Engineering Questions with Answers - Multiple Choice Questions

Computational Fluid Dynamics – Continuity Equation – Infinitesimally Small Element

1 - Question

Which of the equations suit this model?

a) tVρdV+sρV⃗ .dS=0
b) DDtVρdV=0
c) ρt+.(ρV⃗ )=0
d) DρDt+ρ.V⃗ =0

View Answer Answer: c
Explanation: The diagram represents infinitesimally small fluid element fixed in space. This model will give conservative differential equation which is ∂ρ∂t+∇.(ρV⃗ )=0 from the options.



2 - Question

Consider an infinitesimally small fluid element with density ρ (of dimensions dx, dy and dz) fixed in space and fluid is moving across this element with a velocity V⃗ =ui⃗ +vj⃗ +wk⃗ . The net mass flow across the fluid element is given by ______
a) [∂(ρu)∂x+∂(ρv)∂y+∂(ρw)∂z]dxdydz
b) [∂(ρu)∂x+∂(ρv)∂y+∂(ρw)∂z]
c) [ρ]dx dy dz
d) [∂(ρ)∂x+∂(ρ)∂y+∂(ρ)∂z]dxdydz
View Answer Answer: a
Explanation: Mass inflow in x direction=(ρu)dy dz mass outflow in x direction=(ρu+∂(ρu)∂xdx)dydz change in mass flow in x direction=(ρu+∂(ρu)∂xdx)dydz−(ρu)dydz change in mass flow in x direction=∂(ρu)∂xdxdydz Net mass flow across the element = (change in mass flow in x direction + change in mass flow in y direction + change in mass flow in z direction) = ∂(ρu)∂xdxdydz+∂(ρv)∂ydxdydz+∂(ρw)∂zdxdydz Net mass flow across the element = [∂(ρu)∂x+∂(ρv)∂y+∂(ρw)∂z]dxdydz



3 - Question

Consider an infinitesimally small fluid element with density ρ (of dimensions dx, dy and dz) fixed in space and fluid is moving across this element with a velocity V⃗ =ui⃗ +vj⃗ +wk⃗ . The rate of change in mass of the fluid element is given by ____________
a) ∂(ρu)∂x+∂(ρv)∂y+∂(ρw)∂z
b) ∂ρ∂t
c) ∂ρ∂t(dxdydz)
d) [∂(ρu)∂x+∂(ρv)∂y+∂(ρw)∂z]dxdydz
View Answer Answer: c
Explanation: Mass=density × volume mass of fluid element=ρ×dx dy dz rate of change in mass of fluid element=∂ρ∂tdxdydz



4 - Question

According to the conservation law, “Net mass flow across the fluid element is equal to the rate of change of mass inside the element”. But, stating the final equation, “Net mass flow across the fluid element + the rate of change of mass inside the element = 0”. Why is the operation not subtraction?
a) Irrespective of the law, the sum is always zero
b) The two terms are always opposite in sign
c) Change in sign is not considered
d) Rate of change may be increase or decrease
View Answer Answer: b
Explanation: The “net mass flow across the fluid element” being positive means that it is outward flow. If flow is outward, mass inside the fluid element decreases leading to a negative “rate of change of mass inside the element”. Thus, the two terms are always opposite in sign that they can be summed up to get zero.



5 - Question

Consider an infinitesimally small fluid element with density ρ (of dimensions dx, dy and dz) fixed in space and fluid is moving across this element with a velocity V⃗ =ui⃗ +vj⃗ +wk⃗ . What is the final reduced form of net mass flow across the fluid element?
a) ∂ρ∂t
b) ρV⃗ dxdydz
c) ∇.(ρV⃗ )
d) ∇.(ρV⃗ )dx dy dz
View Answer Answer: d
Explanation: Net mass flow across the element = change in mass flow in x direction + change in mass flow in y direction + change in mass flow in z direction = ∂(ρu)∂xdxdydz+∂(ρv)∂ydxdydz+∂(ρw)∂zdxdydz =[(∂∂xi⃗ +∂∂yj⃗ +∂∂zk⃗ )dxdydz.((ρu)i⃗ +(ρv)j⃗ +(ρw)k⃗ )] Net mass flow across the element = [∇.(ρV⃗ )]dxdydz.



6 - Question

Which of the equations suit this model?

a) tVρdV+sρV⃗ .dS=0
b) DDtVρdV=0
c) DρDt+ρ.V⃗ =0
d) ρt+.(ρV⃗ )=0

View Answer Answer: c
Explanation: ∂ρ∂t+∇.(ρV⃗ )=0 is the non-conservative differential equation. Non-conservative differential equation is given by an infinitesimally small fluid element moving along with the flow.



7 - Question

Consider an infinitesimally small fluid element with density ρ (of dimensions dx, dy and dz with mass δ m and volume δ V) moving along with the flow with a velocity V⃗ =ui⃗ +vj⃗ +wk⃗ . What is the time rate of change of mass of this element?
a) D(ρδV)Dt
b) ∂(ρδm)∂t
c) ∂(ρδV)∂t
d) D(ρδm)Dt
View Answer Answer: a
Explanation: Substantial derivative is used as the model is moving. mass = ρ δ V time rate of change of mass=D(ρδV)Dt



8 - Question

Consider an infinitesimally small fluid element with density ρ (of dimensions dx, dy and dz with mass δ m and volume δ V) moving along with the flow with a velocity V⃗ =ui⃗ +vj⃗ +wk⃗ . The continuity equation is DρDt+ρ∇.V⃗ =0. Where does this second term come from?
a) Integral
b) The rate of change of element’s volume
c) Elemental change in mass
d) Local derivative
View Answer Answer: b
Explanation: Applying mass conservation for this element, time rate of change of mass = 0 D(ρδV)Dt=0 δVD(ρ)Dt+ρD(δV)Dt=0 D(ρ)Dt+ρ1δVD(δV)Dt=0 D(ρ)Dt+ρ∇.ρ=0 Thus, the term arises from the rate of change of element’s volume.



9 - Question

Consider the continuity equation ∂ρ∂t+∇.(ρV⃗ )=0. For an incompressible flow, this equation becomes ___________
a) ∇.(ρV⃗ )=0
b) ∂(ρV⃗ )∂t=0
c) div(V⃗ )=0
d) div(ρV⃗ )=0
View Answer Answer: c
Explanation: Taking the continuity equation, ∂ρ∂t+∇.(ρV⃗ )=0 For incompressible flow, ρ is constant ∂ρ∂t+∇.(ρV⃗ )=0 The resulting equation is ∇.(ρV⃗ )=0 ρ∇.(V⃗ )=0 ∇.(V⃗ )=0 Thus, for incompressible flow, divergence of V⃗ =0.



10 - Question

Consider the continuity equation ∂ρ∂t+∇.(ρV⃗ )=0. For a steady flow this equation becomes ___________
a) ∇.(ρV⃗ )=0
b) ∇.(V⃗ )=0
c) div(V⃗ )=0
d) curl(V⃗ )=0
View Answer Answer: a
Explanation: Taking the continuity equation, ∂ρ∂t+∇.(ρV⃗ )=0 For steady flow, flow variables do not vary with time. ∂ρ∂t=0 Thus, for steady flow ∇.(ρV⃗ )=0.

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