environmental systems and facility planning
TRANSCRIPT
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Heat Transfer Review
D. H. WillitsBiological and Agricultural Engineering
North Carolina State University
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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 Rt t
xx x
h k k k h
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Steady-State Ht Conduction
Composite Plane Wall (Fig 6.5)
1 5 1 5
1 21 2 3
1 1 2
1xt t t t q
x xR R R
h k k
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Steady-State Ht Conduction
Composite Plane Wall (Fig 6.5)
4 5 4 5
23
2
x t t t t q xR
k
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Thermal Conductivity ValuesInterpretation of the values in Table 6.2 requiresan understanding of the difference between
resistivity and resistance.
Resistivity = 1/kResistance = x/k
To get resistance from resistivity, you mustmultiply by the thickness of the material.
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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
2
ln( / ) ln( / )ln( / )1 1
W/m of length
r
t tq
R R R R R
t t
r r r r r r
r h k k k r h
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Steady-State Ht Conduction
Composite Cylinder (Fig 6.8)
1 5
1 2 3
1 5
3 22 1
1 1 1 2
2
2
ln( / )ln( / )1
r
t t
q R R R
t t
r rr r
r h k k
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Steady-State Ht Conduction
Composite Sphere (Fig 6.10)
1 2
1 2 3 4
1 2
3 22 1
2 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 t
r rr r
r h k r r k r r r h
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Steady-State Ht Conduction
ProblemConsider a
composite cylindrical tube with an outside diameter of 10 cm. The wallconsists 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 stillair at a temperature of 30 C. Material A is stainless steel, 0.2 cm thick, andmaterial 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 surfacecoefficient can be estimated as ho = 7.36 W/m
2 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 isestimated at 20 W m-2 K-1
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Steady-State Ht Conduction
ProblemAnswers:
a) the heat loss per unit length32 (110 30) 185.1W/mr oq r h
b) the temperature at the interface
int
3 2
2 ( 110)
185.1W/mln( / )
erface
r
b
t
q r r
k
tinterface = 158.2 C.
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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
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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
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Transient Ht TransferCase 1:
1
0.2p
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
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Transient Ht TransferHeisler Charts:
2
1Lines are
o
o
F
B
For non-infinite geometries, TR values are
multiplied together
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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 issuddenly dropped into boiling water at 100 C,determine the temperature at the center after 10min. Assume the following values:
h = 6000 W/m2 Kk = 0.5 W/m K = 1.33 X 10-7 m2/s.
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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 = TR1x TR2= 0.20295;tc= 81.73 C
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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 initialtemperature 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?
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Transient Ht Transfer
Problem 2
Check Bo:
for the cylinder:
Bo = hro/k = (300W/m2
K)(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.
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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.
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Convection
The basic problem is to find the appropriate h
to use inNewtons Law of Cooling. Once thatis done, finding qx is fairly trivial:
xq hA t
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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
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Convection
Free
3 2
2
Nu
Gr
Prp
hD
kD g t
c
k
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ConvectionForced
Nu (Re) (Pr)m nc
Eqs 7.49 7.53
ReVD
where
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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.
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Convection - ProblemUsing 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
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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 t UAt t
t t
a a pa a
b b pb b
q m c t
q m c t
Fig 7.1
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Heat Exchangers
Parallel Flow
ta1
ta2
tb1
tb2
fluid 'a'
fluid 'b'
temperature
distance
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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
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Heat Exchangers Effectiveness
Ratio
1
1 exp 1
11
11 exp 1
1 11 exp 1
p
c
NTU RE
R
NTUR
E
NTUR R
Figs 7.3 and 7.4, or
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Radiation Heat Transfer
Black Body Emissive Power
4
bW T
where Tis absolute temperature and dependsupon the unit system.
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Radiation Heat Transfer
Gray Body Emissive Power
4W T
where is the emissivity.
Note: Gray bodies have constant with wavelength.
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Radiation Heat Transfer
Gray Body Exchange
4 4( )
1
i i j
i ji i i
i j i j j
A T Tq
A
F A
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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 isintercepted 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.