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
increased by 4 times
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.
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:
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:
$$V = \frac{1}{n} \left(\frac{D}{4}\right)^{2/3} S^{1/2}$$
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 (\(S \propto n^2\)).
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.
| 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 (\(S_2 = 4S_1\)) |
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.
Let's review the key concepts:
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