FormulaDen.com

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

Specific Weight of Fluid 1 given Differential Pressure between Two Points Formula

GoSpecific Weight of Inclined Manometer Liquid Formula

GoSpecific Weight of Liquid given Total Hydrostatic Force Formula

Fx Copy

LaTeX Copy

Specific weight 1 is the specific weight of the fluid 1. Check FAQs

γ 1 = \frac{Δp + γ 2 \cdot h 2}{h 1}

γ₁ - Specific Weight 1?Δp - Pressure Changes?γ₂ - Specific Weight 2?h₂ - Height of Column 2?h₁ - Height of Column 1?

Specific Weight of Fluid 1 given Differential Pressure between Two Points Example

With values

With units

Only example

Here is how the Specific Weight of Fluid 1 given Differential Pressure between Two Points equation looks like with Values.

Here is how the Specific Weight of Fluid 1 given Differential Pressure between Two Points equation looks like with Units.

Here is how the Specific Weight of Fluid 1 given Differential Pressure between Two Points equation looks like.

5579.6667 = \frac{3.36 + 1223 \cdot 0.052}{0.012}

5579.6667N/m³ = \frac{3.36Pa + 1223N/m³ \cdot 0.052m}{0.012m}

5579.6667 = \frac{3.36 + 1223 \cdot 0.052}{0.012}

Tip: Use pencil ( Pencil

) icon above to edit Input Values and Units.

Calculate

Recalculate

Solution

Copy

Reset

You are here -

Home » Category

Engineering » Category

Mechanical » Category

Fluid Mechanics » fx Specific Weight of Fluid 1 given Differential Pressure between Two Points

Specific Weight of Fluid 1 given Differential Pressure between Two Points Solution

Follow our step by step solution on how to calculate Specific Weight of Fluid 1 given Differential Pressure between Two Points?

FIRST Step Consider the formula

γ 1 = \frac{Δp + γ 2 \cdot h 2}{h 1}

Next Step Substitute values of Variables

∴

γ 1 = \frac{3.36Pa + 1223N/m³ \cdot 0.052m}{0.012m}

Next Step Prepare to Evaluate

∴

γ 1 = \frac{3.36 + 1223 \cdot 0.052}{0.012}

Next Step Evaluate

∴

γ 1 = 5579.66666666667N/m³

LAST Step Rounding Answer

∴

γ 1 = 5579.6667N/m³

Specific Weight of Fluid 1 given Differential Pressure between Two Points Formula Elements

Variables

Specific Weight 1

Specific weight 1 is the specific weight of the fluid 1.

Symbol: γ₁

Measurement: Specific WeightUnit: N/m³

Note: Value can be positive or negative.

Formulas to find Specific Weight 1

Pressure Changes

Pressure changes is the difference between the pressure inside the liquid droplet and atmospheric pressure.

Symbol: Δp

Measurement: PressureUnit: Pa

Note: Value can be positive or negative.

Formulas to find Pressure Changes

Specific Weight 2

Specific weight 2 is the specific weight of 2nd fluid.

Symbol: γ₂

Measurement: Specific WeightUnit: N/m³

Note: Value can be positive or negative.

Formulas to find Specific Weight 2

Height of Column 2

Height of column 2 is the length of the column2 measured from bottom to Top.

Symbol: h₂

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Height of Column 2

Height of Column 1

Height of column 1 is the length of the column1 measured from bottom to Top.

Symbol: h₁

Measurement: LengthUnit: m

Note: Value can be positive or negative.

Formulas to find Height of Column 1

Other Formulas to find Specific Weight 1

Go Specific Weight of Inclined Manometer Liquid

γ 1 = \frac{P a}{L m \cdot \sin (Θ)}

Go Specific Weight of Liquid given Total Hydrostatic Force

γ 1 = \frac{F hs}{h G \cdot A wet}

Other formulas in Specific Weight category

Go Specific Weight

γ Unit = \frac{W body}{V T}

Go Specific Weight of Fluid 2 given Differential Pressure between Two Points

γ 2 = \frac{γ 1 \cdot h 1 - Δp}{h 2}

Go Specific Weight of Liquid given its Absolute Pressure at Height

γ liq = \frac{P abs - P atm}{h}

Go Specific Weight of Liquid given Buoyancy Force

y = \frac{F b}{V O}

How to Evaluate Specific Weight of Fluid 1 given Differential Pressure between Two Points?

