FormulaDen.com

Physics Chemistry Math Chemical Engineering Civil Electrical Electronics Electronics and Instrumentation Materials Science Mechanical Production Engineering Financial Health

Head2 given Time Required to Lower Liquid for Triangular Notch Formula

GoHead2 given Time Required to Lower Liquid Surface Formula

GoHead2 given Time Required to Lower Liquid Surface using Bazins Formula Formula

Fx Copy

LaTeX Copy

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 (\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}}

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?H_Upstream - Head on Upstream of Weir?

Head2 given Time Required to Lower Liquid for Triangular Notch Example

With values

With units

Only example

Here is how the Head2 given Time Required to Lower Liquid for Triangular Notch equation looks like with Values.

Here is how the Head2 given Time Required to Lower Liquid for Triangular Notch equation looks like with Units.

Here is how the Head2 given Time Required to Lower Liquid for Triangular Notch equation looks like.

4.9291 = {(\frac{1}{(\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{1}{{10.1}^{\frac{3}{2}}})})}^{\frac{2}{3}}

4.9291m = {(\frac{1}{(\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{1}{{10.1m}^{\frac{3}{2}}})})}^{\frac{2}{3}}

4.9291 = {(\frac{1}{(\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{1}{{10.1}^{\frac{3}{2}}})})}^{\frac{2}{3}}

Tip: Use pencil ( Pencil

) icon above to edit Input Values and Units.

Calculate

Recalculate

Solution

Copy

Reset

You are here -

Home » Category

Engineering » Category

Civil » Category

Hydraulics and Waterworks » fx Head2 given Time Required to Lower Liquid for Triangular Notch

Head2 given Time Required to Lower Liquid for Triangular Notch Solution

Follow our step by step solution on how to calculate Head2 given Time Required to Lower Liquid for Triangular Notch?

FIRST Step Consider the formula

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}}

Next Step Substitute values of Variables

∴

h 2 = {(\frac{1}{(\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{1}{{10.1m}^{\frac{3}{2}}})})}^{\frac{2}{3}}

Next Step Convert Units

∴

h 2 = {(\frac{1}{(\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{1}{{10.1m}^{\frac{3}{2}}})})}^{\frac{2}{3}}

Next Step Prepare to Evaluate

∴

h 2 = {(\frac{1}{(\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{1}{{10.1}^{\frac{3}{2}}})})}^{\frac{2}{3}}

Next Step Evaluate

∴

h 2 = 4.9290844130142m

LAST Step Rounding Answer

∴

h 2 = 4.9291m

Head2 given Time Required to Lower Liquid for Triangular Notch 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

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

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 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 Surface using Bazins Formula

h 2 = {(\frac{1}{\frac{Δt \cdot m \cdot \sqrt{2 \cdot g}}{2 \cdot A R} + (\frac{1}{\sqrt{H Upstream}})})}^{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 Head2 given Time Required to Lower Liquid for Triangular Notch?

Head2 given Time Required to Lower Liquid for Triangular Notch evaluator uses 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) to evaluate the Head on Downstream of Weir, Head2 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 Downstream of Weir is denoted by h₂ symbol.

How to evaluate Head2 given Time Required to Lower Liquid for Triangular Notch using this online evaluator? To use this online evaluator for Head2 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), 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 for Triangular Notch

What is the formula to find Head2 given Time Required to Lower Liquid for Triangular Notch?

The formula of Head2 given Time Required to Lower Liquid for Triangular Notch is expressed as 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). Here is an example- 4.929084 = (1/(((1.25*(8/15)*0.66*sqrt(2*9.8)*tan(0.5235987755982/2))/((2/3)*13))+(1/10.1^(3/2))))^(2/3).

How to calculate Head2 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), Cross-Sectional Area of Reservoir (A_R) & Head on Upstream of Weir (H_Upstream) we can find Head2 given Time Required to Lower Liquid for Triangular Notch using the formula - 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). This formula also uses Tangent (tan), 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*Bazins Coefficient*sqrt(2*Acceleration due to Gravity))/(2*Cross-Sectional Area of Reservoir)+(1/sqrt(Head on Upstream of Weir))))^2

Can the Head2 given Time Required to Lower Liquid for Triangular Notch be negative?

Yes, the Head2 given Time Required to Lower Liquid for Triangular Notch, measured in Length can be negative.

Which unit is used to measure Head2 given Time Required to Lower Liquid for Triangular Notch?

Head2 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 Head2 given Time Required to Lower Liquid for Triangular Notch can be measured.

Let Others Know

✖

Facebook

Twitter

Copied!

: Value
: Unit

Head2 given Time Required to Lower Liquid for Triangular Notch Formula

​GoHead2 given Time Required to Lower Liquid Surface Formula

​GoHead2 given Time Required to Lower Liquid Surface using Bazins Formula Formula

Head2 given Time Required to Lower Liquid for Triangular Notch Example

Head2 given Time Required to Lower Liquid for Triangular Notch Solution

Head2 given Time Required to Lower Liquid for Triangular Notch 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 for Triangular Notch?

FAQs on Head2 given Time Required to Lower Liquid for Triangular Notch

Variable Name

GoHead2 given Time Required to Lower Liquid Surface Formula

GoHead2 given Time Required to Lower Liquid Surface using Bazins Formula Formula