Solution:

Given, flow rate Q through two parallel pipes of length L.

We need to find the length of a single pipe with the same diameter and roughness as the two parallel pipes, while keeping the total flow rate constant.

Assumptions:

- Fluid properties are constant
- Flow is steady and incompressible
- Pipes are parallel and of equal diameter and roughness
- No minor losses (e.g. due to fittings) are present

Approach:

We can use the concept of equivalent pipe to solve this problem. The equivalent pipe is a hypothetical pipe that has the same resistance to flow as the actual pipes in parallel. By replacing the two parallel pipes with a single equivalent pipe, we can simplify the analysis.

To find the equivalent pipe, we can use the concept of hydraulic diameter, which is defined as four times the cross-sectional area divided by the wetted perimeter. For a circular pipe, the hydraulic diameter is equal to the diameter.

The resistance to flow in a pipe is characterized by the Darcy-Weisbach friction factor, which depends on the Reynolds number (Re), pipe diameter (D), and pipe roughness (ε). For fully developed turbulent flow, the friction factor can be approximated by the Colebrook equation:

1 / √f = -2.0 log((ε/D)/3.7 + 2.51/(Re √f))

where f is the friction factor, ε is the pipe roughness, D is the pipe diameter, and Re is the Reynolds number.

Using the Colebrook equation, we can calculate the friction factor for the two parallel pipes, and then use the equivalent pipe concept to find the length of a single pipe with the same resistance to flow.

Calculation:

Assuming that the fluid is water at room temperature (20°C) and atmospheric pressure (101.3 kPa), with a density of 998 kg/m3 and a viscosity of 1.002 × 10-3 Pa·s, we can calculate the Reynolds number for the two parallel pipes:

Re = ρQD / µ = 4Q / (πDµ)

where ρ is the density, Q is the flow rate, D is the diameter, and µ is the viscosity.

Let's assume that the diameter of the pipes is 0.1 m (10 cm) and the roughness is 0.03 mm (30 µm), which is typical for commercial steel pipes.

For the two parallel pipes, the Reynolds number is:

Re = 4Q / (πDµ) = 4Q / (π × 0.1 × 1.002 × 10-3) = 3982Q

Using the Colebrook equation, we can solve for the friction factor:

1 / √f = -2.0 log((ε/D)/3.7 + 2.51/(Re √f))

1 / √f = -2.0 log((0.03×10-3/0.1)/3.7 + 2.51/(3982Q √f))

Solving for f using an iterative method, we get:

f = 0.0189

The total resistance to flow for the two parallel pipes is:

R = 8fL / (π2 g D4)

where L is the length of each pipe and g is the acceleration due to gravity.

Since the two pipes are