Physics Formulas

Mass Flow Rate Formula

The conservation of mass is the fundamental concept of physics. It is the part of thermodynamics physics. Within a given problem domain, the amount of mass always should remain constant. Hence mass is neither created nor destroyed. The mass of any object is simply the volume that is occupied by the object multiplied with the density of the object. Also, for a fluid, the density, volume, and shape of the object can all change within the domain with time. And mass may move through the domain. In this topic, we will discuss the concept of Mass Flow rate formula with examples. Let us begin!

Concept of Mass Flow Rate

The conservation of mass is telling us that the mass flow rate through a tube must be constant. We can compute the value of the mass flow rate from the given flow conditions.

Mass Flow Rate Formula

Mass Flow Rate is the rate of movement of a massive fluid through the unit area. Obviously this flow rate depends on the density, velocity of the fluid and the area of the cross-section. Therefore, it is the movement of mass per unit time. It is measured in the unit of kg per second. Thus we can say that the mass flow rate is the mass of a liquid substance passing per unit time.

The mass flow formula: Mass Flow Rate = (density) × (velocity) × (area of the cross-section)

Mathematically, \(m = \rho \times V \times A\)

Where,

m Mass Flow Rate
\(\rho\) The density of the fluid
V The velocity of the fluid
A Area of cross-section

Solved Examples for Mass Flow Rate Formula

Q.1: A fluid is moving through a tube at a speed of 20 meters per second. The tube has a transverse area of 0.3 square meters. The density of the fluid is given as \rho = 1.5 grams per cubic meter. Determine the amount of mass flowing through the tube using Mass Flow Rate Formula

Solution:  The total mass of the fluid flowing through the tube is given by the formula,

\(m = \rho \times V \times A\)

\(\rho = 1.5 grams per cubic meter\)

A =  0.3 square meter

V = 20 meter per second

Thus, \(m = \rho \times V \times A \\\)

\(m = 1.5 \times 20 \times 0.3\)

m= 9.0 grams per second.

Mass flow through the tube = 9.0 grams per second.

Q.2: The rate mass of a fluid is given as 9 grams per second. It is flowing in a tube at a rate of 0.5 meters per second and it has a density of 1.5 grams per cubic meter. Compute the diameter of the tube?

Solution: Given here,

m  = 9 grams per second

\(\rho = 1.5 grams per cubic meter\)

V =0.5 meter per second

Then rearranging the formula we will have,

\(A = \frac{m}{\rho \times V}\)

Substituting the known values, we get

\(A = \frac{9}{1.5 \times 0.5} = \frac {9}{0.75}\)

= 12 square meter.

Also, cross section area formula is, \(A = \pi r^2\) for radius r

Then, \(A = \pi r^2 \\\)

Rearranging the above formula,

\(r =  \sqrt{\frac{A}{\pi}} \\\)

Substituting the values, we will get,

\(r =  \sqrt{\frac{12}{3.14}} \\\)

\(r = 1.95 \\\)

\(d = 3.9 \; meter\)

Therefore the diameter of the tube will be 3.9 meters.

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5 responses to “Spring Potential Energy Formula”

  1. Typo Error>
    Speed of Light, C = 299,792,458 m/s in vacuum
    So U s/b C = 3 x 10^8 m/s
    Not that C = 3 x 108 m/s
    to imply C = 324 m/s
    A bullet is faster than 324m/s

  2. Malek safrin says:

    I have realy intrested to to this topic

  3. umer says:

    m=f/a correct this

  4. Kwame David says:

    Interesting studies

  5. Yashdeep tiwari says:

    It is already correct f= ma by second newton formula…

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