Continuity Equation MCQ Quiz - Objective Question with Answer for Continuity Equation - Download Free PDF

Concept:

For 2D incompressible flow, the continuity equation must hold:

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

Check for Option 3:

$u=x+y,v=x-y$

$\frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial y}=-1$

$\Rightarrow \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=1-1=0$

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 ${\rho }_1{A}_1{V}_1={\rho }_2{A}_2{V}_2$ (Compressible fluid)

Density ρ = C for an incompressible fluid.

∴ Continuity equation for an incompressible fluid A1V1 = A2V2

Additional Information
  

Generalized equation of continuity. 

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

This equation can be written in vector form as,

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

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

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

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

$\mathrm{\nabla }.\overrightarrow{V}=0$

For two dimensional flow

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

Explanation:

Given Data:

• Area at section 1-1: $A_1=5$ m²
• Area at section 2-2: $A_2=2$ m²
• Velocity at section 1-1: $V_1=4$ m/s
• Velocity at section 2-2: $V_2$ (to be determined)

Step 1: Use the Continuity Equation

The continuity equation states that the flow rate remains constant for an incompressible fluid:

$A_1{V}_1=A_2{V}_2$

Step 2: Substitute the Given Values

$5\times 4=2\times V_2$

$20=2V_2$

Step 3: Solve for $V_2$

$V_2=\frac{20}{2}=10$ m/s

Continuity Equation: $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$, Given: $u=x^3+z^6+6,v=y^3+z^4$, Partial Derivatives: $\frac{\partial u}{\partial x}=3x^2,\frac{\partial v}{\partial y}=3y^2$, Substitute: $\frac{\partial w}{\partial z}=-(3x^2+3y^2)$, Integrate: $w=-3z(x^2+y^2)$

​Explanation:

Euler's equation of motion:

The Euler's equation for a steady flow of an ideal fluid along a streamline is a relation between the velocity, pressure and density of a moving fluid. It is based on the Newton's Second Law of Motion which states that if the external force is zero, linear momentum is conserved. The integration of the equation gives Bernoulli's equation in the form of energy per unit weight of the following fluid.

It is based on the following assumptions:

• The fluid is non-viscous (i,e.,inviscid fluid or the frictional losses are zero)
• The fluid is homogeneous and incompressible (i.e., the mass density of the fluid is constant)
• The flow is continuous, steady and along the streamline.
• The velocity of the flow is uniform over the section.
• No energy or force (except gravity and pressure forces) is involved in the flow.

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 = 0.25 m2, A2 = 0.1 m2.

Flow rate: Q = 0.1 m3/s.

Q = A1V1 = A2V2 

$V_1=\frac{Q}{A_1}=\frac{0.1}{0.25}=0.4m/s$

$V_2=\frac{Q}{A_2}=\frac{0.1}{0.1}=1m/s$

∴ The velocity of flow in the reduced pipe is 1 m/s

Explanation:

General Continuity equation:

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

For incompressible and steady flow:

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

$\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{\mathrm{V}}=0$

∴ The flow needs to be steady and incompressible.

Concept:

According to continuity, the liquid flow rate will be conserved.

Q₁ = Q₂ + Q₃

In a pipe, flow rate is given by

Q = AV

Calculation:

Given V₁ = 3 m/s, D₁ = 450 mm, V₂ = 2.5 m/s, D₂ = 300 mm, D₃ = 200 mm;

From continuity,

D₁² V₁ = D₂² V₂ + D₃² V₃

⇒ 450² × 3 = 300² × 2.5 + 200² × V₃

⇒ V₃ = 9.56 m/s 

Explanation:

General Continuity Equation:

The equation is based on the principle of conservation of mass which means mass can neither be created nor be destroyed.

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

$\frac{\partial \rho }{\partial t}+\rho \left(\mathrm{\nabla }.\overrightarrow{V}\right)=0$    (Vector form)

where, ρ = density of fluid, u = velocity in x-direction, v = velocity in y-direction, w = velocity in z-direction

Special Case:

If the flow is steady then $\left(\frac{\partial \rho }{\partial t}=0\right)$.

Thus continuity equation will be

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

For steady, compressible, and two-dimensional flow

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

For steady, incompressible (ρ = constant)

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$​​

Important Points

1-D flow 

​$\frac{\partial u}{\partial x}\ne 0$, $\frac{\partial w}{\partial z}=0$, $\frac{\partial v}{\partial y}=0$

Uniform flow: The velocity vector at all points in the flow is the same at any instant of time.​

Non-Uniform flow: ​​​The velocity vector varies from point to point at any instant of time.​

Explanation:

Continuity Equation

• It is based on the principle of conservation of mass.
• For a fluid flowing through a pipe at all the cross-section, the quantity of fluid per second is constant.
• The continuity equation is given as ${\rho }_1{A}_1{V}_1={\rho }_2{A}_2{V}_2$

Bernoulli’s Equation

• It is based on the principle of conservation of energy.
• For an ideal fluid flowing through a pipe at all the cross-section, the total energy would be constant.

Conservation of momentum

• It states that the net force acting on the fluid mass is equal to the change in momentum of flow per unit time in that direction.

Given Incompressible fluid means ρ = constant

Apply continuity equation

ρAV = constant

So, fluid is incompressible {ρ = constant}

than A₁V₁ = A₂V₂ 

$A_1\times 4={\displaystyle \frac{\pi }{4}}\times (7{)}^2\times 2$

$A_1={\displaystyle \frac{\pi }{4}}\times {\displaystyle \frac{(7{)}^2}{2}}$

A₁ = 19.24 cm²

Concept:

The principle used here is Conservation of mass

∴ Q₁ = Q₂ + Q₃

Calculation:

Q₁ = Q₂ + Q₃

A₁V₁ = A₂V₂ + A₃V₃

⇒ 0.9 = 300 × 10^-3 × V₂ + 200 × 10^-3 × 3

⇒ 0.9 = 0.3 V₂ + 0.6

⇒ V₂ = 1 m/s

Mistake Points 
The question does not mention that the duct is circular. Here in the place of diameter, width has been taken which implies that the cross-section is of rectangular shape.

The total area we can take here as (width*depth). Since the depth is not given and the inlet and outlet lie in the same plane we can take it as unity.

Concept:

Continuity Equation

• It is based on the principle of conservation of mass.
• For a fluid flowing through a pipe at all the cross-section, the quantity of fluid per second is constant.
• The continuity equation is given as ${\rho }_1{A}_1{V}_1={\rho }_2{A}_2{V}_2$

Calculation:

Given:

Q = Q₁ + Q₂

AV = A₁V₁ + A₂V₂

100 × 25  = 50 × 10  + 50 × V₂

∴ V₂ = 40 m/s

Concept:

Using continuity equation

ρ1A1V1 = ρ2A2V2

$A_1{V}_1=A_2{V}_2${∵ ρ₁ = ρ₂}

$\frac{\pi }{4}{d}_1^2{V}_1=\frac{\pi }{4}{d}_2^2{V}_2$

Calculation:

Given:

d₁ = 40 cm, V₁ = 9 m/s, d₂ = 30 cm

$\frac{\pi }{4}{40}^2\times 9=\frac{\pi }{4}{30}^2{V}_2$

V₂ = 16 m/s

Concept:

Flow through pipe is given as, Q = AV

Where, Q = Discharge, A = cross sectional area, V = velocity of flow

Calculation:

Given:

d = 1 m, Q = 1200 ltr/sec = 1.2 m³/sec

$Q=\frac{\pi }{4}\times d^2\times V$

$1.2=\frac{\pi }{4}\times 1^2\times V$

$V=\frac{4.8}{\pi }m/s$