College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
EXAMPLE 1.2. ONE DIMENSIONAL STEADY STATE
Find the heat transfer via the automobile tire as shown in the Figure below, the width of tire is 5
cm, the inside and outside diameter are 20 cm and 40 cm, respectively. The thermal conductivity
of the tire is 50 W/m-K.
T2=27ᵒC
Analytical Solution
0.2
ln ( )
T1 − T2
0.7
0.1
qr =
, R syl =
≈
= 0.0044 K/W
R syl
2π × 0.05 × 50 15.71
qr =
13
= 2888.9 W ≈ 2.888 kW.
0.0045
1.1.
Heat by Convection & Radiation
The applications of the heat transfer by convection can be seen daily in our life, such as natural
convection of oceans and seas evaporation. The sun rays hits the surface of the earth in form of
heat by convection and radiation mode. The convective and radiative heat transfer coefficient are
most important parameter. The heat transfer by convection can be classified as:
1. Natural convection: the heat transits in a medium naturally.
2. Forced convection: the heat transits in a medium by using an external force, such as fans.
Convection:
q = hAs (Ts − T∞ )
1.49
Where,
h is the convective heat transfer coefficient in [W/m2-k,kW/m2-K , Btu/ft2-F….]
As is the surface area exposures the external flow in unit area.
Ts is the surface temperature in [K,or C]
T∞ is the surrounding temperature in [K, or C]
Radiation:
4 )
2 )(T
q = εσAs (Ts4 − Tsur
= εσAs (Ts2 + Tsur
s + Tsur )(Ts − Tsur )
1.50
Where,
ɛ is the emissivity (dimensionless, ɛ <1 for gray body: highly reflective surfaces, ɛ =1 for black
body, highly absorptive surfaces).
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
As is the surface area in a unit area.
Ts is the absolute surface temperature [K].
σ: Stefan-Boltzmann constant = 1.67×10-8 W/m2-K.
The resistance can be measured as:
Convection:
R conv. =
T1 − T2 (Ts − T∞ )
1
=
=
q
q
hAs
1.51
Radiation:
R rad. =
εσAs (Ts2
R rad. ≈
1
1
(Exact)
=
2
+ Tsur )(Ts + Tsur ) As h̅
1
̅̅̅3
As ɛσ4T
̅=
(Approximate) and T
1.52
Ts + Tsur
2
1.5
Table 1.5. Resistance Formulae
Coordinates
Planar Plane
Resistance Formulae
L
kAs
Cylindrical Plane
R cylw =
SPHERICAL PLANE
R Sphw
in
2πLk
1
1
rin − rout
=
4πk
Convection
Resistance
1
hAs
Contact Resistance
R′
Rc =
Ac
Radiation Resistance
r
ln ( rout )
R rad. =
Nomenclature
L is a wall thickness.
As is the area perpendicular to heat
flow.
rout is the outside raduis.
rin inside raduis.
L is the length of the cylinder.
k is the thermal conductivity.
rout is the outside raduis.
rin inside raduis.
h is the convective heat transfer
coefficient.
As is the surface area.
1
2 )(T + T
+ Tsur
s
sur )
1
1
=
=
̅
As h̅ 4εσAs T
εσAs (Ts2
R′ : Contact resistance.
Ac : Contact cross-section area.
Ts and Tsur are the surface and the
surrounding
Type equation here.temperature in
[K].
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
̅=
T
σ is the Steffen Boltzmann
constant.
ɛ is the emissivity constant.
Ts + Tsur
2
EXAMPLE 1.3. COMPOSITE WALL AND RESISTANCES FORMULAE
Shown in (Fig. 1.5) determine:
k =20
2
k=15
W/m-K
k=10W
/m-K
Wall1
Wall2
ho=50 W/m -k
To = 30C
W/m-K
Wall3
L1=0.1 m L2=0.12m
Cylinder, k=5 w/m-K
A. The resistance of each wall “R”?
B. The overall heat transfer coefficient” U”?
C. The heat transfer from high temperature wall to low temperature wall “ q”?
The area of the wall is 0.1m×0.1m, the inner and the outer diameter of the cylinder are 20 cm and
40 cm, respectively.
L3=0.15 m
hi =100 W/m2-K,Ti=100C
r
Fig. 1.5. Furnace wall layers for low temperature
Analytical solution:
From (Table 1.5) the resistance formula can be evaluated as:
To
Ti
R convo .
R wall1 .
R wall2 .
R wall3 .
R cyl .
R convi .
Inside convective resistance:
R convi . =
1
1
K
=
= 0.32
ho As 50 × 2π × 0.1 × 0.1
W
Walls conduction resistances:
Plane Wall1:
R w1 =
L1
0.1
K
=
= 0.67
A1 k1 (0.1 × 0.1) × 15
W
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
Plane wall2:
R w2 =
L2
0.12
K
=
= 1.20
A2 k 2 (0.1 × 0.1) × 10
W
Plane Wall3:
R netw3 =
L3
0.15 − 0.2
K
=
= 0.65
A3 k 3 (0.1 × 0.1) × 20
W
Cylindrical Wall:
R cyl =
r
ln ( rou )
in
2πkL
=
ln(2)
K
= 0.22
2π × 5 × 0.1
W
Outside convective resistance:
R convo . =
1
1
K
=
=1
ho As 100 × 0.1 × 0.1
W
R total = R convo + R w1 + R w2 + R netw3 + R cyl + R convi = 1 + 0.67 + 1.2 + 0.65 + 0.22 +
K
0.32 = 4.06 W
The heat transfer via the composite wall can be evaluated as:
q=
(Ti − To ) 100 − 30
70
=
=
= 17.24 W
R total
4.06
4.06
Note: the heat transfer to the outside is quiet small because the number of wall’s layers. The wall
is placed in a series and it is possible to be placed it in the parallel.
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
CHAPTER 2.0: SECOND SESSION (2ND S): EXTENDED SURFACES (FINS)
Fins are particular relevant to a large number of thermal engineering applications because the fins
that are used to enhance heat transfer in heat exchangers can often be treated as extended surfaces
(fins).
Surrounding Fluid
T∞, h
W
D
x=L
x=0
Energy is conducted axially from the bas at x = 0 (Tb) of the fin. As energy moves along the fin
in the x-direction, it is also conducted laterally to the fin surface where it is finally transferred by
convection to the surrounding fluid when T= T∞ and convective coefficient h.
Analytical Analysis:
The differential energy balance suggested by Figure is:
q x = q conv + q x+dx
1.6
Expanding the higher order term and simplifying leads to:
0 = q conv +
dq
dx
dx
1.7
The convection term is given by:
q conv = hAs (T − T∞ )
1.8
Substituting into Eq.(1.7) yields:
kAc
d2 T
dx = h P dx(T − T∞ )
dx 2
P is the perimeter of the geometry.
Manipulating of Eq. 1.9 leads to:
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
1.9
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
d2 T h P
(T − T∞ )
=
dx 2 kAc
1.20
Rewriting Eq (1.20) becomes:
d2 T h P
hP
−
T=
T
2
dx
kAc
kAc ∞
1.21
The equation 1.21 is the nonhomogeneous 2nd order differential equation and it can be solved as a
particular solution + homogenous solution.
T = Th + Tp
1.22
Tp = C, Th = C1 er1 x + C2 er2 x
1.23
Plugging, the particular and homogenous solution, into Eq.(1.21) yields:
C = T∞ → Tp = T∞
1.24
The homogenous solution can be written as:
r2 −
hP
hP
hP
hP
= 0 → r = ±√
→ r1 = √
and r2 = −√
kAc
kAc
kAc
kAc
1.25
The general solution of the non-homogenous equation becomes:
T = C1
hP
√
x
kA
e c
+ C2 e
−√
hP
x
kAc
+ T∞
1.26
hP
Assume M = √kA ,
c
To evaluate the constants C1 and C2, we should apply the boundary condition:
Assume the tip of the fin is at the fixed temperature: T = Tt
The boundary conditions will be ( at x = 0, T = Tb , and at x = L).
Tb = C1 + C2 + T∞
1.27
Tt = C1 eML + C2 e−ML + T∞
1.28
From Eq.(1.27) C2 = (Tb − T∞ ) − C1 plugging into Eq(1.28)yields:
C1 =
(Tt − T∞ ) + (T∞ − Tb )e−ML
(Tt − T∞ ) + (T∞ − Tb )e−ML
(T
)
,
and
C
=
−
T
−
2
b
∞
eML − e−ML
eML − e−ML
The analytical solution becomes:
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
T(x) =
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
(Tt − T∞ ) + (T∞ − Tb )e−ML Mx
(Tt − T∞ ) + (T∞ − Tb )e−ML −Mx
(T
)
e
+
−
T
−
e
b
∞
eML − e−ML
eML − e−ML
+ T∞
1.29
Numerical Analysis:
The central finite difference scheme of Eq. (30) of the governing equation
d2 T h P
(T − T∞ )
=
dx 2 kAc
1.30
Leads to:
T(i + 1) − 2T(i) + T(i − 1)
hP
hP
=
T(i)
−
T
∆x 2
kAc
kAc ∞
1.31
Manipulating of Eq. 1.31 yields:
T(i) =
1
hP
∆x 2 + 2
kAc
[T(i + 1) + T(i − 1) +
hP∆x 2
T ]
kAc ∞
1.32
The left and the right boundary condition are: LBCS: T(i = 0) = Tb and RBCs: T(i = Nx ) = Tt .
