CHE 314 - Heat Transfer
Midterm Exam (Fall 2021), October 25,
Lecture time
Formula Sheet (10 Pages)
Energy Balance
• Fourier’s law for the one-dimensional plane wall:
00
qx = −k
dT
dx
• Newton’s law of cooling:
00
q = h (Ts − T∞ )
00
q [W/m2 ] is the heat flux, q = As · q
00
• The net heat flux of radiation heat transfer from the surface:
00
4
qrad = ε σ Ts4 − Tsur
σ is the Stefan-Boltzmann constant, σ = 5.67 · 10−8 (W/m2 · K4 ).
ε is emissivity, a radiative property of the surface.
• Energy Conservation Principle for a Control Volume:
Ėst = Ėin − Ėout + Ėg
where
dT
00
00
; Ėin = A · qin
; Ėout = A · qout
; Ėg = q 000 · Vol
dt
where A is the surface area perpendicular to the direction of the heat flux
q 00 , q 000 is the specific heat generation rate [W/m3 ].
Ėst = ρ · V ol · c ·
• Surface Energy Balance for a Control Surface:
Ėin − Ėout = 0
• Example of the heat fluxes balance for the surface, where the heat is
transferred to the surface by the conduction and removed from the surface
by the convection and radiation:
00
00
00
qcond − qconv − qrad = 0
1
Steady State Conductive Heat Transfer
• General Fourier’s law for 3D Cartesian coordinate system
00
00
00
00
~q = −k∇T = i qx + j qy + k qz
where i = ~i, j = ~j, k = ~k
00
qx = −k
∂T
∂T
∂T
00
00
; qy = −k ; qz = −k ;
∂x
∂y
∂z
• The heat diffusion equation written in the general form in Cartezian
coordinates:
∂T
∂
∂T
∂
∂T
∂
∂T
k
+
k
+
k
+ q̇ = ρ cp
∂x
∂x
∂y
∂y
∂z
∂z
∂t
• One-Dimensional (1D) conduction in a planar medium with constant properties and no heat generation:
1 ∂T
∂ 2T
=
α ∂t
∂x2
where α =
k
ρ cp
is the thermal diffusivity [m2 /s].
1D Steady State Conduction-Plane Walls
,1
8
T
Ts,1
qx
q conv
q conv
q cond
T s,2
Hot Fluid
h1
,2
8
T
Cold Fluid
x=L
x
h2
Figure 1: Heat Transfer through a plane wall.
2
• Heat Rate
T∞,1 − T∞,2
Rtot
qx =
where
Rtot =
L
1
1
+
+
A h1 Ak A h2
= Rt,conv,1 + Rt,cond + Rt,conv,2
A is the surface.
• Thermal Resistances
– The thermal resistance for conduction in a plane wall:
Rt,cond =
L
Ak
– The thermal resistance for convection:
Rt,conv =
1
Ah
– The thermal resistance for radiation:
Rt,rad =
1
1
=
2
2 ) (T + T
A hr
A σ ε (Ts + Tsur
s
sur )
– The thermal resistance for convection and radiation (parallel):
Rt,par =
1
Rt,conv
+
1
Rt,rad
−1
1
=
A
1
h + hr
• Circuit representations of heat the conduction problem shown in Fig. (1)
Figure 2: The equivalent thermal circuit for the plane wall with convection
surface conditions shown in Fig. 1.
3
• Thermal Resistance for Unit Surface Area is called as the area-specific
contact resistance:
00
Rtot =
∆T
= A Rtot (K m2 /W )
00
qx
• Overall heat transfer coefficient, U , which is defined by an expression analogous to Newton’s law of cooling:
qx = U · A · ∆T ; =⇒ U =
1
Rtot · A
• Complex Composite Wall, see Fig. 3. The total heat transfer through this
composite system can be expressed as follows:
qx =
Tl − T∞
Rtotal
cond
cond
Rtotal = Rparal
+ R3 + Rconv ; Rparal
=
and
R1 =
1
1
+
R1 R2
−1
L2
L3
1
L1
; R2 =
; R3 =
; Rconv =
K1 A1
K2 A 2
K3 A3
h A3
1D Steady State Conduction-Radial Sysytems
• The heat rate and heat flux at which energy is conducted across any cylindrical surface in the solid may be expressed using Fourier’s law
2π L k(Ts,1 − Ts,2 )
dT
=
qr = −k (2π r L)
| {z } dr
ln(r2 /r1 )
A
00
qr =
qr
k(Ts,1 − Ts,2 )
=
A
r ln(r2 /r1 )
• Conduction Resistance:
Rr,cond =
4
ln(r2 /r1 )
2π L k
X
A1
Tl
insulation
A3
K1
K3
T
h
8
K2
A2
convection
insulation
L3
L 1= L 2
q
R convec
R3
R1
T
8
Tl q
1
R2
q
q2
Figure 3: Complex Composite Wall
• Considering the composite system (see Fig. (4)) the heat transfer rate may
be expressed as
qr =
T∞,1 − T∞,4
1
ln(r2 /r1 ) ln(r3 /r2 ) ln(r4 /r3 )
1
+
+
+
+
L kA} |2π {z
L kB} |2π {z
L kC} |2π r{z
1 L h1}
4 L h4}
|2π r{z
|2π {z
Rt,conv1
Rt,condA
Rt,condB
Rt,condC
Here the interfacial contact resistances was neglected.
