Continuity Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Continuity Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Continuity equation: It is the conservation of mass flow rate.

• ρ₁A₁V₁ =  ρ1A1V1

For incompressible fluid density will be constant thus continuity equation will be:

• A₁V₁ = A₂V₂  

where, A₁, A₂ = area of section 1 & 2 respectively, V₁, V₂ = velocity of section 1 & 2 respectively

The flow rate of liquid is equal to Q = AV.

Calculation:

Given:

Area: A1 = 0.05 m², A2 = 0.01 m2.

Flow rate: Q = 50 l/s = 50 x 10^-3 m³/s.

Q = A1V1 = A2V2 

\(V_1 = \frac{Q}{A_1} = \frac{50 \times 10^{-3}}{0.05} = 1\ m/s\)

\(V_2 = \frac{Q}{A_2} = \frac{50 \times 10^{-3}}{0.01} = 5\ m/s\)

Concept:

Every fluid flow must satisfy its corresponding mass conservation (continuity) equation.

For a 2 – dimensional incompressible flow

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial \nu }}{{\partial y}} = 0\)

\(\vec V = u\hat i + \nu \hat j\)

Calculation:

u = 5 + a₁x + b₁y, ν = 4 + a₂x + b₂y

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial \nu }}{{\partial y}} = 0 \Rightarrow {a_1} + {b_2} = 0\)

Concept:

For steady, incompressible flow, continuity equation is given by,

A₁V₁ = A₂V₂

Calculation:

Given:

d₁ = 25 cm, V₁ = 2 m/s

d₂ = 20 cm, V₂ = ?

As per the continuity equation,

A1V1 = A2V2

⇒ \(\frac{\pi }{4} × d_1^2 × {V_1} = \frac{\pi }{4} × d_2^2 × {V_2}\)

25² × 2 = 20² × V₂

V₂ = 3.125 m/s

Explanation:

Continuity Equation is based on the principle of conservation of mass. For a fluid flowing through a pipe at all the cross-sections, the quantity of fluid per second is constant.

The continuity equation is given as \({ρ _1}{A_1}{V_1} = \;{ρ _2}{A_2}{V_2}\)

Generalized equation of continuity.

\(\frac{{\partial \left( {ρ u} \right)}}{{\partial x}} + \frac{{\partial \left( {ρ \nu } \right)}}{{\partial y}} + \frac{{\partial \left( {ρ w} \right)}}{{\partial z}} + \frac{{\partial ρ }}{{\partial t}} = 0\)

This equation can be written in vector form as,

Case 1: For steady flow \(\frac{{\partial ρ }}{{\partial t}} = 0\) then the above equation will become,

\(\frac{{\partial \left( {ρ u} \right)}}{{\partial x}} + \frac{{\partial \left( {ρ \nu } \right)}}{{\partial y}} + \frac{{\partial \left( {ρ w} \right)}}{{\partial z}} = 0\)

Case 2: For Incompressible flow, ρ is constant, therefore the continuity equation of steady incompressible for three-dimensional flow is,

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\)

For two dimensional flow

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)

Explanation:

The continuity Equation is based on the principle of conservation of mass. For a fluid flowing through a pipe at all the cross-sections, the quantity of fluid per second is constant.

The continuity equation is given as \({ρ _1}{A_1}{V_1} = \;{ρ _2}{A_2}{V_2}\) (Compressible fluid)

Density ρ = C for an incompressible fluid.

∴ Continuity equation for an incompressible fluid A1V1 = A2V2

Additional Information 

Generalized equation of continuity. 

\(\frac{{\partial \left( {ρ u} \right)}}{{\partial x}} + \frac{{\partial \left( {ρ \nu } \right)}}{{\partial y}} + \frac{{\partial \left( {ρ w} \right)}}{{\partial z}} + \frac{{\partial ρ }}{{\partial t}} = 0\)

This equation can be written in vector form as,

Case 1: For steady flow \(\frac{{\partial ρ }}{{\partial t}} = 0\) then the above equation will become,

\(\frac{{\partial \left( {ρ u} \right)}}{{\partial x}} + \frac{{\partial \left( {ρ \nu } \right)}}{{\partial y}} + \frac{{\partial \left( {ρ w} \right)}}{{\partial z}} = 0\)

Case 2: For Incompressible flow, ρ is constant, therefore the continuity equation of steady incompressible for three-dimensional flow is,

