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Question

When the coefficient of rugosity is increased from 0.01 to 0.02, the gradient of a pipe of a given diameter to carry the same flow at the same velocity should be

The correct answer is

increased by 4 times

Understanding Rugosity and Pipe Gradient

The question asks how the gradient of a pipe changes when the coefficient of rugosity increases, while the diameter, flow rate, and velocity remain constant. This problem relates to empirical formulas used for calculating flow in pipes or open channels, where the coefficient of rugosity (often denoted by 'n' in Manning's formula) accounts for the roughness of the pipe material.

Applying Manning's Formula to Pipe Flow

Manning's formula is commonly used to estimate flow velocity in open channels, but it can also be applied to pipes flowing full. The formula for velocity (V) is:

$$V = \frac{1}{n} R^{2/3} S^{1/2}$$

Where:

  • \(V\) is the flow velocity
  • \(n\) is the coefficient of rugosity
  • \(R\) is the hydraulic radius
  • \(S\) is the hydraulic gradient (slope of the energy line or bed slope for uniform flow)

For a circular pipe flowing full, the hydraulic radius \(R\) is given by the ratio of the cross-sectional area to the wetted perimeter. For a pipe of diameter \(D\), the area is \(\frac{\pi D^2}{4}\) and the wetted perimeter is \(\pi D\). Thus, the hydraulic radius is:

$$R = \frac{\text{Area}}{\text{Wetted Perimeter}} = \frac{\frac{\pi D^2}{4}}{\pi D} = \frac{D}{4}$$

Substituting \(R = D/4\) into Manning's formula for velocity:

$$V = \frac{1}{n} \left(\frac{D}{4}\right)^{2/3} S^{1/2}$$

Analyzing the Relationship between Gradient and Rugosity

The question states that the pipe diameter (\(D\)), flow rate (\(Q\)), and velocity (\(V\)) are constant. Since \(Q = A \times V\) and the area \(A\) is constant for a fixed diameter \(D\), keeping \(V\) constant means the left side of the Manning's velocity equation remains unchanged.

Let's rearrange the equation to see how \(S\) relates to \(n\) when \(V\) and \(D\) are constant:

$$V \cdot n = \left(\frac{D}{4}\right)^{2/3} S^{1/2}$$

Since \(V\) and \(D\) are constant, the term \(V \left(\frac{D}{4}\right)^{-2/3}\) is a constant. Let's call this constant \(K\).

$$K \cdot n = S^{1/2}$$

Squaring both sides to solve for \(S\):

$$S = (K \cdot n)^2 = K^2 \cdot n^2$$

This equation shows that the hydraulic gradient \(S\) is directly proportional to the square of the coefficient of rugosity \(n\), assuming velocity and diameter are constant (\(S \propto n^2\)).

Calculating the Change in Gradient

We are given that the coefficient of rugosity is increased from \(n_1 = 0.01\) to \(n_2 = 0.02\).

Let the initial gradient be \(S_1\) corresponding to \(n_1\).

$$S_1 = K^2 \cdot n_1^2$$

Let the new gradient be \(S_2\) corresponding to \(n_2\).

$$S_2 = K^2 \cdot n_2^2$$

We know that \(n_2 = 0.02\) and \(n_1 = 0.01\), so \(n_2 = 2 \cdot n_1\).

Substitute \(n_2 = 2 n_1\) into the equation for \(S_2\):

$$S_2 = K^2 \cdot (2 n_1)^2$$

$$S_2 = K^2 \cdot (4 n_1^2)$$

$$S_2 = 4 \cdot (K^2 n_1^2)$$

Since \(S_1 = K^2 n_1^2\), we can write:

$$S_2 = 4 \cdot S_1$$

This means the new gradient \(S_2\) is 4 times the initial gradient \(S_1\). Therefore, the gradient must be increased by 4 times.

Summary of Changes

Parameter Initial State Final State Change
Coefficient of Rugosity (n) 0.01 0.02 Increased by 2 times
Diameter (D) Constant Constant No Change
Velocity (V) Constant Constant No Change
Flow Rate (Q) Constant Constant No Change
Gradient (S) \(S_1\) \(S_2\) Increased by 4 times (\(S_2 = 4S_1\))

Conclusion

When the coefficient of rugosity of a pipe is increased from 0.01 to 0.02 (doubled), while keeping the diameter, flow rate, and velocity constant, the hydraulic gradient must be increased by 4 times to overcome the increased resistance due to the rougher surface.

Revision Table: Pipe Flow and Rugosity

Let's review the key concepts:

  • Coefficient of Rugosity (n): A measure of the roughness of the pipe or channel surface. Higher 'n' means rougher surface and greater resistance to flow.
  • Hydraulic Gradient (S): Represents the slope of the energy line. It indicates the head loss per unit length of pipe. A steeper gradient means more head loss over a given distance, which is needed to maintain velocity against higher resistance.
  • Manning's Formula: An empirical formula relating velocity, hydraulic radius, hydraulic gradient, and rugosity. \(V \propto \frac{1}{n} S^{1/2}\).
  • Relationship S vs n: If V and R (or D) are constant, \(S^{1/2} \propto n\), which means \(S \propto n^2\).

Additional Information: Factors Affecting Pipe Flow

Several factors influence flow in pipes:

  • Pipe Diameter (D): Larger diameters generally allow for higher flow rates or lower velocities/gradients for the same flow.
  • Pipe Length (L): Longer pipes result in greater total head loss for a given flow rate.
  • Pipe Material and Condition (Rugosity, n): Determines the friction between the fluid and the pipe wall. Smooth materials (like PVC) have low 'n', while rough materials (like concrete or corroded metal) have high 'n'. Increased rugosity increases head loss.
  • Fluid Properties: Density and viscosity affect how the fluid interacts with the pipe walls and itself. These are considered in formulas like the Darcy-Weisbach equation through the friction factor.
  • Hydraulic Gradient (S): The driving force for the flow, related to the available head difference over the pipe length. A steeper gradient provides more potential energy to overcome friction.

While Manning's formula is often used for open channels, for pipes flowing full, the Darcy-Weisbach equation is generally considered more theoretically rigorous, as it directly incorporates the Reynolds number and relative roughness to determine the friction factor. However, empirical formulas like Manning's are also used, especially for large diameter pipes or specific design practices.

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Important Questions from Flow Through Pipes

  1. The velocity of pressure wave in a rigid pipe carrying a fluid of density ‘ρ’, viscosity ‘µ’ varies as

  2. In order to replace a pipe of diameter D by n parallel pipes of diameter d the relation used is

  3. Darcy Weisbach equation is used to find loss of head due to -

  4. To avoid vapourisation, pipe lines are laid over the ridge so that they are not more than _________ above the hydraulic gradient line.

  5. The head of water over the centre of an orifice of diameter 20 mm is 1 m. The actual discharge through the orifice is 0.85 litre/s. Find the coefficient of discharge.

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