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Time Required to Lower Liquid Surface for Triangular Notch Formula

GoTime Required to Lower Liquid Surface Formula

GoTime Required to Lower Liquid Surface using Bazins Formula Formula

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

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

Δt - Time Interval?A_R - Cross-Sectional Area of Reservoir?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?

Time Required to Lower Liquid Surface for Triangular Notch Example

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

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

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

1.1555 = (\frac{(\frac{2}{3}) \cdot 13}{(\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8} \cdot \tan (\frac{30}{2})}) \cdot ((\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1}{{10.1}^{\frac{3}{2}}}))

1.1555s = (\frac{(\frac{2}{3}) \cdot 13m²}{(\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8m/s²} \cdot \tan (\frac{30°}{2})}) \cdot ((\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1}{{10.1m}^{\frac{3}{2}}}))

1.1555 = (\frac{(\frac{2}{3}) \cdot 13}{(\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8} \cdot \tan (\frac{30}{2})}) \cdot ((\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1}{{10.1}^{\frac{3}{2}}}))

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Time Required to Lower Liquid Surface for Triangular Notch Solution

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

Δt = (\frac{(\frac{2}{3}) \cdot 13m²}{(\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8m/s²} \cdot \tan (\frac{30°}{2})}) \cdot ((\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1}{{10.1m}^{\frac{3}{2}}}))

Next Step Convert Units

∴

Δt = (\frac{(\frac{2}{3}) \cdot 13m²}{(\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8m/s²} \cdot \tan (\frac{0.5236rad}{2})}) \cdot ((\frac{1}{{5.1m}^{\frac{3}{2}}}) - (\frac{1}{{10.1m}^{\frac{3}{2}}}))

Next Step Prepare to Evaluate

∴

Δt = (\frac{(\frac{2}{3}) \cdot 13}{(\frac{8}{15}) \cdot 0.66 \cdot \sqrt{2 \cdot 9.8} \cdot \tan (\frac{0.5236}{2})}) \cdot ((\frac{1}{{5.1}^{\frac{3}{2}}}) - (\frac{1}{{10.1}^{\frac{3}{2}}}))

Next Step Evaluate

∴

Δt = 1.1554617380882s

LAST Step Rounding Answer

∴

Δt = 1.1555s

Time Required to Lower Liquid Surface for Triangular Notch Formula Elements

Variables

Functions

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

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

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 Time Interval

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 Time Required to Lower Liquid Surface using Bazins Formula

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

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

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}})

How to Evaluate Time Required to Lower Liquid Surface for Triangular Notch?

Time Required to Lower Liquid Surface for Triangular Notch evaluator uses Time Interval = (((2/3)*Cross-Sectional Area of Reservoir)/((8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2)))*((1/Head on Downstream of Weir^(3/2))-(1/Head on Upstream of Weir^(3/2))) to evaluate the Time Interval, The Time Required to Lower Liquid Surface for Triangular Notch is defined as amount of time taken by lowering water surface from Head at upstream to downstream. Time Interval is denoted by Δt symbol.

How to evaluate Time Required to Lower Liquid Surface for Triangular Notch using this online evaluator? To use this online evaluator for Time Required to Lower Liquid Surface for Triangular Notch, enter Cross-Sectional Area of Reservoir (A_R), 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 Time Required to Lower Liquid Surface for Triangular Notch

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

The formula of Time Required to Lower Liquid Surface for Triangular Notch is expressed as Time Interval = (((2/3)*Cross-Sectional Area of Reservoir)/((8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2)))*((1/Head on Downstream of Weir^(3/2))-(1/Head on Upstream of Weir^(3/2))). Here is an example- 1.155462 = (((2/3)*13)/((8/15)*0.66*sqrt(2*9.8)*tan(0.5235987755982/2)))*((1/5.1^(3/2))-(1/10.1^(3/2))).

How to calculate Time Required to Lower Liquid Surface for Triangular Notch?

With Cross-Sectional Area of Reservoir (A_R), 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 Time Required to Lower Liquid Surface for Triangular Notch using the formula - Time Interval = (((2/3)*Cross-Sectional Area of Reservoir)/((8/15)*Coefficient of Discharge*sqrt(2*Acceleration due to Gravity)*tan(Theta/2)))*((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 Time Interval?

Here are the different ways to Calculate Time Interval-

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

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

No, the Time Required to Lower Liquid Surface for Triangular Notch, measured in Time cannot be negative.

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

Time Required to Lower Liquid Surface for Triangular Notch is usually measured using the Second[s] for Time. Millisecond[s], Microsecond[s], Nanosecond[s] are the few other units in which Time Required to Lower Liquid Surface for Triangular Notch can be measured.

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Time Required to Lower Liquid Surface for Triangular Notch Formula

​GoTime Required to Lower Liquid Surface Formula

​GoTime Required to Lower Liquid Surface using Bazins Formula Formula

Time Required to Lower Liquid Surface for Triangular Notch Example

Time Required to Lower Liquid Surface for Triangular Notch Solution

Time Required to Lower Liquid Surface for Triangular Notch Formula Elements

Other Formulas to find Time Interval

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

How to Evaluate Time Required to Lower Liquid Surface for Triangular Notch?

FAQs on Time Required to Lower Liquid Surface for Triangular Notch

Variable Name

GoTime Required to Lower Liquid Surface Formula

GoTime Required to Lower Liquid Surface using Bazins Formula Formula