Chapter 13
Heat Transfer – Unsteady State
We have shown how heat travels through material. But the movement of heat does not necessarily imply a
movement of temperature changes. In fact, in a steady state, heat flows with no temperature change at all. In
this section, we will examine the unsteady state and see what factors affect the migration of temperature
changes through an object.
1. Equilibrium. Imagine a pot filled with a liquid that has been sitting in a cool water bath for an extended
period. An equilibrium state has been reached in which the water bath, the wall of the pot and the liquid in
the pot are all at the same temperature. Figure 13.1A shows the flat temperature profile that will exist
through the wall of the pot. Clearly, no heat will be flowing through the wall.
2. Unsteady State Heating. Suppose the pot is suddenly plunged into a hot water bath. Heat will begin to
penetrate the pot wall and in a few seconds, the temperature profile will resemble the one shown in Figure
13.1B. The outer surface has heated to the temperature of the hot water while most of the wall is still near
the original temperature. This results in a steep gradient followed by a flatter gradient.
3. Energy Balance. According to Fourier’s first law, heat should move rapidly across the steep gradient, then
slow down when the flatter gradient is reached. An energy balance at the point of gradient change will be
 Rate of   Rate of   Rate of Energy 
 Energy In    Energy Out    Accumulation 

 
 

In other words, energy will accumulate at the point of gradient change with a resulting temperature rise. This
will tend to straighten the temperature profile so that it resembles Figure 13.1C. The rate of temperature rise
with time (dT/dt) at any point in the wall will depend on three factors:
 Thermal conductivity: The greater the conductivity (k), the faster heat will be brought to any given spot
and the faster temperature will rise.
dT
k
dt
 Heat Capacity: The greater the heat capacity (specific heat) of the wall (Cp), the more heat is required to
raise the temperature at any point and the slower the temperature will rise.
dT
1

dt C p
PAGE 325
 Rate of Gradient Change: Since the rate of heat flow is proportional to the slope (gradient) of the
temperature profile, a large change in slope at any point will result in a large difference between the heat
into that point and the heat out of that point. Thus, the faster the slope is changing, the faster the
temperature will rise. We will show below that this change in slope is described by the second derivative
of the temperature profile.
dT
dT d 2T
 Rate of change in

dt
dx dx 2
4. Temperature Rates. In the following derivation, be careful to distinguish between the rate of temperature
change per unit distance (dT/dx, dT/dy and dT/dz) and the rate of change in temperature per unit time
(dT/dt).
5. Steady State Heating. Eventually a linear temperature profile will be established as shown in Figure 13.1D.
Now, as long as the outside and inside temperatures remain the same, all the heat that enters the pot will
flow through and none will accumulate in the wall. Although the temperature will vary at different points in
the wall, at any point in the wall, the temperature will remain constant with time and we say that the wall
will be in a steady state. Thus,
dT
0
dt
GRAPHIC
Figure 13.1
Unsteady State Conduction
6. Thermal Diffusivity. A comparison of successive profiles in Figure 13.1 shows that during the period of
unsteady state, a temperature increase is migrating through the wall of the container. This temperature
migration should not be confused with the heat flow through the wall although the heat flow is necessary for
the temperature migration to take place. This migration of temperature is analogous to the diffusion of
molecules in a solvent. For example, if a cube of sugar is dropped into a cup of coffee without stirring, sugar
molecules will diffuse into the coffee. Initially, a steep concentration gradient will form near the sugar cube
similar to the one in Figure 13.1B. With passing time this sugar concentration will migrate across the cup.
Because of the similarity to molecular diffusion, we call the migration of temperature change thermal
diffusion and in the following sections we will show that the thermal diffusivity depends on thermal
conductivity, heat capacity and the second derivative of the temperature profile.
Page 326
7. A Calculus Review, Functions. The next three sections are provided as a refresher on second derivatives.
A function is any rule that says, give me a number and I will produce a matching number. In general, we
write:
y  f  x
(13.1)
to mean “y is a function of x” or “give me an x and I have a rule called f that will give you a matching y.”
(f(x) does not mean f times x.)
Example:
1. Figure 13.2 is a graph of the function.
y  f  x   2 x  0.008x3
You can compute the following values for the function or read them from the graph:
f(-20) = 24
f(2)
= 3.94
f(-10) = -12
f(10)
= 12
f(0)
etc.
=0
GRAPHIC
Figure 13.2
A Function and its Derivative
Page 327
8. More Calculus Review, The First Derivative. The first derivative of a function is a new function that tells
the slope or rate of change of the original function. It is found by computing the difference between f(x) and
f(x +  x), dividing this difference by  x and taking the limit of this ratio as  x goes to 0. In other words:
f  x 
f  x  x   f  x 
dy
 Lim
dx x0
x
(13.2)
Example:
2. The derivative of the previous function is:
dy
 2  0.024 x 2
dx
In Figure 13.2, this derivative is used to determine the slope of the function at two points:
f  x 
f’(-10) = 2 – 0.024 (-10)2
= -0.40
(A small downward slope)
f’(0)
=2
(A steep upward slope)
=-3.4
(A steep downward slope)
= 2 – 0.024 (0)2
f’(15) = 2 – 0.024 (15)2
9. More Calculus Review, The Second Derivative. A derivative of a first derivative is called a second
derivative. The second derivative tells the slope or rate of change of the first derivative and hence the rate of
change of the slope of the original function. It is computed from the first derivative the same way the first
derivative is computed from the original function.
 dy 
 dy 
 
 
2
d y
 dx  x x  dx  x
f   x   2  Lim
(13.3)

x

0
dx
x
Example:
3. The second derivative of the above example is the derivative of f’ = 2 – 0.024x2 which is
f  
d2y
 0.048 x
dx 2
Study Figure 13.2 and confirm that the function behaves as these representative values of this second
derivative suggest.
Page 328
f”(-10) = –0.048(-10) = 0.48
(The slope is changing from negative to positive at a rate of
0.48 units per unit change in x.)
f”(0)
= –0.048(0) = 0
(For a brief instant the slope is not changing.)
f”(15)
= –0.048(15) = -0.72
(The slope is becoming increasingly negative at a rapid rate.)
10. Another Look at Unsteady State. In part 2 above (Unsteady State Heating), it was pointed out that, when a
temperature profile is non-linear, the temperature tends to rise most rapidly at the point where the
temperature gradient is changing most rapidly. In that case,
 The original function, f(x), describes the temperature profile.
 The first derivative, f’(x), describes the slope or gradient of the profile.
 The second derivative, f”(x), describes the rate of change of the gradient of the profile. Thus, we expect
that temperature will change most rapidly where the second derivative is greatest. We will now derive
the equation that verifies this expectation.
11. The Differential Element. Figure 13.3A shows a cross section of a slab in which we have selected a small
rectangular volume element at some arbitrary location. Figure 13.3B shows a three dimensional enlargement
of this tiny element whose edges have lengths  x,  y and  z. The faces of the element have areas equal
to the products of the edges, i.e.
Ax =  y  z,
Ay =  x  z and
Az =  x  y.
Its volume will be the product of the three dimensions, i.e.
V =  x  y  z.
Page 329
•
12. Three Dimensional Heat Balance. Heat flowing in the x direction ( q x) enters this element through one face
and exits through the opposite face as shown in the figure. Assume that an unsteady state exists and that
there is a nonlinear temperature profile through the slab as shown. This means heat will enter the element
with one gradient and exit with a less steep gradient. By Fourier’s Law then, heat will enter faster than it
leaves, giving the following energy balance:
Energy in – Energy Out = Energy Accumulated
Or
•
 dT 
 dT 
kAx 
  kAx 
  q x  Rate of Heat Accumulation
 dx in
 dx out
So that
•
 dT   dT  
q x  kAx 
 
 
 dx in  dx out 
(13.4)
where the expression in square brackets represents the difference in slope between the two sides of the tiny
element. To make this derivation perfectly general, assume that the same thing is happening in the y and z
directions, so
•
 dT 
 dT  
q y  kAy 
  kA 
 
 dy out 
 dy in
(13.5)
•
 dT 
 dT  
q z  kAz 
  kA 
 
 dz out 
 dz in
(13.6)
The total rate of heat accumulation in the element is the sum of these three heat flows so we can write
•
•
•
•
Total Accumulation = q = q x + q y + q z
(13.7)
Substituting equations (13.4), (13.5), and (13.6) in (13.7) gives the complete three dimensional energy
balance
•

 T   T  
 T   T   
  T   T  

q  k  Ax 


A


A
(13.8)


y 
z 
 

 
 
 
 
 z in  z out  
  x in  x out 

 y in  y out 

Page 330
In this equation the dependent variable (T) is a function of 4 independent variables (x, y, z, and t). We must,
therefore, differentiate T separately for each independent variable and this results in partial derivatives. In
partial derivatives, the d’s are replaced by curly d’s in equation (13.8).
13. Fourier’s Second Law. The rate at which the temperature changes in the differential element depends on
three factors:
 It is inversely proportional to the mass of the element, which equals its volume times its density (
xyz ).
 It is inversely proportional to the heat capacity (specific heat) of the element ( C p ) and
•
 It is proportional to the rate of heat accumulation within the element ( q ).
Thus we can write


T
q
q


t C p m C p  xyz
Or
C p  xyz
T 
q
t
(13.9)
Substituting equation 13.8 in 13.9 gives
 C p xyz

 T   T  
 T   T   
T
  T   T  

 k  Ax 


A


A


y 
z 
 

 
 
 
 
t
 z in  z out  
  x in  x out 

 y in  y out 

If we divide all terms by  x  y  z, noting that Ax =  y  z, etc., we get
Page 331
   T    T     Ty    Ty     T    T   
T
 x out
x in
z in 
out
in
   z out
C p
 k  


t
x
y
z

 





 


Notice that each of the three terms in square brackets is a change in the first derivative divided by  x,  y and
 z respectively. According to equation (13.3), if you take the limit of these ratios as  x,  y and  z go to
zero, the results will be second derivatives of the temperature profile. Thus the equation becomes
T
k

t
C p
  2T  2T  2T 
 x 2  y 2  z 2 


(13.10)
This equation is Fourier’s second law which verifies what we saw earlier, that the rate of temperature change
per unit volume is
 Proportional to conductivity (k).
 Inversely proportional to heat capacity. (Since temperature change per unit mass is proportional to C p ,
change per unit volume is proportional to  C p .)
 Proportional to the rate of gradient change, i.e. the second derivative of the temperature profile.
14. Thermal Diffusivity Coefficient. Since, for any material, k,  and C p are all constants, the expression k/
 C p is also a constant that is frequently called  and is given the name “thermal diffusivity”. Substituting
 , equation (13.5) becomes
  2T  2T  2T 
T
   2  2  2 
t
y
z 
 x
(13.11)
Like thermal conductivity, it is characteristic of the material and reflects the fact that the rate at which a
temperature change migrates through the material depends both on its conductivity and its heat capacity.
Table 11.2 in Chapter 11 gives the thermal conductivity, specific heat (heat capacity), density and thermal
diffusivity for the principal constituents of food products. To summarize:
 A temperature increase will penetrate a food product at a rate that is proportional to  .
 The  for any material is proportional to its thermal conductivity (k).
 The  for any material is inversely proportional to its specific heat ( C p ).
 The  for any material is inversely proportional to its density (  ).
The units of  are:
 watts 


m2
mK 
   

 kg   joules  sec
 3 

 m   kg K 
or
 Btu 
 hr ft o R 
2
  ft
   
 lbm   Btu  hr
 ft 3   lb o R 

 m 
Page 332
13.1
The Biot Number (Bi)
 A dimensionless number, used to evaluate the relative importance of internal & external resistance to
heat transfer.
Bi 
hcx1
k solid
(13.12) GRAPHIC
hc
ksolid
x1 1
= convective heat transfer coefficient
= thermal conductivity of the solid
= characteristic dimension of the solid

Volume
Area
4 3
r
r
3 2 
4 r
3

x1 
 r2L r

2 rL 2
When objects are finite
(13.13a)
For spheres
(13.13b)
For cylinders
(13.13c)
1
thickness of infinite slab & x1  r for infinite cylinder
2
x1
 Bi 
hcx1 ksolid
Internal resistance to heat transfer
(13.14)


1
ksolid
External or boundary resistance to heat flow
h
 In general:
Bi <0.1
Bi = 0.1 to 40
 Internal resistance is negligible (i.e. k is very high) Boundary resistance controls
heat transfer.
 Both the internal and boundary resistances control heat transfer.
 Bi > 40
 Boundary resistance is negligible, internal resistance controls heat transfer (i.e. k
is very small). Surface temperature = heating (or cooling) medium temperature.
13.2 Negligible Internal Resistance & Lumped Parameter Analysis (Bi < 0.1)
t=0
T= To
Area=A
Vol.=V
Density= 
High ksolid
uniform temp

t>0
T=T
Area=A, Volume=V, Density =  ,
hc , T1 (e.g. Steam or hot water
Page 333
At any time t:

Heat given up by the heating medium:
Heat gained by the object:
Thus
VC p
dT
 hc A T1  T  or
dt
 hc A
dT

 T  T  VC p
T T0 1

T T
 In



hA
dT
 c dt
T1  T VC p
t t
t
t 0
hA
T1  T
 c t
T1  T0 VC p
 hc A 

t
 VC 
T1  T
 e p 
T1  T0
 hx
 exp    c 1
  ksolid

  t 
T1  T
 exp    Bi   2   ,
T1  T0
 x1  

T1  T
 exp   Bi  Fo  ,
T1  T0
(13.15)
q  hc A T1  T 

dT
dT
q  mC p
 VC p
dt
dt
  ksolid t  
 
 
   C p x12  
t
x12
 Fo  Fourier no.
T1  T

T1  T0
Dimensionless temp
ratio, 0  1
Page 334
CHECK YOUR UNDERSTANIND:
1. Flat aluminum plates (1m wide x 2m long and 0.5cm thick) are being used as cookie sheets in a bakery.
They have a thermal conductivity of 207 W/mK, a specific heat of 0.9 kJ/kg-K and a density of 2707
kg/m3. Estimate the time required for the temperature of the plates to drop to a handling temperature of
40oC after they are removed from an oven at 230  C and placed in ambient air at 21  C. The convective
heat transfer coefficient is estimated at 6 W/m.2K.
2. Cream with 30% butterfat, is being heated with agitation in a 0.4 m3 jacketted tank. The overall heat
transfer coefficient has been computed to be 300 W/m2  C. The tank has 1.4 m2 of heating surface
and contains 220kg product. If the initial product temperature was 21  C and the heating medium
was 93  C, compute the product temperature after 30 min. The specific heat of the product is 3.35
kJ/kg  C and the density is 996 kg/m3. (From Heldman and Singh)
13.3 Finite Internal & Surface Resistance:
0.1  Bi  40
Solution for an infinite slab (Carlslaw & Jaeger, p.122)

2 Bi Cos  λ n x / d  Sec  λ n 
T1  T

exp  n 2 Fo 
2
T1  T0 n 1
Bi  Bi  1  λ n
(13.16)
First term approximation:
T1  T 2 Bi Cos  λx / d  Sec  λ 

exp   2 Fo 
2
T1  T0
Bi  Bi  1  λ
(13.17)
13.4 Negligible Surface Resistance: Bi > 40
Infinite slab:
 n 2  t 
T1  T 4  1
n x
  Sin
exp  
 2
T1  T0  n 1 n
L
 z  x1 
(13.18)
Expressions of the type shown above have been obtained for different geometries (e.g. cylinder, plate, and
sphere). Utilizing these series solution equations, charts of the type shown below have been obtained.
Page 335
Figure 13.4: Gurney-Lurie chart for unsteady-state heat conduction in a long cylinder.
GRAPHIC
Figure 13.5: Heisler chart for determining center temperature for unsteady-state heat conduction in a long
cylinder.
GRAPHIC
Page 336
Figure 13.6: Gurney-Lurie chart for unsteady-state heat conduction in a large flat plate.
GRAPHIC
Figure 13.7: Heisler chart for determining center temperature for unsteady-state heat conduction in a large flat
plate.
GRAPHIC
.
Page 337
Figure 13.8: Gurney-Lurie chart for unsteady-state heat conduction in a sphere
GRAPHIC
Figure 13.9: Heisler chart for determining center temperature for unsteady-state heat conduction in a sphere.
Page 338
CHECK YOUR UNDERSTANDING.
1
1. A can with the code 301 x 400 is 3 /16” in diameter and 4.00” high. Such a can is filled with tomato paste and
placed in a canning retort in such a position that it is heated by steam from all sides. What is the temperature in
the center of the can? The following are the process parameters:
– k (of paste) = 0.480 Btu/hr ft  F
–  (of paste) = 2.16  106 ft 2 / s  7.78  103 ft 2 hr
– hc (of steam) = 800 Btu / hr ft 2 F
– T0 = initial temperature of paste = 85  F
– T1 = temperature of steam = 240  F
– t = heating time = 45 min
2. An apple is being cooled from an initial temperature of 21  C to 4  C in a high velocity water stream at
2  C. Compute the time required for the center of the apple to reach 4  C when the following properties
are known; density = 800 kg/m3, specific heat = 3.56 kJ/kg  C, thermal conductivity = 0.35 W/m  C and
radius = 0.03m. The water stream produced a convective heat transfer coefficient of 3400 W/m2  C,
compute time required for location at 0.01 m from surface to reach 4  C.
Page 339
Finite Objects
The Gurney-Lurie and Heisler charts allow you to compute temperatures for infinite slabs (edge effects
negligible) and infinite cylinders (end effects negligible) as follows:
m
k
hcx1
Y (from chart) =
n
x
x1
X 
t
x12
T1  T
T1  T0
For a finite cylinder such as a can of radius R and length X,
–Determine Yr for an infinite cylinder of radius R
–Determine Yx for an infinite slab of thickness X
–Solve the following for T:
T1  T
T1  T0
For a rectangular solid of length X, width Y, and
thickness Z,
–Determine Yx for an infinite slab of thickness X
–Determine Yy for an infinite slab of thickness Y
Yxr  Yx Yr  
–Determine Yz for an infinite slab of thickness Z
–Solve the following for T:
Yxyz  Yx  Yy  Yz  
T1  T
T1  T0
Figure 13.10- Approximation of a finite cylinder
Graphic
CHECK YOUR UNDERSTANDING.
1. Two-Dimensional Conduction in a Short Cylinder. A cylindrical can of puree has a diameter of 68.1 mm and
a height of 101.6 mm and is initially at a uniform temperature of 29.4  C. The cans are stacked vertically in a
retort and steam at 115.6  C is admitted. For a heating time of 0.75 h at 115.6ºC, calculate the temperature at
the center of the can. The heat capacity of the metal wall of the can will be neglected. The heat transfer
coefficient of the steam is estimated as 4540 W/ m 2 K. Physical properties of puree are k = 0.830 W/mK and
  2.007  107 m2 / s .
Page 340
Thermal Processing: (Canning/Appertization)
“…a method of food preservation wherein a food and its container are rendered commercially sterile by
application of heat, alone or in combination with pH and /or water activity or other chemicals.”
•
Unsteady-state heating
•
Two methods used to calculate lethality:
– The General method
– Ball’s Formula method
Figure 13.11- Product “cold spot” and retort temperature profiles
GRAPHIC
Page 341
For a sphere, Fourier’s second law is written as:
T
  2T
t
  2T 2  T 
T
   2   
t
r  r 
 r
or
And the series solution is:

2 (sini  i cos i )
T1  T
x
x 
  t

exp  i 2 sin(i ) / ( i ) 
T1  T0 i 1 i  sin i cos i
r
r 
 r
First term approximation gives
 2(sin 1  1 cos 1 )
T1  T
 12t
x
x 

 log {
}sin( i ) /( i )
2
T1  T0 2.303r
r
r 
 1  sin 1 cos 1
And if,
log
 12
1

f 2.303r 2
and
 2(sin 1  1 cos 1 )
x
x 
j  {
}sin( i ) /( i )
r
r 
 1  sin 1 cos 1
T1  T
t
   log j
T1  T0
f
Then,
Or
log(T1  T )  
t
 log[ j (T1  T0 )]
f
A plot of log (T1-T) and time will give a straight line from where f and j parameters can be evaluated, as
illustrated in Figure 13.12.
Page 342
Figure 13.12
Temperature response plot
GRAPHIC
Where,
f = time required for the heating/ cooling line to pass through one log cycle or the time required for the
temperature difference between heating/cooling medium to decrease by 90%.
j = lag factor 
T1  Ta
T1  To
The f and j parameters are used to calculate the lethality (F value) of a thermal process like canning.