Download Summary of Heat Transfer

Heat Transfer
Revision – Examples
Heat transfer: energy transport because of a temperature
difference.
Thermal energy is transferred from one region to another.
Heat transport is the same phenomena like mass transfer,
momentum transfer and electrical conduction. Similar rate
equations, where flux is proportional to a potential
difference.
Potential difference  temperature difference
Three modes of heat transfer:
- Thermal Conduction
- Thermal Convection
- Thermal Radiation
Example 1
Given is a solid brick wall, with the following data
Material:
brick, fired clay, density: 1760 kg/m3,
conductivity (k): 0.8 W/m K
Surface (A):
5 m x 3 m = 15 m2
Interior surface
temperature (t1): 20 °C
Exterior surface
temperature (t2): 0 °C
Thickness of the wall (L):
25 cm = 0.25 m
Calculate:
a) the heat flow through the wall
b) the thermal resistance
Phenomenon: Thermal conduction
The mechanism: energy is transported between parts of
continuum by the transfer of kinetic energy between
particles or groups of particles at the atomic level.
Purely thermal conduction: in solid opaque bodies
(opaque: not permeable for radiation) the thermal
conduction is the significant heat transfer mechanism
because the material doesn’t flow and there is no radiation.
In flowing fluids, thermal conduction dominates in the
region very close to the boundary layer, where
– the flow is laminar
– the flow parallel to the surface
– there is no eddy motion
Steady-state conduction
Steady-state conditions:
t = F(t) = constant
For steady-state heat conduction, in one dimension, the Fourier-law is
q  (kA)
q
k
A
dt/dx
dt
dx
– heat flow rate, W
– thermal conductivity or heat conduction coefficient, W/m·K
– cross sectional area normal to flow, m2
– temperature gradient, K/m
The equation states that the heat flow rate (q) in the x direction is directly
proportional to the temperature gradient dt/dx and the cross sectional area A
normal to the heat flow.
The proportionality factor is the thermal conductivity k.
The minus sign indicates that the heat flow is positive in the direction of
decreasing temperature.
One-layer flat wall
t2  t1
t1  t2
Q  k  A
kA
, W
x2  x1
x2  x1
Or
R – thermal resistance, K/W
x 2  x1 L
R
 ,
k
k
m2  K
W
If A and k are constant (e.g. at a wall) the equation
(t1  t 2 )
qkA
L
L
t1
t2
A
k
–
–
–
–
–
wall thickness
temperature at x = 0
temperature at x = L
surface of the wall
thermal conductivity
and
t
qA , W
R
where
L
R ,
k
m2  K
W
Example 1
Given is a solid brick wall, with the following data:
Material:
brick, fired clay, density: 1760 kg/m3,
conductivity: from table
k = 0.8 W/m K
Surface (A):
5 m x 3 m = 15 m2
Interior surface temperature (t1):
20 °C
Exterior surface temperature (t2):
0 °C
Thickness of the wall (L):
25 cm = 0.25 m
Calculate
a) the heat flow through the wall
b) the thermal resistance
(t1  t 2 )
qkA
L
L
R ,
k
m2  K
W
t
qA , W
R
Example 2
Given is an insulated concrete wall
Layers:
1 – sand and gravel concrete,
2400 kg/m3, k1 = 1.5 W/m K, L1 = 16 cm
2 – expanded polystyrene,
20 kg/m3, k2 = 0.036 W/m K, L2 = 6 cm
3 – sand and gravel concrete,
2400 kg/m3, k3 = 1.5 W/m K, L3 = 8 cm
Surface (A):
5 m x 3 m = 15 m2
Interior surface temperature (t1): 20 °C
Exterior surface temperature (t4): 0 °C
Calculate:
the heat flow through the wall and the
temperature on the surface of the layers
Multi-layer flat wall
t 2  t3
t3  t 4
t1  t 2
q  A·
 A·
 A·
, W
R1
R2
R3
If the number of layers is n:
t1  t n 1
q  A· n
 Ri
i 1
t1  t 4
q  A·
, W
R1  R2  R3
L1
R1  ,
k1
L2
R2 
k2
L3
R3 
k3
m2  K
W
Example 3
Given are a room and a radiator.
Surface temperature:
ts = 85 °C
Parameters of ambient air:
tair = 30 °C, 1 bar, dry air
Natural convection, length in direction of flow: 0,5 m
Phenomenon: Thermal convection
Energy transfer is involved by fluid movement and molecular
conduction.
Heat transfer means energy transfer
• from liquids and gases to the surface of a body or a wall, or
• from the surface of a body to the liquid.
If in the flow the Reynolds number is large enough, three different
flow regions exist:
At the wall:
laminar sublayer (boundary layer)  thermal conduction
Outside the laminar layer:
buffer layer  eddy mixing and conduction
Beyond the buffer layer:
turbulent region  eddy mixing
Convection is divided into two categories:
Free convection, natural convection:
the flow is generated by nonhomogeneous densities caused
temperature difference.
Forced convection:
the flow is produced by external sources (pump, fan).
The heat flow rate is described by the Newton’s formula:
q  hc  A (t s  t f )  hc  A  t
Where
q – heat flow rate, W
hc – convectional heat transfer
coefficient, W/m2K
A – surface of the wall, m2
ts – temperature of the surface, K or °C
tf – temperature of the fluid, K or °C
The heat transfer coefficient can be calculated with dimensionless
numbers, from similarity theory.
Generally
Nu = F (Pr; Gr; Re)
It means, Nusselt number is a function of the product of
Prandtl number,
Grashof number and
Reynolds number.
Dimensionless Numbers
Nu 
hc  L
k
hc – heat transfer coefficient, W/m2·K
L
– characteristic dimension, m
k
– thermal conductivity of the fluid, W/m·K
Dimensionless Numbers
In the equation of Nusselt number
Pr 
 cp
 – absolute viscosity, kg/m·s or Ns/m2
cp – specific heat, J/kg·K
k
k
– thermal conductivity of the fluid, W/m·K
L3   2    g  t L – characteristic dimension, m
Gr 
2
 – density, kg/m3
 – coefficient of thermal expansion, 1/K
g
Re 
vL

