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Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Formula

GoCross Sectional Area given Time required to Lower Liquid Surface Formula

GoCross Sectional Area given time required to Lower Liquid Surface using Bazins Formula Formula

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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. Check FAQs

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

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

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Example

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Here is how the Cross Sectional Area given Time required to Lower Liquid for Triangular Notch equation looks like with Values.

Here is how the Cross Sectional Area given Time required to Lower Liquid for Triangular Notch equation looks like with Units.

Here is how the Cross Sectional Area given Time required to Lower Liquid for Triangular Notch equation looks like.

14.0636 = \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 ((\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1}{{10.1}^{\frac{3}{2}}}))}

14.0636m² = \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 ((\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1}{{10.1m}^{\frac{3}{2}}}))}

14.0636 = \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 ((\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1}{{10.1}^{\frac{3}{2}}}))}

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Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Solution

Follow our step by step solution on how to calculate Cross Sectional Area given Time required to Lower Liquid for Triangular Notch?

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

A R = \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 ((\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1}{{10.1m}^{\frac{3}{2}}}))}

Next Step Convert Units

∴

A R = \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 ((\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1}{{10.1m}^{\frac{3}{2}}}))}

Next Step Prepare to Evaluate

∴

A R = \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 ((\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1}{{10.1}^{\frac{3}{2}}}))}

Next Step Evaluate

∴

A R = 14.0636418016635m²

LAST Step Rounding Answer

∴

A R = 14.0636m²

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Formula Elements

Variables

Functions

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

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

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

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

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 Cross-Sectional Area of Reservoir

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}})}

Go Cross Sectional Area given time required to Lower Liquid Surface using Bazins Formula

A R = \frac{Δt \cdot m \cdot \sqrt{2 \cdot g}}{(\frac{1}{\sqrt{h 2}} - \frac{1}{\sqrt{H Upstream}}) \cdot 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 Cross Sectional Area given Time required to Lower Liquid for Triangular Notch?

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch evaluator uses Cross-Sectional Area of Reservoir = (Time Interval*(8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2))/((2/3)*((1/Head on Downstream of Weir^(3/2))-(1/Head on Upstream of Weir^(3/2)))) to evaluate the Cross-Sectional Area of Reservoir, The Cross Sectional Area given Time required to Lower Liquid for Triangular Notch is area of a two-dimensional shape that is obtained when a three-dimensional object - such as a cylinder. Cross-Sectional Area of Reservoir is denoted by A_R symbol.

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

FAQs on Cross Sectional Area given Time required to Lower Liquid for Triangular Notch

What is the formula to find Cross Sectional Area given Time required to Lower Liquid for Triangular Notch?

The formula of Cross Sectional Area given Time required to Lower Liquid for Triangular Notch is expressed as Cross-Sectional Area of Reservoir = (Time Interval*(8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2))/((2/3)*((1/Head on Downstream of Weir^(3/2))-(1/Head on Upstream of Weir^(3/2)))). Here is an example- 14.06364 = (1.25*(8/15)*0.66*sqrt(2*9.8)*tan(0.5235987755982/2))/((2/3)*((1/5.1^(3/2))-(1/10.1^(3/2)))).

How to calculate Cross Sectional Area given Time required to Lower Liquid for Triangular Notch?

With Time Interval (Δt), Coefficient of Discharge (C_d), Acceleration due to Gravity (g), Theta (θ), Head on Downstream of Weir (h₂) & Head on Upstream of Weir (H_Upstream) we can find Cross Sectional Area given Time required to Lower Liquid for Triangular Notch using the formula - Cross-Sectional Area of Reservoir = (Time Interval*(8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2))/((2/3)*((1/Head on Downstream of Weir^(3/2))-(1/Head on Upstream of Weir^(3/2)))). This formula also uses Tangent (tan), Square Root (sqrt) function(s).

What are the other ways to Calculate Cross-Sectional Area of Reservoir?

Here are the different ways to Calculate Cross-Sectional Area of Reservoir-

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

Can the Cross Sectional Area given Time required to Lower Liquid for Triangular Notch be negative?

No, the Cross Sectional Area given Time required to Lower Liquid for Triangular Notch, measured in Area cannot be negative.

Which unit is used to measure Cross Sectional Area given Time required to Lower Liquid for Triangular Notch?

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch is usually measured using the Square Meter[m²] for Area. Square Kilometer[m²], Square Centimeter[m²], Square Millimeter[m²] are the few other units in which Cross Sectional Area given Time required to Lower Liquid for Triangular Notch can be measured.

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Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Formula

​GoCross Sectional Area given Time required to Lower Liquid Surface Formula

​GoCross Sectional Area given time required to Lower Liquid Surface using Bazins Formula Formula

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Example

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Solution

Cross Sectional Area given Time required to Lower Liquid for Triangular Notch Formula Elements

Other Formulas to find Cross-Sectional Area of Reservoir

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

How to Evaluate Cross Sectional Area given Time required to Lower Liquid for Triangular Notch?

FAQs on Cross Sectional Area given Time required to Lower Liquid for Triangular Notch

Variable Name

GoCross Sectional Area given Time required to Lower Liquid Surface Formula

GoCross Sectional Area given time required to Lower Liquid Surface using Bazins Formula Formula