Specific Energy
Hydraulics
Prof. B.S. Thandaveswara
9.3 Application of Specific Force and Specific Energy
1. Determine the energy Loss in a NHJ Solution: Applying Momentum equation γQ ( V2 − V1 ) = P1 − P2 g
γ Q 2 ( y1 − y 2 ) gb 2
y1y 2
2
2 2 = y1 − y 2
(
)
Q2 2gb Q2 b
2
=
( y1 + y2 ) y1y2
4
= q2
q 2 ( y1 + y 2 ) y1y 2 = (1) 2g 4 Specific energy equation y1 + V12 V2 = y 2 + 2 + ∆E 2g 2g Q2
2 2gy1 b 2
∆E = y1 +
− y2 −
Q2
2 2gy 2 b 2
1 ⎤ q2 ⎡ 1 ∆E = ( y1 − y 2 ) + ⎢ 2 − 2⎥ 2g ⎢ y1 y 2 ⎥ ⎣ ⎦ ⎡ q2 1 ⎤ = ( y1 − y 2 ) ⎢1 + ( y2 + y1 )( −1)⎥ 2 2 ⎢ 2g y1 y 2 ⎥ ⎣ ⎦ Substituting from momentum equation ⎡ ( y + y ) y y ( y + y )( −1) ⎤ = ( y1 − y 2 ) ⎢1 + 2 1 1 2 2 2 1 2 ⎥ 4 y1 y 2 ⎢ ⎥ ⎣ ⎦ = = ∆E =
( y1 − y2 ) ⎡ 4y y ( y1 − y2 ) ⎡ 4y y
4y1y 2 ⎣ 4y1y 2 4y1y 2 ⎣
1 2
2 − ( y1 + y 2 ) ⎤ ⎦ 2 2 − y1 − y 2 − 2y1y 2 ⎤ ⎦
1 2
( y2 - y1 )3
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
2. Problem of specific force Determine the sequent depth in a trapezoidal channel of 3 m width. The initial depth is 0.5 m. The side slope is 1:1. Initial flow velocity is 4 m/s. Solution:
0.5 1 1
3
A= (b+my)y (b+my)y ______________ b+2
Q = A.ν = = (3+0.5) 0.5*4 3.5 = *4 = 7 m3 /s 2 Q2 M1 = + zA gA =
m2+1
y
=R
⎛ 3*0.52 1*0.53 ⎞ +⎜ + ⎟ 3.5 ⎜ 2 3 ⎟ ⎝ ⎠ 9.81* 2 72
= 2.86 + 1.5*0.25 + 0.041666 = 3.2767m3 M1 =M 2 ⎛ 3*y 2 2 2 3 ⎞ +⎜ + y2 ⎟ ⎜ 2 ⎟ 3 ⎝ ⎠ Solve by trial and error y 2 = 1.05m. M1 = 72 9.81* ( 3+y 2 ) y 2 Alternative approach is: z M1 b
3
= =
1*3.2767 33 1*7 2
5
= 0.12135
m2Q2
5
gb 9.81*3 From graph y 2 = 1.05m.
= 2.05×10-2
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
9.3.1 Hydraulic Jump
1. For the case of hydraulic jump in a rectangular channel, complete the following table. y1 ( m) 0.20 1.91
V 1 (m /s)
q (m3 /s)
y2 (m) 1.204 2.50
V 2 (m / s)
Head 1oss ( m )
1.00
26.18
50
2. A hydraulic jump
Prof. B.S. Thandaveswara
9.3 Application of Specific Force and Specific Energy
1. Determine the energy Loss in a NHJ Solution: Applying Momentum equation γQ ( V2 − V1 ) = P1 − P2 g
γ Q 2 ( y1 − y 2 ) gb 2
y1y 2
2
2 2 = y1 − y 2
(
)
Q2 2gb Q2 b
2
=
( y1 + y2 ) y1y2
4
= q2
q 2 ( y1 + y 2 ) y1y 2 = (1) 2g 4 Specific energy equation y1 + V12 V2 = y 2 + 2 + ∆E 2g 2g Q2
2 2gy1 b 2
∆E = y1 +
− y2 −
Q2
2 2gy 2 b 2
1 ⎤ q2 ⎡ 1 ∆E = ( y1 − y 2 ) + ⎢ 2 − 2⎥ 2g ⎢ y1 y 2 ⎥ ⎣ ⎦ ⎡ q2 1 ⎤ = ( y1 − y 2 ) ⎢1 + ( y2 + y1 )( −1)⎥ 2 2 ⎢ 2g y1 y 2 ⎥ ⎣ ⎦ Substituting from momentum equation ⎡ ( y + y ) y y ( y + y )( −1) ⎤ = ( y1 − y 2 ) ⎢1 + 2 1 1 2 2 2 1 2 ⎥ 4 y1 y 2 ⎢ ⎥ ⎣ ⎦ = = ∆E =
( y1 − y2 ) ⎡ 4y y ( y1 − y2 ) ⎡ 4y y
4y1y 2 ⎣ 4y1y 2 4y1y 2 ⎣
1 2
2 − ( y1 + y 2 ) ⎤ ⎦ 2 2 − y1 − y 2 − 2y1y 2 ⎤ ⎦
1 2
( y2 - y1 )3
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
2. Problem of specific force Determine the sequent depth in a trapezoidal channel of 3 m width. The initial depth is 0.5 m. The side slope is 1:1. Initial flow velocity is 4 m/s. Solution:
0.5 1 1
3
A= (b+my)y (b+my)y ______________ b+2
Q = A.ν = = (3+0.5) 0.5*4 3.5 = *4 = 7 m3 /s 2 Q2 M1 = + zA gA =
m2+1
y
=R
⎛ 3*0.52 1*0.53 ⎞ +⎜ + ⎟ 3.5 ⎜ 2 3 ⎟ ⎝ ⎠ 9.81* 2 72
= 2.86 + 1.5*0.25 + 0.041666 = 3.2767m3 M1 =M 2 ⎛ 3*y 2 2 2 3 ⎞ +⎜ + y2 ⎟ ⎜ 2 ⎟ 3 ⎝ ⎠ Solve by trial and error y 2 = 1.05m. M1 = 72 9.81* ( 3+y 2 ) y 2 Alternative approach is: z M1 b
3
= =
1*3.2767 33 1*7 2
5
= 0.12135
m2Q2
5
gb 9.81*3 From graph y 2 = 1.05m.
= 2.05×10-2
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
9.3.1 Hydraulic Jump
1. For the case of hydraulic jump in a rectangular channel, complete the following table. y1 ( m) 0.20 1.91
V 1 (m /s)
q (m3 /s)
y2 (m) 1.204 2.50
V 2 (m / s)
Head 1oss ( m )
1.00
26.18
50
2. A hydraulic jump