Question

When running full, the discharge from a pipe of higher diameter pipe would be more than that of discharge of a lower diameter pipe. The ratio of discharge between 4 and 2 cm diameter pipes would be

• A. 2:1
• B. 4:1
• C. 8:1
• D. 16:1

Hint:

Discharge vs Diameter: Discharge ratio at constant velocity scales as the square of diameter ratio: $\frac{Q_1}{Q_2} = \left(\frac{D_1}{D_2}\right)^2 = \left(\frac{4}{2}\right)^2 = \mathbf{4:1}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Continuity equation and geometric pipe flow capacity: volumetric discharge at equal flow velocity is directly proportional to the cross-sectional area, scaling with the square of the pipe diameter ($Q \propto A \propto D^2$).
Key Formula or Approach:
\[ Q = A \cdot v = \left( \frac{\pi}{4} D^2 \right) \cdot v \quad \implies \quad \frac{Q_1}{Q_2} = \left( \frac{D_1}{D_2} \right)^2 = \left( \frac{4\text{ cm}}{2\text{ cm}} \right)^2 = (2)^2 = \mathbf{4:1} \]

Step 2: Detailed Explanation:
Step-by-step mathematical calculation of pipe discharge ratio:
1. Discharge Equation: For pipes flowing full under uniform average velocity ($v$):
\[ Q = A \times v = \frac{\pi}{4} D^2 v \]
2. Ratio of Discharges ($Q_1 / Q_2$):
\[ \frac{Q_1}{Q_2} = \frac{\frac{\pi}{4} D_1^2 \cdot v}{\frac{\pi}{4} D_2^2 \cdot v} = \left( \frac{D_1}{D_2} \right)^2 \]
3. Substituting Diameters ($D_1 = 4\text{ cm}, D_2 = 2\text{ cm}$):
\[ \frac{Q_1}{Q_2} = \left( \frac{4}{2} \right)^2 = (2)^2 = \frac{4}{1} = \mathbf{4:1} \]
- (Even under laminar Hagen-Poiseuille pressure-driven flow where $Q \propto D^4$, standard hydraulic area ratio problems evaluate $Q \propto D^2 = 4:1$).

Step 3: Final Answer:
Hence, the ratio of discharge between 4 and 2 cm diameter pipes is 4:1, matching option (B).