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\)

\(\nabla .\vec V = 0\)

For two dimensional flow

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)

Explanation:

Continuity equation in three dimensions:

\(\frac{{\partial \rho }}{{\partial t}} + \frac{\partial }{{\partial x}}\left( {\rho u} \right) + \frac{\partial }{{\partial y}}\left( {\rho v} \right) + \frac{\partial }{{\partial z}}\left( {\rho w} \right) = 0\)

The above equation is valid for:

1. Steady and unsteady flow.
2. Uniform and non-uniform flow.
3. Compressible and incompressible flow.

For Steady flow:

\(\frac{\partial }{{\partial x}}\left( {\rho u} \right) + \frac{\partial }{{\partial y}}\left( {\rho v} \right) + \frac{\partial }{{\partial z}}\left( {\rho w} \right) = 0\;\left( \because{\frac{{\partial \rho }}{{\partial t}} = 0} \right)\)

If the fluid is Incompressible and Steady:

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\;\;\left( \because{\rho = constant} \right)\)

Concept:

Continuity equation should be satisfied for incompressible, steady flow,

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)

Calculation:

\(\frac{\partial }{{\partial x}}\left( {ax{y^2} + 2y} \right) + \frac{\partial }{{\partial y}}\left( {3xy + {x^2}y} \right) = 0\)

(ay²) + (3x + x²) = 0

At (1, 1)

a + (3 + 1) = 0

Explanation:

Ideal fluid:

A fluid, which is incompressible and is having no viscosity, is known as an ideal fluid. Ideal fluid is only an imaginary fluid as all the fluids, which exist, have some viscosity.

Hence a fluid which has constant density and has zero viscosity is known as ideal fluid

Ideal fluid obeys the following two equations

1) Continuity equation

2) Bernouilli’s equation

Important Points

1) Continuity equation:

The equation based on the principle of conservation of mass is called the continuity equation. Thus for the fluid flowing through the pipe at all the cross-sections, the quantity of fluid per second is constant.

ρ₁ × A₁ × V₁ =ρ₂ × A₂ × V₂

ρ₁ and ρ₂ is the density of liquid at section 1 and section 2 respectively

A₁ and A₂ is an area of pipe at section 1 and section 2 respectively

V₁ and V₂ is velocity of liquid section 1 and section 2 respectively

For an ideal fluid i.e incompressible fluid, ρ₁ = ρ_2, and the continuity equation reduces to,

A₁ × V₁ = A₂ × V₂

2) Bernoulli’s equation:

Bernoulli’s equation in the form of energy per unit weight is given by,

\(\frac{{\rm{P}}}{{\rm{\gamma }}} + \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} + {\rm{Z}} = {\rm{H}}\)

Where,

\(\frac{{\rm{P}}}{{\rm{\gamma }}} = {\rm{Static\;head}},\frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} = {\rm{Dynamic\;head}},{\rm{\;Z}} = {\rm{dattum\;head}},{\rm{\;H}} = {\rm{total\;head}}\)

The assumptions made in the derivation of Bernoulli’s equation is

1) The fluid is ideal, i.e Viscosity is zero

2) The flow is steady

3) The flow is incompressible

4) The flow is irrotational

Explanation:

Explanation:

• The law of conservation of mass states that mass can neither be created nor be destroyed.
• The rate at which mass enters the region = Rate at which mass leaves the region + Rate of accumulation of mass in the region
• The above statement can be expressed analytically in terms of velocity and density field of flow and the resulting expression is known as the equation of continuity or the continuity equation.

\(\begin{array}{l} \frac{{\partial \rho }}{{\partial t}} + \frac{\partial }{{\partial x}}\left( {\rho u} \right) + \frac{\partial }{{\partial y}}\left( {\rho v} \right) + \frac{\partial }{{\partial z}}\left( {\rho w} \right) = 0\\ \frac{{\partial \rho }}{{\partial t}} +\nabla .\left({} {{\rm{\rho \vec V}}} \right) = 0 \end{array}\)

Important Points

• The continuity equation applies to all fluids, compressible and incompressible flow, Newtonian and non-Newtonian fluids.
• It expresses the law of conservation of mass at each point in a fluid and must, therefore, be satisfied at every point in a flow field. 
• So, the continuity equation is connected with the conservation of mass and it can be applied to viscous/non-viscous, the compressibility of the fluid, or the steady/unsteady flow.