In this article we will discuss about the “ Is Mass flow rate constant” at standard condition, with detailed facts.

**The mass flow rate can be defined as the mass per unit area of a liquid substance at a standard temperature and pressure**.**The unit of the mass flow rate is kilogram per second (kg/s).**

**Mass Flow Rate Equation:**

The formula of the **mass flow** rate is, **Mass flow rate =** change in time/change in mass

Mathematically it can be expressed as,

Where,

=Mass flow rate

dm = Change in the mass

dt = Change in time

From the equation the mass flow rate can be easily describe that, it is **the ratio of the change is mass of the fluid to the change of time**. In the equation the m dot is mainly used to classified the regular m which we are used to denoted the mass in the regular field.

**Read more about Mass flow rate: Its important relations and FAQs**

**Is mass flow rate always constant?**

Yes, the mass flow rate remains always constant, if any external mass is no added or removed.

**The mass flow rate is the rate of movement of the liquid substance, which is passes through in a unit area.**

In the field of chemistry and physics the term mass flow rate comes from the low of **conversion of mass**. For any object the mass is all-time remain constant in a standard pressure and temperature.

If a object is measured in any surface of the earth the total mass of the object is remain constant, there is no change in the mass for the any object. The total amount of mass which is present in an object is cannot be created itself or destroyed itself, only the amount of mass transferred from one to another form just like an energy. By the help of the mass we could recognize the measure of the inertia of a body. If net force applied to a body then it offers to change the resistance to acceleration of that body. The mass of a body is measured by lever balance.

**Why is mass flow rate constant?**

From the field of the thermodynamics we get a clear concept about the mass flow rate.

**The mass flow rate remains constant because it is mainly calculated in a standard condition**.

**When is mass flow rate constant?**

The law of conservation for mass states that, the amount of mass present in a physical body is cannot be created or destroyed by itself.

**The mass flow rate remains constant if the external mass is not added or removed from the system law of conversion states that.**

If a large amount of physical body is transformed in a small physical body then the amount of mass present in the physical body also transformed in the small portion of the new physical body, if they all together added then the total mass for the body remain same.

**Is mass flow rate constant in turbine?**

In turbine steady flow is observed. The **steady flow** can be defined as the amount of liquid flow for every per second in a body. Steady flow can be stays in two form neither uniform nor non uniform.

**In the turbine the mass flow rate is constant, if the flow remains steady.**

The total amount of steam which is present in the reaction turbine is entering mainly through the moving blades.

**Mass flow rate in turbine:**

The steam is spread in the whole circumference of the turbine. Now consider a reaction whose and view of the blade ring.

Let, d= Diameter of the rotor drum

h = Height of blades, and

] *V _{F1}*=Velocity of flow at exit

Now, the total amount of the area from through the steam is flowing,

*A = π(d+h)h*

And, volume of steam flowing =*π(d+h)hV _{F1}*

We know that, volume of 1 kg of steam at the given pressure is * v _{g}* (from the steam table).Therefore mass of steam flowing is,

*m* = *π(d+h)hV _{F1}/v_{g}*

Let, the dryness fraction of the steam is x.

So, the mass of the steam which is flowing is,

**Is mass flow rate constant in a nozzle?**

The nozzle flow can be describing as the liquid or gas moves through a tapering mouthpiece which is attached with a water pipeline without changing the entropy of the system.

**Yes, the mass flow rate in a nozzle is a constant. We get the clear concept about the topic from the law of conversion of mass.**

**Mass of steam discharge through nozzle**:

From the general law of the gas we could know that, the present steam in the nozzle is isentropic.

So, we could write that,

] *pv ^{n}*= Constant

We know that gain in kinematic energy = V^{2}/2

And, Heat drop = Work done during Rankine cycle = n/(n-1) (p_{1}v_{1}-p_{2}v_{2})

Through this process kinetic energy is gain, which is equal to the heat drop,

*n/(n-1)p _{1}v_{1}(1-p_{2}v_{2})/(p_{1}v_{1})*

We know that,

v_{2}/v_{1} = … ^{(}1/n) = … ^{(}-1/n)

v_{2} = v_{1}… ^{(}-1/n)

Substituting the value of v_{2}…v_{1} and we get,

_{Now the volume of steam flowing per second = Cross – sectional area of nozzle} *_{ Velocity of steam =} AV_{2 }

_{And volume of 1 kg of steam specific volume of steam at pressure}, *p _{2} = v_{2}m^{3}/kg*

So, mass of steam discharged through nozzle per second,

m = (volume of steam flowing per second)/(Volume of 1 kg of steam at pressure p_{2})

Substituting the value v_{2 } and we get,

_{Problem: In a nozzle the area is about 200 square in mm.The nozzle which is contains the temperature and pressure are respectively given of 27 degree} _{C and of 20 bar. In the first condition the pressure is enters. Now in the same nozzle the pressure emit at 4 bar. Now calculate the amount of mass of air discharges?}

_{Problem: In a nozzle the area is about 200 square in mm.The nozzle which is contains the temperature and pressure are respectively given of 27 degree}

_{C and of 20 bar. In the first condition the pressure is enters. Now in the same nozzle the pressure emit at 4 bar. Now calculate the amount of mass of air discharges?}

**Solution**: Given data,

T_{1 }= 300K

p_{1 }= 20bar = 20 x 10^{5} Newton per sqauremetre

p_{2} = 4 bar = 4 x 10^{5} Newton per square metre

A = 200squaremm = 200 x 10^{-6}

Let, v_{1} = Specific volume of air in cubic metre per kg

We know that,

*p _{1}v_{1} = mRT_{1}*

*v _{1} = mRT_{1}*/

*p*= 1 x 287 x 300/20 x 10

_{1}^{5}= 0.043cubicmetre/kg

So, the amount of steam is emitted by the nozzle is,

m = …(2n/(n-1)p_{1}/v_{1}…(p_{2}/p_{1})…^{(}2/n) – …(p_{2}/p_{1})…(n+1)/n) = 0.72kg per secong

So,the amount of mass discharge is 0.72 kg/s.

