**Mass flow rate Definition**

**Mass flow rate Equation | Mass flow rate units | Mass flow rate symbol**

It is denoted by **ṁ**, It is Formulated as,

[latex]\dot{m}=\frac{dm}{dt}[/latex]

In Hydrodynamics

[latex]\dot{m}=\rho AV=\rho Q[/latex]

Where,

ρ = Density of the fluid

A = Cross sectional Area

V = Velocity of flow of fluid

Q = Volume flow rate or discharge

It has unit kg/s, lb./min etc.

**Mass flow rate conversion**

**Mass flow rate from volumetric flow rate**

In hydrodynamic, the mass-flow rate can be derived from volume flow rate with help of Continuity Equation.

The Continuity equation is given by

Q=AV

Where,

A = Cross sectional Area

V = Velocity of flow of fluid

Multiplying the continuity equation with density of the fluid we get ,

[latex]\dot{m}=\rho AV=\rho Q[/latex]

Where,

ρ = Density of the fluid

**Mass flow rate to velocity | It’s Relationship with each other**

In Hydrodynamics

[latex]\dot{m}=\rho AV=\rho Q[/latex]

Where,

ρ = Density of the fluid

A = Cross sectional Area

V = Velocity of flow of fluid

Q = Volume flow rate or discharge

For an in-compressible fluid carrying through a fixed cross section, the mass-flow rate is directly proportionate to the velocity of fluid flown.

[latex]\\\dot{m}\propto V\\\\ \frac{\dot{m_1}}{\dot{m_2}}=\frac{V_1}{V_2}[/latex]

**Reynolds number with mass flow rate | Their Generalized relation**

The Reynolds number is given by the equation,

[latex]Re=\frac{\rho VL_c}{\mu}[/latex]

Where,

L_{c} = Characteristic length

V = Velocity of flow of fluid

ρ = Density of the fluid

μ = dynamic viscosity of the fluid

Multiply numerator and denominator by cross sectional Area A

[latex]Re=\frac{\rho AVL_c}{A\mu}[/latex]

But mass-flow rate is

[latex]\dot{m}=\rho AV[/latex]

Thus Reynolds Number becomes

[latex]Re=\frac{\dot{m} L_c}{A\mu}[/latex]

**Mass flow rate problems | Mass flow rate example**

## Q.1] A Turbine operates on a steady flow of air produces 1 kW of Power by expanding air from 300kPa, 350 K, 0.346 m_{3}/kg to 120 kPa. The inlet and outlet velocity are 30 m/s and 50 m/s respectively. The expansion follows the Law PV^{1.4} = C. Determine the mass flow rate of air?

Solution:

[latex]P_1=300 kPa, \;T_1=350 K,\; v_1=0.346\frac{m^3}{kg},\;\dot{W}=1kW=1000W[/latex]

According to Steady Flow energy equation

[latex]q-w=h_2-h_1+\frac{(V_2^2-V_1^2)}{2}+g[Z_2-Z_1][/latex]

Q = 0, Z_{1} = Z_{2}

[latex]W=h_2-h_1+\frac{(V_2^2-V_1^2)}{2}[/latex]

[latex]\dot{W}=\dot{m}w[/latex]

[latex]-w=-\int vdp-\Delta ke[/latex]

PV^{n} = C

[latex]v=\frac{c\frac{1}{n}}{P\frac{1}{n}}[/latex]

[latex]w=-c^\frac{1}{n}\int_{1}^{2}P^\frac{-1}{n}dp-\Delta ke[/latex]

[latex]=-c^\frac{1}{n}*[(P_2^{\frac{-1}{n}+1}-P_1^{\frac{-1}{n}+1}]-\Delta ke[/latex]

[latex]c^{-1/n}=P_1^{1/n} v_1=P_2^{1/n} v_2[/latex]

[latex]w=-\frac{n}{n-1}(P_2 v_2-P_1 v_1 )-\Delta ke[/latex]

[latex]\frac{v_2}{v_1}=[\frac{P_2}{P_1}]^{\frac{1}{n}}[/latex]

We get,

[latex]\\w=-\frac{n}{n-1}P_1v_1[{\frac{P_2}{P_1}}^\frac{n-1}{n}-1]-\Delta ke \\\\w=-\frac{1.4}{1.4-1}300*10^3*0.346*[{\frac{120}{300}}^\frac{1.4-1}{1.4}-1]-\frac{50^2-30^2}{2}\\ \\\\w=82953.18\frac{J}{kg}[/latex]

Mass-Flow rate is

[latex]\dot{m}=\frac{W}{w}=\frac{1000}{82953.18}=0.012\;\frac{kg}{s}[/latex]

## Q.2] Air enters a device at 4 MPa and 300^{o}C with velocity of 150m/s. The inlet area is 10 cm^{2} and Outlet area is 50 cm^{2}.Determine the mass flux if air exits at 0.4 MPa and 100^{o}C?

