Computational Fluid Dynamics Questions and Answers – Energy Equation – Based on Thermal Properties

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This set of Computational Fluid Dynamics Questions and Answers for Freshers focuses on “Energy Equation – Based on Thermal Properties”.

1. The energy equation in terms of total energy is
\(\frac{\partial(\rho e)}{\partial t}+\nabla.(\rho \vec{V}e)=-\nabla.\dot{q}_s-\nabla.(p\vec{V})+\nabla.(τ.\vec{V})+\vec{f_b}.\vec{V}+\dot{q}_v.\)
Where,
t → Time
ρ → Density
e → Specific total energy
\(\vec{V}\) → Velocity vector
\(\dot{q}_s\) → Rate of heat transfer per unit area
\(\vec{f_b}\) → Body force vector
τ → Shear stress
\(\dot{q}_v\) → 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
View Answer

Answer: c
Explanation: 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.
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2. 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
View Answer

Answer: b
Explanation: 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. Expressing \(\tau:\Delta \vec{V}\) in terms of flow variables, we get λφ+μψ. What are φ and ψ?
a) \(\phi=(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})^2 \,and\, \psi=(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y})^2+(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})^2\)
b) \(\psi=(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})^2 \,and\, \phi=(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y})^2+(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})^2\)
c) \(\psi=(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z})^2 \,and\, \phi=2(\frac{\partial u}{\partial x})^2 + 2(\frac{\partial v}{\partial y})^2 +2(\frac{\partial w}{\partial z})^2+(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})^2 +\)
\( (\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y})^2 + (\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x})^2\)
d) \(\phi=(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})^2 \,and\, \psi=2(\frac{\partial u}{\partial x})^2+2(\frac{\partial v}{\partial y})^2+2(\frac{\partial w}{\partial z})^2+(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})^2+\)
\((\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y})^2+(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})^2\)
View Answer

Answer: d
Explanation: \(\tau:\Delta \vec{V} = \lambda(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z})^2 \mu(2(\frac{\partial u}{\partial x})^2+2(\frac{\partial v}{\partial y})^2+2(\frac{\partial w}{\partial z})^2+(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})^2+\)
\((\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y})^2+(\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})^2)\).

4. 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 \(\vec{V}\)), what is the net work done on the system?
a) \(\rho (\nabla .(p\vec{V} )+\nabla .(τ.\vec{V}))\)
b) \(((p\vec{V})+(\tau.\vec{V}))dx \,dy \,dz\)
c) \(\rho(\nabla.(p\vec{V})+\nabla.(\tau.\vec{V})) dx \,dy \,dz\)
d) \((\nabla .(p)+\nabla.(\tau))dx \,dy \,dz\)
View Answer

Answer: c
Explanation: The rate of work done is power which is the product of force and velocity. This can be represented by \((\nabla.(p\vec{V})+\nabla.(\tau.\vec{V})) dx \,dy \,dz\).

5. 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
View Answer

Answer: c
Explanation: 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.
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6. 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
View Answer

Answer: a
Explanation: 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. If \(\vec{f}\) is the body force of an infinitesimally small element (volume dx dy dz and density ρ) moving along with the flow (velocity \(\vec{V}\)), Which term is the work done by the body force?
a) \(\vec{f}.\vec{V}\)dx dy dz
b) \(\rho\vec{f}.\vec{V}\)
c) \(\rho\vec{f}.\vec{V}\)dx dy dz
d) \(\rho\vec{f}\)dx dy dz
View Answer

Answer: c
Explanation: mass = density×volume
mass = ρdx dy dz
rate of work done = force×velocity
rate of work done = \(\rho\vec{f}.\vec{V}\)dx dy dz
\(\rho\vec{f}.\vec{V}\)dx dy dz is the work done by the body force.

