Download - Chapter V-HYDRAULIC JUMP.pdf
A hydraulic jump is a transition flow from supercritical to subcritical flow.
Hydraulic jump is one means of reducing the velocity of flow. It may also be used to separate lighter
solids from heavier ones.
y2>yc
V1 y1 < yc
HYDRAULIC JUMP IN A RECTANGULAR CHANNEL
Consider a freebody of water containing hydraulic jump
F2
F1
W
P1 = Ξ³ y1 P2 = Ξ³ y2 N
y2
y1 V1
y2
y1 Q
V2
Considering the Impulse-Momentum Equation
π΄πΉ = (πππ)ππ’π‘ β (πππ)ππ
π΄πΉπ₯ = (ππππ₯)ππ’π‘ β (ππππ₯)ππ
πΉ1 β πΉ2 β πΈπ = ππ2π2 β ππ1π1
where: Ef = neglected (if distance between sections is relatively small)
πΉ1 = 1
2π1π¦1π =
1
2 Ξ³π¦1π¦1π =
1
2 Ξ³π¦1
2π
πΉ2 = 1
2π2π¦2π =
1
2 Ξ³π¦2π¦2π =
1
2 Ξ³π¦2
2π
Then, 1
2 Ξ³π¦1
2π β1
2 Ξ³π¦2
2π β 0 = Ξ³
π π΄2π2 π2 β
Ξ³
π π΄1π1 π1
1
2 ππ π¦1
2 β π¦22 = π΄2π2
2 β π΄1π12
1
2 ππ π¦1
2 β π¦22 = ( ππ¦2 )π2
2 β ( ππ¦1 )π12
1
2 π π¦1
2 β π¦22 = π¦2π2
2 β π¦1π12
From continuity equation
π1 = π2
π΄1π1 = π΄1π1
ππ¦1π1 = ππ¦2π2
π½π = πππ½π
ππ
Substitute values
1
2 π π¦1
2 β π¦22 = π¦2
π¦12π1
2
π¦22 β π¦1π1
2
1
2 π π¦1
2 β π¦22 = π1
2π¦1 π¦1
π¦2β 1
1
2 π π¦1
2 β π¦22 =
π12π¦1
π¦2 π¦1 β π¦2
1
2 π π¦1 β π¦2 π¦1 + π¦2 =
π12π¦1
π¦2 π¦1 β π¦2
1
2 π π¦1 + π¦2 =
π12π¦1
π¦2
π½ππ =
π
π π
ππ
ππ ππ + ππ
But
π1 = π
π΄=
ππ
ππ¦1=
π
π¦1
π2
π¦12 =
1
2 π
π¦2
π¦1 π¦1 + π¦2
ππ = π
π πππππ ππ + ππ
ENERGY LOST AND POWER LOST IN A JUMP
Energy Equation 1 β 2
π1
πΎ+ π§1 +
π12
2π=
π2
πΎ+ π§2 +
π22
2π+ ππΏ
π¦1 + π1
2
2π= π¦2 +
π22
2π + ππΏ
πΈ1 = πΈ2 + ππΏ
ππ³ = π¬π β π¬π energy head lost
Power Lost: π· = πΈπΈππ³
Depth of Hydraulic Jump
Solve for y2: consider the equation:
π2 = 1
2 ππ¦1π¦2 π¦1 + π¦2
π¦2 π¦1 + π¦2 = 2π2
ππ¦1
π¦22 + π¦1π¦2 =
2π2
ππ¦1
π¦22 + π¦1π¦2 +
1
2π¦1
2=
2π2
ππ¦1 +
1
2π¦1
2
π¦2 + 1
2π¦1
2 =
2π2
ππ¦1 +
1
4π¦1
2
π¦2 + 1
2π¦1
2 =
1
4π¦1
2
8π2
ππ¦13 + 1
Extract the square root:
π¦2 + 1
2π¦1 =
1
2π¦1
8π2
ππ¦13 + 1
π¦2 = β 1
2π¦1 +
1
2π¦1
8π2
ππ¦13 + 1
π¦2 = 1
2π¦1 β1 +
8π2
ππ¦13 + 1
But π2
ππ¦13 =
π2
π2
ππ¦13 =
π΄2π2
π2
ππ¦13 =
π2π¦12π1
2
π2
ππ¦13 =
π12
ππ¦1= ππΉ
2
Hence, ππ = π
πππ βπ + ππ΅ππ
π + π ππΉ1> 1
Likewise,
ππ = π
πππ βπ + ππ΅ππ
π + π ππΉ2< 1
2
y2
y1
1 yc2
y2
y1
yc1
HYDRAULIC JUMP IN A NON-RECTANGULAR SECTION
Consider free body water
y2
y1
F2
Ff
F1
W
section thru 1 - 1 section thru 2 - 2
where:
F1 = Ι£A2yc2
F2 = Ι£A2yc2
Ef = 0