Ans: A_{1} = 10 cm^{2}, P_{1} = 4 MPa, T_{1} = 573 K, V_{1} = 150m/s, A_{2} = 50 cm^{2}, P_{2} = 0.4 MPa, T_{2} = 373 K

[latex]\rho =\frac{P_1}{RT_1}=\frac{4000}{0.287*573}=24.32 kg/m^3[/latex]

[latex]\\\dot {m}=\rho_1 A_1 V_1=24.32*10*10^{-4}*150\\ \\\dot {m}=3.648\frac{kg}{s}[/latex]

## Q.3] A perfect gas having specific heat at constant pressure as 1 kJ/kgK enters and leaves a gas turbine with same velocity. The temperature of the gas at turbine inlet and outlet are 1100, and 400 Kelvin respectively and The power generation is at the rate 4.6 Mega Watt and heat leakages is at the rate of 300 kilo-Joule/seconds through the turbine casing. Compute mass flow rate of the gas through the turbine. (GATE-17-SET-2)

Solution: C_{p} = 1 kJ/kgK, V_{1} = V_{2}, T_{1} = 1100 K, T_{2} = 400 K, Power = 4600 kW

Heat loss from turbine casing is 300 kJ/s = Q

According to Steady Flow energy equation

[latex]\dot{m}h_1+Q=\dot{m}h_2+W[/latex]

[latex]\\\dot{m}h_1+Q=\dot{m}h_2+W\\ \\\dot{m}[h_1-h_2]=W-Q\\ \\\dot{m}C_p[T_1-T_2]=W-Q\\ \\\dot{m}=\frac{W-Q}{C_p[T_1-T_2]}=\frac{4600+300}{1100-400}=7\;\frac{kg}{s}[/latex]

**FAQ**

## Why is mass flow rate important?

Ans: Mass-flow rate is important in the wide range of field which include fluid dynamics, pharmacy, petrochemicals etc. It is important to ensure right fluid possessing desired properties is flowing to the required location. It is important for maintaining and controlling the quality of fluid flowing. Its accurate measurements ensure the safety of workers working in a hazardous and dangerous environment. It is also important for good machine performance and efficiency and environment.

## Mass flow rate of water

Mass-flow rate is given by the equation

[latex]\dot{m}=\rho AV[/latex]

Density of water is 1000 kg/m^{3}

[latex]\dot{m}=1000AV[/latex]

## Mass flow rate of air

Mass-flow rate is given by the equation

[latex]\dot{m}=\rho AV[/latex]

Density of air is 1 kg/m^{3}

[latex]\dot{m}=AV[/latex]

## How to get mass flow rate from enthalpy?

Heat Transfer in fluid and thermodynamics is given by the following equation

[latex]Q=\dot{m}h[/latex]

Where Q = heat transfer, m = mass-flow rate, h = change in enthalpy For constant heat supplied or rejected, enthalpy is inversely proportionate to mass flow rate.

## How to get mass flow rate from Velocity?

In hydrodynamic, the mass-flow rate can be derived from volume flow rate with help of Continuity Equation.

The Continuity equation is given by

Q = AV

Where,

A = Cross sectional Area

V = Velocity of flow of fluid

Multiplying the continuity equation with density of the fluid we get,

[latex]\dot{m}=\rho AV[/latex]

## Can mass flow rate be negative

The magnitude of Mass flow rate cannot be negative. If we are provided the mass-flow rate with negative sign it generally indicates the direction of mass flux is reversed than the direction taken into consideration.

## Mass flow rate for an ideal compressible gas

Air is assumed to be an Ideal compressible gas with C_{p} = 1 kJ/kg. K.

Mass flow rate is given by the equation

[latex]\dot{m}=\rho AV[/latex]

Density of air is 1 kg/m^{3}

[latex]\dot{m}=AV[/latex]

## How can I find the mass flow of a refrigeration fluid R 134a and its temperatures in a domestic freezer How can I find them?

Assuming the Domestic freezer works on a vapor compression cycle, in order to find out mass-flow rate of the coolant R-134a we are required to find:

- Net refrigeration capacity or effect – generally given for that particular model of freezer.
- Compressor Inlet Pressure and Temperature
- Compressor outlet Pressure and Temperature
- Temperature and pressure at the inlet of evaporator
- Temperature and pressure at the outlet of condenser
- For P-h chart find enthalpy at all the above points.
- Net Refrigeration effect = mass-flow rate * [h
_{1}– h_{2}]

## What is the relationship between pressure and mass flow rate Does the mass flow rate increase if there’s a pressure increase and does the mass flow rate decrease if there’s a pressure decrease ?

Let,

L = length of pipe

V = Velocity of flow of fluid

μ = dynamic viscosity of the fluid

d = diameter of pipe

According to Hagen Poiseuille equation

[latex]\Delta P=\frac{32\mu lV}{d^2}[/latex]

Multiplying numerator and denominator by ρA

[latex]\Delta P=\frac{32\mu lV\rho A}{\rho Ad^2}[/latex]

[latex]\Delta P=\frac{32\nu \dot{m}l}{\frac{\pi}{4}d^2*d^2}[/latex]

[latex]\Delta P=\frac{40.743\nu \dot{m}l}{d^4}[/latex]

where, ν = kinematic viscosity = μ/ρ

Thus, as pressure difference increases, mass-flow rate increases and vice versa.

