Question

Applying a pressure drop across a capillary results in a volumetric flow rate ‘Q' under laminar flow conditions. The flow rate, for the same pressure drop, in a capillary of the same length but half the radius is:

• A. \(\frac{Q}{2}\)
• B. \(\frac{Q}{4}\)
• C. \(\frac{Q}{8}\)
• D. \(\frac{Q}{16}\)

Hint:

For fluid flow problem types involving circular channels, always remember that volumetric flow rate is highly sensitive to radius changes: - Flow rate is directly proportional to the fourth power of the radius (\( Q \propto R^4 \)). - Halving the radius (\( \frac{1}{2} \)) drops flow capacity by factor of \( (\frac{1}{2})^4 = \frac{1}{16} \). - Doubling the radius (\( 2 \)) multiplies flow capacity by factor of \( 2^4 = 16 \).

Answer 1

The Correct Option is D

Concept: The steady, laminar flow of an incompressible fluid through a long cylindrical pipe or capillary tube of uniform cross-section is governed precisely by Hagen-Poiseuille's Law. This fundamental law describes how physical attributes like pressure difference, geometry of the tube (length and radius), and fluid properties (viscosity) affect the total volume of fluid passing through a cross-section per unit time. Key mathematical variables used in Hagen-Poiseuille's equation:
• Volumetric flow rate represented by \( Q \)
• Pressure drop or pressure difference along the length of the capillary represented by \( \Delta P \)
• Internal radius of the capillary tube represented by \( R \)
• Absolute or dynamic fluid viscosity represented by \( \mu \)
• Length of the capillary tube along the axis of flow represented by \( L \) The direct equation formulation establishing this relationship is: \[ Q = \frac{\pi \cdot \Delta P \cdot R^4}{8 \cdot \mu \cdot L} \] By evaluating this proportional structure, we can clearly understand how modifications to any geometrical factor dramatically alter the total system behavior.

Step 1: Establishing the initial configuration parameter state.
Let us consider the initial capillary setup where the physical attributes are defined precisely as follows:
• Initial radius of the original capillary tube = \( R_1 = R \)
• Initial length of the original capillary tube = \( L_1 = L \)
• Given pressure drop across this system = \( \Delta P_1 = \Delta P \)
• Dynamic viscosity of the running fluid = \( \mu \) Substituting these direct initial state variables explicitly into Hagen-Poiseuille's standard governing equation gives us our baseline reference flow rate, \( Q_1 \): \[ Q_1 = Q = \frac{\pi \cdot \Delta P \cdot R^4}{8 \cdot \mu \cdot L} \quad \cdots (1) \]

Step 2: Defining the second modified system parameter conditions.
Now, we look closely at the problem statement to define the variables for the altered capillary configuration. The question specifies that we are operating under the exact same pressure drop, utilizing a capillary of the exact same length, but containing a radius that is cut exactly in half. Thus, we can express the new secondary configuration variables mathematically as:
• New operational radius = \( R_2 = \frac{R}{2} \)
• New operational length = \( L_2 = L \) (since length remains identical)
• New operational pressure drop = \( \Delta P_2 = \Delta P \) (since pressure drop remains identical) Let the corresponding new volumetric flow rate resulting from these physical modifications be denoted as \( Q_2 \).

Step 3: Formulating the new expression for modified volumetric flow rate.
By systematically substituting our updated parameters from Step 2 directly back into the core governing physics equation, we establish the explicit relationship for \( Q_2 \): \[ Q_2 = \frac{\pi \cdot \Delta P_2 \cdot (R_2)^4}{8 \cdot \mu \cdot L_2} \] Now substitute the exact proportional values relative to the original state (\( \Delta P_2 = \Delta P \), \( L_2 = L \), and \( R_2 = \frac{R}{2} \)): \[ Q_2 = \frac{\pi \cdot \Delta P \cdot \left(\frac{R}{2}\right)^4}{8 \cdot \mu \cdot L} \]

Step 4: Expanding the geometric exponential term mathematically.
Let us carefully apply algebraic distribution rules to expand the fourth-power term containing the modified capillary radius: \[ \left(\frac{R}{2}\right)^4 = \frac{R^4}{2^4} = \frac{R^4}{16} \] Substitute this newly expanded exponential fraction back directly into our equation framework for the second flow rate \( Q_2 \): \[ Q_2 = \frac{\pi \cdot \Delta P \cdot \left(\frac{R^4}{16}\right)}{8 \cdot \mu \cdot L} \] By moving the scalar denominator value of 16 cleanly outside the main fraction group, we can rewrite the expression as: \[ Q_2 = \frac{1}{16} \times \left( \frac{\pi \cdot \Delta P \cdot R^4}{8 \cdot \mu \cdot L} \right) \quad \cdots (2) \]

Step 5: Substituting the initial base reference equation to find the final value.
If we carefully compare equation (2) with our baseline reference layout established in equation (1), we can immediately observe that the complex algebraic grouping inside the parentheses perfectly matches our original flow rate \( Q \): \[ Q = \frac{\pi \cdot \Delta P \cdot R^4}{8 \cdot \mu \cdot L} \] Replacing that parenthetical term in equation (2) directly with the variable symbol \( Q \) gives us: \[ Q_2 = \frac{1}{16} \cdot Q = \frac{Q}{16} \] This mathematically demonstrates that shrinking the radius to half its original scale reduces the fluid volume flow rate down to precisely one-sixteenth of its initial value, which corresponds directly with Option (D).