Topic 9
Transient Heat Transfer
FOOD1360/1577
Principles of Food Engineering
Robert Driscoll
FST 2010
1/97
Introduction

We need a method for calculating processing
times, for example in cooking, pasteurising or
sterilising a product.

For example, retorting:
• A can is filled with the product (leaving a small
headspace)
• The can is placed into a retort (a long metal
cylinder which can hold pressure).
• The can is steam-heated.
FST 2010
2/97
Types of questions

How long until centre reaches a specified
temperature?

How much heat is added/removed?

Find final average product temperature.

Calculations of lethality.
FST 2010
3/97
Combined modes
Can example
 Assume solid food.
 External: convection.
 Internal: conduction
 Need to calculate
thermal effect.
 How do we combine
convection and
conduction?

FST 2010
4/97
Steady State

Steady state means state variables don’t
change with time.
• This means that in heat transfer, T stays constant
with time, but may change with location.
• Example: heat exchangers - milk pasteuriser
T
hold
cool
heat
FST 2010
x
5/97
Unsteady State

‘Unsteady state’ means variables change with
time (and position!)

Examples:
• Heating a can in a retort.
• cooling of fruits and vegetables.
FST 2010
6/97
General Conduction Equation

We have learnt about the 3 modes of heat transfer
(conduction, convection and radiation)

conservation of energy
dQ  
 Q  Qin  Q out
dt

energy accumulates as internal energy:
Q  mcT
FST 2010
7/97
Combining

We can combine these equations into one
powerful equation called the General
Conduction Equation (GCE).

The GCE can be used to solve common heat
processing problems.

Graphical solutions are the easiest – so that’s
what we will use!
FST 2010
8/97
Deriving the GCE

Heat flow along a bar:
q
xo
A
dx
FST 2010
9/97
Some definitions

Sink – a point where
heat is lost from a
system

Source – a point
where heat enters a
system
FST 2010
10/97
Setting up the solution

Now apply a heat balance to the element dx:
HEAT
ENERGY
IN
+
HEAT
GENERATED IN
ELEMENT

CHANGE
IN INTERNAL
ENERGY
FST 2010
+
HEAT
ENERGY
OUT
11/97
TERM 1: Heat Energy In

Fourier’s equation at x=xo
T

Qin  kA
x
FST 2010
x  xo
12/97
TERM 2: Heat Generated

Let qv be heat generated per unit volume:
Q gen  Q v Adx
Useful for nonconduction terms,
eg cooking, heat
losses.
volume
FST 2010
13/97
TERM 3: Change in Internal
Energy

Assume no phase changes!
T
T

Q  mc
 cAdx
t
t

Compare with definition of sensible heat.
FST 2010
14/97
TERM 4: Heat Energy Out

Fourier’s equation again …
T

Qout  kA
x
 T
 kA
 x
x  xo + dx
T
+ dx 2
x
2
x  xo


x  xo 

using a Taylor series expansion.
FST 2010
15/97
Adding terms

Adding and dividing by the element volume:
 T

T
k 2 + Q v  c
t
x
2
Each term is a power density (W/m3)
FST 2010
16/97
In 3 directions

This equation can be generalised to 3D as
follows:
T

k T + Qv  c
t
2

where the  symbol indicates differentiation
in three dimensions:
2
2
2

T

T

T
2
T 2+ 2+ 2
x
y
z
FST 2010
17/97
Meaning of equation

T
k T + Q v  c
t
2
Heat Flow
in 3D by
Conduction
Heat
Sources or
Sinks
FST 2010
Sensible
heat
store
18/97
Some examples

This equation is easy to solve for certain
situations, for example:
• Steady state (last term=0)
• One-dimensional (which term changes?)
• No sources or sinks (which term is 0?)
FST 2010
19/97
Example: Steady State, 1D

For steady state, 1D flow, T/t=0
k
 2T
x 2
+ Q v  0
FST 2010
20/97
Problem

If there are no heat sources or sinks, simplify
the general equation.

NOTE: The ratio k/c is often called the
thermal diffusivity, .
FST 2010
21/97
Unsteady state

But the equation is hard to solve for realistic
situations.