**Is mass flow rate constant in compressible flow?**

Compressible flow: The compressible flow can be defined as the, during the flow of the liquid, the velocity and the density will changes. Compressible flow is present in all gasses in the atmosphere.

**The compressible flow means the flow in the ideal gas the mass flow rate is constant under some fixed unique function.**

The functions which are dependent on the mass flow rate constant in the compressible flow is given below,

**Mach number****Pressure****Temperature****Property of the gas**

**Frequent ask questions:**

**Question: What is the Mach number and write down the importance of it?**

**Question: What is the Mach number and write down the importance of it?**

**Solution:**With the help of the Mach number we get a clear concept about the type of flow of any fluid.

**Mach number: Mach number can be defined as the, the ratio between the velocity of the fluid and the velocity of sound wave.**

In general, the Mach number divided in **four categories** for the flow of fluid. They are,

**Subsonic flow**: It can be describe as the when the Mach number is **less from the unity.**

**Sonic flow**: It can be defined as the when the Mach number is **equal to the unity.**

**Supersonic flow**: It can be defined as the when the Mach number is** between the unity number of 1 and 6.**

**Hyper sonic flow**: It can be defined as the when the Mach number is **more than unity number of 6**.

**Question: In a tank the water capacity is 90 cubic meters, which fill the brim. Calculate the mass flow rate. To fill up the tank it takes time about 3 hours.**

**Question: In a tank the water capacity is 90 cubic meters, which fill the brim. Calculate the mass flow rate. To fill up the tank it takes time about 3 hours.**

**Solution**: Given data are, v = 90 cubic metre

We know that,

= 1000 kg per cubic metre

So, m = 90 x 1000 = 90000 kg

So, the mass flow rate of the brim is 8.3 kg/s.

**Question: From a cylinder water is drained about 700 grams. After the drainage of the water found that the remaining water in the cylinder is about 300 grams. The whole process takes time 60 second. Calculate the mass flow rate.**

**Question: From a cylinder water is drained about 700 grams. After the drainage of the water found that the remaining water in the cylinder is about 300 grams. The whole process takes time 60 second. Calculate the mass flow rate.**

**Solution**: We know that, the mass flow rate,

The mass flow rate is -6.66 kg/s.

The negative sign means the water is emitted from the system.

**Question: In the direction of the motion for a reaction turbine which has the blades tips in the inclination angles. The tips angles of the reaction turbines are respectively are** **20 degree** **and** **35 degree** **.Now the guide blades which are contained by the reaction turbine, have the exactly same shape just like the moving blades. The direction of the moving blade and the guide blades are the opposite. In a fixed place the reaction turbine has the drum diameter is about 1 metre and the height is 100 mm. In this condition the reaction turbine has the pressure of the steam is 1.7 bar. Now the dryness factor of the steam is 0.935.The steam of the reaction turbine passes through the blades without any shock. The speed in the reaction turbine is 250 r.p.m. **

**Now calculate the**,

**1. Mass of the steam flow**

**2. How much power is developed?**

**Solution**: Given data,

θ = β = 35 degree

d = 1m

h = 100m = 0.1m

p = 1.7 bar

x = 0.937

N = 250 r.p.m

We know that blade speed,

V_{b }= π (d+h)N/60 = π (1+0.1)250/60 = 14.4m/s

1. First of all, draw a horizontal line, and cut off AB equal to 14.4 m/s to some suitable scale to represent the velocity of the blade.

2. Now draw inlet velocity triangle ABC the base AB with α = 20 degree and β= 35 degree .

3. Similarly draw outlet velocity triangle ABD on the same base AB with

= 20 degree and = 35 degree

4. From C and D draw perpendiculars to meet the line AB produced at E and F.

By measuring from velocity triangle, we find that,

Change in the velocity of whirl, (Vω+Vω_{1} )= EF = 42.5 m/s

And velocity of flow at outlet, V_{f1} = DF = 10 m/s

Mass flow of steam,

If we go through of the steam tables, the pressure of 1.7 bars, we could easily find that specific volume for the steam,

*v _{g}*= 1.031

From the equation of mass flow of steam we could write,

*m = π(d+h)h x V _{f1}/xv_{g }= 3.58 kg/s*

Now, power developed

P = m(Vω+Vω_{1} )V_{b} = 3,58 x 42.5 x 14.4 = 2191W

**Question: Where the mass flow rate is applied?**

**Solution**: The mass flow rate devices are uses in many purposes in the engineering field.

They are discuss in below,

**To check the high accuracy of the instruments****Controlling the molecules of the flowing gas.****Precision measurement.**