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Fig. 6 Nomograph for Estimating Value of M12 from Reciprocalof Biot Number and Smith’s (1966) Geometry Index

Fig. 6 Nomograph for Estimating Value of M12 from Reciprocalof Biot Number and Smith’s (1966) Geometry Index

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Cooling and Freezing Times of Foods

20.5

For two-dimensional, irregularly shaped foods, E0 (the equivalent heat transfer dimensionality for Bi = 0) is given by
1 – 1  2

1-  1 +  -----------------
E 0 = 1 + -----

1 

 2 1 + 2

(18)

For three-dimensional, irregularly shaped foods, E0 is
2

2 0.4

2

1 + 2 + 1  1 + 2  + 2  1 + 1    1 – 2  
- – ------------------------------------ (19)
E0 = 1.5 ---------------------------------------------------------------------------------1 2  1 + 1 + 2 
15
For finite cylinders, bricks, and infinite rectangular rods, E0 may
be determined as follows:
1- + -----1E 0 = 1 + -----1 2

(20)

For spheres, infinite cylinders, and infinite slabs, E0 = 3, 2, and 1,
respectively.
For both two-dimensional and three-dimensional food items, the
general form for E at Bi   is given as

Licensed for single user. © 2010 ASHRAE, Inc.

E = 0.75 + p1 f (1) + p2 f (2)

(21)

where
2
1

f () = ------- + 0.01p 3 exp  – ----2
6


(22)

with 1 and 2 as previously defined. The geometric parameters p1,
p2, and p3 are given in Table 4 for various geometries.
Lin et al. (1993, 1996a, 1996b) also developed an expression for
the lag factor jc applicable to the thermal center of a food as
1
1.35
+ ---Bi

j c = ------------------------1.35
Bi
1
-------------- + ---j


(23)

where j is as follows:
j = 1.271 + 0.305 exp(0.1721 – 0.11512)
+ 0.425exp(0.092 – 0.12822)

(24)

and the geometric parameters , 1, and 2 are given in Table 4.
For the mass average temperature, Lin et al. gave the lag factor jm
as follows:
jm =  jc

(25)

where
 1.5 + 0.69 Bi 
 =  ------------------------------- 
 1.5 + Bi 

N

(26)

and N is the number of dimensions of a food in which heat transfer
is significant (see Table 4).

Algorithms for Estimating Cooling Time
The following suggested algorithm for estimating cooling time
of foods and beverages is based on the equivalent heat transfer
dimensionality method by Lin et al. (1993, 1996a, 1996b).
1. Determine thermal properties of the food (see Chapter 19).
2. Determine surface heat transfer coefficient for cooling (see Chapter 19).
3. Determine characteristic dimension L and dimensional ratios 1
and 2 using Equations (16) and (17).
4. Calculate Biot number using Equation (1).

Table 4 Geometric Parameters
Shape
Infinite slab
(1 = 2 = )
Infinite rectangular rod
(1 1, 2 = )
Brick
(1 1, 2 1)
Infinite cylinder
(1 = 1, 2 = )
Infinite ellipse
(1 > 1, 2 = )
Squat cylinder
(1 = 2, 1 1)
Short cylinder
(1 = 1, 2 1)
Sphere
(1 = 2 = 1)
Ellipsoid
(1 1, 2 1)

N

p1

p2

p3

γ1

γ2

λ

1

0

0

0





1

2

0.75

0

–1

41/



1

3

0.75

0.75

–1

41/

1.52

1

2

1.01

0

0

1



1

2

1.01

0

1

1



1

3

1.01

0.75

–1

1.2251

3

1.01

0.75

–1

1

1.52

1

3

1.01

1.24

0

1

1

1

3

1.01

1.24

1

1

2

1

1.2252 1

Source: Lin et al. (1996b)