• The Resistance for convection and radiation (parallel):
−1
1
1
Rt,par =
+
Rt,conv Rt,rad
• The Resistance for radiation:
1
1
Rt,rad =
=
2 ) (T + T
As hr
As σ ε (Ts2 + Tsur
s
sur )
• Critical insulation radius:
rcr =
5
k
h
Rt,conv4
,1
T s,1 A B C
Ts,2
Ts,3
Ts,4
r1
r2
r3
T
,4
8
symmetry line
8
T
r4
h4
h1
Hot
fluid
Cold
fluid
Figure 4: Cylindrical composite system
• Heat Transfer Rate for a Spherical Wall:
qr = −k(4πr2 )
4π k(Ts,1 − Ts,2 )
dT
=
1
1
dr
r1 − r2
• Conduction Resistance for a Spherical Wall:
1
1
1
−
Rt,cond =
4π k r1 r2
• Convection Resistance for a Spherical Wall:
Rt,conv =
1
h As
• The Resistance for convection and radiation (parallel) for a Spherical Wall::
−1
1
1
Rt,par =
+
Rt,conv Rt,rad
• The Resistance for radiation for a Spherical Wall::
Rt,rad =
1
1
=
2 ) (T + T
As hr
As σ ε (Ts2 + Tsur
s
sur )
6
• The composite system for a Spherical Wall:
qr =
Ts − T∞
, Rt,total = Rt,cond1 + Rt,cond2 + ... + Rt,conv
Rt,total
Conduction with Thermal Energy Generation
• The heat diffusion equation for a solid cylinder
dT
q̇
1 d
r
+ =0
r dr
dr
k
• If q̇ = const., the general solution is
T (r) = −
q̇ 2
r + C1 ln(r) + C2
4k
• The heat diffusion equation for a solid sphere
dT
q̇
1 d
r2
+ =0
2
r dr
dr
k
• If q̇ = const., the general solution is
T (r) = −
q̇ 2 C1
r −
+ C2
6k
r
• Boundary conditions are
dT
= 0 ; T (r = r1 ) = Ts or
r=0
dr
| {z }
−k
dT
dr
due to the symmetry
Transient Heat Transfer
• The Lumped Capacitance Method - LCM
t
T − T∞
= exp −
Ti − T∞
τt
where τt is the thermal time constant
τt =
7
ρ · V ol · c
h As
r=r1
= h(T − T∞ )
• The Biot number
h Lc
k
where Lc is the characteristic length, Lc =
Bi =
V ol
As .
• LCM model is valid if Bi < 0.1.
• The total energy transfer Q occurring up to some time t:
t
Q = Qmax 1 − exp −
τt
Qmax = Ct · Θi = ρ · V ol · c · Θi
Θi = Ti − T∞
Convective Heat Transfer: External Flow
Boundary Layer & Dimensional Analysis
• The total heat transfer rate:
q = h · As · (Ts − T∞ )
• An average convection heat transfer coefficient
1
h=
As
Z
h dAs
As
The special case of flow over a flat plate
1
h=
L
ZL
h dx
0
• Local convection heat transfer coefficient
h=
• Biot number: Bi =
−kf
∂T
(Ts − T∞ ) ∂y
hL
ks
8
y=0
• Nusselt number: N u =
hL
kf
• Prandtl number: P r =
cp µ
k
• Reynolds number: Re =
• Fourier number: F o =
ρ V∞ L
µ
αt
L2
Flat Plate
• Mean boundary layer temperature Tf , termed the film temperature:
Tf ≡
Ts + T∞
2
• The laminar boundary layer thickness δ:
5x
u∞ · x
δ=√
; Rex =
ν
Rex
• For P r ≥ 0.6 the ratio of the velocity to thermal boundary layer thickness
is
δ
≈ P r1/3
δt
The velocity tubrbulent boundary layer thickness may be expressed as
δ = 0.37 · x · Re−1/5
x
Isothermal plate
• For laminar flow (Rex < 5·105 ) over an isothermal plate the local Nusselt
number has the form:
hx x
= 0.332 · Re1/2 · P r1/3 f or P r ≥ 0.6 (gase & water)
N ux ≡
k
N ux = 0.564 · Re1/2 · P r1/2 f or P r ≤ 0.5(liquid metals), P ex ≥ 100
P ex ≡ Rex · P r is the Peclet number.
• For laminar flow over an isothermal plate the average Nusselt number
has the form:
N ux ≡
hx · x
1/3
= 0.664 · Re1/2
f or P r ≥ 0.6
x · Pr
k
9
• The local Nusselt number for turbulent flow is
1/2
N ux = 0.0296 · Re4/5
f or 0.6 ≤ P r ≤ 60
x · Pr
Uniform surface heat flux
• For a uniform surface heat flux imposed at the plate and for laminar
flow the local Nusselt number can be estimated as follows:
1/3
N ux = 0.453 · Re1/2
f or P r ≥ 0.6
x · Pr
• For turbulent flow:
1/3
N ux = 0.0308 · Re4/5
f or 0.6 ≤ P r ≤ 60
x · Pr
• The average Nusselt number for laminar flow:
1/2
N uL = 0.68 · ReL P r1/3
• An average surface temperature from
00
Ts − T∞
q ·L
hL L
= s
; N uL =
k
k · N uL
Cylinder & Sphere
Re =
V∞ D
ρ V∞ D
=
µ
ν
where D is the diameter.
• Cylinder: Overall average Nusselt number according to Whitaker (Re >
0):
1/2
2/3
N u ≈ 0.4Re + 0.06Re
P r0.4
• Sphere: Average Nusselt number according to the Ranz and Marshall
correlation:
N u = 2 + 0.6 · Re1/2 · P r1/3
10