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LECTURE 3
Governing Equations for
Fluid Flow and Heat Transfer
Lecturer:
Dr Mohd Faizal Mohamad
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Governing Equations
The governing equations represent the mathematical statements
of the conservation laws of physics;
 The mass of the fluid is conserved
 The rate of change of momentum
= the sum of the forces on a fluid particle (Newton’s 2nd Law)
 The rate of change of energy
= the sum of the rate of heat addition to + the rate of work
done on a fluid particle (1st law of thermodynamics)
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Fluid Element
Fluid behaviour is described in terms of macroscopic properties
and are functions of space and time:
• Velocity, 𝑢 (𝑥, 𝑦, 𝑧, 𝑡)
z
y
• Pressure, 𝑝 (𝑥, 𝑦, 𝑧, 𝑡)
x
(x, y, z)
• Density, 𝜌 (𝑥, 𝑦, 𝑧, 𝑡)
δz
• Temperature, 𝑇 (𝑥, 𝑦, 𝑧, 𝑡)
δy
Fluid properties at the faces can be expressed
by the first two terms of Taylor series
expansion;
e.g. 𝑝 −
𝛿𝑥
and
𝑝+
δx
Fluid element
𝛿𝑥
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Mass Conservation
Mass balance of the fluid element
Rate of increase of mass
in fluid element
𝜌𝑣 +
Velocity comp.;
u; x-direction
v; y-direction
w; z-direction
ρ; density
𝜌𝑢 −
y
𝜕 𝜌𝑣
𝜕𝑦
𝜕 𝜌𝑢
𝜕𝑥
1
𝛿𝑦
2
Net rate of flow of mass into
fluid element
=
𝜌𝑤 +
1
𝛿𝑥
2
1
𝛿𝑧
2
𝜌𝑢 +
(x, y, z)
𝜕 𝜌𝑢
𝜕𝑥
1
𝛿𝑥
2
z
x
𝜌𝑣 −
𝜌𝑤 −
4
𝜕 𝜌𝑤
𝜕𝑧
𝜕 𝜌𝑤
𝜕𝑧
1
𝛿𝑧
2
𝜕 𝜌𝑣
𝜕𝑦
1
𝛿𝑦
2
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Mass Conservation
Mass balance of the fluid element
Rate of increase of mass
in fluid element
𝜕
𝜕𝑡
=
𝜌𝛿𝑥𝛿𝑦𝛿𝑧
Net rate of flow of mass into
fluid element
𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑖𝑛
− 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑜𝑢𝑡
