# Hydrosystems Hydrulics open channel flow iit nptel _19_ by hemlalb

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```									Hydraulics                                                                                                Prof. B.S. Thandaveswara

6.3 Comparison Between Momentum and Energy Equation
Theoretically when the flow is gradually varied, energy and momentum equation should

yield same results. Consider a gradually varied flow. The pressure distribution in the

sections is taken as hydrostatic, the channel bed slope as small. For a rectangular

channel of small slope and width b, in a short reach the expression for pressure forces

can be written as

1
P = γ by12
1
2
1
and P2 = γ by2 2
2
If Force due to friction can be written as Pf = γ h′f by

in which h 'f is the friction head and y is the average depth, or ( y1 +y2 ) / 2. The

discharge through the reach is equal to

Q=
1
2
(
V1 + V 2 by     )
Also, the weight of the body of water is
W = γ byL
z1 − z2
and            sin θ =
L
Then the momentum equation, after substituting these expressions simplifies (see box) as
2                    2
V1                  V2   '
z1 + y1 + β1    = z 2 + y2 + β 2    +h f
2g                  2g
2                   2
V1                  V2
z1 + y1 + α1    = z 2 + y2 + α 2    + hf
2g                  2g     1-2

This equation appears to be practically the same as the energy equation (Bernoulli

equation). However, the energy loss given by momentum equation is due to external

forces whereas the loss given by energy equation is due to internal forces. One is a

vector quantity and other is scalar quantity. However, if the flow is uniform, then hf = h'f if

the difference between α and ρ is ignored. Similarity ends here. There are cases where

either momentum equation or energy equation can be used with the continuity equation.

Indian Institute of Technology Madras
Hydraulics                                                                                                                            Prof. B.S. Thandaveswara

Momentum Application
⎛ V1 +V 2 ⎞ ⎛ y1 +y 2 ⎞     ⎛ V1 +V 2 ⎞
Q= ⎜         ⎟b⎜         ⎟ =by ⎜         ⎟
⎝    2 ⎠ ⎝ 2 ⎠              ⎝ 2 ⎠
W = γbyL= Specific weight * (Volume)
z1 -z 2
sinθ =
L
γ ⎛ V1 +V 2 ⎞                    1      1
⎟ ⎡ β 2 V2 -β1 V1 ⎤ = γby1 - γby 2 + γby Lsinθ - γh f by
2       2                  '
by ⎜
g ⎝ 2 ⎠⎣                      ⎦ 2       2
γby ⎡                                    1      1             z -z
β V V + β V2 − β1 V1 − β1 V1 V2 ⎤ = γby1 - γby 2 + γby L 1 2 - γh 'f by
2       2                   2       2
2g ⎢⎣ 2 1 2 2                        ⎥ 2
⎦         2               L
divided by γb
y ⎡                                     1 2 1       y +y     y +y             ⎛ y1 +y 2 ⎞
β V − β V + β2 V1 V2 − β1 V1 V 2 ⎤ = y1 - y 2 + 1 2 z1 - 1 2 z 2 − h 'f
2     2
⎢ 2 2 1 1                          ⎥ 2        2                             ⎜         ⎟
2g ⎣                                  ⎦       2        2        2              ⎝ 2 ⎠
y1 +y 2 ⎡ V2                        V V2 ⎤ 1 2 1
2       2
V1     V V2                       y     y      y       y          '       ⎛ y1 +y 2 ⎞
⎢β 2   − β1    + β2 1   − β1 1 ⎥ = y1 - y 2 + 1 z1 + 2 z1 − 1 z 2 − 2 z 2 + h f
2                                           ⎜         ⎟
2 ⎢ 2g           2g      2g       2g ⎥ 2    2      2      2      2       2                 ⎝ 2 ⎠
⎣                                ⎦
Simplifying
2
V2     V2 β V V 2 − β1 V1 V 2                              '
β2      − β1 1 + 2 1               = ( y1 -y 2 ) − z1 − z 2 − h f
2g     2g        2g
β 2 V1 V 2 − β1 V1 V 2
If β1 ≈ β2 we can neglect                            ≈0
2g
2              2
V1                  V2    '
z1 + y1 + β1    = z 2 + y2 + β 2    + hf
2g                  2g

Indian Institute of Technology Madras

```
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