$ in fully turbulent flow under a constant pressure gradient ($\frac{\Delta P}{L} = \text{constant}$), we use the Darcy-Weisbach equation: \[ \Delta P = \frac{4 \cdot f \cdot L \cdot \rho \cdot v^2}{2 \cdot d} \] where $f$ is the Fanning friction factor, $L$ is the pipe length, $\rho$ is the fluid density, and $v$ is the average fluid velocity. The volumetric flow rate is related to velocity by the continuity equation: \[ Q = A \cdot v = \left( \frac{\pi}{4} d^2 \right) v \implies v = \frac{4Q}{\pi d^2} \]

Step 1: Analyze the behavior of the friction factor in fully turbulent flow.
For fully turbulent flow at very high Reynolds numbers ($\text{Re} > 10^5$), the flow enters the "fully rough" regime on the Moody diagram. In this regime, the friction factor $f$ becomes independent of the Reynolds number and depends only on the relative roughness ($\frac{\varepsilon}{d}$). For a smooth pipe or a constant roughness profile, $f$ can be treated as a constant.

Step 2: Express the pressure gradient in terms of volumetric flow rate $Q$.
Substitute the velocity expression $v \propto \frac{Q}{d^2}$ into the Darcy-Weisbach equation: \[ \Delta P \propto \frac{f \cdot L \cdot \rho \cdot \left( \frac{Q}{d^2} \right)^2}{d} \propto \frac{Q^2}{d^5} \] Rearranging this gives the pressure gradient expression: \[ \frac{\Delta P}{L} \propto \frac{Q^2}{d^5} \]

Step 3: Determine the scaling relationship for $Q$ with respect to $d$.
Since the pressure gradient $\frac{\Delta P}{L}$ is held constant: \[ \frac{Q^2}{d^5} = \text{constant} \implies Q^2 \propto d^5 \] Taking the square root of both sides gives: \[ Q \propto d^{5/2} \implies Q \propto d^{2.5} \] Thus, the volumetric flow rate scales with the diameter raised to the power of $2.5$.