Chapter V-HYDRAULIC JUMP.pdf

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Transcript of Chapter V-HYDRAULIC JUMP.pdf

Page 1: Chapter V-HYDRAULIC JUMP.pdf
Page 2: 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

Page 3: Chapter V-HYDRAULIC JUMP.pdf

HYDRAULIC JUMP IN A RECTANGULAR CHANNEL

Consider a freebody of water containing hydraulic jump

F2

F1

W

P1 = γ y1 P2 = γ y2 N

y2

y1 V1

y2

y1 Q

V2

Page 4: Chapter V-HYDRAULIC JUMP.pdf

Considering the Impulse-Momentum Equation

𝛴𝐹 = (𝜌𝑄𝑉)𝑜𝑢𝑡 − (𝜌𝑄𝑉)𝑖𝑛

𝛴𝐹𝑥 = (𝜌𝑄𝑉𝑥)𝑜𝑢𝑡 − (𝜌𝑄𝑉𝑥)𝑖𝑛

𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1

where: Ef = neglected (if distance between sections is relatively small)

𝐹1 = 1

2𝑃1𝑦1𝑏 =

1

2 γ𝑦1𝑦1𝑏 =

1

2 γ𝑦1

2𝑏

𝐹2 = 1

2𝑃2𝑦2𝑏 =

1

2 γ𝑦2𝑦2𝑏 =

1

2 γ𝑦2

2𝑏

Then, 1

2 γ𝑦1

2𝑏 −1

2 γ𝑦2

2𝑏 − 0 = γ

𝑔 𝐴2𝑉2 𝑉2 −

γ

𝑔 𝐴1𝑉1 𝑉1

1

2 𝑔𝑏 𝑦1

2 − 𝑦22 = 𝐴2𝑉2

2 − 𝐴1𝑉12

1

2 𝑔𝑏 𝑦1

2 − 𝑦22 = ( 𝑏𝑦2 )𝑉2

2 − ( 𝑏𝑦1 )𝑉12

1

2 𝑔 𝑦1

2 − 𝑦22 = 𝑦2𝑉2

2 − 𝑦1𝑉12

Page 5: Chapter V-HYDRAULIC JUMP.pdf

From continuity equation

𝑄1 = 𝑄2

𝐴1𝑉1 = 𝐴1𝑉1

𝑏𝑦1𝑉1 = 𝑏𝑦2𝑉2

𝑽𝟐 = 𝒚𝟏𝑽𝟏

𝒚𝟐

Substitute values

1

2 𝑔 𝑦1

2 − 𝑦22 = 𝑦2

𝑦12𝑉1

2

𝑦22 − 𝑦1𝑉1

2

1

2 𝑔 𝑦1

2 − 𝑦22 = 𝑉1

2𝑦1 𝑦1

𝑦2− 1

1

2 𝑔 𝑦1

2 − 𝑦22 =

𝑉12𝑦1

𝑦2 𝑦1 − 𝑦2

1

2 𝑔 𝑦1 − 𝑦2 𝑦1 + 𝑦2 =

𝑉12𝑦1

𝑦2 𝑦1 − 𝑦2

1

2 𝑔 𝑦1 + 𝑦2 =

𝑉12𝑦1

𝑦2

𝑽𝟏𝟐 =

𝟏

𝟐 𝒈

𝒚𝟐

𝒚𝟏 𝒚𝟏 + 𝒚𝟐

But

𝑉1 = 𝑄

𝐴=

𝑏𝑞

𝑏𝑦1=

𝑞

𝑦1

𝑞2

𝑦12 =

1

2 𝑔

𝑦2

𝑦1 𝑦1 + 𝑦2

𝒒𝟐 = 𝟏

𝟐 𝒈𝒚𝟏𝒚𝟐 𝒚𝟏 + 𝒚𝟐

Page 6: Chapter V-HYDRAULIC JUMP.pdf

ENERGY LOST AND POWER LOST IN A JUMP

Energy Equation 1 – 2

𝑃1

𝛾+ 𝑧1 +

𝑉12

2𝑔=

𝑃2

𝛾+ 𝑧2 +

𝑉22

2𝑔+ 𝑕𝐿

𝑦1 + 𝑉1

2

2𝑔= 𝑦2 +

𝑉22

2𝑔 + 𝑕𝐿

𝐸1 = 𝐸2 + 𝑕𝐿

𝒉𝑳 = 𝑬𝟏 − 𝑬𝟐 energy head lost

Power Lost: 𝑷 = 𝜸𝑸𝒉𝑳

Page 7: Chapter V-HYDRAULIC JUMP.pdf

Depth of Hydraulic Jump

Solve for y2: consider the equation:

𝑞2 = 1

2 𝑔𝑦1𝑦2 𝑦1 + 𝑦2

𝑦2 𝑦1 + 𝑦2 = 2𝑞2

𝑔𝑦1

𝑦22 + 𝑦1𝑦2 =

2𝑞2

𝑔𝑦1

𝑦22 + 𝑦1𝑦2 +

1

2𝑦1

2=

2𝑞2

𝑔𝑦1 +

1

2𝑦1

2

𝑦2 + 1

2𝑦1

2 =

2𝑞2

𝑔𝑦1 +

1

4𝑦1

2

𝑦2 + 1

2𝑦1

2 =

1

4𝑦1

2

8𝑞2

𝑔𝑦13 + 1

Extract the square root:

𝑦2 + 1

2𝑦1 =

1

2𝑦1

8𝑞2

𝑔𝑦13 + 1

𝑦2 = − 1

2𝑦1 +

1

2𝑦1

8𝑞2

𝑔𝑦13 + 1

𝑦2 = 1

2𝑦1 −1 +

8𝑞2

𝑔𝑦13 + 1

But 𝑞2

𝑔𝑦13 =

𝑄2

𝑏2

𝑔𝑦13 =

𝐴2𝑉2

𝑏2

𝑔𝑦13 =

𝑏2𝑦12𝑉1

2

𝑏2

𝑔𝑦13 =

𝑉12

𝑔𝑦1= 𝑁𝐹

2

Hence, 𝒚𝟐 = 𝟏

𝟐𝒚𝟏 −𝟏 + 𝟖𝑵𝑭𝟏

𝟐 + 𝟏 𝑁𝐹1> 1

Likewise,

𝒚𝟏 = 𝟏

𝟐𝒚𝟐 −𝟏 + 𝟖𝑵𝑭𝟐

𝟐 + 𝟏 𝑁𝐹2< 1

Page 8: Chapter V-HYDRAULIC JUMP.pdf

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 = ɣA2yc2

F2 = ɣA2yc2

Ef = 0

Page 9: Chapter V-HYDRAULIC JUMP.pdf

Impulse-Momentum Equation:

𝛴𝐹𝑥 = (𝑝𝑄𝑉)𝑜𝑢𝑡 − (𝑝𝑄𝑉)𝑖𝑛

𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1

𝛾𝐴1𝑦𝑐1− 𝛾𝐴2𝑦𝑐2

− 0 = 𝛾

𝑔 𝐴2𝑉2𝑉2 − 𝐴1𝑉1𝑉1

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴2𝑉22 − 𝐴1𝑉1

2

Continuity Equation:

𝑄1 = 𝑄2

𝐴1𝑉1 = 𝐴1𝑉1

𝑉2 = 𝐴1𝑉1

𝐴2

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴2𝐴1

2𝑉12

𝐴22 − 𝐴1𝑉1

2

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴1𝑉12

𝐴1

𝐴2− 1

𝑔 𝐴1𝑦𝑐1− 𝐴2𝑦𝑐2

= 𝐴1𝑉12

𝐴1−𝐴2

𝐴2

𝑽𝟏𝟐 =

𝒈𝑨𝟐

𝑨𝟏 𝑨𝟏𝒚𝒄𝟏

−𝑨𝟐𝒚𝒄𝟐

𝑨𝟏− 𝑨𝟐 or 𝑽𝟏

𝟐 = 𝒈𝑨𝟐

𝑨𝟏 𝑨𝟐𝒚𝒄𝟐

−𝑨𝟏𝒚𝒄𝟏

𝑨𝟐− 𝑨𝟏

Page 10: Chapter V-HYDRAULIC JUMP.pdf

*Another solution

𝛴𝐹𝑥 = (𝜌𝑄𝑉𝑥)𝑜𝑢𝑡 − (𝜌𝑄𝑉𝑥)𝑖𝑛

𝐹1 − 𝐹2 − 𝐸𝑓 = 𝜌𝑄2𝑉2 − 𝜌𝑄1𝑉1

γ𝐴1𝑦𝑐1 − γ𝐴2𝑦𝑐2 =γ

𝑔 𝐴2𝑉2

2 − 𝐴1𝑉12

𝐴1𝑦𝑐1 − 𝐴2𝑦𝑐2 =1

𝑔 𝐴2

𝑄2

𝐴22 − 𝐴1

𝑄2

𝐴12

𝐴1𝑦𝑐1 − 𝐴2𝑦𝑐2 =𝑄2

𝑔

1

𝐴2−

1

𝐴1

𝑸𝟐

𝒈=

𝑨𝟏𝒚𝒄𝟏−𝑨𝟐𝒚𝒄𝟐

𝟏

𝑨𝟐−

𝟏

𝑨𝟏

Page 11: Chapter V-HYDRAULIC JUMP.pdf

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

Page 12: Chapter V-HYDRAULIC JUMP.pdf

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

𝐴23 𝐴22

𝐴13

𝐴12

1.80

1.20 𝐴11

𝐴21

2

y2 = 1.80m

y1 = 1.20m

1

Page 13: Chapter V-HYDRAULIC JUMP.pdf

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

Δx

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