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Head1 given Time Required to Lower Liquid for Triangular Notch Formula

GoHead1 given Time Required to Lower Liquid Surface Formula

GoHead1 given Time Required to Lower Liquid Surface using Bazins Formula 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{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}}

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

Head1 given Time Required to Lower Liquid for Triangular Notch Example

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Here is how the Head1 given Time Required to Lower Liquid for Triangular Notch equation looks like with Values.

Here is how the Head1 given Time Required to Lower Liquid for Triangular Notch equation looks like with Units.

Here is how the Head1 given Time Required to Lower Liquid for Triangular Notch equation looks like.

11.2224 = {(\frac{1}{(\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1.25 \cdot (\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8} \cdot \tan (\frac{30}{2})}{(\frac{2}{3}) \cdot 13})})}^{\frac{2}{3}}

11.2224m = {(\frac{1}{(\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1.25s \cdot (\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8m/s²} \cdot \tan (\frac{30°}{2})}{(\frac{2}{3}) \cdot 13m²})})}^{\frac{2}{3}}

11.2224 = {(\frac{1}{(\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1.25 \cdot (\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8} \cdot \tan (\frac{30}{2})}{(\frac{2}{3}) \cdot 13})})}^{\frac{2}{3}}

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Head1 given Time Required to Lower Liquid for Triangular Notch Solution

Follow our step by step solution on how to calculate Head1 given Time Required to Lower Liquid for Triangular Notch?

FIRST Step Consider the formula

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}}

Next Step Substitute values of Variables

∴

H Upstream = {(\frac{1}{(\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1.25s \cdot (\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8m/s²} \cdot \tan (\frac{30°}{2})}{(\frac{2}{3}) \cdot 13m²})})}^{\frac{2}{3}}

Next Step Convert Units

∴

H Upstream = {(\frac{1}{(\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1.25s \cdot (\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8m/s²} \cdot \tan (\frac{0.5236rad}{2})}{(\frac{2}{3}) \cdot 13m²})})}^{\frac{2}{3}}

Next Step Prepare to Evaluate

∴

H Upstream = {(\frac{1}{(\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1.25 \cdot (\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8} \cdot \tan (\frac{0.5236}{2})}{(\frac{2}{3}) \cdot 13})})}^{\frac{2}{3}}

Next Step Evaluate

∴

H Upstream = 11.2223927927199m

LAST Step Rounding Answer

∴

H Upstream = 11.2224m

Head1 given Time Required to Lower Liquid for Triangular Notch 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

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

Coefficient of Discharge

The Coefficient of Discharge is ratio of actual discharge to theoretical discharge.

Symbol: C_d

Measurement: NAUnit: Unitless

Note: Value should be between 0 to 1.2.

Formulas to find Coefficient of Discharge

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

Theta

Theta is an angle that can be defined as the figure formed by two rays meeting at a common endpoint.

Symbol: θ

Measurement: AngleUnit: °

Note: Value can be positive or negative.

Formulas to find Theta

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

tan

The tangent of an angle is a trigonometric ratio of the length of the side opposite an angle to the length of the side adjacent to an angle in a right triangle.

Syntax: tan(Angle)

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 Surface using Bazins Formula

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

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}})

How to Evaluate Head1 given Time Required to Lower Liquid for Triangular Notch?

Head1 given Time Required to Lower Liquid for Triangular Notch evaluator uses 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) to evaluate the Head on Upstream of Weir, Head1 given Time Required to Lower Liquid for Triangular Notch 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 for Triangular Notch using this online evaluator? To use this online evaluator for Head1 given Time Required to Lower Liquid for Triangular Notch, enter Head on Downstream of Weir (h₂), Time Interval (Δt), Coefficient of Discharge (C_d), Acceleration due to Gravity (g), Theta (θ) & Cross-Sectional Area of Reservoir (A_R) and hit the calculate button.

FAQs on Head1 given Time Required to Lower Liquid for Triangular Notch

What is the formula to find Head1 given Time Required to Lower Liquid for Triangular Notch?

The formula of Head1 given Time Required to Lower Liquid for Triangular Notch is expressed as 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). Here is an example- 11.22239 = (1/((1/5.1^(3/2))-((1.25*(8/15)*0.66*sqrt(2*9.8)*tan(0.5235987755982/2))/((2/3)*13))))^(2/3).

How to calculate Head1 given Time Required to Lower Liquid for Triangular Notch?

With Head on Downstream of Weir (h₂), Time Interval (Δt), Coefficient of Discharge (C_d), Acceleration due to Gravity (g), Theta (θ) & Cross-Sectional Area of Reservoir (A_R) we can find Head1 given Time Required to Lower Liquid for Triangular Notch using the formula - 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). This formula also uses Tangent (tan), Square Root (sqrt) 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/((Time Interval*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)-(1/sqrt(Head on Downstream of Weir))))^2)

Can the Head1 given Time Required to Lower Liquid for Triangular Notch be negative?

Yes, the Head1 given Time Required to Lower Liquid for Triangular Notch, measured in Length can be negative.

Which unit is used to measure Head1 given Time Required to Lower Liquid for Triangular Notch?

Head1 given Time Required to Lower Liquid for Triangular Notch 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 for Triangular Notch can be measured.

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Head1 given Time Required to Lower Liquid for Triangular Notch Formula

​GoHead1 given Time Required to Lower Liquid Surface Formula

​GoHead1 given Time Required to Lower Liquid Surface using Bazins Formula Formula

Head1 given Time Required to Lower Liquid for Triangular Notch Example

Head1 given Time Required to Lower Liquid for Triangular Notch Solution

Head1 given Time Required to Lower Liquid for Triangular Notch 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 for Triangular Notch?

FAQs on Head1 given Time Required to Lower Liquid for Triangular Notch

Variable Name

GoHead1 given Time Required to Lower Liquid Surface Formula

GoHead1 given Time Required to Lower Liquid Surface using Bazins Formula Formula