Basic Concepts and Definitions

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Basic Concepts and Definitions

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𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐹𝑙𝑜𝑤 𝑣𝑠. 𝑂𝑝𝑒𝑛 𝐶ℎ𝑎𝑛𝑛𝑒𝑙 𝐹𝑙𝑜𝑤: 𝐹𝑟𝑖𝑐𝑖𝑡𝑜𝑛 𝐹𝑜𝑟𝑐𝑒 𝑣𝑠. 𝐺𝑟𝑎𝑣𝑖𝑡𝑦

Simplifying Assumptions in an Open Channel Flow:

1- One dimensional 2- Steady

3- Flow at each section: Uniform: 𝐿𝑎𝑟𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑐𝑎𝑙𝑒 𝑎𝑛𝑑 𝑅𝑒

∴ 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 (𝛼 = 1)

4- The pressure distribution is Hydrostatic In fluid low dir.: ∆𝑝 = 0

Major pressure gradient across each section

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Basic Concepts and Definitions

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Channel Geometry

𝑦: 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑓𝑙𝑜𝑤: 𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑏𝑒𝑑 𝑡𝑜 𝑓𝑟𝑒𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐴:𝐹𝑙𝑜𝑤 𝑎𝑟𝑒𝑎 ⊥ 𝐹𝑙𝑜𝑤 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛

𝑃: 𝑊𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟: 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑤𝑖𝑡ℎ 𝑙𝑖𝑞𝑢𝑖𝑑 𝑅_ℎ: 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑟𝑎𝑑𝑖𝑢𝑠

𝑹_𝒉 = 𝑨 𝑷

𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐹𝑙𝑜𝑤: 𝐷_ℎ = 4𝐴 𝑝

𝐹𝑜𝑟 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛: 𝐷_ℎ = 2𝑅_ℎ = 𝐷 2

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Channel Geometry

𝐹𝑜𝑟 𝑛𝑜𝑛 − 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑐ℎ𝑎𝑛𝑛𝑒𝑙𝑠: 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑑𝑒𝑝𝑡ℎ:𝑦_ℎ = 𝐴 𝑏_𝑠 𝑏_𝑠: 𝑤𝑖𝑑𝑡ℎ 𝑎𝑡 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

𝐴 = 𝑏_𝑠𝑦_ℎ

the hydraulic depth represents the average depth of the channel at any cross section. It gives the depth of an equivalent rectangular channel.

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Speed of Surface Waves and the Froude Number

Effect of Gravitational Disturbance on Surface Wave تسدلااب اب تسد نییاپ رد لایس راتفر قیبطت

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Speed of Surface Waves and the Froude Number

1. Steady ﬂow.

2. Incompressible ﬂow.

3. Uniform velocity at each section.

4. Hydrostatic pressure distribution at each section.

5. Frictionless ﬂow.

Continuity:

→ ∆𝑉 = 𝑐 ∆𝑦 𝑦 + ∆𝑦

𝑎𝑠𝑠𝑢𝑚𝑒 𝑐 > ∆𝑉

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Momentum:

𝑭_𝑺𝒙? ? ? ^𝑝𝑐: 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

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ො𝑛 ො𝑛

𝑐 − ∆𝑉 𝑐

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𝑢𝑠𝑒 ∆𝑉 = 𝑐 ∆𝑦 𝑦 + ∆𝑦

∆𝑦 ≪ 𝑦 𝒄 = 𝒈𝒚 :کچوک هنماد اب جاوما یارب سین هتسباو لایس صاوخ هب جوم تکرح تعرس ت

.

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Speed of Surface Waves and the Froude Number

Example

11.1 SPEED OF FREE SURFACE WAVES

You are enjoying a summer’s afternoon relaxing in a rowboat on a pond. You decide to find out how deep the water is by splashing your oar and timing how long it takes the wave you produce to reach the edge of the pond. (The pond is artificial; so it has approximately the same depth even to the shore.) From floats installed in the pond, you know you’re 20 ft from shore, and you measure the time for the wave to reach the edge to be 1.5 s. Estimate the pond depth. Does it matter if it’s a freshwater pond or if it’s filled with seawater?

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Speed of Surface Waves and the Froude Number

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تعرس اب لایس دوش یم لامعا نآ هب تسد نییاپ رد شاشتغا کی هک دراد نایرج 𝑉

.

کی طسوت دناوت یم شاشتغا نیا 𝑏𝑢𝑚𝑝 𝑖𝑛 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑓𝑙𝑜𝑜𝑟

عنام کی طسوت ای 𝑏𝑎𝑟𝑟𝑖𝑒𝑟

دوش داجیا .

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تعرس اب شاشتغا نیا دنک یم تکرح تسدلااب رد لایس هب تبسن 𝑐

.

رگا  تعرس لایس

مک

،دشاب 𝑉 < 𝑐

شاشتغا اب

تعرس 𝑐 − 𝑉 قلطم

رد تسدلااب

تکرح یم

دنک .

رگا  تعرس لایس

دایز

،دشاب 𝑉 > 𝑐

شاشتغا یمن

دناوت رد تسدلااب تکرح

دنک و طقف

رد نییاپ تسد

تکرح یم

دنک .

دورف ددع فیرعت

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𝑭𝒓 = 𝑽 𝒈𝒚

Instead of the rather loose terms “slow” and “fast,” we now have the following criteria:

• 𝐹𝑟 < 1 Flow is

subcritical

. Disturbances can travel upstream; downstream conditions can affect the ﬂow upstream. The ﬂow can gradually adjust to the disturbance.

• 𝐹𝑟 = 1 Flow is

critical

.

• 𝐹𝑟 > 1 Flow is

supercritical

, rapid, or shooting. No disturbance can travel upstream; downstream conditions cannot be felt upstream. The ﬂow may

“violently” respond to the disturbance because the ﬂow has no chance to adjust to the disturbance before encountering it.

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For nonrectangular channels: 𝑭𝒓 = 𝑽

𝒈𝒚_𝒉 _{(𝐴 = 𝑏}

𝑠𝑦_ℎ)

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Energy Equation for Open-Channel Flows

assumptions:

1. Steady ﬂow

2. Incompressible ﬂow

3. Uniform velocity at a section

4. Pressure distribution is hydrostatic 5. Small bed slope

6. 𝑊ሶ_𝑆 = ሶ𝑊_{𝑠ℎ𝑎𝑓𝑡} = ሶ𝑊_{𝑜𝑡ℎ𝑒𝑟} = 0

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Energy Equation for Open-Channel Flows

𝐶ℎ𝑎𝑛𝑛𝑒𝑙 𝑏𝑒𝑑

𝐶ℎ𝑎𝑛𝑛𝑒𝑙 𝑏𝑒𝑑 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑑𝑒𝑝𝑡ℎ

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𝑑𝐴 = 𝑏𝑑𝑦

_

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Notice that:

4. Pressure distribution is hydrostatic

For section 1:

Therefore,

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Total head

Same for section 2, therefore, the energy equation becomes:

The total head or energy head, H, at any location in an open-channel flow:

𝑉₁²

2 + 𝑔𝑦₁ + 𝑔𝑧₁ − 𝑉₂²

2 + 𝑔𝑦₂ + 𝑔𝑧₂ = ℎ_{𝑙 𝑇}

÷ 𝑔 𝑉₁²

2𝑔 + 𝑦₁ + 𝑧₁ = 𝑉₂²

2𝑔 + 𝑦₂ + 𝑧₂ + 𝐻_𝑙

𝑯 = 𝑽^𝟐

𝟐𝒈 + 𝒚 + 𝒛 𝑦 & 𝑧: 𝑙𝑜𝑐𝑎𝑙 𝑑𝑒𝑝𝑡ℎ 𝑎𝑛𝑑 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑏𝑒𝑑 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛

𝐻₁ − 𝐻₂ = 𝐻_𝑙 & 𝐻_𝑙 > 0 → 𝐻₁ > 𝐻₂

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Specific Energy

𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 − 𝑧 (𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 ℎ𝑒𝑎𝑑)

𝑬 = 𝑽^𝟐

𝟐𝒈 + 𝒚 → 𝐸₁ − 𝐸₂ + 𝑧₁ − 𝑧₂ = 𝐻_𝑙

𝑬 = 𝑸^𝟐

𝟐𝒈𝑨^𝟐 + 𝒚

𝐾𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑉 = 𝑄 𝐴 :

𝐹𝑜𝑟 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑠 𝑦 ↑: 𝐴 ↑→ 𝑄²

2𝑔𝐴² ↓ 𝐴 𝑦

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Specific Energy

Critical depth (𝑦_𝑐): For a given flow rate, at a specified depth, the specific energy would be minimum.

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Specific Energy

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Specific Energy

Example 11.2 SPECIFIC ENERGY CURVES FOR A RECTANGULAR CHANNEL

For a rectangular channel of width b = 10 m, construct a family of speciﬁc energy curves for Q = 0, 2, 5, and 10 ^𝑚

3

𝑠 . What are the minimum speciﬁc energies for these curves?

𝑬 = 𝑸^𝟐

𝟐𝒈𝑨^𝟐 + 𝒚

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Specific Energy

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Specific Energy

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Specific Energy

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Specific Energy

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Critical Depth: Minimum Speciﬁc Energy

Example 11.2 treated the case of a rectangular channel. We now consider channels of general cross section.