5. Calculate equivalent heat transfer dimensionality E for food
geometry using Equation (15). This calculation requires evaluation of E0 and E using Equations (18) to (22).
6. Calculate lag factor corresponding to thermal center and/or mass
average of food using Equations (23) to (26).
7. Calculate root of transcendental equation given in Equation (14).
8. Calculate cooling time using Equation (13).
The following alternative algorithm for estimating the cooling
time of foods and beverages is based on the use of f and j factors.
1. Determine thermal properties of food (see Chapter 19).
2. Determine surface heat transfer coefficient for cooling process
(see Chapter 19).
3. Determine characteristic dimension L of food.
4. Calculate Biot number using Equation (1).
5. Calculate f and j factors by one of the following methods:
(a) Method of Pflug et al. (1965): Figures 2 to 5.
(b) Method of Lacroix and Castaigne (1987a): Tables 1, 2, and 3.
(c) Method of Smith et al. (1968): Equations (7) to (10) and
Figure 6.
(d) Method of Hayakawa and Villalobos (1989): Equations (11)
and (12) in conjunction with Equations (7) to (10).
6. Calculate cooling time using Equation (4).

SAMPLE PROBLEMS FOR ESTIMATING
COOLING TIME
Example 1. A piece of ham, initially at 70°C, is to be cooled in a blast
freezer. The air temperature within the freezer is –1°C and the surface
heat transfer coefficient is estimated to be 48.0 W/(m2 ·K). The overall
dimension of the ham is 0.102 by 0.165 by 0.279 m. Estimate the time
required for the mass average temperature of the ham to reach 10°C.
Thermophysical properties for ham are given as follows:
c = 3740 J/(kg·K)
k = 0.379 W/(m·K)
 = 1080 kg/m3
Solution: Use the algorithm based on the method of Lin et al. (1993,
1996a, 1996b).
Step 1: Determine the ham’s thermal properties (c, k, ).
These were given in the problem statement.
Step 2: Determine the heat transfer coefficient h.
The heat transfer coefficient is given as h = 48.0 W/(m2 ·K).
Step 3: Determine the characteristic dimension L and dimensional ratios
1 and 2.

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20.6

2010 ASHRAE Handbook—Refrigeration (SI)

For cooling time problems, the characteristic dimension is the shortest distance from the thermal center of a food to its surface. Assuming
that the thermal center of the ham coincides with its geometric center,
the characteristic dimension becomes

 = 2.68
Step 8: Calculate cooling time.
The unaccomplished temperature difference is
Tm – T
–1 – 10
- = ------------------- = 0.1549
Y = ---------------T m – T i –1 – 70

L = 0.102 m/2 = 0.051 m
The dimensional ratios then become [Equations (16) and (17)]

Using Equation (13), the cooling time becomes

m- = 1.62
------------------1 = 0.165
0.102 m

2

 3   1080   3740   0.051   0.721 
 = --------------------------------------------------------------ln  ---------------- = 12 228 s = 3.41 h
2
 0.1549
 2.68   0.379   1.45 

0.279 m
2 = -------------------- = 2.74
0.102 m
Step 4: Calculate the Biot number.
Bi = hL/k = (48.0)(0.051)/0.379 = 6.46
Step 5: Calculate the heat transfer dimensionality.
Using Equation (19), E0 becomes
2

Solution.
2

1.62 + 2.74 + 1.62  1 + 2.74  + 2.74  1 + 1.62 
E 0 = 1.5 ------------------------------------------------------------------------------------------------------------------- 1.62   2.74   1 + 1.62 + 2.74 
2 0.4

Licensed for single user. © 2010 ASHRAE, Inc.

