**1 Answer**

written 2.2 years ago by |

**Derivation :** Length of Back Water Curve

Consider the flow of water through a channel in which depth of water is rising as shown in Figure.

Let the two sections A and B are at such a distance that the distance between them represents the length of back water curve.

**Notations :**

$h_{1}=$ depth of flow at section A,

$V_{1}=$ velocity of flow at section A,

$h_{2}=$ depth of flow at section B,

$V_{2}=$ velocity of flow at section B,

$i_{b}=$ bed slope,

$i_{e}=$ energy line slope, and

$L=$ length of back water curve.

Applying Bernoulli's equation at sections A and B,

$$ z_{1}+h_{1}+\frac{V_{1}^{2}}{2 g}=z_{2}+h_{2}+\frac{V_{2}^{2}}{2 g}+h_{L} ...(i) $$

where $h_{L}=$ Loss of energy due to friction $=i_{e} \times L$

Also taking datum line passing through the bed of the channel at section B. Then $z_{2}=0$

Equation (i) becomes as $$z_{1}+h_{1}+\frac{V_{1}^{2}}{2 g}=h_{2}+\frac{V_{2}^{2}}{2 g}+i_{e} \times L$$

From Figure, $z_{1}=i_{b} \times L$

$$ i_{b} ~L+h_{1}+\frac{v_{1}^{2}}{2 g}=h_{2}+\frac{v_{2}^{2}}{2 g}+i_{e} ~ L$$

$$ i_{b} ~ L-i_{e} ~L=\left(h_{2}+\frac{v_{2}^{2}}{2 g}\right)-\left(h_{1}+\frac{v_{1}^{2}}{2 g}\right) $$

$$ L\left(i_{b}-i_{e}\right)=E_{2}-E_{1}, $$

Where $$\left (E_{2}=h_{2}+\frac{V_{2}^{2}}{2 g}, E_{1}=h_{1}+\frac{V_{1}^{2}}{2 g}\right) $$

$$\therefore L=\frac{E_{2}-E_{1}}{i_{b}-i_{e}} $$