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Heat Transfer Review
D. H. WillitsBiological and Agricultural EngineeringNorth Carolina State University
Steady-State Ht Conduction – Composite Plane Wall (Fig 6.5)
1 2
1 2 3 4 5
1 2
31 2
1 1 2 3 2
2
1 1
W/m of surface area
x
t tq
R R R R R
t txx x
h k k k h
Steady-State Ht Conduction – Composite Plane Wall (Fig 6.5)
1 5 1 5
1 21 2 3
1 1 2
1x
t t t tq
x xR R Rh k k
Steady-State Ht Conduction – Composite Plane Wall (Fig 6.5)
4 5 4 5
23
2
x
t t t tq
xRk
Thermal Conductivity Values
Interpretation of the values in Table 6.2 requiresan understanding of the difference betweenresistivity and resistance.
Resistivity = 1/kResistance = x/k
To get resistance from resistivity, you mustmultiply by the thickness of the material.
Steady-State Ht Conduction – Composite Cylinder (Fig 6.8)
1 2
1 2 3 4 5
1 2
3 2 4 32 1
1 1 1 2 3 4 2
2
2ln( / ) ln( / )ln( / )1 1
W/m of length
r
t tq
R R R R R
t tr r r rr r
r h k k k r h
Steady-State Ht Conduction – Composite Cylinder (Fig 6.8)
1 5
1 2 3
1 5
3 22 1
1 1 1 2
2
2ln( / )ln( / )1
r
t tq
R R R
t tr rr r
r h k k
Steady-State Ht Conduction – Composite Sphere (Fig 6.10)
1 2
1 2 3 4
1 2
3 22 12 2
1 1 1 1 2 2 2 3 3 2
4
4( )( )1 1
W
rs s s s
t tq
R R R R
t tr rr r
r h k r r k r r r h
Steady-State Ht Conduction – Problem
Consider a composite cylindrical tube with an outside diameter of 10 cm. The wall consists of two layers of different materials, A and B. The inner material A is in contact with a hot fluid and the outer material B is in contact with still air at a temperature of 30 C. Material A is stainless steel, 0.2 cm thick, and material B is 0.3 cm thick with a thermal conductivity of 0.0378 W m-1 K-1. If the outer surface temperature is 110 C, and the outside surface coefficient can be estimated as ho = 7.36 W/m2 K, determine: a) the heat transfer through the wall, in W/m of lengthb) the temperature of the interface between the two materialsc) the temperature of the hot fluid if the inside surface coefficient is estimated at 20 W m-2 K-1
Steady-State Ht Conduction – Problem
Answers:
a) the heat loss per unit length–
32 (110 30) 185.1W/mr oq r h
b) the temperature at the interface –
int
3 2
2 ( 110)185.1W/m
ln( / )erface
r
b
tq
r r
k
tinterface = 158.2 C.
Steady-State Ht Conduction – Problem
Answers:
(c) the temperature of the hot fluid
-1
-2 -1 -1 -1
2 ( 158.2)185.1Wm
1 ln(0.047 / 0.045 )(0.045m)(20Wm K ) 21.5Wm K
191.0C
fluid
fluid
t
m m
t
Critical Radii for Cylinders and Spheres
Radius at which maximum heat transfer occurs:
Cylinder Biot No. = 1.0 = roho/k
Sphere Biot No. = 2.0
Transient Ht Transfer
Case 1:
1
0.2 p
hA
c Vo
o
t thBo e
k t t
Case 2: 100h
k
Case 3: 0.2 100h
k
Heisler ChartsFigs 6.11-6.13
Transient Ht Transfer
Heisler Charts:
2
1Lines are
o
o
F
B
For non-infinite geometries, TR values are multiplied together
Transient Ht Transfer – Problem 1
A 3 cm diameter hot dog with a length of 10 cm has an initial uniform temperature of 10 C. If it is suddenly dropped into boiling water at 100 C, determine the temperature at the center after 10 min. Assume the following values:
h = 6000 W/m2 K k = 0.5 W/m Kα = 1.33 X 10-7 m2/s.
Transient Ht Transfer – Problem 1
Intersection of cylinder and slabfor cylinder:
Bo = hr/k = 180 ; 1/Bo = 0.0056 (Case 2);= /r2 = 0.3547;TR1 = 0.205 (from Fig 6.12;)
for slab:Bo = hL/k = 600; 1/Bo = 0.0016665 (Case 2);
= /r2 = 0.0.03192;TR2 = 0.99 (from Fig. 6.11);
TR = TR1 x TR2 = 0.20295; tc = 81.73 C
Transient Ht Transfer – Problem 2
An aluminum cylinder (thermal conductivity = 160 W/m K, density = 2790 kg/m3, specific heat = 0.88 kJ/kg K) of radius r = 5 cm, length L = 0.5 m, and a uniform initial temperature of 200 C is suddenly immersed at time zero in a well-stirred fluid maintained at a constant temperature of 25 C. The heat transfer coefficient between the cylinder and the fluid is h = 300 W/m2 K. Determine the time required for the center of the cylinder to reach 50 C. What will the surface temperature be at that time?
Transient Ht Transfer – Problem 2
Check Bo:for the cylinder: Bo = hro/k = (300W/m2K)(0.05m)/(160W/mK) = 0.094
This is Case 1
Note: we do not have to check Bo for the slab because Case 1 says that the internal temperature gradient for the cylinder is already negligible, which says that the heat is conducted to the edge of the material faster than it can be convected away by the water. The slab case will not change that.
Transient Ht Transfer – Problem 2
For Case 1, use Eq 6.120 in the text:
)exp(1 Vc
hA
tt
tt
po
o
)))(J/kgK880)(kg/m2790(
)m2m)(KW/m300(exp(
25200
255023
2222
Lr
rDL
solving for gives 6.03 min or 361.8 s
The surface is the same as the center because Case 1 assumes no internal gradient.
Convection
The basic problem is to find the appropriate h to use in Newton’s Law of Cooling. Once that is done, finding qx is fairly trivial:
xq hA t
Convection
Free ( ) (Pr)m nNu c Gr
Horizontal cylinder, laminar flow – 1x104 < GrPr < 1x108
0.250.56( Pr)Nu Gr
Horizontal cylinder, turbulent flow – 1x108 < GrPr < 1x1012
1/30.13( Pr)Nu Gr
Convection
Free
3 2
2
Nu
Gr
Pr p
hD
k
D g t
c
k
Convection
Forced
Nu (Re) (Pr)m nc
Eqs 7.49 – 7.53
ReVD
where
Convection - Problem
Air at 60 C if flowing normal to a cylindrical copper tube with a diameter of 15 cm at a velocity of 2 m/s.
Estimate the convective heat transfer coefficient at the surface.
Convection - Problem
Using Eq 7.51, with properties determined at Tfilm
rho 1.0604 kg/m^3 Density of air
mu .00002005 Pa*s Viscosity of air
cp 1007 J/kg*K Specific Heat of air
k .02856 W/m*K Thermal conductivity of air
Tfilm 60 C Film temperature
Pr .706945028011204 - Prandtl No.
Nu 77.6381180236944 - Nusselt No.
h 14.7822976717114 W/m^2 Conv heat transfer coefficient
Re 15866.3341645885 - Reynold's Number
Heat Exchangers – Basic Eqns
ln
1 1 2 2
1 1
2 2
ln
a b a b
a b
a b
q UA t
t t t tUA
t t
t t
a a pa a
b b pb b
q m c t
q m c t
Fig 7.1
Heat Exchangers – Parallel Flow
ta 1
ta 2
tb 1
tb 2
f luid 'a '
f luid 'b '
tem
pera
ture
d is tanc e
Heat Exchangers – Effectiveness Ratio
max min
max min
a a
a b
b b
a a
a a
t tE
t t
w cR
w c
UANTU
c w
Heat Exchangers – Effectiveness Ratio
11 exp 1
11
11 exp 1
1 11 exp 1
p
c
NTUR
E
R
NTUR
E
NTUR R
Figs 7.3 and 7.4, or
Radiation Heat Transfer
Black Body Emissive Power
4bW T
where T is absolute temperature and depends upon the unit system.
Radiation Heat Transfer
Gray Body Emissive Power
4W Twhere is the emissivity.
Note: Gray bodies have constant with wavelength.
Radiation Heat Transfer
Gray Body Exchange
4 4( )
1i i j
i ji i i
i j i j j
A T Tq
A
F A
Radiation Heat Transfer
The first problem is determining the shape factor Fij. For simple geometries, Table 8.3 may suffice. Fij is the fraction of the energy leaving i that is intercepted by j.
For infinite parallel planes, the value is 1.0.
For small bodies enclosed by a larger body (where the smaller body cannot see itself), thevalue is also 1.0.