  1.62 – 2.74   = 2.06
– -------------------------------------------15
Assuming the ham to be ellipsoidal, the geometric factors can be
obtained from Table 4:
p1 = 1.01

p2 = 1.24

Example 2. Repeat the cooling time calculation of Example 1, but use
Hayakawa and Villalobos’ (1989) estimation algorithm based on the
use of f and j factors.

p3 = 1

From Equation (22),
2
1 +  0.01   1  exp  1.62 – -----------1.62  = 0.414
f (1) = -----------

2
6


1.62
2

2.74 
1
f (2) = ------------ +  0.01   1  exp  2.74 – ------------ = 0.178
2
6 

2.74

Step 1: Determine the thermal properties of the ham.
The thermal properties of ham are given in Example 1.
Step 2: Determine the heat transfer coefficient.
From Example 1, h = 48.0 W/(m2 ·K).
Step 3: Determine the characteristic dimension L and the dimensional
ratios 1 and 2.
From Example 1, L = 0.051 m, 1 = 1.62 2 = 2.74
Step 4: Calculate the Biot number.
From Example 1, Bi = 6.46
Step 5: Calculate the f and j factors using the method of Hayakawa and
Villalobos (1989).
For simplicity, assume the cross sections of the ham to be ellipsoidal. The area of an ellipse is the product of  times half the minor axis
times half the major axis, or
A1 = L21

A2 = L22

Using Equations (7) and (8), calculate the geometry index G:

From Equation (21),

2

A1
L  1
- = --------------- = 1 = 1.62
B1 = -------2
2
L
L

E = 0.75 + (1.01)0.414 + (1.24)(0.178) = 1.39
Thus, using Equation (15), the equivalent heat transfer dimensionality becomes

2

A2
L  2
- = --------------- = 2 = 2.74
B2 = -------2
2
L
L

43

6.46
+ 1.85
- = 1.45
E = ---------------------------------43
1.85
6.46
------------------ + ---------1.39
2.06
Step 6: Calculate the lag factor applicable to the mass average temperature.
From Table 4,  = 1, 1 = 1, and 2 = 2. Using Equation (24), j
becomes
j = 1.271 + 0.305 exp[(0.172)(1.62) – (0.115)(1.62)2]
+ 0.425 exp[(0.09)(2.74) – (0.128)(2.74)2] = 1.78
Using Equation (23), the lag factor applicable to the center temperature becomes
1
6.46
+ ---------1.62
- = 1.72
jc = ----------------------------------1.35
1
6.46
------------------ + ---------1.78
1.62
1.35

Using Equations (25) and (26), the lag factor for the mass average
temperature becomes

3
3
G = 0.25 + -------------------------- + -------------------------- = 0.443
2
2
 8   2.74 
 8   1.62 
Using Equation (12), determine the characteristic value M12:
Xg = ln(G) = ln(0.443) = –0.814
Xb = ln(1/Bi) = ln(1/6.46) = –1.87
ln (M12 )

= 0.92083090 + 0.83409615  –0.814  – 0.78765739  – 1.87 
– 0.04821784  – 0.814   – 1.87  – 0.04088987  – 0.814 
2

– 0.10045526  – 1.87  + 0.01521388  – 0.814 
3

3

(1.72) = 0.721

Step 7: Find the root of transcendental Equation (14):
cot + Bi – 1 = 0
cot + 6.46– 1 = 0

3

+ 0.00119941  – 0.814   – 1.87  + 0.00129982  – 1.87 
= 1.28
M12 = 3.60
From Equation (9), the f factor becomes
2

1.5 +  0.69   6.46 
jm = --------------------------------------------1.5 + 6.46

2

2
2.303L c
f = 2.303L
------------------- = -------------------------2
2
M1 k
M1 
2

 2.303   0.051   1080   3740 
f = -------------------------------------------------------------------------- = 17 700 s
 3.60   0.379 
From Equation (10), the j factor becomes

4

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Cooling and Freezing Times of Foods

20.7

jm = 0.892e(–0.0388)(3.60) = 0.776
Step 6: Calculate cooling time.
From Example 1, the unaccomplished temperature difference was
found to be Y = 0.1549 Using Equation (4), the cooling time becomes

spheres, and rectangular bricks. In these regression equations, the
effects of surface heat transfer, precooling, and final subcooling are
accounted for by the Biot, Plank, and Stefan numbers, respectively.
In this section, the Biot number is defined as
hD
Bi = ------ks

17 700  0.1549
 = – ---------------- ln  ---------------- = 12 400 s = 3.44 h
2.303  0.776 

FREEZING TIMES OF FOODS AND BEVERAGES
As discussed at the beginning of this chapter, freezing of foods
and beverages is not an isothermal process but rather occurs over a
range of temperatures. This section discusses Plank’s basic freezing
time estimation method and its modifications; methods that calculate freezing time as the sum of the precooling, phase change, and
subcooling times; and methods for irregularly shaped foods. These
methods are divided into three subgroups: (1) equivalent heat transfer dimensionality, (2) mean conducting path, and (3) equivalent
sphere diameter. All of these freezing time estimation methods use
thermal properties of foods (Chapter 19).

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Plank’s Equation
One of the most widely known simple methods for estimating
freezing times of foods and beverages was developed by Plank
(1913, 1941). Convective heat transfer is assumed to occur between
the food and the surrounding cooling medium. The temperature of
the food is assumed to be at its initial freezing temperature, which is
constant throughout the freezing process. Furthermore, constant
thermal conductivity for the frozen region is assumed. Plank’s
freezing time estimation is as follows:
Lf 
2
 = ------------------ PD
-------- + RD
---------- 
Tf – T m  h
ks 

(27)

where Lf is the volumetric latent heat of fusion (see Chapter 19), Tf
is the initial freezing temperature of the food, Tm is the freezing
medium temperature, D is the thickness of the slab or diameter of
the sphere or infinite cylinder, h is the convective heat transfer
coefficient, ks is the thermal conductivity of the fully frozen food,
and P and R are geometric factors. For an infinite slab, P = 1/2 and
R = 1/8. For a sphere, P = 1/6 and R = 1/24; for an infinite cylinder,
P = 1/4 and R = 1/16.
Plank’s geometric factors indicate that an infinite slab of thickness D, an infinite cylinder of diameter D, and a sphere of diameter
D, if exposed to the same conditions, would have freezing times in
the ratio of 6:3:2. Hence, a cylinder freezes in half the time of a slab
and a sphere freezes in one-third the time of a slab.

Modifications to Plank’s Equation
Various researchers have noted that Plank’s method does not
accurately predict freezing times of foods and beverages. This is
because, in part, Plank’s method assumes that foods freeze at a
constant temperature, and not over a range of temperatures as is the
case in actual food freezing processes. In addition, the frozen food’s
thermal conductivity is assumed to be constant; in reality, thermal
conductivity varies greatly during freezing. Another limitation of
Plank’s equation is that it neglects precooling and subcooling, the
removal of sensible heat above and below the freezing point. Consequently, researchers have developed improved semianalytical/
empirical cooling and freezing time estimation methods that account for these factors.
Cleland and Earle (1977, 1979a, 1979b) incorporated corrections
to account for removal of sensible heat both above and below the
food’s initial freezing point as well as temperature variation during
freezing. Regression equations were developed to estimate the geometric parameters P and R for infinite slabs, infinite cylinders,

(28)

where h is the convective heat transfer coefficient, D is the characteristic dimension, and ks is the thermal conductivity of the fully frozen food. In freezing time calculations, the characteristic dimension
D is defined to be twice the shortest distance from the thermal center
of a food to its surface: the thickness of a slab or the diameter of a
cylinder or a sphere.
In general, the Plank number is defined as follows:
Cl  Ti – Tf 
Pk = ---------------------------H

(29)

where Cl is the volumetric specific heat of the unfrozen phase and
H is the food’s volumetric enthalpy change between Tf and the
final food temperature (see Chapter 19). The Stefan number is similarly defined as
Cs  Tf – Tm 
Ste = -----------------------------H