Specific Weight of Fluid 1 given Differential Pressure between Two Points evaluator uses Specific Weight 1 = (Pressure Changes+Specific Weight 2*Height of Column 2)/Height of Column 1 to evaluate the Specific Weight 1, The Specific Weight of Fluid 1 given Differential Pressure between Two Points formula is defined as the force exerted by gravity on a unit volume of a fluid. Specific Weight 1 is denoted by γ₁ symbol.

How to evaluate Specific Weight of Fluid 1 given Differential Pressure between Two Points using this online evaluator? To use this online evaluator for Specific Weight of Fluid 1 given Differential Pressure between Two Points, enter Pressure Changes (Δp), Specific Weight 2 (γ₂), Height of Column 2 (h₂) & Height of Column 1 (h₁) and hit the calculate button.

FAQs on Specific Weight of Fluid 1 given Differential Pressure between Two Points

What is the formula to find Specific Weight of Fluid 1 given Differential Pressure between Two Points?

The formula of Specific Weight of Fluid 1 given Differential Pressure between Two Points is expressed as Specific Weight 1 = (Pressure Changes+Specific Weight 2*Height of Column 2)/Height of Column 1. Here is an example- 5579.667 = (3.36+1223*0.052)/0.012.

How to calculate Specific Weight of Fluid 1 given Differential Pressure between Two Points?

With Pressure Changes (Δp), Specific Weight 2 (γ₂), Height of Column 2 (h₂) & Height of Column 1 (h₁) we can find Specific Weight of Fluid 1 given Differential Pressure between Two Points using the formula - Specific Weight 1 = (Pressure Changes+Specific Weight 2*Height of Column 2)/Height of Column 1.

What are the other ways to Calculate Specific Weight 1?

Here are the different ways to Calculate Specific Weight 1-

Specific Weight 1=Pressure A/(Length of Manometer*sin(Angle))
Specific Weight 1=Hydrostatic Force/(Depth of Centroid*Wet Surface Area)

Can the Specific Weight of Fluid 1 given Differential Pressure between Two Points be negative?

Yes, the Specific Weight of Fluid 1 given Differential Pressure between Two Points, measured in Specific Weight can be negative.

Which unit is used to measure Specific Weight of Fluid 1 given Differential Pressure between Two Points?

Specific Weight of Fluid 1 given Differential Pressure between Two Points is usually measured using the Newton per Cubic Meter[N/m³] for Specific Weight. Newton per Cubic Centimeter[N/m³], Newton per Cubic Millimeter[N/m³], Kilonewton per Cubic Meter[N/m³] are the few other units in which Specific Weight of Fluid 1 given Differential Pressure between Two Points can be measured.

Let Others Know

✖

Facebook

Twitter

Copied!

Specific Weight of Fluid 1 given Differential Pressure between Two Points Formula

​GoSpecific Weight of Inclined Manometer Liquid Formula

​GoSpecific Weight of Liquid given Total Hydrostatic Force Formula

Specific Weight of Fluid 1 given Differential Pressure between Two Points Example

Specific Weight of Fluid 1 given Differential Pressure between Two Points Solution

Specific Weight of Fluid 1 given Differential Pressure between Two Points Formula Elements

Other Formulas to find Specific Weight 1

Other formulas in Specific Weight category

How to Evaluate Specific Weight of Fluid 1 given Differential Pressure between Two Points?

FAQs on Specific Weight of Fluid 1 given Differential Pressure between Two Points

Variable Name

GoSpecific Weight of Inclined Manometer Liquid Formula

GoSpecific Weight of Liquid given Total Hydrostatic Force Formula