CHAPTER 1
BASIC CONCEPTS
1.1 Convection Heat Transfer
•
Examine thermal interaction between a surface and
an adjacent moving fluid
1.2 Important Factors in Convection Heat
Ts
Transfer
•
• Surface temperature is too high.
How to reduce it?
(1) Use a fan
(2) Change the fluid
(3) Increase surface area
q′s′
V∞
T∞
+
Fig. 1.1
−
1
• Conclusion: Three important factors in convection
(1) fluid motion
(2) fluid nature
(3)
surface
geometryof the role of fluid motion in convection:
• Examples
• Fanning to feel cool
• Stirring a mixture of ice and water
• Blowing on the surface of coffee in a cup
• Orienting a car radiator to face air flow
1.3 Focal Point in Convection Heat Transfer
Determination of temperature distribution in a
moving fluid
T = T ( x, y, z, t )
(1.1)
2
1.3 Fourier’s Law of Conduction
A (T − T )
q∝
L
A (Tsi − Tso )
qx = k
L
si
Tso
Tsi
so
A
x
qx
dx
(1.2)
k = thermal conductivity
0
L
x
Fig. 1.2
• Valid for:
(1) steady state
(2) constant k
(3) one-dimensional conduction
3
• Reformulate to relax restrictions. Consider element dx
T ( x ) − T ( x + dx )
T ( x + dx ) − T ( x )
q =k A
=−k A
dx
dx
dT
(1.3)
qx = − k A
dx
q′x′ = Heat flux
qx
(1.4)
q′x′ =
A
dT
(1.5)
q ′x′ = − k
dx
x
Generalize (1.5):
∂T
q ′x′ = − k
,
∂x
∂T
q ′y′ = − k
,
∂y
∂T
q ′z′ = − k
∂z
(1.6)
4
(1) Why negative sign?
(2) k ≠ constant
(3) Find T(x,y,z,t), use (1.6) to obtain q′′
(4) Changing fluid motion changes T(x,y,z,t)
1.5 Newton's Law of Cooling
q′s′ ∝ (Ts − T∞ )
q′s′ = surface flux
Ts = surface temperature
q′s′ = h(Ts − T∞ )
(1.7)
5
• Eq. (1.7) is Newton's law of cooling
• h is called the heat transfer
coefficient
h = f (geometry, motion, properties, ∆T )
(1.8)
1.6 The Heat Transfer Coefficient h
• Is h a property?
• Does h depend on temperature
distribution?
Apply Fourier’s law
∂ T ( x ,0 , z )
q′s′ = − k
∂y
(1.9)
Combine (1.7) and (1.9)
∂T ( x ,0, z )
∂y
h = −k
(Ts − T∞ )
(1.10)
Temperature distribution in needed to determine h
6
Ts•
• Apply Newton’s law to the bulb:
q ′s′
Ts = T∞ +
h
(1.11)
• Increase V∞ to increase h and lower Ts
qs′′
V∞
T∞
Fig. 1.1
+
Table 1.1 Typical values of h
h ( W/m 2 − o C )
Process
Free convection
Gases
5-30
Liquids
20-1000
Forced convection
Gases
20-300
Liquids
50-20,000
Liquid metals
5,000-50,000
Phase change
2,000-100,000
Boiling
5,000-100,000
Condensation
−
7
1.7 Differential Formulation of Basic Laws
• Three basic laws:
conservation of mass,
momentum, and energy
y
v
u
• Formulation
w
• Differential
• Integral
• Finite difference
• Key assumption: continuum
x
z
1.8 Mathematical Background
Fig. 1.4
r
(a) Velocity Vector V
r
V = ui + v j + wk
(1.12)
8
(b) Velocity Derivative
r
∂V ∂u ∂v
∂w
i+
j+
k
=
∂x ∂x
∂x
∂x
(1.13)
(c) The Operator ∇
Cartesian:
∂
∂
∂
∇≡
i+
j+ k
∂x
∂x
∂x
(1.14)
Cylindrical:
∂
1 ∂
∂
∇ ≡ ir +
iθ + i z
∂r
r ∂θ
∂z
(1.15)
Spherical: ∇ ≡
∂
1 ∂
1
∂
ir +
iθ +
iφ
∂r
r ∂θ
r sin θ ∂φ
(1.16)
9
(d) Divergence of a Vector
r
r ∂u ∂v ∂w
div .V ≡ ∇ ⋅ V = + +
∂x ∂y ∂z
(1.17)
(e) Derivative of the Divergence
r
∂
∂  ∂u ∂v ∂w 
(∇ ⋅ V ) =  + + 
∂x
∂x  ∂x ∂y ∂z 
(1.18)
or
or
r
∂
∂
(∇ ⋅V ) = ∇ ⋅ (u i + v j + w k )
∂x
∂x
r
r
∂
∂V
(∇ ⋅ V ) = ∇ ⋅
∂x
∂x
(1.19)
10
(f) Gradient of Scalar
∂T
∂T
∂T
Grad T = ∇T =
i+
j+
k
∂x
∂y
∂z
(1.22)
(g) Total Differential and Total Derivative
f = flow field dependent variable such as u, v, p, etc.
Cartesian Coordinates:
f = f ( x, y, z, t )
(a)
Total differential of f:
∂f
∂f
∂f
∂f
df = dx + dy + dz + dt
∂x
∂y
∂z
∂t
or
df Df ∂f dx ∂f dy ∂f dz ∂f
+
+
+
≡
=
dt Dt ∂x dt ∂y dt ∂z dt ∂t
(b)
11
But
dx
dy
dz
= u,
= v,
=w
dt
dt
dt
(c)
Substitute (c) into (b)
df Df
∂f
∂f
∂ f ∂f
+w +
=
=u
+v
dt Dt
∂x
∂ z ∂t
∂y
Total derivative:
df Df
=
dt Dt
Convective derivative:
∂f
∂f
∂f
u
+w
+v
∂x
∂y
∂z
Local derivative:
∂f
∂t
(1.21)
(d)
(e)
12
Apply to velocity component u. Set f = u
du Du
∂u
∂u
∂u ∂u
=
+w +
=u
+v
dt Dt
∂x
∂z ∂t
∂y
(1.22)
(1.22) represents
∂u
∂u
∂u
= convective acceleration in the x-direction
u +v +w
∂x
∂y
∂z
(f)
∂u
= local acceleration
∂t
(g)
Cylindrical coordinates : r ,θ , z
dv r Dv r
∂v r v θ ∂v θ v θ2
∂v r ∂v r
= vr
+vz
=
+
−
+
dt
Dt
∂r
r ∂θ
r
∂z
∂t
(1.23a)
13
dv θ Dv θ
=
dt
Dt
∂v θ v θ ∂v θ v r v θ
∂v θ ∂v θ
= vr
+vz
+
+
+
∂r
r ∂θ
r
∂z
∂t
dv z Dv z
∂v z ∂v z
∂v z v θ ∂v z
= vr
+vz
=
+
+
dt
Dt
∂r
r ∂θ
∂z
∂t
(1.23b)
(1.23c)
• Total derivative of temperature:
set f = T in (1.21)
dT DT
∂T
∂ T ∂T
∂T
=u
+v
+w
+
=
dt
Dt
∂x
∂y
∂z ∂t
(1.24)
14
1.9 Problem Solving Format
Solve problems in stages:
(1)
(2)
(3)
(4)
Observations
Problem Definition
Solution Plan
Plan Execution
(i) Assumptions
(ii) Analysis
(iii) Computations
(iv) Checking
(5) Comments
15
1.10 Units
SI units
Length (L): meter (m)
Time (t): second (s)
Mass (m): kilogram (kg)
Temperature (T): kelvin (K)
• Celsius and kelvin scales
T(oC) = T(K) - 273.15
(1.25)
Derived units:
• Force: newton (N)
One newton = force to accelerate one kilogram
one meter per sec per sec:
Force = mass × acceleration
N = kg . m /s2
16
• Energy: joules (J)
One joule = energy due to a force of one newton
moving a distance of one meter
J = N. m = kg . m2 /s2
• Power: watts (W)
One watt = one joule per second
W = J/s = N. m/s = kg . m2 /s3
17
Example 1.1: Heat Loss from Identical Triangles
• Surface temperature: Ts
• Variable h:
C
h( x ) =
x
q1
• Determine:
q2
(1)
•
•
•
Observations
Newton’s law givesq
h varies with x
Integration is required
(2) Problem Definition.
Determine dq for dx of each triangle
18
(3) Solution Plan.
Apply Newton's law to element and integrate
(4) Plan Execution
(i) Assumptions
• steady state
•
one• dimensional
uniform T∞
• uniform Ts
• negligible radiation
(ii) Analysis
Apply Newton’s law
dq = h( x )(Ts − T∞ )dA
(a)
C
h=
x
(b)
19
Triangle 1:
dA1 = y1 ( x )dx
(c)
Triangle 2:
dA2 = y 2 ( x )dx
(d)
Geometry:
H
y1 ( x ) = ( L − x )
L
H
y2 ( x) =
x
L
(e)
(f)
(e) into (c), (f) into (d):
H
dA1 = ( L − x )dx
L
(g)
H
dA2 = xdx
L
(h)
(b) and (g) into (a), integrate
20
H L− x
q1 = ∫ dq1 = ∫ C (Ts − T∞ )
dx
1/ 2
0
L x
L
q1 = ( 4 / 3)C (T − T∞ ) HL1 / 2
s
Similarly
(i)
H x
q2 = ∫ dq2 = ∫ C (T − T∞ )
dx
1
/
2
0
L x
L
s
q 2 = ( 2 / 3)C (T − T∞ ) HL1 / 2
s
Ratio of (i) and (j)
q1
=2
q2
(j)
(k)
(iii) Checking
Dimensional check:
(b) gives units of C:
C = W/m 3/2 − o C
21
(i) gives units of q1
q1 = C(W/m3/2-oC)( Ts - T∞ )(oC)H(m)L1/2 (m1/2) = W
Qualitative check:
q1 > q2 because base of 1 is at x = 0 where h = ∞ .
(5) Comments
• Orientation is important.
Same area triangles but different q
• Use same approach for other geometries
22