(30)

where Cs is the volumetric specific heat of the frozen phase.
In Cleland and Earle’s method, Plank’s original geometric factors
P and R are replaced with the modified values given in Table 5, and
the latent heat Lf is replaced with the volumetric enthalpy change of
the food H10 between the freezing temperature Tf and the final
center temperature, assumed to be –10°C. As shown in Table 5, P
and R are functions of the Plank and Stefan numbers. Both parameters should be evaluated using the enthalpy change H10. Thus, the
modified Plank equation takes the form
 H 10
 = -----------------Tf – T m

 PD RD 2
 -------- + ----------
ks 
 h

(31)

where ks is the thermal conductivity of the fully frozen food.
Equation (31) is based on curve-fitting of experimental data in
which the product final center temperature was –10°C. Cleland and
Earle (1984) noted that this prediction formula does not perform as
well in situations with final center temperatures other than –10°C.
Cleland and Earle proposed the following modified form of Equation (31) to account for different final center temperatures:
 H 10
 = -----------------Tf – T m

 PD RD 2
1.65 Ste  T c – T m 
 -------- + ---------- 1 – -------------------- ln  ----------------------
ks 
ks
 h
 T ref – T m

(32)

where Tref is –10°C, Tc is the final product center temperature, and
H10 is the volumetric enthalpy difference between the initial freezing temperature Tf and –10°C. The values of P, R, Pk, and Ste should
be evaluated using H10, as previously discussed.
Hung and Thompson (1983) also improved on Plank’s equation to
develop an alternative freezing time estimation method for infinite
slabs. Their equation incorporates the volumetric change in enthalpy
H18 for freezing as well as a weighted average temperature difference between the food’s initial temperature and the freezing medium
temperature. This weighted average temperature difference T is
given as follows:

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20.8

2010 ASHRAE Handbook—Refrigeration (SI)
2

2

Cl 
C
 T – T  ----- – Tf – T c ------s
f
 i

 2
2
T = (Tf – Tm) + --------------------------------------------------------------------- H 18

(33)

where Tc is the food’s final center temperature and H18 is its
enthalpy change between initial and final center temperatures; the
latter is assumed to be –18°C. Empirical equations were developed
to estimate P and R for infinite slabs as follows:

Ste 
P = 0.7306 – 1.083 Pk + Ste 15.40U – 15.43 + 0.01329 --------- (34)
Bi 

R = 0.2079 – 0.2656U(Ste)

(35)

where U = T/(Tf – Tm). In these expressions, Pk and Ste should
be evaluated using the enthalpy change H18. The freezing time prediction model is

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H 18
 = -----------T

PD RD 2
-------- + ----------
ks 
 h

 Tm – Tf 
3 = f3log  j 3 --------------------
 Tm – Tc 



(37)

2

Lf D
 P

2 = ----------------------------  ---------- + R
 Tf – T m k c  2Bi c


where Lf is the food’s volumetric latent heat of fusion, P and R are
the original Plank geometric shape factors, kc is the frozen food’s
thermal conductivity at (Tf + Tm)/2, and Bic is the Biot number for
the subcooling period (Bic = hL/kc).
Lacroix and Castaigne (1987a, 1987b) adjusted P and R to obtain
better agreement between predicted freezing times and experimental data. Using regression analysis, Lacroix and Castaigne suggested the following geometric factors:

Total freezing time  is as follows:

P = 0.51233

(42)

R = 0.15396

(43)

P = 0.27553

(44)

R = 0.07212

(45)

P = 0.19665

(46)

R = 0.03939

(47)

