### Manometer Calculator:

This manometer calculator will help you find the hydrostatic pressure in a liquid using a manometer. Read on to learn about Pascal’s principle and find out what manometers and u-tube manometers are. To help you easily understand the physics behind this simple device, we will also show some examples with the manometer equation.

The free manometer pressure calculator tool is one of the best tools to help you find the gauge pressure of any type of manometer. Also, it contains a simple step-by-step procedure for finding the pressure of a manometer with manometer definition and formulas.

We have also given solved sample questions at the end for better understanding of the topic.

A manometer is a simple instrument used to measure fluid pressure. It consists of a uniform glass tube connected to a tube. It is nothing but a tube used to measure pressure differences due to the interaction of liquids and atmospheric pressure.

A manometer contains liquid mercury of high density and can contain water or any other liquid. A manometer works following Pascal’s principle. The pressure exerted on a region of an enclosed incompressible fluid,

Pascal’s principle states that conduction through a fluid is equal in all directions.

One of the best examples of a manometer is toothpaste. Formulas for calculating manometer pressure or hydrostatic pressure equations are along the lines

### Manometer Formula:

Manometer formula is pressure of the liquid P_{(pa)} in pascal is equal to the liuid density ρ_{(kg/m3)} in kilogram per meter^{3} and multiply the acceleration due to gravity g_{(m/s2)} in meter per second^{2} and multiply the height h_{(m)} in meter.Hence the manometer formula can be written as

P_{(pa)}=ρ_{(kg/m3)}*g_{(m/s2)}*h_{(m)}

The pressure equation for u tube manometer attached to pipe

P=ρ_{1}*g*h_{1}+(-)+ρ_{2}*g*h_{2}

Where

P=is the pressure of the liquid

H=is the height

Ρ=is the liquid density

G=is the acceleration due to gravity

### Calculation:1

In a single column monomer, one end is attached to the atmosphere and the other is an open tank.If the atmospheric pressure is 1525 pa and liquid , liquid density is 1500 kg /m^{3} and the liquid height is 0.4m,find the gauge pressure and absolute pressure?

### Answer:

Given that

Atmospheric pressure =1525 pa

Liquid density=1500 kg/m3

Liquid height=0.4m

Gauge pressure is

P_{(pa)}=ρ_{(kg/m3)}*g_{(m/s2)}*h_{(m)}

P=1500*9.8*0.4

P=5880pa

Absolute pressure =5880-1525

=4355pa

Therefore,gauge pressure is 5880pa and absolute pressure 4355pa

### Example:2

In a single column monomer, one end is attached to the atmosphere and the other is an open tank.If the atmospheric pressure is 1025 pa and liquid , liquid density is 1050kg /m^{3} and the liquid height is 0.2m,find the gauge pressure and absolute pressure?

### Answer:

Given that

Atmospheric pressure =1025 pa

Liquid density=1050 kg/m3

Liquid height=0.2m

Gauge pressure is

P_{(pa)}=ρ_{(kg/m3)}*g_{(m/s2)}*h_{(m)}

P=1050*9.8*0.2

P=2058pa

Absolute pressure =2058-1025

=1033pa

Therefore,gauge pressure is 5880pa and absolute pressure 4355pa