Transform the statement into mathematical expression
Rate of increase of mass is given by
𝜕
𝜕𝜌
𝜌𝛿𝑥𝛿𝑦𝛿𝑧 =
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑡
𝜕𝑡
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Mass Conservation
Mass balance of the fluid element
Rate of increase of mass
in fluid element
𝜕
𝜕𝑡
𝜌𝛿𝑥𝛿𝑦𝛿𝑧
=
Net rate of flow of mass into
fluid element
𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑖𝑛
− 𝑀𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑜𝑢𝑡
Mass flow rate = product of density, area and
velocity component normal to the face
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Mass Conservation
Net rate flow of mass
x-direction;
𝜌𝑢 −
𝜕 𝜌𝑢 1
𝜕 𝜌𝑢 1
𝜕 𝜌𝑢
𝛿𝑥 𝛿𝑦𝛿𝑧 − 𝜌𝑢 +
𝛿𝑥 𝛿𝑦𝛿𝑧 = −
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥
y-direction;
𝜌𝑣 −
𝜕 𝜌𝑣 1
𝜕 𝜌𝑣 1
𝜕 𝜌𝑣
𝛿𝑦 𝛿𝑥𝛿𝑧 − 𝜌𝑣 +
𝛿𝑦 𝛿𝑥𝛿𝑧 = −
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
z-direction;
𝜌𝑤 −
𝜕 𝜌𝑤 1
𝜕 𝜌𝑤 1
𝜕 𝜌𝑤
𝛿𝑧 𝛿𝑥𝛿𝑦 − 𝜌𝑤 +
𝛿𝑧 𝛿𝑥𝛿𝑦 = −
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
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Mass Conservation
Rewritten the equation
𝜕𝜌
𝜕 𝜌𝑢
𝜕 𝜌𝑣
𝜕 𝜌𝑤
𝛿𝑥𝛿𝑦𝛿𝑧 = −
𝛿𝑥𝛿𝑦𝛿𝑧 −
𝛿𝑥𝛿𝑦𝛿𝑧 −
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜌 𝜕 𝜌𝑢
𝜕 𝜌𝑣
𝜕 𝜌𝑤
+
+
+
=0
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
Continuity equation
𝜕𝜌
+ 𝑑𝑖𝑣 𝜌𝑢 = 0
𝜕𝑡
𝑑𝑖𝑣 𝑢 = 0
Unsteady, compressible fluid
Incompressible fluid
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Rates of Change of Fluid Particle & Element
Each property of fluid particle is described based on the function of:
1. Time, t
2. Position, (x, y, z)
The total/substantive derivative of φ (property per unit mass) is written as:
𝐷∅ 𝜕∅ 𝜕∅ 𝑑𝑥 𝜕∅ 𝑑𝑦 𝜕∅ 𝑑𝑧
=
+
+
+
𝐷𝑡
𝜕𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑧 𝑑𝑡
𝑢
𝑣
𝑤
𝐷∅ 𝜕∅
𝜕∅
𝜕∅
𝜕∅
=
+𝑢
+𝑣
+𝑤