𝑬 = 𝑸^𝟐

𝟐𝒈𝑨^𝟐 + 𝒚

For a given flow rate Q, to find the depth for minimum specific energy,

𝑑𝐸

𝑑𝑦 = 0 = − 𝑄² 𝑔𝐴³

𝑑𝐴

𝑑𝑦 + 1 & 𝑑𝐴 = 𝑏_𝑠𝑑𝑦 𝑑𝐸

𝑑𝑦 = − 𝑄²

𝑔𝐴³ 𝑏_𝑠 + 1 = 0

→ 𝑄² = 𝑔𝐴³ 𝑏_𝑠

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Critical Depth: Minimum Speciﬁc Energy

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𝐹𝑟𝑜𝑚 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑉 = 𝑄 𝐴 𝑉 = 𝑄

𝐴 = 1 𝐴

𝑔𝐴³ 𝑏_𝑠

12

= 𝑔𝐴 𝑏_𝑠

& 𝐴 = 𝑏_𝑠𝑦_ℎ 𝑉 = 𝑔𝑦_ℎ

𝑭𝒓 = 𝑽 𝒈𝒚_𝒉 But the Froude number is given by:

• For minimum speciﬁc energy, Fr = 1, which corresponds to critical ﬂow.

• We obtain the important result that, for ﬂow in any open channel, the speciﬁc energy is at its minimum at critical conditions.

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Critical Depth: Minimum Speciﬁc Energy

For critical ﬂow:

𝑸^𝟐 = 𝒈𝑨^𝟑 𝒃_𝒔 𝑽_𝒄 = 𝒈𝒚_𝒉_𝒄

𝐴_𝑐, 𝑉_𝑐, 𝑏_𝑠𝑐 and 𝑦_ℎ𝑐 are the critical ﬂow area, velocity, channel surface width, and hydraulic depth, respectively.

For rectangular channel: 𝑏_𝑠 = 𝑏 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 & 𝐴 = 𝑏𝑦

𝑄² = 𝑔𝐴³_𝑐

𝑏_𝑠_𝑐 = 𝑔𝑏³𝑦_𝑐³

𝑏 = 𝑔𝑏²𝑦_𝑐³ 𝑏 𝑦

→ 𝒚_𝒄 = 𝑸^𝟐 𝒈𝒃^𝟐

𝟏𝟑

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𝑽_𝒄 = 𝒈𝒚_𝒄 = 𝒈𝑸 𝒃

𝟏𝟑

With:

And minimum energy:

𝐸 = 𝐸_𝑚𝑖𝑛 = 𝑉_𝑐²

2𝑔 + 𝑦_𝑐 = 𝑔𝑦_𝑐

2𝑔 + 𝑦_𝑐

𝑬_𝒎𝒊𝒏 = 𝟑 𝟐 𝒚_𝒄

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Example 11.3 CRITICAL DEPTH FOR TRIANGULAR SECTION

A steep-sided triangular section channel (𝛼 = 60°) has a ﬂow rate of 300 ^𝑚

3

𝑠 . Find the critical depth for this ﬂow rate. Verify that the Froude number is unity.

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Critical Depth: Minimum Speciﬁc Energy

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Localized Effect of Area Change (Frictionless Flow)

a simple ﬂow case in which the channel bed is horizontal and for which the effects of channel cross section (area change) predominate:

Flow Over a Bump

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Localized Effect of Area Change (Frictionless Flow)

𝑉₁²

2𝑔 + 𝑦₁ + 𝑧₁ = 𝑉₂²

2𝑔 + 𝑦₂ + 𝑧₂ = 𝑉²

2𝑔 + 𝑦 + 𝑧 = 𝑐𝑜𝑛𝑠𝑡 𝐸₁ + 𝑧₁ = 𝐸₂ + 𝑧₂ = 𝐸 + 𝑧 = 𝑐𝑜𝑛𝑠𝑡

it takes place over a short distance, the effects of friction (on either momentum or energy) may be reasonably neglected.

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Flow over a Bump

تباث ضرع اب یلیطتسم لاناک دیریگب رظن رد 𝑏

.

یگدمآرب هلداعم 𝑧 = ℎ(𝑥) :

𝑦 𝑥 : 𝑓𝑟𝑜𝑚 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑏𝑒𝑑 ،

1: 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚 𝑝𝑜𝑖𝑛𝑡 & 2: 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑢𝑚𝑝:

𝑉₁²

2𝑔+ 𝑦₁ = 𝐸₁ = ^𝑉_2𝑔² + 𝑦 + ℎ = 𝐸 + ℎ 𝑥 = 𝑐𝑜𝑛𝑠𝑡 → 𝐸 𝑥 = 𝐸₁ − ℎ 𝑥

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Flow over a Bump

𝐹𝑟𝑜𝑚 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦: 𝑄 = 𝑏𝑉₁𝑦₁ = 𝑏𝑉𝑦

𝑄²

2𝑔𝑏²𝑦₁² + 𝑦₁ = 𝑄²

2𝑔𝑏²𝑦² + 𝑦 + ℎ = 𝑐𝑜𝑛𝑠𝑡

هب تبسن لااب هطبار زا دوجو لیلد هب دازآ حطس عافترا تارییغت یارب یا هطبار ات میریگ یم قتشم 𝑥

𝑏𝑢𝑚𝑝

دیایب تسد هب .

− 𝑄² 𝑔𝑏²𝑦³

𝑑𝑦

𝑑𝑥 + 𝑑𝑦

𝑑𝑥 + 𝑑ℎ

𝑑𝑥 = 0 𝑑𝑦

𝑑𝑥 =

𝑑ℎ𝑑𝑥

[ 𝑄𝑔𝑏²²𝑦³ − 1]

=

𝑑ℎ𝑑𝑥 [𝑉𝑔𝑦 − 1]²

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Flow over a Bump

𝒅𝒚

𝒅𝒙 = 𝟏 𝑭𝒓^𝟐 − 𝟏

𝒅𝒉 𝒅𝒙

response to a bump very much depends on the local Froude number, Fr

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Flow over a Bump

رگا 𝐹𝑟 < 1: 𝑆𝑢𝑏𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐹𝑙𝑜𝑤: 𝐹𝑟² − 1 < 0:

𝑑𝑦 بیش بیش فلاخم دازآ حطس رد 𝑑𝑥

𝑑ℎ

رد 𝑑𝑥

تسا 𝑏𝑢𝑚𝑝 .

عافترا رگا 𝑏𝑢𝑚𝑝

:↑ لایس عافترا سکعلاب و ↓

.

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Flow over a Bump

رگا 𝐹𝑟 = 1: C𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐹𝑙𝑜𝑤: 𝐹𝑟² − 1 = 0: 𝑑𝑦

𝑑𝑥 → ∞: 𝐼𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒

→ 𝑑ℎ

𝑑𝑥 = 0

هک یلحم رد دناوت یم طقف ینارحب نایرج نیاربانب

𝑑ℎ

𝑑𝑥 = 0 دتفیب قافتا ،تسا

.

ینعی : 𝑎𝑡 𝑏𝑢𝑚𝑝 𝑐𝑟𝑒𝑠𝑡 𝑜𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑖𝑠 𝑓𝑙𝑎𝑡

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Flow over a Bump

رگا 𝐹𝑟 > 1: 𝑆𝑢𝑝𝑒𝑟𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐹𝑙𝑜𝑤: 𝐹𝑟² − 1 > 0:

𝑑𝑦 بیش بیش قفاوم دازآ حطس رد 𝑑𝑥

𝑑ℎ

رد 𝑑𝑥

تسا 𝑏𝑢𝑚𝑝 .

عافترا رگا 𝑏𝑢𝑚𝑝

:↑ لایس عافترا سکعلاب و ↑

.

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Flow over a Bump

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Flow over a Bump

Example 11.4 FLOW IN A RECTANGULAR CHANNEL WITH A BUMP OR A NARROWING

A rectangular channel 2 m wide has a ﬂow of 2.4 𝑚³/𝑠 at a depth of 1.0 m.

Determine whether critical depth occurs at:

(a) a section where a bump of height h = 0.20 m is installed across the channel bed,

(b) a side wall constriction (with no bumps) reducing the channel width to 1.7 m, and

(c) both the bump and side wall constrictions combined.

Neglect head losses of the bump and constriction caused by friction, expansion, and contraction.

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Flow over a Bump

Solution: Compare the specific energy to the minimum specific energy for the given flow rate in each case to establish whether critical depth occurs.

(56)

Flow over a Bump

(57)

Flow over a Bump

(58)

Flow over a Bump

(59)

Flow over a Bump

(60)

Hydraulic Jump

When ﬂow at a section is supercritical,

and downstream conditions will require a change to subcritical ﬂow, the need for this change cannot be communicated upstream;

the ﬂow speed exceeds the speed of surface waves, which are the mechanism for transmitting changes.

Thus a gradual change with a smooth transition through the critical point is not possible.

The transition from supercritical to subcritical ﬂow occurs abruptly through a hydraulic jump.

(61)

Hydraulic Jump

(62)

Hydraulic Jump

Unlike the changes due to phenomena such as a bump,

the abrupt change in depth involves a signiﬁcant loss of mechanical energy through turbulent mixing.

فا یم قافتا هاتوک اتبسن هزاب کی رد یکیلوردیه شرپ هک دنهد یم ناشن یبرجت یاه شیامزآ دت

.

شرپ لوط ممیزکام :

6 رت گرزب قمع ربارب (𝑦₂)

: زا رظنفرص 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒

هلداعم رد طقف

یژرنا هلداعم رد هن و موتنموم (

سنلاوبروت هدیدپ لیلد هب )

Although hydraulic jumps can occur on inclined surfaces, for simplicity we assume a horizontal bed, and rectangular channel of width b; the results we obtain will apply generally to hydraulic jumps.

(63)

Hydraulic Jump

(64)

Hydraulic Jump

(65)

Hydraulic Jump

assumptions:

1. Steady ﬂow

3. Uniform velocity at each section

4. Hydrostatic pressure distribution at each section 5. Frictionless ﬂow (for the momentum equation)

Continuity: ෍

𝐶𝑆𝑉. Ԧ𝐴 = 0 → −𝑉₁𝑏𝑦₁ + 𝑉₂𝑏𝑦₂ = 0 𝑉₁𝑦₁ = 𝑉₂𝑦₂

(66)

Hydraulic Jump

x-mom: 𝐹_𝑥 = 𝐹_𝑠𝑥 + 𝐹_𝐵𝑥 = 𝜕

𝜕𝑡න

𝐶𝑉

𝑢𝜌 𝑑𝑉 + ෍

𝐶𝑆𝑢𝜌𝑉. Ԧ𝐴

→ 𝐹_𝑠𝑥 = ෍

𝐶𝑆𝑢𝜌𝑉. Ԧ𝐴

𝐹_𝑅 = 𝑝_𝑐𝐴

𝑭_𝒔𝒙? ?

𝐹_𝑠𝑥 = 𝐹_𝑅1 − 𝐹_𝑅2 = 𝑝_𝑐𝐴 ₁ − 𝑝_𝑐𝐴 ₂ = 𝜌𝑔𝑦₁

2 𝑦₁𝑏 − 𝜌𝑔𝑦₂

2 𝑦₂𝑏 = 𝐹_𝑠𝑥 = 𝜌𝑔𝑏

2 (𝑦₁² − 𝑦₂²) 𝐹_𝑠𝑥 = 𝜌𝑔𝑏

2 𝑦₁² − 𝑦₂² = ෍

𝐶𝑆𝑢𝜌𝑉. Ԧ𝐴 = 𝑉₁𝜌 −𝑉₁𝑏𝑦₁ + 𝑉₂𝜌 𝑉₂𝑏𝑦₂

(67)

Hydraulic Jump

Momentum equation for the hydraulic jump:

Momentum Eq. for open channel flow:

For horizontal hydraulic jump, 𝑧₁ = 𝑧₂, so:

𝑬_𝟏 = 𝑽_𝟏^𝟐

𝟐𝒈 + 𝒚_𝟏 = 𝑽_𝟐^𝟐

𝟐𝒈 + 𝒚_𝟐 + 𝑯_𝒍 = 𝑬_𝟐 + 𝑯_𝒍

𝑉₁²

2𝑔 + 𝑦₁ + 𝑧₁ = 𝑉₂²

2𝑔 + 𝑦₂ + 𝑧₂ + 𝐻_𝑙

𝑽_𝟏^𝟐𝒚_𝟏

𝒈 + 𝒚_𝟏^𝟐

𝟐 = 𝑽_𝟐^𝟐𝒚_𝟐

𝒈 + 𝒚_𝟐^𝟐 𝟐

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Hydraulic Jump

Loss of mechanical energy for hydraulic jump:

∆𝐸 = 𝐸₁ − 𝐸₂ = 𝐻_𝑙

𝑽_𝟏𝒚_𝟏 = 𝑽_𝟐𝒚_𝟐 𝑽_𝟏^𝟐𝒚_𝟏

𝒈 + 𝒚_𝟏^𝟐

𝟐 = 𝑽_𝟐^𝟐𝒚_𝟐

𝒈 + 𝒚_𝟐^𝟐 𝟐 𝑬_𝟏 = 𝑽_𝟏^𝟐

𝟐𝒈 + 𝒚_𝟏 = 𝑽_𝟐^𝟐

𝟐𝒈 + 𝒚_𝟐 + 𝑯_𝒍 = 𝑬_𝟐 + 𝑯_𝒍

یکیلوردیه شرپ تلاداعم

(69)

Depth Increase Across a Hydraulic Jump

(𝑉₁𝑦₁ = 𝑉₂𝑦₂) ∴ 𝑉₁²𝑦₁

𝑔 + 𝑦₁²

2 = 𝑉₁²𝑦₁ 𝑔 (𝑦₁

𝑦₂) + 𝑦₂² 2 Rearranging:

Dividing by 𝑦₂ − 𝑦₁:

× 𝑦₂ ÷ 𝑦₁² gives:

(70)

