-1-
KURSES / COURSE: INGENIEURSWESE / ENGINEERING
VAK/ SUBJECT: WARMTE OORDRAG 4A / HEAT TRANSFER 4A
UNIVERSITEIT VAN JOHANNESBURG/
UNIVERSITY OF JOHANNESBURG
SEMESTER TEST 3 – 17 June 2014
COURSE:
ENGINEERING
SUBJECT:
HEAT TRANSFER 4A
TIME:
1.0 hours
MARKS:
50
----------------------------------------------------------------------------------------------------------------------------------Requirements : Calculator
Answer all questions.
Attachments: Information Sheet
This is a closed book test.
________________________________________________________________
QUESTION 1
1.1
Prove Kirchoff’s Identity.
1.2
Derive the reciprocity relation
1.3
From first principles show that the radiation heat exchange between non-black bodies is given
by:
𝐸𝑏 − 𝐽
𝑞=
1−𝜖
𝜖𝐴
explaining the significance of each variable
1.4
If this equation is interpreted as an electrical analogy explain the relevance of the various
terms.
1.5
The 𝐹𝑖𝑗 terms in question 1.2 may be interpreted as shape factors. Explain briefly
(mathematical derivation not required):
1.5.1
Their relevance to general radiation heat transfer calculations, and
1.5.2
How they could be derived from basic geometrical considerations.
A j F ji = Ai Fij explaining the reasoning behind the derivation.
(25 marks)
QUESTION 2
A long furnace has inside dimensions of 3 m x 3 m. The walls, roof, and floor have zero transmissivity.
The temperature of the roof is 1400 K, the side walls 1700 K, and the floor 600 K. The emissivity of the
roof is 0.6, the side walls 0.5, and the floor 0.4. Calculate the radiant heat flux of the floor. Use the
component numbering system given in the figure. For the geometry of the furnace 𝐹13 = 0.414. StefanBoltzmann constant = 5.67 × 10 −8 W/ 𝑚2 𝐾4
3m
3
2
2
3m
1
(25 marks)
HEAT TRANSFER WAO4A
MEMORANDUM – TEST 3 JUNE 2014
QUESTION 1
1.1
Prove Kirchoff’s Identity.
𝒒𝒊 𝑨 𝜶
EA
Black Enclosure
Consider a body with surface area 𝐴 located in a black enclosure. When the black body and
enclosure are in equilibrium; no heat transfer between the two:
For the body in the enclosure - Energy Outflow = Energy absorbed, i.e.:
𝐸𝐴 = 𝑞𝑖 𝐴 𝛼
Where:
(1)
𝐸 =Emissive power of body
𝐴 = surface area of body
𝑞𝑖 = radiant flux from enclosure received by body
𝛼 = absorptivity of body
If the body is replaced with a black body then 𝛼 = 1 aqnd:
Divide equation (1) by equation (2):
𝐸𝑏 𝐴 = 𝑞𝑖 𝐴
𝑬
𝑬𝒃
=𝜶
(2)
(3)
Thus, the ratio of the emissive power on a body to that of a similar black body at the same
temperature is equal to the absorptivity of the body. Thus the emissivity is given by:
and hence:
𝜖=
𝐸
(4)
𝐸𝑏
𝜖=𝛼
(5)
Which is referred to as Kirchoff’s law or Kirchoff’s identity.
1.2
Derive the reciprocity relation
A j F ji = Ai Fij explaining the reasoning behind the
derivation.
The net radiation energy exchange between two bodies labelled 1 and 2 is given by:
If:
𝐸𝑏1 𝐴1 𝐹12 − 𝐸𝑏2 𝐴2 𝐹12 = 𝑄12
1
(1)
then:
𝑇1 = 𝑇2
(2)
𝐸𝑏1 = 𝐸𝑏2
(3)
(4)
and equation (1) may be written:
𝑄12 = 0
(5)
i.e.
𝐴1 𝐹12 − 𝐴2 𝐹12 = 0
𝐴1 𝐹12 = 𝐴2 𝐹12
(6)
𝐴𝑖 𝐹𝑖𝑗 = 𝐴𝑗 𝐹𝑗𝑖
(7)
This is the reciprocity equation and may be written generally as:
1.3
From first principles show that the radiation heat exchange between non-black bodies
is given by:
𝑞=
𝐸𝑏 − 𝐽
1−𝜖
𝜖𝐴
The surface energy components on an opaque material are shown in the following figure:
𝑱 = 𝝐𝑬𝒃 + 𝝆G
G
𝒒
= 𝑱−𝑮
𝑨
where:
𝐺 = Incident radiation
∈ =emissivity of material
𝐸𝑏 = emissive power of surface (a function of its absolute temperature)
𝜌 = reflectivity of material
𝑞
=energy per unit area supplied to surface to balance energies
𝐴
𝐽= Radiosity
Energy radiated from surface (Radiosity) is:
Balance of energy supplied to surface:
𝐽 = 𝜖𝐸𝑏 + 𝜌𝐺
𝑞
𝐴
= 𝐽−𝐺
(1)
(2)
For an opaque material where the transmissivity is zero, by Kirchoff’s identity:
𝜌 = 1 − 𝛼 = 1−∈
2
(3)
Equation (2) may be written:
𝑞
𝐴
= 𝐽 − 𝐺 = 𝜖𝐸𝑏 + 𝜌𝐺 − 𝐺 = 𝜖𝐸𝑏 + (1−∈) 𝐺 − 𝐺 = 𝜖𝐸𝑏 −∈ 𝐺
Substituting for 𝐺 using equation (1) gives:
𝑞
𝐴
𝑞
𝐴
which may be written:
=∈ 𝐸𝑏 −∈
𝐽−∈𝐸𝑏
𝜌
=∈ 𝐸𝑏 −∈
𝐸𝑏 −∈𝐸𝑏−𝐽+∈𝐸𝑏
=∈ �
𝑞=
(5)
1−∈
𝐸𝑏 −𝐽
� =∈ �
1−∈
𝐽−∈𝐸𝑏
1−∈
(4)
�
(6)
𝐸𝑏−𝑗
(7)
1−∈
∈𝐴
1.4 If this equation is interpreted as an electrical analogy explain the relevance of the various terms
In terms of an electrical analogy:
𝑞 = electrical current
𝐸𝑏 − 𝑗 = Potential difference; voltage, and
1−∈
∈𝐴
= resistance, in this case a surface resistance
1.5
The 𝑭𝒊𝒋 terms in question 1.2 may be interpreted as shape factors. Explain briefly
(mathematical derivation not required):
1.5.1
Their relevance to general radiation heat transfer calculations, and
•
•
•
•
1.5.2
•
The 𝐹𝑖𝑗 terms, shape factors or view factors define the rate of radiation heat transfer from a
surface 1 to a surface 2 or between a number of surfaces.
It defines the rate of radiation heat transfer between surfaces each of which has a constant
surface temperature and hence a constant emissive power.
It takes into account the geometrical orientation of the two or more surfaces.
The shape factor which is defined for surfaces of common orientations and are available as
standards.
How they could be derived from basic geometrical considerations.
Shape factors, which define the radiation between surfaces are obtained by integration of the
radiative heat transfer between two surfaces. Formulations are of the form:
𝑞𝑛𝑒𝑡 1−2 = (𝐸𝑏1 − 𝐸𝑏2 ) �
𝐴2
•
� cos �φ1 �cos (φ2 )
𝐴1
𝑟2
which defines the total net exchange in radiation energy. The shape factor is the integral term.
Shape factors are obtained through double integrals od the transfer of radiation energy
between elemental areas of the surfaces involved.
QUESTION 2
2.1
𝑑 𝐴1 𝑑𝐴2
Glass properties: (Not asked)
Reflectivity
Transmissivity
= 0.1
= 0.75
3
Glass thickness
Irradiance
The absorptivity
= 3 mm
2
= 1000 W/m
= 1 − 0.1 − 0.75 = 0.15
𝐶𝑃 𝑔𝑙𝑎𝑠𝑠 = 800 𝐽/ 𝑘𝑔 𝐾
𝜌𝑔𝑙𝑎𝑠𝑠 = 2640 𝑘𝑔/ 𝑚3
Heat absorbed by the glass plate = 0.15 × 1000 = 150 𝑊
Mass of glass with a surface area of 1 m
= 0.003 × 2460 = 7.38 𝑘𝑔/ 𝑚2
2
The rate of increase in temperature of the glass at its initial condition is given by:
𝑑𝑇
𝑑𝜏
and:
𝑑𝑇
𝑑𝜏
2.2
=
𝑄
𝐶𝑃 𝑚 = 𝑄
=
𝐶𝑃 𝑚
1000
800 ×7.38
(1)
= 0.169 𝐾/𝑠
(3)
Furnace dimensions given in figure.
