Question

Two pipe lines of equal length and diameters of 10 cm and 40 cm are connected in parallel between two reservoirs. If friction factor f is same for both the pipes, the ratio of the discharges in the larger to the smaller pipe is
1. 4
2. 16
3. 32
4. 64

Answer 1

Correct Answer - Option 3 : 32

Concept:

Flow-through Pipes: In flow-through pipe different type of losses takes place.

Major losses are due to viscous (friction) resistance through the wall, and head loss due to friction is given by Darcy's Weichback equation: 

​ \(h_f =\frac{fLV^2}{2gD} = \frac{fLQ^2}{12.1\ D^5}\) 

where, f = friction factor, L = length of pipe, D = diameter of the pipe, Q = pipe flow

Minor losses these losses are due to change in momentum of the fluid due to the sudden expansion or contraction, change in the shape of the pipe.

Compound Pipes: Types of pipe connection are:

In Series Connection of pipe discharge is same and losses are different in each pipe.

• Q₁ = Q₂ = Q₃ = ................ = Q
• Equivalent head h_f = h_f1 + h_f2 + hf2 ....

In Parallel Connection of pipe discharge is different and losses are same in each pipe.

• Q = Q₁ + Q2 + Q₃ ..... + Q
• Equivalent head hf = hf1 = hf2 = hf2 ......

Calculation:

Given:

D₁ = 10 cm, D₂ = 40 cm , same length L₁ = L₂ = L, same friction factor f₁ = f₂ = f

In parallel Connection Head loss in pipe are equal:

h_f1 = hf2

\(\frac{fLQ_1^2}{12.1\ D_1^5} = \frac{fLQ_2^2}{12.1\ D_2^5}\) ⇒ \(\frac{Q_1^2}{(10)^5} = \frac{Q_2^2}{(40)^5}\)

\(\frac{Q_2^2}{Q_1^2} = \frac{(40)^5}{(10)^5}\)

\(\frac{Q_2}{Q_1} = 32\)