To make life a little easier, solutions have
been published as graphs that we can use.
FST 2010
22/97
Heisler Charts
A common situation is as follows:
An object at uniform initial temperature Ti is
dropped suddenly into a convective
environment at temperature T
 Some typical questions:

• How long until average temperature is T?
• How long until centre temperature is T?
• How much heat enters object?
FST 2010
23/97
Diagram and TR

T
T  T
TR 
T  Ti
Ti

Convective Environment
Define a temperature
ratio, TR:
FST 2010
where T is any
temperature of interest.
24/97
Thermal Resistance

Outside the object is a convective
environment, so R=1/hA

Inside the object is a conduction environment,
so R=D/kA, D=size

Ratio tells us which is important:
D
internal
hD
kA
Bi 


external 1
k
hA
FST 2010
Called BIOT
Number
25/97
Large Biot number

If Bi is large, then internal resistance is large,
and we can ignore the convective resistance.

Standard solutions exist for “regular” solids –
spheres, cylinders etc.

These solutions are based on diffusion
theory.
FST 2010
26/97
Small Biot Number

If Bi is small, then external resistance is high,
and we can ignore heat diffusion inside the
solid.

Then we can use a method of predicting
temperatures called the LUMPED HEAT
CAPACITY method: assume all of solid is at
constant temperature T
FST 2010
27/97
Lumped Heat Capacity

If all solid is at T, then
T

Q  hA(T  T )  mc p
t

Solving,
T
hA
dT
0 mcp dt  Ti T  T
t

T  T

e
Ti  T

FST 2010
hA
t
mc p
28/97
Diagram for LHC
T
tc 
mcp
hA
Ti
tc
FST 2010
time
29/97
Example:

A can of chunky soup at 1000C is left to cool
on a bench in air at 20oC. Estimate how long
until the soup cools to 60oC.

Data:
k=1.35 W/m.K
A=0.02 m2
L=0.015 m
h=5 W/m2.K
m=0.250 kg
cp=3800 W/kg.K
Note: k is high, because convection in the can
increases the effective rate of heat transfer.
FST 2010
30/97
Check for type of problem
hL 5  0.015
Bi 

 0.056
k
1.35

First check Bi:

Since this is small, convection is slow
compared to conduction. So treat as lumped
heat capacity problem.
FST 2010
31/97
Solution
T  T
e
Ti  T

60  20

e
100  20
 t  1.8 hrs
FST 2010
hA
t
mc p

5 ( 0.02)
t
0.2503800
32/97
Bi  1
So we can solve high and low Biot number
problems.
 For Bi1, we use the charts.
 Charts exist for 4 regular shapes:

•
•
•
•

Semi-infinite solids
Thin slabs
Cylinders
Spheres
Use these to approximate food shapes.
FST 2010
33/97
1) Semi-infinite solids
Definition: changes in T
close to surface
 Or mathematically, x<<W,
L, H where

x
• x is depth below surface
• W,L,H are dimensions of
object
•  is thermal diffusivity
W
FST 2010
34/97
Solution
1-TR
h
t
k
x 2 t
FST 2010
35/97
Example

A beef carcase at 38oC enters a cold room at
5oC. Meat density is 1042 kg/m3 thermal
conductivity 0.44 W/mK, h is 20 W/m2.K,
specific heat 3558 J/kgK. How cold is the
meat 2 cm below the surface after 20 min?
Assume “thick body response”
FST 2010
36/97
Solving

Thermal diffusivity is
k
0.44


 1.187  10 7
c p 1042  3558

Horizontal axis:
x
2 t

0.02
2 1.187  10 7  20  60
FST 2010
 0.838
37/97
ctd

function value
h
 20 
7
t  
 1.187  10  20  60  0.542
k
 0.44 

reading solution on vertical axis
1  TR  0.05  T  36 C
o
(see definition of TR)
FST 2010
38/97
2) Sphere

r
Geometry of sphere
r0
To
FST 2010
39/97
Graphs
sphere
TRo
k/hr
t/r2
FST 2010
40/97
This is in the form …

Dimensionless centre temperature, TRo

Dimensionless time (Fourier number Fo)

Reciprocal Biot number Bi.

So the graph allows prediction of centre
temperature at varying times.
FST 2010
41/97
Example
Estimate the time required for To at the centre of
a 6 cm dia apple dipped in 2oC water to reach
3oC. The apple Ti is 15oC.
DATA:
h=50 W/m2K
k=0.355 W/mK
cp=3.6 kJ/kg.K
=820 kg/m3
FST 2010
42/97
Method

First find Bi and TR:
hr 50  0.03
Bi 

 4.23
k
0.355
T  T
23
TR 

 0.077
T  Ti 2  15

Now read Fo from graph
FST 2010
43/97
Solving


Fo=0.5
So
k
0.355


 1.2  10 7
c p 820  3600
t  r Fo /   1.1 hrs
2

Is this a realistic value?
FST 2010
44/97
Can we calculate other
temperatures?

Define  as the temperature relative to T:
=T-T

o=To-T
i=Ti-T
Following plot is
X axis: 1/Bi
versus
Y Axis: /o
showing lines of constant r/ro:
FST 2010
45/97
Simplified
sphere
1
/o
0
r/ro
1/Bi
FST 2010
46/97
Example:
To find temperature at radius r:

Calculate Bi and r/ro

Use graph to find /o

Solve for: /i=/o x o/i
(where o/i is the temperature ratio TR)

Solve for  = T-T= i x /o x TR
(and so solve for T)
FST 2010
47/97
Problem

Peas (diameter = 6 mm) are blanched in hot
water at 95oC. If the peas enter at 15oC,
calculate the time until the centre temperature
is 85oC.

Data:
h=1200 W/m2K, k=0.35 W/mK,
cp=3.3 kJ/kgK, =980 kg/m3.
FST 2010
48/97
Draw Diagram, Calcs.
1.
Pea
Ti=15oC
T0=85oC
hL 1200  31000
Bi 

 10.3
k
0.35
T0  T 10
  0.125
2. TR 
Ti  T 80
3.
From graph, Fo=0.37
T∞=95oC
FST 2010
49/97
Solving
4.
k
0.35
7
2




1
.
08

10
m
/s
Diffusivity:
c 980  3300
0.37  

 31 s
5. Since Fo  2  t 
r
1.08  10
r 1.5

 0.5
6. Find T at r=1.5 mm:
ro
3
t
7.
So from graph
8.
Solving, T=88o.
2
3
1000
7

 0 .7
o
FST 2010
50/97
Heat loss by sphere

Calculating amount of heat lost/gained.

Final graph on ‘sphere’ handout.

This is a plot of
X Axis: Fourier number
Y Axis: Q/Qi
Lines: Reciprocal Biot number
FST 2010
51/97
Simplified graph
1
Q/Qi
1/Bi
0
Fo
FST 2010
52/97
Meaning of graph?

Qo is the amount of heat to add/remove to
achieve equilibrium with the fluid:
Q0 = mc(T-Ti)

Q is the amount of heat actually
added/removed at time t:
Q = mc(Tavg-Ti)
FST 2010
53/97
So for spheres:

First graph relates Fo, o/i and Bi
(time ... centre T ... Bi)

Second graph relates Bi, /o and r/ro
(Bi ... T ... radius)

Third graph relates Fo, Bi and Q/Qi
(time ... Bi ... heat)
Given 2 of 3, use graph to find 3rd.
FST 2010
54/97
Other shapes

Thin plate (also called infinite plate)
Q
x
2L
Q
L << W1,W2
FST 2010
55/97

Infinite cylinder
Q
r << H
r
H
FST 2010
56/97
Some uses

Thin plate:
• heating a steak
• cooling a product which is thin in one dimension

Cylinder:
• cooking a sausage
• cooling an extruded product
FST 2010
57/97
Tutorial Problem
A 5 kg fish at 70% mc is cooled by brine at
-5oC from its catch temperature of 16oC. Find
the temperature history of the fish at 1, 2 and
3 cm below the surface.
 Assume the fish is a plate of 6 cm thickness
(L=3 cm):

•
•
•
•
density 1000 kg/m3
thermal conductivity 0.5 W/m.K
specific heat 3800 W/kg.K
h=50W/m2.K
FST 2010
58/97
Combining shapes

Three of the four basic shapes can be
combined:
• ‘combined’ means intersection volume of two
shapes
• sphere cannot be combined!
• this allows true 2D and 3D shapes
FST 2010
59/97
Example
infinite
cylinder
thin plate
FST 2010
finite
cylinder
60/97
Uses

We can choose the cylinder radius
and the thickness of the thin plate to
match the dimensions of a can.

We use the
• cylinder solution - for heat transfer
through can walls
• thin plate solution - for heat transfer
through can ends.
An example might make this clearer.
FST 2010
61/97
Example

Problem: A can is suddenly dropped into a
steam environment. Find its temperature at a
specific point.

Solution:
• Find heat flow through can walls, (/i) infinite cylinder
• Find heat flow through ends, (/i)thin plate
• Find combined solution: (/i)=(/i)cyl x (/i)plate
FST 2010
62/97
Second example
Two plates: intersection
gives ‘chip’ shape
FST 2010
63/97
Practise

Do tutorial example: cylinder
(eg Holman p 163)

If I combine the three basic shapes, what
other shapes are possible?

How could I model a chip being heated?
FST 2010
64/97
Problem
Can Heating
A can of OD 7.5 cm and height 11.2 cm is
filled with product at 25oC, sealed, and placed
in a steam environment (retort) at 121 oC,
construct a curve at 5 minute time intervals of
the temperature at the centre of the can.
 Data:

•
•
•
•
htc from the steam to the can is 2000 W/m2K.
density 900 kg/m3.
thermal conductivity 0.34 W/m.K and
specific heat 3.5 kJ/kg.K,
FST 2010
65/97
Solution Outline

Define temperature differences o = T - To
and i = T - Ti .

Divide the problem into a cylinder and a thin
plate.

Fill in the tables (guided solution) on the
following pages.
FST 2010
66/97
Looking at can as cylinder

Infinite cylinder of radius 3.75 cm:
heat flow
in
centre
T=To
7.5 cm
FST 2010
can67/97
Cylinder






Cylinder: radius = 3.75 cm.
Biot number = hr/k = 2000 x 0.0375 / 0.34
= …………………………..
Thermal diffusivity = k/c = 0.34 /(900 x 3500)
= ………………………..…
Fourier number = t/r2 = ………… x 300/(.0375)2
= …………………………..
Temperature difference i = T - Ti = 121 – 25
= …………………………..
Now use Fig 4.9 (cylinder) to calculate the
temperature ratio o/i.
FST 2010
68/97
Looking at can as thin plate

Width and length of plate very large
H=11.2 cm
centre
T=To
FST 2010
can69/97
Thin Plate






Thin Plate: thickness 2L = 11.2 cm.
Biot number = hL/k = 2000 x 0.056 / 0.34
= …………………………..
Thermal diffusivity = k/c = 0.34 /(900 x 3500)
= ………………………..…
Fourier number = t/L2 = ………… x 300/(.056)2
= …………………………..
Temperature difference i = T - Ti = 121 – 25
= …………………………..
Now use Fig 4.8 (plate) to calculate the temperature
ratio o/i.
FST 2010
70/97
Combine Solutions

Combine solutions:

Find the product (o/i) cyl.x (o/i) plate.
= …………………………..

Since o is T - To and i is T - Ti , solve for
the centre temperature To.

Repeat at five minute intervals (t = 300 s),
plotting a graph of centre temperature versus
time.
FST 2010
71/97
Solution: To for can

Your values may differ slightly from these.
min
Fo
o/i
Fo
0
5
15
30
60
90
120
0.07
0.14
0.28
0.42
0.55
1
0.95
0.7
0.27
0.15
0.07
0 0.01
0.03
0.06
0.12
0.19
0.25
0.0 0.02
1
o/i
1
1
1
0.9
0.9
0.8
0.7
T
25
25
30
60
98
109
116
FST 2010
72/97
As a graph:
T, C
Centre Temperature
140
120
100
80
60
40
20
0
0
50
100
150
Time, min
FST 2010
73/97
Relevance

Thermal processing is the heart of the food
industry:
• heating to pasteurise/sterilise
• cooling to slow reaction rates
• modifying flavours
• modifying water activity (drying)

Reasonable approximate solutions can be
found graphically using the Heisler charts.
FST 2010
74/97
Types of questions

How long until centre reaches a specified
temperature?

How much heat is added/removed?

Find final average product temperature.

Calculations of lethality.
FST 2010
75/97
What you have learnt

It is important to develop a feel for
cooking/processing times.

The charts can help do this!

Most thermal processing is transient.

Now, how can we efficiently heat on an
industrial scale?
Ans: STEAM !
FST 2010
76/97
TUTORIAL PROBLEMS
Covering all of heat transfer
FST 2010
77/97
Problem 1

A large plate of Al at 200oC is suddenly
immersed in water at 70oC. The plate is
5 cm thick.

Find temperature T at 1.25 cm below
surface after 1 min.

Assume
• h=525 W/m2.K, k=215 W/m.K
• =8.4x10-5 m2/s
FST 2010
78/97
Solution 1

DATA:
• Ti=200oC
T=70oC
• i=200-70=130
• 2L=5 cm so L=2.5 cm
T=?
• T=1 min = 60 s
• x=2.5 – 1.25 = 1.25 cm

First find Bi and Fo.
FST 2010
79/97
Solution 1 (ctd)


Fo, Bi:
t 8.4  105  60

 8.064
2
2
L
0.025
k
215

 16.38
hL 525  0.025
x 1.25

 0.5
L 2.5
From centre T graph (thin slab):
o
 0.61
i
 o  To  T  0.61 130  79.3 oC
FST 2010
80/97
Solution 1 (ctd)

From T ratio graph
for a thin slab,

 0.98
o
   T  T  0.98  79.3  77.7 oC
 T  77.7 + 70  147.7 oC

Energy loss: =2700
kg/m3, c=0.9 kJ/kg.K
h 2 5252  8.4  105  60

 0.03
2
2
k
215
hL 525  0.025

 0.061
k
215
Q
  0.41
Q0
FST 2010
81/97
Solution 1 (ctd)

Qo cVi

 c(2 L)i
A
A
 2700  900  0.05  130
For unit area:
 15.8  106 J/m2

So heat removed is:
Q
 15.8  106  0.41  6.48  106 J/m2
A
FST 2010
82/97
Problem 2

A large cylinder of Al at 200oC is suddenly
immersed in water at 70oC. The cylinder is 5
cm in diameter.

Find temperature T at 1.25 cm below surface
after 1 min.

Assume
• h=525 W/m2.K, k=215 W/m.K
• =8.4x10-5 m2/s
FST 2010
83/97
Solution 2

DATA:
• Ti=200oC
T=70oC
• i=200-70=130
• 2ro=5 cm so ro=2.5 cm
T=?
• T=1 min = 60 s
• r=2.5 – 1.25 = 1.25 cm

First find Bi and Fo.
FST 2010
84/97
Solution 2 (ctd)

Fo, Bi:
t 8.4  105  60

 8.064
2
2
L
0.025
k
215

 16.38
hro 525  0.025
r 1.25

 0.5
ro 2.5

From centre T graph (cylinder):
o
 0.38
i
 o  To  T  0.38  130  49.4 oC
FST 2010
85/97
Solution 2 (ctd)

From T ratio graph
for a cylinder,

 0.98
o
   T  T  0.98  49.4  48.4 oC
 T  48.4 + 70  118.4 oC

Energy loss: =2700
kg/m3, c=0.9 kJ/kg.K
h 2 5252  8.4  105  60

 0.03
2
2
k
215
hro 525  0.025

 0.061
k
215
Q
  0.65
Q0
FST 2010
86/97
Solution 2 (ctd)

Qo cVi

 c(ro2 )i
A
A
 2700  900  0.0252  130
For unit area:
 6.203  105 J/m2

So heat removed is:
Q
 6.203 105  0.65  4.0  105 J/m2
A
FST 2010
87/97
End of Notes
FST 2010
88/97
Solutions to Heat Transfer
Assignment 2009
Q1a. Find the rate of heat flow per square
metre through an aluminium plate of
thickness 1.8 cm if the temperature
difference across the plate is 21.4oC.
A1b. Simple test of Fourier’s equation:
q  kA
T q
21.4

 204 
 243 kW/m2
A
x
0.018
FST 2010
89/97
Q1.
Q1b. Find the rate of heat flow per square
metre if 2 cm of insulation (k = 7.3 W/m.K) is
added to the aluminium plate.
A1b. Fourier’s equation: series resistance
q
T
21.4


 7.56 kW/m2
A
x1 x2 0.018 0.02
+
+
204
7.3
k1
k2
FST 2010
90/97
Q2.
Q2a. Steam is transported by a steel pipe (OD=5.1 cm)
from a boiler to a double-jacketted kettle at 1 kg/min.
The length of pipe is 26.4 m, and the steam
temperature is 121.1oC. Assume h is 20 W/m2.K:

Estimate the heat losses per unit length (Ta=25oC)
A2a. Convection equation: cylinder
q / L  hAT / L  20    0.051  121.6  25  308 W/m
FST 2010
91/97
Q2b Find the steam dryness factor at the kettle.
A2b. Dryness is defined as proportion of steam to total
(total includes liquid condensate):
htotal  d .hsteam + 1  d .hwater  hwater + d
Equate energy loss in steam to pipe wall loss:
qloss  qloss per unit length  L pipe  m steam h  m steamd
 308  26.4 W  1 / 60  2707.44  507.97  d
 d  0.22
This gives a steam dryness of 0.78.
FST 2010
92/97
Q2c How much insulation (at k=0.14 W/m.K) would be
required to obtain steam at a dryness of 0.90?
A2c. This tests resistances in series, insulation and
dryness equation.
(i) Addition of insulation of thickness x:
lnrnew rold 
1
Rtotal 
+
h2Lrnew 
2Lkins
Energy loss  m steam d .  T Rtotal
 Rtotal  T m steam d .  121.1  25 /
601  0.1 2199.47  0.02622
Second term in Rt is most important : solve by iteration.
FST 2010
93/97

Rewrite Rt equation to solve for rnew:


1
0.00700
  0.6089 
lnrnew rold   2Lkins  Rtotal 
h2Lrnew  
rnew


Try rnew=0.04 then iterate:
rnew  0.04  0.0394  0.03924  0.03924
 thickness  0.03924  0.0255  1.4cm
FST 2010
94/97
Q3.
Q3a. Heat flow per unit length through Cu pipe, thickness 1.2 cm,
internal diameter 0.8 cm, surrounded by 4.3 cm of fibreglass, if
the fluid is at 350oC, ambient is 20oC, h is 17 W/m2.K
A3a. Geometry shown opposite.
 1 
 lnr3 r2  
 lnr2 r1  

Rtotal  
 + 
 +
 2k  Cu  2k  Ins  2hr3  Conv
 5.7  10 4 + 5.92 + 0.16 K/W
 q 
0.8
1.2
350  20
T

 54 W /m
Rtotal
6.08
FST 2010
4.3
95/97
Q3b. What is the critical thickness of insulation?
rcrit
k 0.0351
3
 
 2.06 10 m
h
17
FST 2010
96/97
Q4.
Q4. Find the centre temperature of a 2cm thick beef
steak at 20oC suddenly exposed to an environment at
500oC, h=500 W/m2.K, five minutes after exposure.
A4. Typical meat values:
k=0.4-0.5 W/m.K, =1050 kg/m3, c=3500 kJ/kg
Solving: =1.3x10-7
Bi=0.095
Fo=0.38
Chart (use thin slab): 0/i = 0.63
Solving, To=160-200oC
FST 2010
97/97
Q5. German sausage

A cylindrical sausage of meat is hung in an air-drying
room at 45oC.
• Plot centre T at 1 hourly intervals.
• Plot T at 1 cm from centre.
• Plot average T.

DATA:
• sausage: r=5 cm, h=45 cm
• density: 980 kg/m3
• therm cond: k=0.47 W/m.K
Ti=25oC
htc: h=10 W/m2.K
spec heat: c=3.55 W/kg.K
Solution (1 hour)

1.351x10-7 m2/s
Thermal diffusivity =k/c = …………

Biot number

0.195
Fourier number Fo=t/ro2= ………….

Use Fig 4.8 (check!):
1.064
Bi=hr0/k= ………….
• Centre T at 1 hour:
• So To = 27oC
o/i= ………….
0.90
FST 2010
99/97

Use Fig 4.11 at 1 cm from centre:
• r/r0=0.20
0.98
• So /o= …………….
27.4oC
• So T1 = …………….

Use Fig 4.15 to find Tavg:
0.22
• FoBi2 = …………….
0.40
• So Q/Qo= …………….
33oC
• So T1 = …………….
FST 2010
Qo=mc(T-Ti)
Q mcTavg  Ti 

Q0 mcT  Ti 
 Tavg  Ti +
Q
T  Ti 
Q0
100/97

Now can you find out the three T’s at 2
hours? (easier!).
FST 2010
101/97