## For a convergent nozzle if the exit pressure is less than the critical pressure then what will be the mass flow rate?

As per described situation, nozzle’s outlet velocity is

[latex]C_2=\sqrt{\frac{2n}{n+1}P_1V_1}[/latex]

Mass-flow rate will be

[latex]\dot{m}=\frac{A_2C_2r^\frac{1}{n}}{V_2}[/latex]

Where

A_{1}, A_{2} = Inlet and Outlet Area of nozzle

C_{1}, C_{2} = Inlet and exit velocity of nozzle

P_{1}, P_{2} = Inlet and Outlet Pressure

V_{1}, V_{2} = Volume at Inlet and Outlet of nozzle

r = Pressure ratio =P_{2}/P_{1}

n = Index of expansion

## Why is mass flow rate is ρVA but volumetric flow rate is AV?

In hydrodynamic, the mass flux can be derived from volume flow rate with help of Continuity Equation.

The Continuity equation is given by

Q =AV

Where,

A = Cross sectional Area

V = Velocity of flow of fluid

Multiplying the continuity equation with density of the fluid we get the mass flow rate,

[latex]\dot{m}=\rho AV=\rho Q[/latex]

Where,

ρ = Density of the fluid

## How is the Coriolis principle used to measure mass flow?

A Coriolis mass flowmeter works on the principle of the **Coriolis Effect** and this are true mass meter because they measure the mass rate of flow directly rather than measuring the volumetric flow rate and converting it into the mass flow rate.

Coriolis meter operates linearly, In the meantime no adjustments are essential for changing fluid characteristic. It is independent of fluid characteristics.

Operating Principle:

The fluid is allowed to flow through a U-shaped tube. An oscillation-based excitation force is utilized to the tube, causing it to oscillate. The vibration causes the fluid to induce twist or rotation to the pipe because of Coriolis acceleration. Coriolis acceleration is acting opposite to applied excitation force. The generated twist results in a time lag in flow between the entry and exit-side of the tube, and this Lag or phase difference is proportionate to the mass flow rate.

## What is the relationship between mass flow rate and volume flow rate?

In hydrodynamic, the mass flow rate can be derived from volume flow rate with help of Continuity Equation.

The Continuity equation is given by

Q = AV

Where,

A = Cross sectional Area

V = Velocity of flow of fluid

Multiplying the continuity equation with density of the fluid we get,

[latex]\dot{m}=\rho AV=\rho Q[/latex]

Where,

ρ = Density of the fluid

## What is the formula for finding mass flow rate in a water cooled condenser?

Let,

h_{1} = enthalpy of water at inlet of the condenser

T_{1} = Temperature of water at inlet of the condenser

h_{2} = enthalpy of water at exit of the condenser

T_{2} = Temperature of water at exit of the condenser

C_{p} = Specific heat of water at constant pressure

Power of the condenser,

[latex]\\P=\dot{m}[h_1-h_2]\\ \\\dot{m}=\frac{P}{h_1-h_2}\\ \\\dot{m}=\frac{P}{C_p[T_1-T_2]}[/latex]

## How do you find mass flow with temperature and pressure?

Let,

L = length of pipe

V = Velocity of flow of fluid

μ = dynamic viscosity of the fluid

d = diameter of pipe

According to Hagen Poiseuille equation

[latex]\Delta P=\frac{32\mu lV}{d^2}[/latex]

Multiplying numerator and denominator by ρA

[latex]\Delta P=\frac{32\mu lV\rho A}{\rho Ad^2}[/latex]

[latex]\Delta P=\frac{32\nu \dot{m}l}{\frac{\pi}{4}d^2*d^2}[/latex]

[latex]\Delta P=\frac{40.743\nu \dot{m}l}{d^4}[/latex]

where, ν = kinematic viscosity = μ/ρ

Thus, as pressure difference increases, m increases.

According to Steady Flow energy equation

[latex]\\\dot{m}h_1\pm Q=\dot{m}h_2\pm W\\ \\\dot{m}(h_1-h_2)=W\pm Q\\ \\\dot{m}C_p(T_1-T_2)=W\pm Q[/latex]

## Why in choked flow we always control downstream pressure while the maximum mass flow rate is dependent on upstream pressure

It is impossible to regulate Choked mass flows by changing the downstream pressure. When sonic conditions reach the throat, Pressure disturbances caused due to regulated downstream pressure cannot propagate upstream. Thus, you cannot control the maximum flow rate by regulating the downstream backpressure for a choked flow.

## What is the average fluid mass flow rate of water in pipes with diameter 10cm, velocity of flow is 20 m/s.

In Hydrodynamics

[latex]\\\dot{m}=\rho AV \\\dot{m}=1000*\frac{\pi}{4}*0.1^2*20\\ \\\dot{m}=157.08\;\frac{kg}{s}[/latex]

**To know about Polytropic Process (click here**)**and Prandtl Number (Click here)**