8. Energy equation in terms of specific internal energy is
\(\frac{\partial(\rho \hat{u})}{\partial t}+\nabla.(\rho\vec{V}\hat{u})=-\nabla.\dot{q_s} – p\nabla .\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
Where,
t → Time
ρ → Density
\(\hat{u}\) → Specific internal energy
\(\vec{V}\) → Velocity vector
\(\dot{q_s}\)→ Rate of heat transfer per unit area
τ → Shear stress
\(\dot{q}_v\) → Rate of heat source or sink per unit volume
Convert this equations in terms of specific enthalpy \(\hat{h}\).
a) \(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla .(\rho \vec{V}\hat{h})=-\nabla .\dot{q_s}+\frac{Dp}{Dt} -\tau:\nabla \vec{V}+\dot{q_v}\)
b) \(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla .(\rho \vec{V}\hat{h})=-\nabla .\dot{q_s}-p\nabla .\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
c) \(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla .(\rho \vec{V}\hat{h})=-\nabla .\dot{q_s}-p\nabla .\vec{V}+\nabla .(p\vec{V})-\tau:\nabla \vec{V}+\dot{q_v}\)
d) \(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla .(\rho \vec{V}\hat{h})=-\nabla .\dot{q_s}+\vec{V}.\nabla p-\tau:\nabla \vec{V}+\dot{q_v}\)
View Answer

Answer: a
Explanation: Take the given equation.
\(\frac{\partial (\rho \hat{u})}{\partial t}+\nabla.(\rho \vec{V}\hat{u})=-\nabla.\dot{q_s}-p\nabla.\vec{V}-\tau:\nabla \vec{V}+\dot{q_v} \)
The relation between internal energy and enthalpy is
\(\hat{u}=\hat{h}-\frac{p}{\rho}\)
Substituting this in the equation,
\(\frac{\partial(\rho \hat{h})}{\partial t}-\frac{\partial(p)}{\partial t}+\nabla .(\rho\vec{V}\hat{h})-\nabla.(p\vec{V}) =-\nabla.\dot{q_s}-p\nabla .\vec{V}-\tau:\nabla \vec{V}+\dot{q_v} \)
\(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla.(\rho \vec{V}\hat{h}) = -\nabla.\dot{q}_s-\frac{\partial(p)}{\partial t}-\nabla.(p\vec{V})-p\nabla.\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
\(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla.(\rho \vec{V}\hat{h}) = -\nabla.\dot{q_s} +\frac{\partial(p)}{\partial t}+\nabla.(p\vec{V})-p\nabla.\vec{V}-\tau:\nabla \vec{V}+\dot{q_v}\)
\(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla.(\rho \vec{V}\hat{h}) = -\nabla.\dot{q_s} +\frac{\partial(p)}{\partial t}+\vec{V}.\nabla p-\tau:\nabla \vec{V}+\dot{q_v}\)
\(\frac{\partial(\rho \hat{h})}{\partial t}+\nabla.(\rho\vec{V}\hat{h}) = -\nabla.\dot{q_s} +\frac{D(p)}{Dt}-\tau:\nabla \vec{V}+\dot{q_v}\)
This is the energy equation in specific enthalpy terms.

9. Let \(\hat{u}\) be the specific internal energy of a system moving along with the flow with a velocity \(\vec{V}\). What is the time rate of change of the total energy of the system per unit mass?
a) \(\hat{u}+\frac{1}{2}\vec{V}.\vec{V}\)
b) \(\frac{D}{Dt}(\hat{u}+\frac{1}{2}\vec{V}.\vec{V})\)
c) \(\frac{\partial}{\partial t}(\hat{u}+\frac{1}{2}\vec{V}.\vec{V})\)
d) \(\frac{D}{Dt}(\hat{u}+\vec{V}.\vec{V})\)
View Answer

Answer: b
Explanation: The total energy of the system is
\(E=\hat{u}+\frac{1}{2}\vec{V}.\vec{V}\)
Take its substantial derivative as the model is not stationary.
rate of change of total energy=\(\frac{DE}{Dt}=\frac{D}{Dt}(\hat{u}+\frac{1}{2} \vec{V}.\vec{V})\).
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10. \(\frac{\partial(\rho\hat{u})}{\partial t}+\nabla.(\rho\vec{V}\hat{u}) = -\nabla.\dot{q_s}-p\nabla.\vec{V}-\tau:\nabla\vec{V}+\dot{q_v}\). 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
View Answer

Answer: a
Explanation: 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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn