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 q s
T

Transfer
s
 Surface temperature is too high.
How to reduce it?
(1) Use a fan
(2) Change the fluid
(3) Increase surface area
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
Ts
Ts
so
i
A
x
qx
d
(1.2)
0
k = thermal conductivity
L
x
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
(
g
m
p
,

T
)
e
(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
 T ( x ,0 , z )
y
h  k
Ts  T 
(1.9)
Combine (1.7) and (1.9)
(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
qs
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
F 1
i
1.8 Mathematical Background

(a) Velocity Vector V

V  ui  v j  w k
(1.12)
8
(b) Velocity Derivative

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 
i  i z
r
r 
z
(1.15)

1 
1

Spherical:  
ir 
i 
i
r
r 
r sin  
(1.16)
9
(d) Divergence of a Vector

 u v w
div .V    V 


x y z
(1.17)
(e) Derivative of the Divergence


  u v w 
  V      
x
x  x y z 
or
or



  V     u i  v j  w k 
x
x



V
  V    
x
x
(1.18)
(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)
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But
dx
dy
dz
 u,
 v,
w
dt
dt
dt
(c)
Substitute (c) into (b)
df Df
f
f
f f

u
v
w 
dt Dt
x
y
z t
Total derivative:
df Df

dt
Dt
Convective derivative:
f
f
f
u
v
w
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

u
v
w 
dt Dt
x
y
z t
(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 , , 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 θ v z
v z 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)
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1.9 Problem Solving Format
Solve problems in stages:
(1) Observations
(2) Problem Definition
(3) Solution Plan
(4) 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
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