Understanding Pipe Replacement with Parallel Pipes

When designing fluid systems, sometimes a single large pipe needs to be replaced by multiple smaller pipes running in parallel. This is often done to maintain the same flow capacity and pressure drop characteristics as the original system. The core of the problem is to find the relationship between the diameter of the original large pipe (\(D\)) and the required diameter (\(d\)) of each of the \(n\) smaller parallel pipes.

Deriving the Pipe Diameter Relation

To derive this relationship, we assume that the total flow rate (\(Q_{total}\)) through the system remains the same, and importantly, the head loss (\(h_f\)) across the length of the pipe(s) is also the same in both configurations. A standard engineering approach uses the Darcy-Weisbach equation to model head loss due to friction in pipes.

The Darcy-Weisbach equation is given by:

\[h_f = f \frac{L}{D_h} \frac{V^2}{2g}\]

Where:

• \(h_f\) is the head loss due to friction.
• \(f\) is the Darcy friction factor.
• \(L\) is the length of the pipe.
• \(D_h\) is the hydraulic diameter (which is simply the pipe diameter \(D\) or \(d\) for circular pipes).
• \(V\) is the average flow velocity in the pipe.
• \(g\) is the acceleration due to gravity.

The flow rate \(Q\) through a pipe is related to velocity by \(Q = A \cdot V\), where \(A\) is the cross-sectional area (\(A = \frac{\pi D_h^2}{4}\) for a circular pipe). Thus, velocity \(V = \frac{Q}{A} = \frac{4Q}{\pi D_h^2}\).

Substituting the expression for \(V\) into the Darcy-Weisbach equation, we get head loss in terms of flow rate and diameter:

\[h_f = f \frac{L}{D_h} \frac{\left(\frac{4Q}{\pi D_h^2}\right)^2}{2g} = f \frac{L}{D_h} \frac{16Q^2}{2g \pi^2 D_h^4} = \frac{8 f L Q^2}{\pi^2 g D_h^5}\]

Applying to the Problem: Single Pipe vs. Parallel Pipes

Case 1: Single Large Pipe

Let the original pipe have diameter \(D\). The total flow rate through this pipe is \(Q_{total}\). The head loss is \(h_f\). Using the formula with \(D_h = D\) and \(Q = Q_{total}\):

\[h_f = \frac{8 f_D L Q_{total}^2}{\pi^2 g D^5}\]

(We use \(f_D\) to denote the friction factor for the large pipe).

Case 2: \(n\) Parallel Small Pipes

Each of the \(n\) parallel pipes has diameter \(d\). Since the pipes are in parallel and have the same length \(L\), the head loss \(h_f\) across each small pipe is the same as the head loss across the original single pipe. The total flow rate \(Q_{total}\) is distributed among the \(n\) pipes. Assuming equal distribution, the flow rate through each small pipe is \(Q_d = \frac{Q_{total}}{n}\).

Using the head loss formula for one small pipe with \(D_h = d\) and \(Q = Q_d = \frac{Q_{total}}{n}\):

\[h_f = \frac{8 f_d L Q_d^2}{\pi^2 g d^5} = \frac{8 f_d L \left(\frac{Q_{total}}{n}\right)^2}{\pi^2 g d^5} = \frac{8 f_d L Q_{total}^2}{n^2 \pi^2 g d^5}\]

(We use \(f_d\) for the friction factor of the small pipes).

Equating Head Loss and Solving for \(d\)

For the systems to be equivalent in terms of head loss for the same total flow, we equate the head loss expressions from Case 1 and Case 2:

\[\frac{8 f_D L Q_{total}^2}{\pi^2 g D^5} = \frac{8 f_d L Q_{total}^2}{n^2 \pi^2 g d^5}\]

In typical simplified scenarios for such questions, it is assumed that the friction factor \(f\) is the same for both the large pipe and the smaller pipes (\(f_D \approx f_d \approx f\)). This assumption simplifies the derivation significantly, although in reality, \(f\) depends on the flow conditions (Reynolds number) and pipe roughness, which would change with diameter and velocity.

Assuming \(f_D = f_d = f\), we can cancel the common terms \(\frac{8 f L Q_{total}^2}{\pi^2 g}\) from both sides:

\[\frac{1}{D^5} = \frac{1}{n^2 d^5}\]

Now, we rearrange the equation to solve for \(d\):

\[n^2 d^5 = D^5\]

\[d^5 = \frac{D^5}{n^2}\]

To find \(d\), we take the fifth root of both sides:

\[d = \left(\frac{D^5}{n^2}\right)^{\frac{1}{5}} = \frac{(D^5)^{\frac{1}{5}}}{(n^2)^{\frac{1}{5}}} = \frac{D^{5 \cdot \frac{1}{5}}}{n^{2 \cdot \frac{1}{5}}} = \frac{D^1}{n^{\frac{2}{5}}}\]

So, the relation is:

\[d = \frac{D}{n^{\frac{2}{5}}}\]

Comparing with Options

Let's check the derived formula against the given options:

• Option 1: \(d~=~\frac{D}{n^{\frac{2}{5}}}\) - This matches our derived formula.
• Option 2: \(d~=~\frac{D}{n^{{2}}}\) - This is incorrect.
• Option 3: \(d~=~\frac{D}{n^{\frac{3}{2}}}\) - This is incorrect.
• Option 4: \(d~=~\frac{D}{n}\) - This is incorrect.

The relation \(d~=~\frac{D}{n^{\frac{2}{5}}}\) is based on the principle of maintaining equivalent head loss for the same total flow rate, typically derived using the Darcy-Weisbach equation under the assumption of a constant friction factor.

Revision Table: Pipe Diameter Relations

Additional Information: Pipe Flow Assumptions

The derivation \(d = \frac{D}{n^{\frac{2}{5}}}\) relies on simplifying assumptions, primarily that the friction factor \(f\) remains constant regardless of the change in pipe diameter and velocity. In reality, the friction factor depends on the flow regime (laminar or turbulent) and the pipe's roughness. For turbulent flow, \(f\) is determined from the Moody chart or correlations like the Colebrook-White equation, which require calculating the Reynolds number (\(Re = \rho V D_h / \mu\)) and the relative roughness (pipe roughness \(\epsilon\) divided by hydraulic diameter \(D_h\)).

Since the velocity (\(V = Q/A\)) changes with diameter for a given flow rate per pipe, and the diameter itself changes, the Reynolds number will differ between the large pipe and the small pipes. This difference in Reynolds number and relative roughness means the friction factor \(f\) will likely not be identical. However, for many practical engineering problems and for the purpose of standard exam questions, the assumption of a constant or nearly constant friction factor is made to obtain a clear power-law relationship between the diameters and the number of pipes.