Temperatures of surfaces:
Roof
Walls
Floor
= 1400 K
= 1700 K
= 600 K
Emissivities:
Floor
Walls
Roof
𝜖1 = 0.4
𝜖2 = 0.5
𝜖3 = 0.6
Stefan-Boltzmann constant = 5.67 × 10−8 W/ 𝑚2 𝐾4
𝐹13 = 0.414
3m
3
2
2
1
4
3m
Consider the heat balance at a surface
G
𝑱 = 𝝐𝑬𝒃 + 𝝆G
𝒒
= 𝑱−𝑮
𝑨
The radiosity 𝐽 of a surface, as shown in the diagramme, is given by the sum of emissive power of the
surface multiplied by its emissivity plus the incident radiation 𝐺 on the surface multiplied by its
reflectivity.
For the opaque surfaces of the furnace:
𝝆 = 𝟏 − 𝜶; 𝝐 = 𝜶; 𝝆 = 𝟏 − 𝝐
𝑱 = 𝝐𝑬 𝒃 + (𝟏 − 𝝐) 𝑮
(1)
(2)
The incident radiation on the surface will be the sum of the radiations from all surfaces and hence for
surface 𝑖 the radiosity is given by:
𝑱𝒊 = 𝝐𝒊 𝑬 𝒃 + (𝟏 − 𝝐) ∑𝒏𝒌=𝟏 𝑱𝒌 𝑭𝒊𝒌
(3)
𝑱𝟏 = 𝝐𝟏 𝑬 𝒃𝟏 + (𝟏 − 𝝐𝟏 )(𝑱𝟏 𝑭𝟏𝟏 + 𝑱𝟐 𝑭𝟏𝟐 + 𝑱𝟑 𝑭𝟏𝟑 )
(4)
𝑱𝟑 = 𝝐𝟑 𝑬 𝒃𝟑 + (𝟏 − 𝝐𝟑 )(𝑱𝟏 𝑭𝟑𝟏 + 𝑱𝟐 𝑭𝟑𝟐 + 𝑱𝟑 𝑭𝟐𝟑 )
(6)
For the three different surfaces in the furnaces:
𝑱𝟐 = 𝝐𝟐 𝑬 𝒃𝟐 + (𝟏 − 𝝐𝟐 )(𝑱𝟏 𝑭𝟐𝟏 + 𝑱𝟐 𝑭𝟐𝟐 + 𝑱𝟑 𝑭𝟐𝟑 )
(5)
A shape factor 𝐹𝑙𝑙 will be zero if the surface is flat but will non-zero if radiation from the surface is
incident on itself.
The shape factors need to be quantified.
Given:
From geometry of the furnace:
𝑭𝟏𝟑 = 𝟎. 𝟒𝟏𝟒
(7)
𝑭𝟏𝟏 = 𝟎. 𝟎
(8)
𝑭𝟑𝟑 = 𝟎. 𝟎
(9)
𝑭𝟑𝟏 = 𝑭𝟏𝟑 = 𝟎. 𝟒𝟏𝟒
From the summation rule:
𝑭𝟐𝟐 = 𝑭𝟏𝟑 = 𝟎. 𝟒𝟏𝟒
𝑭𝟏𝟐 = 𝟏 − 𝑭𝟏𝟏 − 𝑭𝟏𝟑 = 𝟏 − 𝟎 − 𝟎. 𝟒𝟏𝟒 = 𝟎. 𝟓𝟖𝟔
5
(10)
(11)
(12)
From the reciprocal rule:
From symmetry:
𝑭𝟐𝟏 =
𝑨𝟏
𝑨𝟐
𝑭𝟏𝟐 = 𝟎. 𝟓 × 𝟎. 𝟓𝟖𝟔 = 𝟎. 𝟐𝟗𝟑
𝑭𝟐𝟑 = 𝑭𝟐𝟏 = 𝟎. 𝟐𝟗𝟑
𝑭𝟑𝟐 = 𝑭𝟏𝟐 = 𝟎. 𝟓𝟖𝟔
(13)
(14)
(15)
Emissive power of the surfaces is obtained from the Stefan-Boltzmann law:
𝑬 𝒃𝟏 = 𝟓. 𝟔𝟕 × 𝟏𝟎−𝟖 × 𝟔𝟎𝟎𝟒 = 𝟕. 𝟑𝟒𝟖𝟑 𝒌𝑾/𝒎𝟐
(16)
𝑬 𝒃𝟑 = 𝟓. 𝟔𝟕 × 𝟏𝟎−𝟖 × 𝟔𝟎𝟎𝟒 = 𝟐𝟏𝟕. 𝟖𝟐 𝒌𝑾/𝒎𝟐
(18)
𝑱𝟏 = 𝟎. 𝟒 × 𝟕. 𝟑𝟒𝟖𝟑 + 𝟎. 𝟔 × (𝟎 + 𝟎. 𝟓𝟖𝟔 𝑱𝟐 + 𝟎. 𝟒𝟏𝟒 𝑱𝟑 )
(19)
𝑱𝟑 = 𝟎. 𝟔 × 𝟐𝟏𝟕. 𝟖𝟐 + 𝟎. 𝟒 × (𝟎. 𝟒𝟏𝟒 𝑱𝟏 + 𝟎. 𝟓𝟖𝟔 𝑱𝟐 + 𝟎)
(21)
𝑬 𝒃𝟐 = 𝟓. 𝟔𝟕 × 𝟏𝟎−𝟖 × 𝟔𝟎𝟎𝟒 = 𝟒𝟕𝟑. 𝟓𝟔 𝒌𝑾/𝒎𝟐
(17)
Substituting the known values in equations (4), (5), and (6) gives:
𝑱𝟐 = 𝟎. 𝟓 × 𝟒𝟕𝟑. 𝟓𝟔 + 𝟎. 𝟓 × (𝟎. 𝟐𝟗𝟑 𝑱𝟏 + 𝟎. 𝟒𝟏𝟒 𝑱𝟐 + 𝟎. 𝟐𝟗𝟑 𝑱𝟑 )
i.e.:
(20)
𝑱𝟏 = 𝟐. 𝟗𝟑𝟗 + 𝟎. 𝟑𝟓𝟏𝟔 𝑱𝟐 + 𝟎. 𝟐𝟒𝟖𝟒 𝑱𝟑
(22)
𝑱𝟑 = 𝟏𝟑𝟎. 𝟕 + 𝟎. 𝟏𝟔𝟓𝟔 𝑱𝟏 + 𝟎. 𝟐𝟑𝟒𝟒 𝑱𝟐
(24)
𝑱𝟐 = 𝟐𝟑𝟔. 𝟖 + 𝟎. 𝟏𝟒𝟔𝟓 𝑱𝟏 + 𝟎. 𝟐𝟏𝟐 𝑱𝟐 + 𝟎. 𝟏𝟒𝟔𝟓 𝑱𝟑
(23)
Substitute equation (24) into equation (22):
i.e.:
𝑱𝟏 = 𝟐. 𝟗𝟑𝟗 + 𝟎. 𝟑𝟓𝟏𝟔 𝑱𝟐 + 𝟎. 𝟐𝟒𝟖𝟒 (𝟏𝟑𝟎. 𝟕 + 𝟎. 𝟏𝟔𝟓𝟔 𝑱𝟏 + 𝟎. 𝟐𝟑𝟒𝟒 𝑱𝟐 )
𝟎. 𝟗𝟓𝟖𝟗 𝑱𝟏 = 𝟑𝟓. 𝟒𝟎𝟓 + 𝟎. 𝟒𝟎𝟗𝟖 𝑱𝟐 )
(25)
(26)
Substitute equation (24) into equation (23):
i.e.:
𝑱𝟐 = 𝟐𝟑𝟔. 𝟖 + 𝟎. 𝟏𝟒𝟔𝟓 𝑱𝟏 + 𝟎. 𝟐𝟏𝟐 𝑱𝟐 + 𝟎. 𝟏𝟒𝟔𝟓 (𝟏𝟑𝟎. 𝟕 + 𝟎. 𝟏𝟔𝟓𝟔 𝑱𝟏 + 𝟎. 𝟐𝟑𝟒𝟒 𝑱𝟐 )
(27)
𝟎. 𝟕𝟓𝟑𝟔 𝑱𝟐 = 𝟐𝟓𝟔. 𝟎 + 𝟎. 𝟏𝟕𝟎𝟖 𝑱𝟏
(28)
Equation (26):
(29)
Equation (28):
𝟐. 𝟑𝟒𝟎 𝑱𝟏 = 𝟖𝟔. 𝟒𝟎 + 𝑱𝟐
(30)
(29)-(30):
𝟎. 𝟐𝟐𝟔𝟕 𝑱𝟏 = −𝟑𝟑𝟗. 𝟕 + 𝑱𝟐
𝟐. 𝟏𝟏𝟑𝟑 𝑱𝟏 = 𝟒𝟐𝟔. 𝟏
(31)
6
𝑱𝟏 = 𝟐𝟎𝟏. 𝟔
(29):
(24):
𝑱𝟐 = 𝟐. 𝟑𝟒𝟎 × 𝟐𝟎𝟏. 𝟔 − 𝟖𝟔. 𝟒 = 𝟑𝟖𝟓. 𝟑𝟒
𝑱𝟑 = 𝟏𝟑𝟎. 𝟕 + 𝟎. 𝟏𝟔𝟓𝟏 × 𝟐𝟎𝟏. 𝟔 + 𝟎. 𝟐𝟑𝟒𝟒 × 𝟑𝟖𝟓. 𝟑𝟒 = 𝟐𝟓𝟒. 𝟑
(32)
(33)
(34)
The radiant heat transfer from surface, as shown above, is given by:
𝒒=
𝒒=
𝑬𝒃𝟏−𝑱𝟏
𝟏−𝝐𝟏
𝜺𝟏𝑨
=
𝑬𝒃−𝒋
(35)
𝟏−∈
∈𝑨
𝟎.𝟒(𝟕.𝟑𝟒𝟖 −𝟐𝟎𝟏.𝟔)×𝟏
𝟏−𝟎.𝟒
7
= −𝟏𝟐𝟗. 𝟓 𝒌𝑾/ 𝒎𝟐
(36)
-2-
KURSES / COURSE: INGENIEURSWESE / ENGINEERING
VAK/ SUBJECT: WARMTE OORDRAG 4A / HEAT TRANSFER 4A
INLIGTINGSBLAD / INFORMATION SHEET
∇ 2T +
q ''' 1 ∂T
;
=
k α ∂t
qx = −kA
α=
k
;
ρC p
k= −
q ''
∂T
∂x
dT
dT
; q r = −kAr