For infinite cylinders

Precooling, Phase Change, and Subcooling
Time Calculations
For spheres
(38)

where 1, 2, and 3 are the precooling, phase change, and subcooling times, respectively.
DeMichelis and Calvelo (1983) suggested using Cleland and
Earle’s (1982a) equivalent heat transfer dimensionality method, discussed in the Cooling Times of Foods and Beverages section of this
chapter, to estimate precooling and subcooling times. They also
suggested that the phase change time be calculated with Plank’s
equation, but with the thermal conductivity of the frozen food evaluated at temperature (Tf + Tm)/2, where Tf is the food’s initial freezing temperature and Tm is the temperature of the cooling medium.
Lacroix and Castaigne (1987a, 1987b, 1988) suggested the use of
f and j factors to determine precooling and subcooling times of foods
and beverages. They presented equations (see Tables 1 to 3) for estimating the values of f and j for infinite slabs, infinite cylinders, and
spheres. Note that Lacroix and Castaigne based the Biot number on
the shortest distance between the thermal center of the food and its
surface, not twice that distance.
Lacroix and Castaigne (1987a, 1987b, 1988) gave the following
expression for estimating precooling time 1:
 Tm – Ti 
1 = f1log  j 1 ------------------
 Tm – Tf 

(41)

For infinite slabs

where Tref is –18°C, Tc is the product final center temperature, and
H18 is the volumetric enthalpy change between the initial temperature Ti and –18°C. The weighted average temperature difference
T, Pk, and Ste should be evaluated using H18.

 = 1 + 2 + 3

(40)

where Tc is the final temperature at the center of the food. The f3 and
j3 factors are determined from a Biot number calculated using the
thermal conductivity of the frozen food evaluated at the temperature
(Tf + Tm)/2.
Lacroix and Castaigne model the phase change time 2 with
Plank’s equation:

(36)

Cleland and Earle (1984) applied a correction factor to the Hung
and Thompson model [Equation (36)] and improved the prediction
accuracy of the model for final temperatures other than –18°C. The
correction to Equation (36) is as follows:
H 18
 = -----------T

where Tm is the coolant temperature, Ti is the food’s initial temperature, and Tf is the initial freezing point of the food. The f1 and j1 factors are determined from a Biot number calculated using an average
thermal conductivity, which is based on the frozen and unfrozen
food’s thermal conductivity evaluated at (Tf + Tm)/2. See Chapter 19
for the evaluation of food thermal properties.
The expression for estimating subcooling time 3 is

(39)

For rectangular bricks

1 - – 0.01956 -------1 - – 1.69657
P = P  – 0.02175 -------
Bi c
Ste



(48)


1 - + 0.02932 -------1 - + 1.58247
R = R  5.57519 -------
Bi
Ste


c

(49)

For rectangular bricks, P and R are calculated using the expressions given in Table 5 for the P and R of bricks.
Pham (1984) also devised a freezing time estimation method,
similar to Plank’s equation, in which sensible heat effects were considered by calculating precooling, phase-change, and subcooling
times separately. In addition, Pham suggested using a mean freezing
point, assumed to be 1.5 K below the initial freezing point of the
food, to account for freezing that takes place over a range of temperatures. Pham’s freezing time estimation method is stated in terms
of the volume and surface area of the food and is, therefore, applicable to foods of any shape. This method is given as
Qi 
Bi 
i = --------------------- 1 + -------i 
hA s T mi 
ki 

i = 1, 2, 3

(50)

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Cooling and Freezing Times of Foods

20.9
Table 5 Expressions for P and R

Shape

P and R Expressions

Infinite
slab

P = 0.5072 + 0.2018 Pk + Ste  0.3224 Pk + 0.0105
---------------- + 0.0681


Bi

Applicability
10  h  500 W/(m2 ·K)
0  D  0.12 m
Ti  40°C
–45 Tm  –15°C

R = 0.1684 + Ste  0.2740 Pk – 0.0135 
Infinite
cylinder

0.155  Ste  0.345
0.5  Bi  4.5
0  Pk  0.55

P = 0.3751 + 0.0999 Pk + Ste  0.4008 Pk + 0.0710
---------------- – 0.5865


Bi
R = 0.0133 + Ste  0.0415 Pk + 0.3957 

Sphere

0.155  Ste  0.345
0.5  Bi  4.5
0  Pk  0.55

0.3114
P = 0.1084 + 0.0924 Pk + Ste  0.231 Pk – ---------------- + 0.6739


Bi
R = 0.0784 + Ste  0.0386 Pk – 0.1694 
P = P 2 + P 1  0.1136 + Ste  5.766P 1 – 1.242  