𝐷𝑡
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝐷∅ 𝜕∅
=
+ 𝑢 . 𝑔𝑟𝑎𝑑∅
𝐷𝑡
𝜕𝑡
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Rates of Change of Fluid Particle & Element
Most of the cases we are interested in developing equations for rates of
change per unit volume. Therefore, total derivative of property φ per unit
volume is given by;
𝜌
𝐷∅
𝜕∅
=𝜌
+ 𝑢 . 𝑔𝑟𝑎𝑑 ∅
𝐷𝑡
𝜕𝑡
Generalization of arbitrary conserved property φ;
𝜕𝜌
+ 𝑑𝑖𝑣 𝜌𝑢
𝜕𝑡
Continuity equation
𝜕 𝜌∅
+ 𝑑𝑖𝑣 𝜌∅𝑢
𝜕𝑡
Arbitrary property
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Rates of Change of Fluid Particle & Element
Re-written to show the relationship with the substantive derivative of φ
𝜕 𝜌∅
𝜕∅
𝜕𝜌
+ 𝑑𝑖𝑣 𝜌∅𝑢 = 𝜌
+ 𝑢 . 𝑔𝑟𝑎𝑑 ∅ + ∅
+ 𝑑𝑖𝑣 𝜌𝑢
𝜕𝑡
𝜕𝑡
𝜕𝑡
=𝜌
𝐷∅
𝐷𝑡
0
Continuity equation
𝜕 𝜌∅
𝐷∅
+ 𝑑𝑖𝑣 𝜌∅𝑢 = 𝜌
𝜕𝑡
𝐷𝑡
Rate of increase of φ
of fluid element
Rate of increase of φ
for a fluid particle
Net rate of flow of φ
out of fluid element
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Momentum Equation
Statement of the momentum is stated by Newton’s 2nd Law as below;
Rate
of
increase
of
momentum of fluid particle
𝐷
𝐷𝑡
=
Sum of forces on fluid
particle
𝐹
𝜌∅
where φ is u, v and w
Types of forces on fluid particles;
 Surface forces – pressure & viscous forces
 Body forces – gravity, centrifugal, Coriolis,
electromagnetic forces
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Momentum Equation
Previously we have derived the momentum and energy equations for relevant
entries of φ;
Rate of increase of momentum
x - momentum
u
𝜌
𝐷𝑢
𝐷𝑡
𝜕 𝜌𝑢
+ 𝑑𝑖𝑣 𝜌𝑢𝑢
𝜕𝑡
y - momentum
v
𝜌
𝐷𝑣
𝐷𝑡
𝜕 𝜌𝑣
+ 𝑑𝑖𝑣 𝜌𝑣𝑢
𝜕𝑡
z - momentum
w
𝜌
𝐷𝑤
𝐷𝑡
𝜕 𝜌𝑤
+ 𝑑𝑖𝑣 𝜌𝑤𝑢
𝜕𝑡
Energy
E
𝜌
𝐷𝐸
𝐷𝑡
𝜕 𝜌𝐸
+ 𝑑𝑖𝑣 𝜌𝐸𝑢
𝜕𝑡
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Momentum Equation
Stress of fluid element is defined in terms of;