Depth Increase Across a Hydraulic Jump

𝒚_𝟐

𝒚_𝟏 = 𝟏

𝟐 𝟏 + 𝟖𝑭𝒓_𝟏^𝟐 − 𝟏

تسدلااب هب تسد نییاپ رد قمع نایم تبسن تسدلااب رد دورف ددع ∝

𝒚_𝟏 & 𝒚_𝟐: 𝑪𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒅𝒆𝒑𝒕𝒉𝒔

(71)

Head Loss Across a Hydraulic Jump

𝑯_𝒍 = 𝑬_𝟏 − 𝑬_𝟐 = 𝑽_𝟏^𝟐

𝟐𝒈 + 𝒚_𝟏 − (𝑽_𝟐^𝟐

𝟐𝒈 + 𝒚_𝟐)

𝑉₁𝑦₁ = 𝑉₂𝑦₂ ∴ 𝐻_𝑙 = 𝑉₁²

2𝑔 1 − 𝑦₁ 𝑦₂

2

+ (𝑦₁ − 𝑦₂)

→ 𝐻_𝑙

𝑦₁ = 𝑉₁²

2𝑔𝑦₁ 1 − 𝑦₁ 𝑦₂

2

+ [1 − 𝑦₂ 𝑦₁]

→ 𝐻_𝑙

𝑦₁ = 𝐹𝑟₁²

2 1 − 𝑦₁ 𝑦₂

2

+ [1 − 𝑦₂ 𝑦₁]

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𝑯_𝒍

𝒚_𝟏 = 𝟏 𝟒

𝒚_𝟐

𝒚_𝟏 − 𝟏 ^𝟑 𝒚_𝟐 𝒚_𝟏

𝒐𝒓 𝑯_𝒍 = 𝒚_𝟐 − 𝒚_𝟏 ^𝟑 𝟒𝒚_𝟏𝒚_𝟐

𝐻_𝑙 > 0^{𝑝𝑟𝑜𝑜𝑓}𝑦₂

𝑦₁ > 1

یزاس دعب یب 𝐻_𝑙

طسوت 𝐸₁

:

𝐸₁ = 𝑉₁²

2𝑔 + 𝑦₁ = 𝑦₁ 𝑉₁²

2𝑔𝑦₁ + 1 = 𝑦₁ (𝐹𝑟₁² + 2) 2

(73)

→ 𝐻_𝑙

𝐸₁ = 1 2

𝑦₂

𝑦₁ − 1 ³ 𝑦₂

𝑦₁ [𝐹𝑟₁² + 2]

𝒚_𝟐

𝒚_𝟏 = 𝟏

𝟐 𝟏 + 𝟖𝑭𝒓_𝟏^𝟐 − 𝟏

𝑯_𝒍 𝑬_𝟏 =

𝟏 + 𝟖𝑭𝒓_𝟏^𝟐 − 𝟑

𝟑

𝟖 𝟏 + 𝟖𝑭𝒓_𝟏^𝟐 − 𝟏 [𝑭𝒓_𝟏^𝟐 + 𝟐]

𝐹𝑟₁ ↑→ 𝐻_𝑙 ↑ 𝐹𝑟₁ = 1 → 𝐻_𝑙 = 0

(74)

Hydraulic Jump

Example

11.5 HYDRAULIC JUMP IN A RECTANGULAR CHANNEL FLOW

A hydraulic jump occurs in a rectangular channel 3 m wide. The water depth before the jump is 0.6 m, and after the jump is 1.6 m. Compute (a) the ﬂow rate in the channel (b) the critical depth (c) the head loss in the jump.

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Hydraulic Jump

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Hydraulic Jump

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Hydraulic Jump

(78)

Hydraulic Jump

(79)

Steady Uniform Flow

Steady uniform ﬂow is something that is to be expected to occur for channels of constant slope and cross section: Prismatic channel

We have Uniform Flow.

Fully Developed Flow

Constant depth: Normal depth (𝒚_𝒏)

𝐴₁ = 𝐴₂ = 𝐴 (cross-section areas), 𝑄₁ = 𝑄₂ = 𝑄 (ﬂow rates),

𝑉₁ = 𝑉₂ = 𝑉 (average velocity, 𝑉 = ^𝑄

𝐴), 𝑦₁ = 𝑦₂ = 𝑦_𝑛 (ﬂow depth)

(80)

Steady Uniform Flow

Assumptions:

1. Steady ﬂow

4. Pressure distribution is hydrostatic 5. Small bed slope

6. 𝑊ሶ_𝑆 = ሶ𝑊_{𝑠ℎ𝑎𝑓𝑡} = ሶ𝑊_{𝑜𝑡ℎ𝑒𝑟} = 0

ضرف 5

: قفا یاتسار رد لایس تعرس ،دومع یاتسار رد لایس قمع

Continuity: 𝑄 = 𝑉₁𝐴₁ = 𝑉₂𝐴₂ = 𝑉𝐴

(81)

