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

Step 1: Set up the head loss condition.
Both pipes connect the same two reservoirs, so both pipes span the same total head difference \(H\) between the reservoirs. Since the pipes are in parallel between the same two points, the head loss due to friction is the same in both pipes: \(h_{f1} = h_{f2}\).

Step 2: Write the Darcy-Weisbach head loss in terms of discharge.
The Darcy-Weisbach equation gives head loss as
\[ h_f = \frac{f L V^2}{2 g D} \]
where \(f\) is the friction factor, \(L\) the pipe length, \(V\) the mean velocity, \(g\) gravity, and \(D\) the pipe diameter. Since \(V = \frac{Q}{A} = \frac{4Q}{\pi D^2}\), substitute for \(V\):
\[ h_f = \frac{f L}{2gD}\left(\frac{4Q}{\pi D^2}\right)^2 = \frac{8 f L Q^2}{\pi^2 g D^5} \]
So for a given \(f\) and \(L\), \(h_f\) is proportional to \(\frac{Q^2}{D^5}\).

Step 3: Apply the equal head loss condition.
Both pipes have the same length and same material (so take \(f\) as equal for both, as is standard for this type of problem), and \(h_{f1} = h_{f2}\). So:
\[ \frac{Q_1^2}{D_1^5} = \frac{Q_2^2}{D_2^5} \]
Rearranged, this gives:
\[ \frac{Q_1}{Q_2} = \left(\frac{D_1}{D_2}\right)^{5/2} \]

Step 4: Substitute the diameters.
Let pipe 1 be the bigger pipe, \(D_1 = 600\) mm, and pipe 2 the smaller pipe, \(D_2 = 400\) mm.
\[ \frac{Q_1}{Q_2} = \left(\frac{600}{400}\right)^{5/2} = (1.5)^{2.5} \]
Compute \((1.5)^{2.5}\): \((1.5)^2 = 2.25\), and \(\sqrt{1.5} = 1.2247\).
\[ (1.5)^{2.5} = 2.25 \times 1.2247 = 2.76 \]

Final Answer:
The flowrate ratio of the bigger pipe to the smaller pipe is 2.76. \[ \boxed{\frac{Q_1}{Q_2} = 2.76} \]