1) Pressure, p
2) Viscous stresses, 𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
𝜏
y
z
x
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𝜏
𝜏
𝜏
𝜏
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Momentum Equation
First, consider the forces acting on the x-direction;
𝜏
𝜏
𝑝−
𝜏
−
+
+
𝜕𝜏 1
𝛿𝑧
𝜕𝑧 2
𝜕𝜏 1
𝛿𝑦
𝜕𝑦 2
𝜕𝑝 1
𝛿𝑥
𝜕𝑥 2
𝑝+
𝜕𝜏 1
𝛿𝑥
𝜕𝑥 2
𝜏
𝜏
y
z
𝜏
−
−
𝜕𝑝 1
𝛿𝑥
𝜕𝑥 2
+
𝜕𝜏 1
𝛿𝑥
𝜕𝑥 2
𝜕𝜏 1
𝛿𝑦
𝜕𝑦 2
𝜕𝜏 1
𝛿𝑧
𝜕𝑧 2
x
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Momentum Equation
Sum all the forces;
Left-right faces
𝑝−
𝜕𝑝 1
𝛿𝑥 − 𝜏
𝜕𝑥 2
−
𝜕𝜏 1
𝛿𝑥
𝜕𝑥 2
𝛿𝑦𝛿𝑧 + − 𝑝 +
= −
𝜕𝑝 𝜕𝜏
+
𝜕𝑥
𝜕𝑥
𝜕𝑝 1
𝛿𝑥 + 𝜏
𝜕𝑥 2
+
𝜕𝜏 1
𝛿𝑥
𝜕𝑥 2
𝛿𝑦𝛿𝑧
𝛿𝑥𝛿𝑦𝛿𝑧
Front-back faces
− 𝜏
−
𝜕𝜏 1
𝛿𝑦 𝛿𝑥𝛿𝑧 + 𝜏
𝜕𝑦 2
+
𝜕𝜏 1
𝜕𝜏
𝛿𝑦 𝛿𝑥𝛿𝑧 =
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑦 2
𝜕𝑦
Bottom-top faces
− 𝜏
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−
𝜕𝜏 1
𝛿𝑧 𝛿𝑥𝛿𝑦 + 𝜏
𝜕𝑧 2
+
𝜕𝜏 1
𝜕𝜏
𝛿𝑧 𝛿𝑥𝛿𝑦 =
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑧 2
𝜕𝑧
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Momentum Equation
Total force per unit volume due to surface stresses is;
𝜕 −𝑝 + 𝜏
𝜕𝑥
+
𝜕𝜏
𝜕𝜏
+
𝜕𝑦
𝜕𝑧
x-component momentum equation
𝜌
𝐷𝑢 𝜕 −𝑝 + 𝜏
=
𝐷𝑡
𝜕𝑥
+
𝜕𝜏
𝜕𝜏
+
+𝑆
𝜕𝑦
𝜕𝑧
y-component momentum equation
𝜌
𝜕 −𝑝 + 𝜏
𝐷𝑣 𝜕𝜏
=
+
𝐷𝑡
𝜕𝑥
𝜕𝑦
+
𝜕𝜏
+𝑆
𝜕𝑧
z-component momentum equation
𝜌
𝜕𝜏
𝐷𝑤 𝜕𝜏
𝜕 −𝑝 + 𝜏
=
+
+
𝐷𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
+𝑆
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Derive the momentum equations for y and z directions
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Energy Equation
The energy equation is derived from the 1st Law of Thermodynamics
Rate of increase
of energy of fluid
particle
=
Net rate of heat
added to fluid
particle
𝐷𝐸
𝐷𝑡
𝜌
Net rate of work
done on fluid
particle
+
𝑞 −𝑞
𝑤𝑜𝑟𝑘 − 𝑤𝑜𝑟𝑘
