Question

Two reservoirs having different water levels are connected by two long parallel pipelines of same length and same material but having diameters of 600 mm and 400 mm. Using Darcy-Weisbach equation, the ratio of flowrate of water in the bigger diameter pipe to that in the smaller diameter pipe is

• A. 0.54
• B. 1.22
• C. 1.84
• D. 2.76

Hint:

Both pipes see the same head difference between the two reservoirs, so equate their Darcy-Weisbach head losses and solve for the discharge ratio in terms of the diameter ratio.

Answer 1

The Correct Option is D

Two parallel pipes joining the same pair of reservoirs must produce identical head loss, because both ends sit at the same two water levels. We use this equal head loss rule with the Darcy-Weisbach formula to compare the two discharges directly, without ever finding the actual head difference.

Write the Darcy-Weisbach head loss as $h_f = \frac{fLV^2}{2gD}$, and replace the velocity $V$ with discharge using $V = Q / A = 4Q/(\pi D^2)$:

\[ h_f = \frac{8fLQ^2}{\pi^2 g D^5} \]

For a fixed pair of reservoirs, both pipes are "long" (so minor losses are ignored), made of the "same material" (so the friction factor $f$ can be treated as identical for both), and have the "same length" $L$. That leaves $h_f \propto Q^2/D^5$ as the only quantity that varies from pipe to pipe.

Since $h_{f,1} = h_{f,2}$ for the two parallel pipes:

\[ \frac{Q_1^2}{D_1^5} = \frac{Q_2^2}{D_2^5} \quad \Rightarrow \quad \frac{Q_1}{Q_2} = \left(\frac{D_1}{D_2}\right)^{5/2} \]

Now plug in the diameters, taking pipe 1 as the 600 mm (bigger) pipe and pipe 2 as the 400 mm (smaller) pipe:

\[ \frac{D_1}{D_2} = \frac{600}{400} = 1.5 \]

Raise this ratio to the power $5/2$. Splitting the exponent as $2 + 0.5$ makes the arithmetic easy: $1.5^2 = 2.25$, and $1.5^{0.5} = \sqrt{1.5} \approx 1.2247$. Multiplying these:

\[ \frac{Q_1}{Q_2} = 2.25 \times 1.2247 \approx 2.76 \]

So the bigger 600 mm pipe carries about 2.76 times the discharge of the smaller 400 mm pipe, purely because a larger diameter gives a much bigger flow area and much lower resistance per unit flow, even under the same driving head. This matches option (D).

\[ \boxed{2.76} \]