dr
dx
∂ 2T ∂ 2T ∂ 2T q 1 ∂T
+
+
+ =
∂x 2 ∂y 2 ∂z 2 k α ∂τ
∂ 2T 1 ∂T 1 ∂ 2T ∂ 2T q 1 ∂T
+
+
+
+ =
∂r 2 r ∂r r 2 ∂φ 2 ∂z 2 k α ∂τ
d dT
d 2T dT
(r
)=r 2 +
dr dr
dr
dr
1 ∂ 2 (rT )
1
1
∂
∂T
∂ 2T q 1 ∂T
(sin
)
+
+
θ
+ =
r ∂r 2
r 2sin θ ∂θ
r 2sin 2 θ ∂φ 2 k α ∂τ
∂θ
q = h1 A(TA − T1 ) =
q=
kA
(T1 − T2 )
∆x
TA − TB
r0
1
ri
+
hi Ai 2πkL
ln
Rt =
1
;
hA
Rt =
1
;
UA
θ=
( x ) C1emx + C2e− mx ;
Tm +1, n + Tm −1, n − 2Tm , n
(∆x) 2
+
S wall =
q= kS ∆T ;
θ
= e − mx
θ0
q f = hPkAc θ 0 ;
Tm , n +1 + Tm , n −1 − 2Tm , n
(∆y ) 2
A
; Sedge = 0.54 D ; Scorner = 0.15 L
L
qi + ∑ (
q
+ =0
k
j
Ti =
∑(
j
nπ
θ T − T1 4 ∞ 1 −[ 2 L ]
=
= ∑ e
θi Ti − T1 π n =1 n
T ( x, t ) − T0
x
= erf
Ti − T0
2 αt
2
ατ
Sin
nπx
2L
erf
Tj
Rij
1
)
Rij
n = 1,3,5
x
2 αt
=
x / 2 αt
∫
e −η dη
2
hx h 2ατ
h ατ
T − Ti
)];
= 1 − erf X − [exp( + 2 )][1 − erf ( X +
T∞ − Ti
k
k
k
∂ 2T1
∂ 2T2 1
∂ 2T2
∂ 2T1
T2 2 + T1 2 = (αT1 2 + αT2 2 )
∂x
∂z
∂z
∂x
α
X =
x
2 ατ
)
-3-
KURSES / COURSE: INGENIEURSWESE / ENGINEERING
VAK/ SUBJECT: WARMTE OORDRAG 4A / HEAT TRANSFER 4A
θ
T − Tw 3 y 1 y
=
=
− ( )
θ ∞ T∞ − Tw 2 δ t 2 δ t
u
∂T
)w
3k 3 k
∂y
h=
=
=
Tw − T∞
2 δ t 2 ςδ
− k(
3
∂T
∂T
∂ 2T
µ ∂u 2
+v
=α 2 +
( )
∂x
∂y
∂y
ρc p ∂y
u=
d
Red Pr
L
= 3,66 +
2
d
1 + 0,04[( ) Red Pr ] 3
L
0,0668
1 τw
ln y + C
K ρ
Hausen : N ud
1
Knudsen and Katz :
hd
u d
=
= C ( ∞ ) n Prf3
νf
kf
N udf
Redf
0,4 – 4
4 – 40
40 – 4000
4000 – 40000
40000 - 400000
ρ (u
C
0,989
0,911
0,683
0,193
0,0266
n
0,330
0,385
0,466
0,618
0,805
∂T
∂u
∂ 2u
+ v ) = gρβ (T − T∞ ) + µ 2
∂x
∂y
∂y
1
0,670 Ra 4
Churchill and Chu : N u = 0,68 +
9 4
0,492 16 9