Steady Uniform Flow

(82)

Steady Uniform Flow

x-mom: 𝐹_𝑥 = 𝐹_𝑠𝑥 + 𝐹_𝐵𝑥 = 𝜕

𝜕𝑡න

𝐶𝑉

𝑢𝜌 𝑑𝑉 + ෍

𝐶𝑆𝑢𝜌𝑉. Ԧ𝐴 𝐹_𝐵𝑥 = 𝑊 sin 𝜃 & 𝐹_𝑠𝑥: 𝐹_𝑓: 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛

𝐹_𝑓 = 𝑊 sin 𝜃 = 𝜏_𝑤𝑃𝐿

𝑊 sin 𝜃 = 𝜌𝑔𝐴𝐿 sin 𝜃 ≈ 𝜌𝑔𝐴𝐿𝜃 ≈ 𝜌𝑔𝐴𝐿𝑆_𝑏

→ 𝜏_𝑤𝑃𝐿 = 𝜌𝑔𝐴𝐿𝑆_𝑏

(83)

Steady Uniform Flow

→ 𝜏_𝑤 = 𝜌𝑔𝐴𝑆_𝑏

𝑃 = 𝜌𝑔𝑅_ℎ𝑆_𝑏 𝑹_𝒉 = 𝑨 𝑷 میدوب هدرک فیرعت لابق 𝐶_𝑓 = 𝜏_𝑤 :

12 𝜌𝑉²

→ 1

2𝐶_𝑓𝜌𝑉² = 𝜌𝑔𝑅_ℎ𝑆_𝑏

𝑉 = 2𝑔

𝐶_𝑓 𝑅_ℎ𝑆_𝑏

نینچمه و لاناک یسدنه تاصخشم ساسا رب تعرس 𝑪_𝒇

𝐶_𝑓 ( : لکشم )

هب هتسباو

𝑏𝑒𝑑 𝑟𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠, 𝑓𝑙𝑢𝑖𝑑 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠, 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦:

(84)

Manning Equation for Uniform Flow

یاج هب 𝑪_𝒇

تیمک کی دیدج

فیرعت یم

دوش :

بیرض 𝑪𝒉𝒆𝒛𝒚

طسوت یبرجت تروص هب 𝑀𝑎𝑛𝑛𝑖𝑛𝑔

تسا هدش هئارا .

𝐶 = 2𝑔

𝐶_𝑓 : 𝐶ℎ𝑒𝑧𝑦 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑽 = 𝑪 𝑹_𝒉𝑺_𝒃: 𝑪𝒉𝒆𝒛𝒚 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏

𝑪 = 𝟏 𝒏 𝑹_𝒉

𝟏𝟔 𝒏: 𝒓𝒐𝒖𝒈𝒉𝒏𝒆𝒔𝒔 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕

𝑽 = 𝟏 𝒏 𝑹_𝒉

𝟐𝟑𝑺_𝒃

𝟏𝟐: 𝑴𝒂𝒏𝒏𝒊𝒏𝒈 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏

(85)

Manning Equation for Uniform Flow

(86)

Manning Equation for Uniform Flow

Manning’s Eq. for Flow Rate in SI:

𝑸 = 𝟏

𝒏 𝑨 𝑹_𝒉

𝟐𝟑𝑺_𝒃

𝟏𝟐

Manning’s Eq. for Flow Rate in English Engineering unit (𝑉 𝑖𝑛^𝑓𝑡

𝑠 & 𝑅_ℎ 𝑖𝑛 𝑓𝑒𝑒𝑡):

𝑽 = 𝟏. 𝟒𝟗 𝒏 𝑹_𝒉

𝟐𝟑𝑺_𝒃

𝟏𝟐 & 𝑸 = 𝟏. 𝟒𝟗

𝒏 𝑨 𝑹_𝒉

𝟐𝟑𝑺_𝒃

𝟏𝟐

𝑸 = 𝟏

𝒏 𝑨 𝑹_𝒉

𝟐𝟑𝑺_𝒃

𝟏𝟐 یازا هب طقف ،صخشم بیش اب لاناک کی رد صخشم یبد کی یازا هب 𝒚 کی یم تخاونکی نایرج

دوش 𝒚_𝒏 :

.

(87)

Steady Uniform Flow

Example 11.6 FLOW RATE IN A RECTANGULAR CHANNEL

An 8-ft-wide rectangular channel with a bed slope of 0.0004 ft/ft has a depth of flow of 2 ft. Assuming steady uniform flow, determine the discharge in the channel. The Manning roughness coefficient is n = 0.015.

