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# Computational Fluid Dynamics – Energy Equation – Based on Thermal Properties

1 - Question

The energy equation in terms of total energy is

∂(ρe)∂t+∇.(ρV⃗ e)=−∇.q˙s−∇.(pV⃗ )+∇.(τ.V⃗ )+fb→.V⃗ +q˙v.

Where,

t → Time

ρ → Density

e → Specific total energy

V⃗ → Velocity vector

q˙s → Rate of heat transfer per unit area

fb→ → Body force vector

τ → Shear stress

q˙v → Rate of heat source or sink per unit volume

While converting this equation in terms of internal energy, which of these terms lose its explicit presence?

a) Pressure term

b) Shear stress term

c) Body force term

d) Heat transfer term

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Answer: cExplanation: While converting the energy equation from total energy terms to internal energy terms, an equation with kinetic energy is subtracted from total energy. In this process, the body force term loses its explicit presence.

2 - Question

While converting the energy equation from one form to another, which of the following happens?

a) Either the left-hand side or the right-hand side of the equation changes

b) Both the left-hand side and the right-hand side of the equation change

c) The right-hand side of the equation changes

d) The left-hand side of the equation changes

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Answer: bExplanation: Changes are applied to only left-hand side terms in the equations, but they affect both the left and right-hand sides of the equation.

3 - Question

Expressing τ:ΔV⃗ in terms of flow variables, we get λφ+μψ. What are φ and ψ?

a) ϕ=(∂u∂x+∂v∂y+∂w∂z)2andψ=(∂u∂y+∂v∂x)2+(∂v∂z+∂w∂y)2+(∂u∂z+∂w∂x)2

b) ψ=(∂u∂x+∂v∂y+∂w∂z)2andϕ=(∂u∂y+∂v∂x)2+(∂v∂z+∂w∂y)2+(∂u∂z+∂w∂x)2

c) ψ=(∂u∂x+∂v∂y+∂w∂z)2andϕ=2(∂u∂x)2+2(∂v∂y)2+2(∂w∂z)2+(∂u∂y+∂v∂x)2+

(∂v∂z+∂w∂y)2+(∂u∂z+∂w∂x)2

d) ϕ=(∂u∂x+∂v∂y+∂w∂z)2andψ=2(∂u∂x)2+2(∂v∂y)2+2(∂w∂z)2+(∂u∂y+∂v∂x)2+

(∂v∂z+∂w∂y)2+(∂u∂z+∂w∂x)2

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Answer: dExplanation: τ:ΔV⃗ =λ(∂u∂x+∂v∂y+∂w∂z)2μ(2(∂u∂x)2+2(∂v∂y)2+2(∂w∂z)2+(∂u∂y+∂v∂x)2+ (∂v∂z+∂w∂y)2+(∂u∂z+∂w∂x)2).

4 - Question

If p and τ are the net pressure and net shear stress acting on an infinitesimally small element (volume dx dy dz) moving along with the flow (velocity V⃗ ), what is the net work done on the system?

a) ρ(∇.(pV⃗ )+∇.(τ.V⃗ ))

b) ((pV⃗ )+(τ.V⃗ ))dxdydz

c) ρ(∇.(pV⃗ )+∇.(τ.V⃗ ))dxdydz

d) (∇.(p)+∇.(τ))dxdydz

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Answer: cExplanation: The rate of work done is power which is the product of force and velocity. This can be represented by (∇.(pV⃗ )+∇.(τ.V⃗ ))dxdydz.

5 - Question

The energy equation which is in terms of temperature can be changed to terms of internal energy using ___________

a) momentum equation

b) stress-strain relations

c) equations of state

d) continuity equation

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Answer: cExplanation: Equations of state give the relationship between temperature and internal energy which is i=Cv T. Using this relation, one can obtain the energy equation in internal energy terms.