[1 + (
) ]
Pr
1
3
1
Nu x = 0,332 Pr 3 Re x2 [1 − (
N ud = 0,023 Rd0,8 Prn
1
x0 4 − 3
) ]
x
RaL < 109
1
1
Nu L =
hL
= 0,664 Re L2 Pr 3
k
Re L =
ρu∞ L
µ
n = 0,4: heating ; = 0,3 : cooling
Eckert and Drake:
N u = (0,43 + 0,50 Re ) Pr
0,5
N uf = C (Grf Prf ) m
0 , 38
(
Prf
Prw
)0, 25 for 1 < Re < 103 0,25 Re Pr
0, 6
0 , 38
(
Prf
Prw
)0, 25 or 103 < Re < 2 × 105
-4-
KURSES / COURSE: INGENIEURSWESE / ENGINEERING
VAK/ SUBJECT: WARMTE OORDRAG 4A / HEAT TRANSFER 4A
Constants for Isothermal Surfaces
Raf
4
Vertical planes and cylinders
10 - 109
109 – 1013
109 – 1013
Horizontal cylinders
0 – 10-5
104 – 109
109 - 1012
Upper surface of heated plates or lower 2x104 – 8x106
surface of cooled plates
8x106 - 1011
Lower surface of heated plates or upper
105 - 1011
surface of cooled plates
Vertical cylinder
104 - 106
Height = diameter
Characteristic length = diameter
1
N ue = 0,56(Gre PreCosθ ) 4
θ < 88o ; 105 < Gre PreCosθ < 1011
1
1
C
0,59
0,021
0,10
0,4
0,53
0,13
0,54
0,15
0,27
m
0,25
0,4
0,333
0
0,25
0,333
0,25
0,333
0,25
0,775
0,21
Te = Tw − 0,25(Tw − T∞ )
1
N ue = 0,14[(Gre Pre ) 3 − (Grc Pre ) 3 ] + 0,56(Gre PreCosθ ) 4 − 75 < θ < −10; 105 < Gre PreCosθ <1011
θ(deg)
-15
-30
-60
-75
q = UA
Grc
5x109
2x109
108
106
− UA
m c
)(1 + c c )]
m c cc
m h ch

mc
1+ c c
m h ch
1 − exp[(
(Th 2 − Tc 2 ) − (Th1 − Tc1 )
(T − T )
ln[ h 2 c 2 ]
(Th1 − Tc1 )
ε=
− UA
C
)(1 + min )]
Cmin
Cmax
C
1 + min
Cmax
1 − exp[(
ε=
Re =
ρUD
µ
Ra = Gr Pr
Pr =
Cpµ
k
Pe = Re Pr
Nu =
hx
k
Gr = Re Pr
Gr =
d
L
gβ (Tw − T∞ ) x 3
ϑ
Bi =
2
hs
k
St =
Fo =
h
ρC pu
ατ
s
2
=
kτ
ρcs 2