(88)

Steady Uniform Flow

(89)

Steady Uniform Flow

Example 11.7 FLOW VERSUS AREA THROUGH TWO CHANNEL SHAPES Open channels, of square and semicircular shapes, are being considered for carrying flow on a slope of Sb = 0.001; the channel walls are to be poured concrete with n = 0.015. Evaluate the flow rate delivered by the channels for maximum dimensions between 0.5 and 2.0 m. Compare the channels on the basis of volume flow rate for given cross-sectional area.

(90)

Steady Uniform Flow

(91)

Steady Uniform Flow

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Steady Uniform Flow

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Steady Uniform Flow

Rectangular Channel

Semi-circular Channel

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Steady Uniform Flow

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Steady Uniform Flow

Example 11.8 NORMAL DEPTH IN A RECTANGULAR CHANNEL Determine the normal depth (for uniform ﬂow) if the channel described in Example 11.6 has a ﬂow rate of 100 cfs.

(96)

Steady Uniform Flow

(97)

Steady Uniform Flow

Example 11.9 DETERMINATION OF FLUME SIZE

An above-ground ﬂume, built from timber, is to convey water from a mountain lake to a small hydroelectric plant. The ﬂume is to deliver water at Q =2 ^𝑚

3

𝑠 ; the slope is 𝑆_𝑏

= 0.002 and n = 0.013. Evaluate the required ﬂume size for (a) a rectangular section with y/b = 0.5 and (b) an equilateral triangular section.

(98)

Steady Uniform Flow

(99)

Steady Uniform Flow

(100)

Steady Uniform Flow

(101)

Steady Uniform Flow

(102)

Steady Uniform Flow

(103)

Energy Equation for Uniform Flow

𝑉₁²

2𝑔 + 𝑦₁ + 𝑧₁ = 𝑉₂²

2𝑔 + 𝑦₂ + 𝑧₂ + 𝐻_𝑙

(𝑉₁= 𝑉₂ = 𝑉 & 𝑦₁ = 𝑦₂ = 𝑦_𝑛)

→ 𝑧₁ = 𝑧₂ + 𝐻_𝑙 → 𝐻_𝑙 = 𝑧₁ − 𝑧₂ = 𝐿𝑆_𝑏 𝐸 = 𝐸₁ = 𝑉₁²

2𝑔 + 𝑦₁ = 𝐸₂ = 𝑉₂²

2𝑔 + 𝑦₂ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝐸𝐺𝐿 = 𝑝

𝜌𝑔 + 𝑉²

2𝑔 + 𝑧_{𝑡𝑜𝑡𝑎𝑙} 𝐸𝐺𝐿 = 𝑝

𝜌𝑔 + 𝑧_{𝑡𝑜𝑡𝑎𝑙}

𝑎𝑛𝑑 𝑧_{𝑡𝑜𝑡𝑎𝑙} = 𝑧 + 𝑦

(104)

Energy Equation for Uniform Flow

𝐸𝐺𝐿 = 𝑉²

2𝑔 + 𝑧 + 𝑦

𝑝

𝜌𝑔 + 𝑉²

2𝑔 + 𝑧_{𝑡𝑜𝑡𝑎𝑙} = 𝜌𝑔(𝑦₁ − 𝑦)

𝜌𝑔 + 𝑉²

2𝑔 + 𝑧₁ + 𝑦

𝐻𝐺𝐿 = 𝑧 + 𝑦

𝐸𝐺𝐿₁ − 𝐸𝐺𝐿₂ = 𝑧₁ − 𝑧₂ = 𝐻_𝑙 𝐻𝐺𝐿₁ − 𝐻𝐺𝐿₂ = 𝑧₁ − 𝑧₂ = 𝐻_𝑙

For normal flow: channel bed || HGL || EGL

(105)

Energy Equation for Uniform Flow

(106)

Optimum Channel Cross Section

هنیهب تلاح :

ای دشاب ممیزکام عطقم حطس هب یبد تبسن لاناک صخشم یربز و بیش یازا هب

اب نیرتمک حطس

عطقم نیرتشیب

روبع نایرج

قافتا دتفیب

.

𝑸

𝑨 = 𝟏 𝒏𝑹_𝒉

𝟐𝟑𝑺_𝒃

𝟏𝟐

𝑄

𝐴 : 𝑚𝑎𝑥 → 𝑅_ℎ: 𝑚𝑎𝑥, 𝑅_ℎ = 𝐴

𝑃 → 𝑃: 𝑚𝑖𝑛 ∴ 𝐹𝑜𝑟 min𝑖𝑚𝑢𝑚 𝐴 : 𝑃 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑚𝑖𝑛𝑖𝑚𝑖𝑧𝑒𝑑

𝑨 = 𝒏𝑸 𝑺_𝒃

𝟏𝟐 𝟑𝟓

𝑷^𝟐^𝟓

(107)