6 - Question

The energy equation which is in terms of total energy can be changed to terms of internal energy using ___________

a) momentum equation

b) stress-strain relations

c) equations of state

d) continuity equation

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Answer: aExplanation: Total energy is the sum of internal energy and kinetic energy. By some manipulation in the momentum equation, we can get the kinetic energy terms. If this is subtracted from the energy equation written in total energy terms, we can get the same in terms of internal energy.

7 - Question

If f⃗ is the body force of an infinitesimally small element (volume dx dy dz and density ρ) moving along with the flow (velocity V⃗ ), Which term is the work done by the body force?

a) f⃗ .V⃗ dx dy dz

b) ρf⃗ .V⃗

c) ρf⃗ .V⃗ dx dy dz

d) ρf⃗ dx dy dz

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Answer: cExplanation: mass = density×volume mass = ρdx dy dz rate of work done = force×velocity rate of work done = ρf⃗ .V⃗ dx dy dz ρf⃗ .V⃗ dx dy dz is the work done by the body force.

8 - Question

Energy equation in terms of specific internal energy is

∂(ρu^)∂t+∇.(ρV⃗ u^)=−∇.qs˙–p∇.V⃗ −τ:∇V⃗ +qv˙

Where,

t → Time

ρ → Density

u^ → Specific internal energy

V⃗ → Velocity vector

qs˙→ Rate of heat transfer per unit area

τ → Shear stress

q˙v → Rate of heat source or sink per unit volume

Convert this equations in terms of specific enthalpy h^

. a) ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙+DpDt−τ:∇V⃗ +qv˙

b) ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙−p∇.V⃗ −τ:∇V⃗ +qv˙

c) ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙−p∇.V⃗ +∇.(pV⃗ )−τ:∇V⃗ +qv˙

d) ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙+V⃗ .∇p−τ:∇V⃗ +qv˙

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Answer: aExplanation: Take the given equation. ∂(ρu^)∂t+∇.(ρV⃗ u^)=−∇.qs˙−p∇.V⃗ −τ:∇V⃗ +qv˙ The relation between internal energy and enthalpy is u^=h^−pρ Substituting this in the equation, ∂(ρh^)∂t−∂(p)∂t+∇.(ρV⃗ h^)−∇.(pV⃗ )=−∇.qs˙−p∇.V⃗ −τ:∇V⃗ +qv˙ ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.q˙s−∂(p)∂t−∇.(pV⃗ )−p∇.V⃗ −τ:∇V⃗ +qv˙ ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙+∂(p)∂t+∇.(pV⃗ )−p∇.V⃗ −τ:∇V⃗ +qv˙ ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙+∂(p)∂t+V⃗ .∇p−τ:∇V⃗ +qv˙ ∂(ρh^)∂t+∇.(ρV⃗ h^)=−∇.qs˙+D(p)Dt−τ:∇V⃗ +qv˙ This is the energy equation in specific enthalpy terms.

9 - Question

Let u^ be the specific internal energy of a system moving along with the flow with a velocity V⃗ . What is the time rate of change of the total energy of the system per unit mass?

a) u^+12V⃗ .V⃗

b) DDt(u^+12V⃗ .V⃗ )

c) ∂∂t(u^+12V⃗ .V⃗ )

d) DDt(u^+V⃗ .V⃗ )

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Answer: bExplanation: The total energy of the system is E=u^+12V⃗ .V⃗ Take its substantial derivative as the model is not stationary. rate of change of total energy=DEDt=DDt(u^+12V⃗ .V⃗ ).

10 - Question

∂(ρu^)∂t+∇.(ρV⃗ u^)=−∇.qs˙−p∇.V⃗ −τ:∇V⃗ +qv˙ . This form of the energy equation is applicable to _________

a) Both Newtonian and non-Newtonian fluids

b) Newtonian fluids

c) Non-Newtonian fluids

d) Pseudo-plastics

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Answer: aExplanation: The energy equation given here is in terms of shear stresses. So, no restrictions based on the viscosity of the fluid. It is applicable to both Newtonian and Non-Newtonian fluids.