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HEAT TRANSFER OPERATION
MIDTERM NOTE
I. LAW OF HEAT TRANSFER:
- Fundamental law:
Define: ALWAYS HAVE TO SATISFY, REGARDLESS OF SITUATION.
(Xài lúc nào cũng đúng)
a. Types of system consider:
_ Closed system:
Fix amount of matter (total moles=constant ¿
No mass flows across the system boundaries (mass=constant ¿
Energy flow may occur across the boundaries (exchange heat, work to surrounding)
The boundaries may change with time (
V change ¿
_Open System - Control Volume
(mass change )
b. Law of Conservation of Mass:
_For closed system :mass=constant(vô TH này thì mass trước = mass sau)
_ Fluid flow through a control volume:
rate mass
¿
−rate mass
out
−mass acumulation
system
With:
rate mass=
dm
dt
=ρ
dV
dt
=ρdA
dx
dt
=ρAv( 1 −Dimesion , small A )
rate mass
¿
∬
control surface
❑
ρ v
¿
d A
¿
(intergerated above for hold surface)
rate mass
¿
∬
control surface
❑
ρ v
out
d A
out
mass acumulation
system
∂ t
∭
Control volume
❑
ρdV
Equation (1)
∬
control surface
❑
ρ v
¿
d A
¿
∬
control surface
❑
ρ v
out
d A
out
∂t
{
∭
Control volume
❑
ρdV
}
(Description : Rate of mass tại 1 điểm là ρAv vói v theo phương x, do xét trong hệ vật
diện tich cố định, Rate of mass trên cả mặt của vật thì lấy nguyên hàm 2 phương y, z.
Tương tự với mass accumulation nếu V đổi theo 3 phương thì ta nguyên hàm theo 3
phương đêr có đụowc tổng thay đổi của V)
Types of fluids Steady state
(No accumulation)
Non-steady state
Compressible fluid
∬
C .V
❑
ρ v
¿
d A
¿
∬
C. V
❑
ρ v
out
d A
out
Equation (1)
Incompressible fluid
(ρ=const ¿
∬
C .V
❑
v
¿
d A
¿
∬
C .V
❑
v
out
d A
out
Equation (1)
_In 1 dimensional : Remove term rate mass out of the integrate (Do A không đổi theo
phương y,z; A tại điểm vào và ra của mass là hằng số)
Types of fluids Steady state
(No accumulation)
Non-steady state
Compressible fluid
ρ
¿
v
¿
A
¿
=ρ
out
v
out
A
out
ρ
¿
v
¿
A
¿
−ρ
out
v
out
A
out
=acumulate
Incompressible fluid
ρ=const ¿
v
¿
A
¿
=v
out
A
out
v
¿
A
¿
−v
out
A
out
=acumulate
c. Newton’s Second law of motion:
_For close system: ∑
F=
d
dt
( m ⃗v)( folow thedirection you chose)
_Fluid flow through a control volume:
∑
force=rate of moment out−rate of moment∈+ Acumulate
With:
rate momentum=
dF
dt
v
dm
dt
v ρdA
dx
dt
v ρA v
n
( 1 −Dimesion , small A)
T at certain point distance x form hot face:
T
H
−T
T
H
−T
C
x
b
+Infinite long hollow cylinder
r
o
r
i
:radius outside (inside) , L:lengh of the cylinder
T
o
T
i
:temperature outside(inside )
q
A
q
2 πrL
=−K
∂T
∂ r
q
2 πL
∫
r o
r i
∂ r
r
=−K
∫
T o
T i
∂ T ↔
q
2 πL
ln
(
r
i
r
o
)
=−K (T
i
−T
o
↔ q=
− 2 πKL
T
i
−T
o
ln
(
r
i
r
o
)
T at certain r :
T −T
i
T
i
−T
o
ln(r /r
i
ln (r
o
/r
i
_At not steady state:
Coordinate x y z
Cartesian (slab)
x=x
∂ T
∂n
∂T
∂ x
y= y
∂ T
∂n
∂T
∂ y
z=z
∂ T
∂n
∂T
∂ z
Cylindrical
x=r cosθ=r
↔ ∂ x =∂ r
∂ T
∂ x
∂T
∂ r
(radial heat transfer)
y=r sinθ=rθ
↔ ∂ y=r ∂ θ
∂ T
∂ y
r
∂ T
∂ θ
z=z
∂ T
∂n
∂T
∂ z
At small θ :sinθ ≈ θ , cosθ ≈ 1
Sphere
x=r sin θ cos φ=rθ
↔ ∂ x =r ∂θ
∂ T
∂ x
r
∂ T
∂ θ
y=r sin θ sin φ=rθφ
↔ ∂ y=rθ ∂ φ
∂ T
∂ y
rθ
∂ T
∂ φ
z=r cosθ=r
∂ T
∂ z
∂T
∂ r
b. Newton’s law of cooling
q
A
=h
T
W
−T
f
T
W
: Heat of object ,T
f
:heat of fluid , h : Heat transfer coefficient
(
W
m
2
K
)
c. Stefan-Boltzmann Law:
q
A
=σ T
4
σ =5.67 × 10
− 8
(
W
m
2
K
4
)
(
Btu
h. ft
2
4
)
: Stefan−Boltzman constant
d. Thermal resistance: (nhiệt trở, na ná điện trở)
Infinite slab Infinite hollow
Cylinder
Surface of liquid
Thermal resistance (
Z
th
b
KA
ln (r
o
/r
i
2 πRKL
hA
If the wall have only 1 way to transfer heat (many layers wall): similar to resistance put
in series:
Z
th ,total
∑
Z
th
If the wall have more than 1 way to transfer heat similar to resistance put in parallel:
(Ex: The wall have 1 nail stick through its, heat can transfer in two paths: Through nail or
through wall)
Z
th ,total
∑
Z
th , pathi
( 1 pathmay have many resistance∈ series)
We can calculate the resistance of the path to find temperature of cold side knowing hot
side and heat rate:
q=
∆ T
Z
th ,total
Ex: Calculate the temp at intersection of 2, 3 layers in 3 layers wall
Z
th ,total
=Z
1
+Z
2
Z1 Z2 Z
heat generation
Steady state
∂T
∂ t
Into equation in I.2.b
K ∇
2
T = 0 ¿
In 1 direction Ex: 3D cylindrical: Assume temperature
change only on r coordinate (radial heat
transfer)
∂T
∂ θ
∂ T
∂ z
r
∂r
(
r
∂ T
∂r
)
q
K
α
∂T
∂ t
r
(
∂T
∂ r
2
T
∂ r
2
)
With
( uv ) ' =u ' v+uv '
In (*)
∂ r
:is differental with r ,
∂ T
∂r
∧r is function of r
Step 3: find
T ∧∂T
∂ x
¿ applied∈boudary condition :
Ex: for 1D slab (condition 4), steady state (condition 2) and K constant (condition 1)
General equation become:
2
T
∂ x
2
q
K
2
T
∂ x
2
−q
K
∂ T
∂ x
−q
K
x +C
1
→ T (x )=
−q
K
x
2
+C
1
x +C
2
Ex: for 1D hollow cylinder (condition 4), steady state (condition 2) and K constant
(condition 1)
r
∂r
(
K. r
∂ T
∂ r
)
q
K
r
(
∂ T
∂ r
+r
2
T
∂r
2
)
q
K
r
∂ T
∂ r
2
T
∂r
2
−q
K
∂ T
∂ r
+r
2
T
∂r
2
−q
K
r (multiply r for each side)
↔ r
∂ T
∂ r
−q
K
r
2
( integral 2 side) ↔
∂ T
∂ r
−q
K
r
→T (r )=
−q
K
r
2
+C
1
Step 4: Find boundary condition for solving C1 and C2 (normally maximum 2 is
require):
Boundary condition
Temperature of surface is maintain at
T
sur
x=b ,r =r
o
T (b)=T
sur
T
sur
is a number, position of surface
depend on how you put the origin and
coordinate)
Knowing heat flux (q/A) at surface
x=b ,r =r
o
q / A=
(
∂T
∂ x
)
x=b
Special case for heat
flux
Insulation surface
(
∂ T
∂ x
)
x=b
Thermal symmetry
(Heat flow in/out from
2 side to center)
(
∂ T
∂ x
)
x=b / 2
(b is the thickness of wall, if chose x =0 at
center change the x=b/2 by x=0)
Convection Boundary Condition
(Have fluid flow to cool/heat surface with
temperature
T
f
and h)
−K
(
∂ T
∂ x
)
x=b
=−h(T
f
−T
x=b
Step 5: Put C1 and C2 to equation.
(TO SEE HOLD SOLVING CAN SEE TAKE HOME PRACTICE)
III. Unsteady State Heat Conduction:
- Some dimension for use:
Biot number : Bi=
hL
K
(Ratio of Z
th
of solid∧Z
th
of surface )
Fourier Number : X=
αt
L
2
Kt
ρC
p
L
2
(dimensionless of time)
α=
K
ρ C
p
; thermal diffusity (m
2
/s)
Step 5: Calculate Biot, Fourier, m, n for each component and find for
θ
θ
o
in Appendix
F:
If there are no line near with you’re calculate m:
Ex: m = 0.6, but in chart only have 0.5 and 0.75:
You can find
θ
θ
o
for 0.5 and 0.75, assume m and
θ
θ
o
are linear:
θ
θ
o
=am+b
And use 2 point at 0.5 and 0.75 you are found to calculate a and b
Step 6: Found the
θ
θ
o
of the shape and calculate T
There are some problem like give you T and make you find t:
Step 1: You will calculate
θ
θ
o
of the shape and write it component:
Ex:
θ
θ
o
ficyl
θ
θ
o
infi cyl
θ
θ
o
infi slab
Step 2: Then you calculate the Fourier number of component through time:
Ex:
X
infi slab
=a× t , X
infi cyl
=b × t
Step 3: Then you will guess :)))))) any relation you want: (CHOSE ONLY 1 COMPONENT)
Ex
θ
θ
o
infi cyl
θ
θ
o
fi cyl
θ
θ
o
infi cyl
θ
θ
o
fi cyl
(You can have the¿chose bay consider the ratio of each L of each componet )
Step 4: Then put it in
θ
θ
o
infi cyl
of the shape to calculate each of component, put in appendix of
find time by each component.
Step 5Take that time to find other component
θ
θ
o
. Then cal
θ
θ
o
shape
Step 6: If
θ
θ
o
of the shape not fit, adjust t and try again
- Fin:
a. Some equation:
We have the fin with:
extended L , A
tip
is area of the tip∧Pis perimeter of the tip
M =
√
hPK A
tip
, m=
√
hP
K A
tip
fin effectiveness:ε
f
q
f
q
heat transfer with fin
heat transfer without the fin
ε
f
√
KP
h A
surface
tanh(mL)(adiabatic fin)
fin efficiency : n
f
q
real
q
ideal
(heat at tip equal at base)
q
real
h A
sur
(T −T
L
b. Calculation:
_For only pin fin:
_ For circular and other strainght fin:
