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Head1 given Time Required to Lower Liquid Surface using Bazins Formula Formula

GoHead1 given Time Required to Lower Liquid Surface Formula

GoHead1 given Time Required to Lower Liquid for Triangular Notch Formula

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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. Check FAQs

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

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

Head1 given Time Required to Lower Liquid Surface using Bazins Formula Example

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Here is how the Head1 given Time Required to Lower Liquid Surface using Bazins Formula equation looks like with Values.

Here is how the Head1 given Time Required to Lower Liquid Surface using Bazins Formula equation looks like with Units.

Here is how the Head1 given Time Required to Lower Liquid Surface using Bazins Formula equation looks like.

7.8825 = ({(\frac{1}{\frac{1.25 \cdot 0.407 \cdot \sqrt{2 \cdot 9.8}}{2 \cdot 13} - (\frac{1}{\sqrt{5.1}})})}^{2})

7.8825m = ({(\frac{1}{\frac{1.25s \cdot 0.407 \cdot \sqrt{2 \cdot 9.8m/s²}}{2 \cdot 13m²} - (\frac{1}{\sqrt{5.1m}})})}^{2})

7.8825 = ({(\frac{1}{\frac{1.25 \cdot 0.407 \cdot \sqrt{2 \cdot 9.8}}{2 \cdot 13} - (\frac{1}{\sqrt{5.1}})})}^{2})

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Head1 given Time Required to Lower Liquid Surface using Bazins Formula Solution

Follow our step by step solution on how to calculate Head1 given Time Required to Lower Liquid Surface using Bazins Formula?

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

H Upstream = ({(\frac{1}{\frac{1.25s \cdot 0.407 \cdot \sqrt{2 \cdot 9.8m/s²}}{2 \cdot 13m²} - (\frac{1}{\sqrt{5.1m}})})}^{2})

Next Step Prepare to Evaluate

∴

H Upstream = ({(\frac{1}{\frac{1.25 \cdot 0.407 \cdot \sqrt{2 \cdot 9.8}}{2 \cdot 13} - (\frac{1}{\sqrt{5.1}})})}^{2})

Next Step Evaluate

∴

H Upstream = 7.88247677128312m

LAST Step Rounding Answer

∴

H Upstream = 7.8825m

Head1 given Time Required to Lower Liquid Surface using Bazins Formula Formula Elements

Variables

Functions

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

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 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

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 Upstream of Weir

Go Head1 given Time Required to Lower Liquid Surface

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

Go Head1 given Time Required to Lower Liquid for Triangular Notch

H Upstream = {(\frac{1}{(\frac{1}{{h 2}^{\frac{3}{2}}}) - (\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{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 Head1 given Time Required to Lower Liquid Surface using Bazins Formula?

Head1 given Time Required to Lower Liquid Surface using Bazins Formula evaluator uses Head on Upstream of Weir = ((1/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)-(1/sqrt(Head on Downstream of Weir))))^2) to evaluate the Head on Upstream of Weir, Head1 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 Upstream of Weir is denoted by H_Upstream symbol.

How to evaluate Head1 given Time Required to Lower Liquid Surface using Bazins Formula using this online evaluator? To use this online evaluator for Head1 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 Downstream of Weir (h₂) and hit the calculate button.

FAQs on Head1 given Time Required to Lower Liquid Surface using Bazins Formula

What is the formula to find Head1 given Time Required to Lower Liquid Surface using Bazins Formula?

The formula of Head1 given Time Required to Lower Liquid Surface using Bazins Formula is expressed as Head on Upstream of Weir = ((1/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)-(1/sqrt(Head on Downstream of Weir))))^2). Here is an example- 7.882477 = ((1/((1.25*0.407*sqrt(2*9.8))/(2*13)-(1/sqrt(5.1))))^2).

How to calculate Head1 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 Downstream of Weir (h₂) we can find Head1 given Time Required to Lower Liquid Surface using Bazins Formula using the formula - Head on Upstream of Weir = ((1/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)-(1/sqrt(Head on Downstream of Weir))))^2). This formula also uses Square Root Function function(s).

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

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

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

Can the Head1 given Time Required to Lower Liquid Surface using Bazins Formula be negative?

Yes, the Head1 given Time Required to Lower Liquid Surface using Bazins Formula, measured in Length can be negative.

Which unit is used to measure Head1 given Time Required to Lower Liquid Surface using Bazins Formula?

Head1 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 Head1 given Time Required to Lower Liquid Surface using Bazins Formula can be measured.

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Head1 given Time Required to Lower Liquid Surface using Bazins Formula Formula

​GoHead1 given Time Required to Lower Liquid Surface Formula

​GoHead1 given Time Required to Lower Liquid for Triangular Notch Formula

Head1 given Time Required to Lower Liquid Surface using Bazins Formula Example

Head1 given Time Required to Lower Liquid Surface using Bazins Formula Solution

Head1 given Time Required to Lower Liquid Surface using Bazins Formula Formula Elements

Other Formulas to find Head on Upstream of Weir

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

How to Evaluate Head1 given Time Required to Lower Liquid Surface using Bazins Formula?

FAQs on Head1 given Time Required to Lower Liquid Surface using Bazins Formula

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GoHead1 given Time Required to Lower Liquid Surface Formula

GoHead1 given Time Required to Lower Liquid for Triangular Notch Formula