EXAMPLE 2.1. RECTANGULAR FINS
Find the temperature distribution of the rectangular fin as shown below analytically and
numerically. The physical properties of the fin are (k = 20 W/m-K, c =1000 J/kg-K, ρ = 2700
kg/m3). The surrounding air has (h = 25 W/m2-k, T∞ =25 C). The length of the fin (L) is 10.0 m
and the depth (D) is 0.02 m and the width (W) is 0.02m.
Tb=100ᵒC
T∞, h, air
Tt=25ᵒC
D
W
X
Analytical solution:
hP
Ac = W × D = 0.02 × 0.02 = m2 , M = √kA , P = 2(W + D) = 2(0.02 + 0.02) = 0.08 m
c
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
25 × 0.08
1
M=√
= 0.5
0.4 × 2300
m
T(x) =
(25 − 100)e−(0.5×10) 0.5x
(25 − 100)e−(0.5×10) −0.5x
(100
e
+
−
25)
−
e
+ 25
e0.5×10 − e−0.5×10
e0.5×10 − e−(0.5×10)
−75 × e−(0.5×10) 0.5x
75 × e−(0.5×10)
T(x) = 0.5×10
e
+ 0.5×10
e−0.5x + 100
e
− e−0.5×10
e
− e−(0.5×10)
T(x) =
e0.5×10
75
(e−0.5x − e0.5x ) + 100
− e−(0.5×10)
Note: for any position the temperature distribution along the fin can be determined.
Numerical solution:
Tb=100C
i=1
i−1
𝑖
i+1
i = Nx + 1, Tt=25C
For central nodes: 2 ≤ i ≥ Nx
T(i + 1) − 2T(i) + T(i − 1)
hP
hP
=
T(i) −
T
2
∆x
kAc
kAc ∞
1.32
Manipulating of Eq. 1.32 yields:
T(i) =
1
hP
∆x 2 + 2
kAc
[T(i + 1) + T(i − 1) +
hP∆x 2
T ]
kAc ∞
clear all
clc
% Fin Dimensions
L=10;
W=0.02;
D=0.02;
Ac=W*D;
P=2*(W+D);
% Physical Properties of the fin
k=0.01;
% Surrounding fluid properties
Tinf=25;
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
1.33
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
h=25;
% Number of nodes and Distance between two adjacent nodes
Nx=100;
dx=L/Nx;
% Given Boundary conditions
Tb=100;
Tinf=25;
Tt=25;
T(1:Nx+1)=0;
% Boundray Conditions
T(1)=Tb;
T(Nx+1)=Tt;
Niter=20000;
% Interior nodes
for k =1:Niter
for i=2:Nx
T(i)=1/((h*P)/(k*Ac)*dx^2+2)*(T(i+1)+T(i-1)+(h*P)/(k*Ac)*dx^2* Tinf);
end
end
x=0:dx:L;
% Analytical Solution
M=sqrt((h*P)/(k*Ac));
Ta(1:Nx+1)=0;
for i=1:Nx+1
Ta(i)=((Tt-Tinf)+(Tinf-Tb)*exp(-M*L))/(exp(M*L)-exp(-M*L))*exp(M*x(i))+((TbTinf)-((Tt-Tinf)+(Tinf-Tb)*exp(-M*L))/(exp(M*L)-exp(-M*L)))*exp(M*x(i))+Tinf;
end
plot (x,Ta,'o',x,T,'r')
grid on
xlabel('L in meter');ylabel('Temperature Profile along the Rectangular Fin in
Degree C')
legend('Analytical’, ‘Prediction')
Note: the analytical and the numerical solution are based on the boundary conditions which are
given in the problem. The different solutions are listed in the Table 2.1.
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]
College of Engineering
Summer Session- 2015
Heat Transfer - ME 372
Dr. Saeed J. Almalowi, [email protected]
Table 2.1. Soutions f or different tip conditions for constant cross-section area
Note: Perimeter is per = 2(W+D) for rectangular base area, per = 4πD for circular fin, m = M =
hP
√kA , P = per, L is the length of the fin, Ac= WD (rectnagular or square bas)or Ac =
c
πD2
4
(circular bas).
Mathematical Relations: sinsh(x) = (ex − e−x )/2 , sinsh(mx) = (emx − e−mx )/2 ,
cosh(x) = (ex + e−x )/2, and cosh(mx) = (emx + e−mx )/2.
d
d
[cosh(x)] = sinsh(x) and [sinsh(x)] = cosh(x).
dx
dx
Copyright@2015, Dr. Saeed J. Almalowi,E-mail: [email protected]