Impulse-Momentum Equation:
π΄πΉπ₯ = (πππ)ππ’π‘ β (πππ)ππ
πΉ1 β πΉ2 β πΈπ = ππ2π2 β ππ1π1
πΎπ΄1π¦π1β πΎπ΄2π¦π2
β 0 = πΎ
π π΄2π2π2 β π΄1π1π1
π π΄1π¦π1β π΄2π¦π2
= π΄2π22 β π΄1π1
2
Continuity Equation:
π1 = π2
π΄1π1 = π΄1π1
π2 = π΄1π1
π΄2
π π΄1π¦π1β π΄2π¦π2
= π΄2π΄1
2π12
π΄22 β π΄1π1
2
π π΄1π¦π1β π΄2π¦π2
= π΄1π12
π΄1
π΄2β 1
π π΄1π¦π1β π΄2π¦π2
= π΄1π12
π΄1βπ΄2
π΄2
π½ππ =
ππ¨π
π¨π π¨ππππ
βπ¨ππππ
π¨πβ π¨π or π½π
π = ππ¨π
π¨π π¨ππππ
βπ¨ππππ
π¨πβ π¨π
*Another solution
π΄πΉπ₯ = (ππππ₯)ππ’π‘ β (ππππ₯)ππ
πΉ1 β πΉ2 β πΈπ = ππ2π2 β ππ1π1
Ξ³π΄1π¦π1 β Ξ³π΄2π¦π2 =Ξ³
π π΄2π2
2 β π΄1π12
π΄1π¦π1 β π΄2π¦π2 =1
π π΄2
π2
π΄22 β π΄1
π2
π΄12
π΄1π¦π1 β π΄2π¦π2 =π2
π
1
π΄2β
1
π΄1
πΈπ
π=
π¨ππππβπ¨ππππ
π
π¨πβ
π
π¨π
1. Water flows in a rectangular channel with a width of 4.0 m at a uniform depth of 1.2 m. Adjustment is made downstream to raise water level to 2.0 m. consequently causing hydraulic jump. a. Calculate the discharge in the canal. b. Determine the power lost in a jump.
1.2 m
2.0 m
1.2 m
2. A hydraulic jump occurs in a 5 m wide rectangular canal carrying 6 m3/s on a slope of 0.005. the depth
after the jump is 1.4m.
a.) Calculate the depth before the jump.
b.) Calculate the power lost in a jump.
y2 = 1.4 m
y1 =?
3. A hydraulic jump occurs in a trapezoidal section with bottom width of 4m and side slope of 1:2. The
depth before the jump is 1.20m and after the jump is 1.80m.
a.) Calculate the flow rate in the canal.
b.) Calculate the power lost.
4m 4m
π΄23 π΄22
π΄13
π΄12
1.80
1.20 π΄11
π΄21
2
y2 = 1.80m
y1 = 1.20m
1
4. A rectangular canal has a width of 4.0m and carries water at the rate of 12m3/s. its bed slope is 0.0003 and roughness is 0.02. To control the flow, a sluice gate is provided at the entrance to the canal.
a. Determine whether a hydraulic jump would occur when the sluice gate is adjusted so that minimum depth after the gate is 0.40 m.
b. If a hydraulic jump would occur in letter (a), how far from the sluice gate will it occur?
y1
yo=1.60
m ys=0.50m
Ξx
5. A rectangular channel has a width of 5m, so=0.0009 and n=0.012. its uniform flow depth is 1.60m. if a sluice gate is adjusted such that a min. depth immediately downstream of the gate is 0.50m.
a. Determine whether a hydraulic jump would occur, and if it occurs b. how far downstream will it occur c. type of profile