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Energy Equation
Work done by surface forces;
force x velocity
𝑢𝜏
𝑢𝜏
𝜕 𝑝𝑢
𝜕𝑥
𝜕𝑢𝜏
−
𝜕𝑥
𝑝𝑢 −
𝑢𝜏
+
+
𝜕𝑢𝜏 1
𝛿𝑧
𝜕𝑧 2
𝜕𝑢𝜏 1
𝛿𝑦
𝜕𝑦 2
1
𝛿𝑥
2
1
𝛿𝑥
2
𝑢𝜏
y
z
𝑢𝜏
x
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−
−
𝑝𝑢 +
𝜕 𝑝𝑢 1
𝛿𝑥
𝜕𝑥 2
𝑢𝜏
+
𝜕𝑢𝜏 1
𝛿𝑥
𝜕𝑥 2
𝜕𝑢𝜏 1
𝛿𝑦
𝜕𝑦 2
𝜕𝑢𝜏 1
𝛿𝑧
𝜕𝑧 2
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Energy Equation
Sum all the forces;
Left-right faces
𝑝𝑢 −
𝜕 𝑝𝑢 1
𝜕(𝜏 𝑢) 1
𝛿𝑥 − (𝜏 𝑢) −
𝛿𝑥
𝜕𝑥 2
𝜕𝑥 2
𝛿𝑦𝛿𝑧 + − 𝑝𝑢 +
= −
𝜕(𝑝𝑢) 1
𝜕 𝜏 𝑢 1
𝛿𝑥 + 𝜏 𝑢 +
𝛿𝑥
𝜕𝑥 2
𝜕𝑥 2
𝛿𝑦𝛿𝑧
𝜕(𝑝𝑢) 𝜕(𝜏 𝑢)
+
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑥
𝜕𝑥
Front-back faces
− 𝜏 𝑢−
𝜕(𝜏 𝑢) 1
𝜕(𝜏 𝑢) 1
𝜕(𝜏 𝑢)
𝛿𝑦 𝛿𝑥𝛿𝑧 + 𝜏 𝑢 +
𝛿𝑦 𝛿𝑥𝛿𝑧 =
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
Bottom-top faces
− 𝜏 𝑢−
𝜕(𝜏 𝑢) 1
𝜕(𝜏 𝑢) 1
𝜕(𝜏 𝑢)
𝛿𝑧 𝛿𝑥𝛿𝑦 + 𝜏 𝑢 +
𝛿𝑧 𝛿𝑥𝛿𝑦 =
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
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Energy Equation
Net rate of work in x-direction
𝜕 𝑢 −𝑝 + 𝜏
𝜕𝑥
+
𝜕 𝑢𝜏
𝜕𝑦
+
𝜕 𝑢𝜏
𝜕𝑧
𝛿𝑥𝛿𝑦𝛿𝑧
Net rate of work in y-direction
𝜕 𝑣𝜏
𝜕𝑥
+
𝜕 𝑣 −𝑝 + 𝜏
𝜕𝑦
+
𝜕 𝑣𝜏
𝜕𝑧
𝛿𝑥𝛿𝑦𝛿𝑧
Net rate of work in z-direction
𝜕 𝑤𝜏
𝜕𝑥
+
𝜕 𝑤𝜏
𝜕𝑦
+
𝜕 𝑤 −𝑝 + 𝜏
𝜕𝑧
𝛿𝑥𝛿𝑦𝛿𝑧
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Energy Equation
Total rate of work done per unit volume on the fluid particle
𝜕 𝑢 −𝑝 + 𝜏
𝜕𝑥
+
+
𝜕 𝑢𝜏
𝜕𝑦
+
𝜕 𝑢𝜏
𝜕𝑧
𝜕 𝑣𝜏
𝜕𝑥
+
𝜕 𝑣 −𝑝 + 𝜏
𝜕𝑦
+
= −𝑑𝑖𝑣 𝑝𝑢
+
𝜕 𝑢𝜏
𝜕𝑥
+
𝜕 𝑢𝜏
𝜕𝑦
+
𝛿𝑥𝛿𝑦𝛿𝑧
+
𝜕 𝑤𝜏
𝜕𝑥
𝜕 𝑢𝜏
𝜕𝑧
+
+
𝜕 𝑣𝜏
𝜕𝑧
𝜕 𝑤𝜏
𝜕𝑦
𝜕 𝑣𝜏
𝜕𝑥
+
𝛿𝑥𝛿𝑦𝛿𝑧
+
𝜕 𝑣𝜏
𝜕𝑦
𝜕 𝑤 −𝑝 + 𝜏
𝜕𝑧
+
𝜕 𝑣𝜏
𝜕𝑧
+
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕 𝑤𝜏
𝜕𝑥
+
𝜕 𝑤𝜏
𝜕𝑦
+
Task: Derive the term −𝑑𝑖𝑣 𝑝𝑢
𝜕 𝑤𝜏
𝜕𝑧
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Energy Equation
Energy flux due to heat conduction;
𝑞 +
𝑞 −
y
𝜕𝑞 1
𝛿𝑦
𝜕𝑦 2
𝜕𝑞 1
𝛿𝑧
𝜕𝑧 2
𝜕𝑞 1
𝛿𝑥
𝜕𝑥 2
𝑞 +
z
𝑞 −
x
24
𝑞 +
𝑞 −
𝜕𝑞 1
𝛿𝑧
𝜕𝑧 2
𝜕𝑞 1
𝛿𝑥
𝜕𝑥 2
𝜕𝑞 1
𝛿𝑦