– gravitational acceleration, m/s2
t – temperature difference, K or °C
 – absolute viscosity, kg/m·s or Ns/m2
v
– velocity, m/s
L
– characteristic dimension, m  D, diameter

– kinematic viscosity = /, m2/s
Practical cases of convection
Natural or free convection: effect of temperature difference.
General relationship:
Nu 
hc  L
 C  (Gr  Pr) n
k
n
k  L      g  t 
hc  C  

L 
2
f
3
2
  c p

 k

n



f
Open spaces:
hc = F(t; direction of heat flow)
Direction of heat flow:
Ceiling  down
Wall
 horizontal
Floor
 up
t = ts – tair > 0
hc up >hc hor > hc down
Some values for hc heat transfer coefficient
Air, gas
4 … 10 W/m2 K
Water, liquid
100 … 500 W/m2 K
Forced convection: fluid stream derived from outer force
Nu = F(Pr, Re) = F(v)
Some values for hc heat transfer coefficient
Air, gas
Water, liquid
10 …
50 W/m2 K
500 … 5000 W/m2 K
Example 3
Given are a room and a radiator.
Surface temperature:
ts = 85 °C
Parameters of ambient air:
tair = 30 °C, 1 bar, dry air
Natural convection, length in direction of flow: 0,5 m
Solution:
h L
Nu  c
 C  (Gr  Pr) n 
k
n
k  L3   2    g  t 
hc  C  

L 
 av2
f
  cp

 k

n



f
Properties of dry air at tair = 30 °C
 30 = 1,1270 kg/m3
density,
coefficient of thermal expansion,  30 = 3,255·10-3 1/K
= 9,806 m/s2
gravitational acceleration,
g
absolute viscosity,
 30 = 1,8682 ·10-5 Ns/m2
specific heat,
cp30 = 1013 kJ/kg·K
thermal conductivity of the fluid, k30 = 0.0258 W/m·K
t = 85 – 30 = 55 °C
Temperature difference
Average temperature
tav = (85 + 30)/2 = 115/2 ~ 60 °C
absolute viscosity at 60 °C
 60 = 1,9907 ·10-5 Ns/m2
ASHRAE Fundamentals 2005, Chapter 3, Page 3.17 Table 10
A special problem of convection:
Convection in closed spaces: hollows, air layer between surfaces
e.g. window construction
t1 > t2
A1 = A2
Complex process of heat transfer:
conduction
convection
radiation
Equivalent heat conduction coefficient: ke
Equivalent conduction resistance of air
layer:
Requi 
L
kequi
From table, e.g. ASHRAE Fundamentals 2001,
Chapter 25, Table 3
Thermal Resistances of Plane Air spaces, m2 K/W
Phenomenon: Thermal radiation
The radiation energy transfer is through energy-carrying
electromagnetic waves that are emitted by atoms and
molecules due to change in their energy content.
It means: does not depend on an intermediate material.
The rate of thermal energy emitted by a surface depends on
its quantity and its absolute temperature.
A black surface absorbs all incident radiation.
The total energy emitted per unit time and unit area is given
by the Stefan-Boltzman law:
Eblack    T 4
where
 – Stefan-Boltzmann constant: 5.670·10-8 W/m2 K4
For nonblack surfaces
E    Eblack      T 4
where
 – hemispherical emittance or emissivity.
 is a function of the material, condition of its surface.
Lambert’s cosine law:
Lambert's cosine law is the statement that the total power
observed from a "Lambertian" surface is directly proportional to
the cosine of the angle Φ made by the observer's line of sight
and the line normal to the surface.
E  En  cos 
Utilising the Lambert’s law the total energy
radiated to the hemisphere is:
ETotal  En  
The heat flow rate between the surfaces
Given are two surfaces:
First let’s examine two general plane elements!
The energy flux which leaves dA1 element and which is absorbed by
dA2 element is quadraticly small:
d2E1 = E1n · cos 1 · dA1 ·dW1
From this, the radiation absorbed by dA2 is:
d2E1-2 = 2 ·d2E1
The heat exchange through radiation between the two plane elements is:
d2q1-2 = d2E1-2 – d2E2-1
Introducing the angle factor, which is affected by
geometrical parameters, the equation is:


q  A1  1   2    T  T F12
4
1
4
2
A – surface m2
– hemispherical emittance or emissivity
– Stefan-Boltzmann constant: 5.670·10-8 W/m2 K4
F12 – angle factor, shape factor,
Definition of Angle Factor
The fraction of all radiant energy leaving a surface i
that is directly incident on surface k is the angle
factor Fik (also known as view factor, shape
factor, and configuration factor).
Example 4:
Calculation of a U-value
Given is a threelayer wall
The wall construction is
the following:
Material
1 – Brick, fired clay
Density
Conductivity
Thickness
, kg/m3
k, W/m,K
L, m
1760
0.8
0.25
2 – Mineral fibreboard
260
0.049
0.06
3 – Wood wallboard, pine
660
0.16
0.02
Phenomenon: Overall Heat Transfer
Given is a multilayer wall. The number of layers is 3.
From inside air to outside air there is a complex process:
Inside
 convection
In the wall construction
 conduction
Outside
 also convection
The heat flow rate with overall heat transfer coefficient:
q  U  A (t i  t o )
Example 4
Given is a multilayer wall, where the wall construction data are the following:
Material
1 – Brick, fired clay
Density
Conductivity
Thickness
, kg/m3
k, W/m,K
L, m
1760
0.8
0.25
2 – Mineral fibreboard
260
0.049
0.06
3 – Wood wallboard, pine
660
0.16
0.02
Heat transfer coeff./surface resistance on the laminar air layer
ASHRAE F.25.2
EN 12831
Still air (inside)
Horizontal surface (upward)
9.26/0.11*
10.0/0.1
Horizontal surface (downward)
6.13/0.16*
5.9/0.17
Vertical surface
8.29/0.12*
7.7/0.13
* non reflective surfaces
Moving air (outside)
Any position
34/0.03*
25/0.04
* wind velocity 6,7 m/s (24 km/h)
Calculations
1. Overall heat transfer coefficient:
With overall thermal resistances:
U
U
1
1 n Li 1
 
hi j 1 ki ho
1
n
Ris   R j  Ros
j 1
2. The heat flow, if
Inside temperature: 20 °C
Outside temperature: 4 °C
Area of the surface: 15 m2
q  U  A (ti  to )
3. The temperatures of the layers
1 1
tis  ti  q 
A hi
1 L1
t1  tis  q
A k1
1 1
tos  q   to
A ho