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Head2 given Time Required to Lower Liquid Surface using Bazins Formula Formula

GoHead2 given Time Required to Lower Liquid Surface Formula

GoHead2 given Time Required to Lower Liquid for Triangular Notch Formula

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Head on Downstream of Weir pertains to the energy status of water in water flow systems and is useful for describing flow in hydraulic structures. Check FAQs

h 2 = {(\frac{1}{\frac{Δt \cdot m \cdot \sqrt{2 \cdot g}}{2 \cdot A R} + (\frac{1}{\sqrt{H Upstream}})})}^{2}

h₂ - Head on Downstream of Weir?Δt - Time Interval?m - Bazins Coefficient?g - Acceleration due to Gravity?A_R - Cross-Sectional Area of Reservoir?H_Upstream - Head on Upstream of Weir?

Head2 given Time Required to Lower Liquid Surface using Bazins Formula Example

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Here is how the Head2 given Time Required to Lower Liquid Surface using Bazins Formula equation looks like with Values.

Here is how the Head2 given Time Required to Lower Liquid Surface using Bazins Formula equation looks like with Units.

Here is how the Head2 given Time Required to Lower Liquid Surface using Bazins Formula equation looks like.

6.21 = {(\frac{1}{\frac{1.25 \cdot 0.407 \cdot \sqrt{2 \cdot 9.8}}{2 \cdot 13} + (\frac{1}{\sqrt{10.1}})})}^{2}

6.21m = {(\frac{1}{\frac{1.25s \cdot 0.407 \cdot \sqrt{2 \cdot 9.8m/s²}}{2 \cdot 13m²} + (\frac{1}{\sqrt{10.1m}})})}^{2}

6.21 = {(\frac{1}{\frac{1.25 \cdot 0.407 \cdot \sqrt{2 \cdot 9.8}}{2 \cdot 13} + (\frac{1}{\sqrt{10.1}})})}^{2}

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Head2 given Time Required to Lower Liquid Surface using Bazins Formula Solution

Follow our step by step solution on how to calculate Head2 given Time Required to Lower Liquid Surface using Bazins Formula?

FIRST Step Consider the formula

h 2 = {(\frac{1}{\frac{Δt \cdot m \cdot \sqrt{2 \cdot g}}{2 \cdot A R} + (\frac{1}{\sqrt{H Upstream}})})}^{2}

Next Step Substitute values of Variables

∴

h 2 = {(\frac{1}{\frac{1.25s \cdot 0.407 \cdot \sqrt{2 \cdot 9.8m/s²}}{2 \cdot 13m²} + (\frac{1}{\sqrt{10.1m}})})}^{2}

Next Step Prepare to Evaluate

∴

h 2 = {(\frac{1}{\frac{1.25 \cdot 0.407 \cdot \sqrt{2 \cdot 9.8}}{2 \cdot 13} + (\frac{1}{\sqrt{10.1}})})}^{2}

Next Step Evaluate

∴

h 2 = 6.20998840007798m

LAST Step Rounding Answer

∴

h 2 = 6.21m

Head2 given Time Required to Lower Liquid Surface using Bazins Formula Formula Elements

Variables

Functions

Head on Downstream of Weir

Head on Downstream of Weir pertains to the energy status of water in water flow systems and is useful for describing flow in hydraulic structures.

Symbol: h₂

Measurement: LengthUnit: m

Note: Value can be positive or negative.

Formulas to find Head on Downstream of Weir

Time Interval

Time interval is the time duration between two events/entities of interest.

Symbol: Δt

Measurement: TimeUnit: s

Note: Value should be greater than 0.

Formulas to find Time Interval

Bazins Coefficient

Bazins Coefficient is the constant value obtained by Head.

Symbol: m

Measurement: NAUnit: Unitless

Note: Value should be greater than 0.

Formulas to find Bazins Coefficient

Acceleration due to Gravity

The Acceleration due to Gravity is acceleration gained by an object because of gravitational force.

Symbol: g

Measurement: AccelerationUnit: m/s²

Note: Value should be greater than 0.

Formulas to find Acceleration due to Gravity

Cross-Sectional Area of Reservoir

Cross-Sectional Area of Reservoir is the area of a reservoir that is obtained when a three-dimensional reservoir shape is sliced perpendicular to some specified axis at a point.

Symbol: A_R

Measurement: AreaUnit: m²

Note: Value should be greater than 0.

Formulas to find Cross-Sectional Area of Reservoir

Head on Upstream of Weir

Head on Upstream of Weirr pertains to the energy status of water in water flow systems and is useful for describing flow in hydraulic structures.

Symbol: H_Upstream

Measurement: LengthUnit: m

Note: Value can be positive or negative.

Formulas to find Head on Upstream of Weir

sqrt

A square root function is a function that takes a non-negative number as an input and returns the square root of the given input number.

Syntax: sqrt(Number)

Other Formulas to find Head on Downstream of Weir

Go Head2 given Time Required to Lower Liquid Surface

h 2 = {(\frac{1}{\frac{Δt \cdot (\frac{2}{3}) \cdot C d \cdot \sqrt{2 \cdot g} \cdot L w}{2 \cdot A R} + (\frac{1}{\sqrt{H Upstream}})})}^{2}

Go Head2 given Time Required to Lower Liquid for Triangular Notch

h 2 = {(\frac{1}{(\frac{Δt \cdot (\frac{8}{15}) \cdot C d \cdot \sqrt{2 \cdot g} \cdot \tan (\frac{θ}{2})}{(\frac{2}{3}) \cdot A R}) + (\frac{1}{{H Upstream}^{\frac{3}{2}}})})}^{\frac{2}{3}}

Other formulas in Time Required to Empty a Reservoir with Rectangular Weir category

Go Time Required to Lower Liquid Surface

Δt = (\frac{2 \cdot A R}{(\frac{2}{3}) \cdot C d \cdot \sqrt{2 \cdot g} \cdot L w}) \cdot (\frac{1}{\sqrt{h 2}} - \frac{1}{\sqrt{H Upstream}})

Go Coefficient of Discharge for Time Required to Lower Liquid Surface

C d = (\frac{2 \cdot A R}{(\frac{2}{3}) \cdot Δt \cdot \sqrt{2 \cdot g} \cdot L w}) \cdot (\frac{1}{\sqrt{h 2}} - \frac{1}{\sqrt{H Upstream}})

Go Length of Crest for time required to Lower Liquid Surface

L w = (\frac{2 \cdot A R}{(\frac{2}{3}) \cdot C d \cdot \sqrt{2 \cdot g} \cdot Δt}) \cdot (\frac{1}{\sqrt{h 2}} - \frac{1}{\sqrt{H Upstream}})

Go Cross Sectional Area given Time required to Lower Liquid Surface

A R = \frac{Δt \cdot (\frac{2}{3}) \cdot C d \cdot \sqrt{2 \cdot g} \cdot L w}{2 \cdot (\frac{1}{\sqrt{h 2}} - \frac{1}{\sqrt{H Upstream}})}

How to Evaluate Head2 given Time Required to Lower Liquid Surface using Bazins Formula?

Head2 given Time Required to Lower Liquid Surface using Bazins Formula evaluator uses Head on Downstream of Weir = (1/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)+(1/sqrt(Head on Upstream of Weir))))^2 to evaluate the Head on Downstream of Weir, Head2 given Time Required to Lower Liquid Surface using Bazins Formula in fluid dynamics, head is concept that relates energy in incompressible fluid to height of equivalent static column. Head on Downstream of Weir is denoted by h₂ symbol.

How to evaluate Head2 given Time Required to Lower Liquid Surface using Bazins Formula using this online evaluator? To use this online evaluator for Head2 given Time Required to Lower Liquid Surface using Bazins Formula, enter Time Interval (Δt), Bazins Coefficient (m), Acceleration due to Gravity (g), Cross-Sectional Area of Reservoir (A_R) & Head on Upstream of Weir (H_Upstream) and hit the calculate button.

FAQs on Head2 given Time Required to Lower Liquid Surface using Bazins Formula

What is the formula to find Head2 given Time Required to Lower Liquid Surface using Bazins Formula?

The formula of Head2 given Time Required to Lower Liquid Surface using Bazins Formula is expressed as Head on Downstream of Weir = (1/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)+(1/sqrt(Head on Upstream of Weir))))^2. Here is an example- 6.209988 = (1/((1.25*0.407*sqrt(2*9.8))/(2*13)+(1/sqrt(10.1))))^2.

How to calculate Head2 given Time Required to Lower Liquid Surface using Bazins Formula?

With Time Interval (Δt), Bazins Coefficient (m), Acceleration due to Gravity (g), Cross-Sectional Area of Reservoir (A_R) & Head on Upstream of Weir (H_Upstream) we can find Head2 given Time Required to Lower Liquid Surface using Bazins Formula using the formula - Head on Downstream of Weir = (1/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)+(1/sqrt(Head on Upstream of Weir))))^2. This formula also uses Square Root (sqrt) function(s).

What are the other ways to Calculate Head on Downstream of Weir?

Here are the different ways to Calculate Head on Downstream of Weir-

Head on Downstream of Weir=(1/((Time Interval*(2/3)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*Length of Weir Crest)/(2*Cross-Sectional Area of Reservoir)+(1/sqrt(Head on Upstream of Weir))))^2
Head on Downstream of Weir=(1/(((Time Interval*(8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2))/((2/3)*Cross-Sectional Area of Reservoir))+(1/Head on Upstream of Weir^(3/2))))^(2/3)

Can the Head2 given Time Required to Lower Liquid Surface using Bazins Formula be negative?

Yes, the Head2 given Time Required to Lower Liquid Surface using Bazins Formula, measured in Length can be negative.

Which unit is used to measure Head2 given Time Required to Lower Liquid Surface using Bazins Formula?

Head2 given Time Required to Lower Liquid Surface using Bazins Formula is usually measured using the Meter[m] for Length. Millimeter[m], Kilometer[m], Decimeter[m] are the few other units in which Head2 given Time Required to Lower Liquid Surface using Bazins Formula can be measured.

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Head2 given Time Required to Lower Liquid Surface using Bazins Formula Formula

​GoHead2 given Time Required to Lower Liquid Surface Formula

​GoHead2 given Time Required to Lower Liquid for Triangular Notch Formula

Head2 given Time Required to Lower Liquid Surface using Bazins Formula Example

Head2 given Time Required to Lower Liquid Surface using Bazins Formula Solution

Head2 given Time Required to Lower Liquid Surface using Bazins Formula Formula Elements

Other Formulas to find Head on Downstream of Weir

Other formulas in Time Required to Empty a Reservoir with Rectangular Weir category

How to Evaluate Head2 given Time Required to Lower Liquid Surface using Bazins Formula?

FAQs on Head2 given Time Required to Lower Liquid Surface using Bazins Formula

Variable Name

GoHead2 given Time Required to Lower Liquid Surface Formula

GoHead2 given Time Required to Lower Liquid for Triangular Notch Formula