𝜕𝑦 2
24
12
18/5/2020
Energy Equation
Net rate of heat transfer to the fluid particle;
x- direction;
𝑞 −
𝜕𝑞 1
𝜕𝑞 1
𝛿𝑥 − 𝑞 +
𝛿𝑥
𝜕𝑥 2
𝜕𝑥 2
𝛿𝑦𝛿𝑧 = −
𝜕𝑞
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑥
y- direction;
𝑞 −
𝜕𝑞 1
𝜕𝑞 1
𝛿𝑦 − 𝑞 +
𝛿𝑦
𝜕𝑦 2
𝜕𝑦 2
𝛿𝑥𝛿𝑧 = −
𝜕𝑞
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑦
z- direction;
𝑞 −
𝜕𝑞 1
𝜕𝑞 1
𝛿𝑧 − 𝑞 +
𝛿𝑧
𝜕𝑧 2
𝜕𝑧 2
𝛿𝑥𝛿𝑦 = −
𝜕𝑞
𝛿𝑥𝛿𝑦𝛿𝑧
𝜕𝑧
25
25
Energy Equation
Therefore, total rate of heat added to the fluid particle per unit volume;
−
𝜕𝑞
𝜕𝑞
𝜕𝑞
−
−
= −𝑑𝑖𝑣𝑞⃗
𝜕𝑥
𝜕𝑦
𝜕𝑧
From the Fourier’s law of heat conduction;
𝑞 = −𝑘
𝜕𝑇
𝜕𝑥
𝑞 = −𝑘
𝜕𝑇
𝜕𝑦
𝑞 = −𝑘
𝜕𝑇
𝜕𝑧
𝑞⃗ = −𝑘 𝑔𝑟𝑎𝑑 𝑇
Substitute
into
;
−𝑑𝑖𝑣 𝑞⃗ = 𝑑𝑖𝑣 𝑘 𝑔𝑟𝑎𝑑 𝑇
26
26
13
18/5/2020
Energy Equation
Total derivative for energy is given by;
𝜌
𝐸=𝑖+
𝐷𝐸
𝜕 𝑢𝜏
= −𝑑𝑖𝑣 𝑝𝑢 +
𝐷𝑡
𝜕𝑥
1
𝑢 +𝑣 +𝑤
2
+
Internal energy
𝜕 𝑣𝜏
𝜕𝑥
+
Kinetic energy
SE; source term
o Potential energy
o Heat production
𝜕 𝑢𝜏
𝜕𝑦
+
+
𝜕 𝑤𝜏
𝜕𝑥
𝜕 𝑣𝜏
𝜕𝑦
+
𝜕 𝑢𝜏
𝜕𝑧
+
+
𝜕 𝑤𝜏
𝜕𝑦
𝜕 𝑣𝜏
𝜕𝑧
+
𝜕 𝑤𝜏
𝜕𝑧
+ 𝑑𝑖𝑣 𝑘 𝑔𝑟𝑎𝑑 𝑇 + 𝑆
27
27
Kinetic Energy Equation
Multiplying u, v and w the equations in Slide 17 yielding ;
𝜌
𝐷
1
𝑢 +𝑣 +𝑤
2
𝐷𝑡
= −𝑢. 𝑔𝑟𝑎𝑑 𝑝 + 𝑢
𝜕𝜏
𝜕𝜏
𝜕𝜏
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
+𝑣
𝜕𝜏
𝜕𝜏
𝜕𝜏
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
+𝑤
𝜕𝜏
𝜕𝜏
𝜕𝜏
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
+ 𝑢. 𝑆
28
28
14
18/5/2020
Internal Energy Equation
Subtracting kinetic energy from energy the equation in Slide 27 yielding ;
𝜌
𝐷𝑖
= −𝑝 𝑑𝑖𝑣 𝑢 + 𝑑𝑖𝑣 𝑘 𝑔𝑟𝑎𝑑 𝑇 + 𝜏
𝐷𝑡
𝜕𝑢
+𝜏
𝜕𝑥
+ 𝜏
+ 𝜏
Where
𝜕𝑢
+𝜏
𝜕𝑦
𝜕𝑣
+𝜏
𝜕𝑥
𝜕𝑤
+𝜏
𝜕𝑥
𝜕𝑢
𝜕𝑧
𝜕𝑣
+𝜏
𝜕𝑦
𝜕𝑣
𝜕𝑧
𝜕𝑤
𝜕𝑤
+𝜏
+𝑆
𝜕𝑦
𝜕𝑧
𝑆 = 𝑆 − 𝑢. 𝑆
29
29
Temperature Equation
Incompressible fluid, i = cT (c is specific heat) and 𝑑𝑖𝑣 𝑢 = 0 ;
𝜌𝑐
𝐷𝑇
= 𝑑𝑖𝑣 𝑘 𝑔𝑟𝑎𝑑 𝑇 + 𝜏
𝐷𝑡
𝜕𝑢
+𝜏
𝜕𝑥
𝜕𝑢
+𝜏
𝜕𝑦
+ 𝜏
+ 𝜏
Where
𝜕𝑢
𝜕𝑧
𝜕𝑣
+𝜏
𝜕𝑥
𝜕𝑤
+𝜏
𝜕𝑥
𝜕𝑣
+𝜏
𝜕𝑦
𝜕𝑣
𝜕𝑧
𝜕𝑤
𝜕𝑤
+𝜏
+𝑆
𝜕𝑦
𝜕𝑧
𝑆 = 𝑆 − 𝑢. 𝑆
30
30
15