Brick

0.155  Ste  0.345
0  Pk  0.55
0  Bi  22
1  1  4
1  2  4

R = R 2 + R 1  0.7344 + Ste  49.89R 1 – 2.900  

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where

and

P 2 = P 1 1.026 + 0.5808 Pk + Ste  0.2296 Pk + 0.0182
---------------- + 0.1050


Bi
R 2 = R 1  1.202 + Ste  3.410 Pk + 0.7336  
1 2
P 1 = ------------------------------------------2  1 2 + 1 + 2 

Q
r -  –  s – 1    – s    – s  ln  ----------s -  + ----1-  2 + 2 – 1 
R 1 = ----  r – 1    1 – r    2 – r  ln  ---------1
2
1
2
 r – 1
 s – 1
2
72
in which
1- = 4   –     – 1  +   – 1  2 1  2
--1
2
1
2
Q

2 1  2
1
r = ---   1 +  2 + 1 +    1 –  2    1 – 1  +   2 – 1   
3


and

2 1  2
1
s = ---   1 +  2 + 1 –    1 –  2    1 – 1  +   2 – 1   
3


Second shortest dimension of food
 1 = ----------------------------------------------------------------------------------Shortest dimension of food
Longest dimension of food
 2 = ----------------------------------------------------------------Shortest dimension of food
Source: Cleland and Earle (1977, 1979a, 1979b)

where 1 is the precooling time, 2 is the phase change time, 3 is the
subcooling time, and the remaining variables are defined as shown
in Table 6.
Pham (1986) significantly simplified the previous freezing time
estimation method to yield
Bi 
V   H1  H2  
 = ---------  ---------- + ----------  1 + -------s- 
hA s   T1 T 2  
4 

(51)

H2 = Lf + Cs(Tfm – Tc)
T i + Tf m
T1 = ------------------- – Tm
2
T2 = Tfm – Tm

Tfm = 1.8 + 0.263Tc + 0.105Tm

(54)

where all temperatures are in °C.

in which
H1 = Cl (Ti – Tfm)

Pham suggested that the mean freezing temperature Tfm used in
Equations (52) and (53) mainly depended on the cooling medium
temperature Tm and product center temperature Tc. By curve fitting
to existing experimental data, Pham (1986) proposed the following
equation to determine the mean freezing temperature for use in
Equations (52) and (53):

Geometric Considerations
(52)

(53)

where Cl and Cs are volumetric specific heats above and below
freezing, respectively, Ti is the initial food temperature, Lf is the
volumetric latent heat of freezing, and V is the volume of the food.

Equivalent Heat Transfer Dimensionality. Similar to their
work involving cooling times of foods, Cleland and Earle (1982b)
also introduced a geometric correction factor, called the equivalent
heat transfer dimensionality E, to calculate the freezing times of
irregularly shaped foods. The freezing time of an irregularly shaped
object shape was related to the freezing time of an infinite slab slab
using the equivalent heat transfer dimensionality:
shape = slab /E

(55)

Freezing time of the infinite slab is then calculated from one of the
many suitable freezing time estimation methods.

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2010 ASHRAE Handbook—Refrigeration (SI)

Table 6 Definition of Variables for Freezing Time
Estimation Method
Process

Variables

Precooling

i=1
k1 = 6
Q1 = Cl (Ti – Tf m )V

 Tm1

Phase change

Bi1 = (Bil + Bis)/2
 T i – T m  –  T fm – T m 
= ----------------------------------------------------- Ti – Tm 
ln  ---------------------- 
 Tf m – T m 

T m3

G2

G3

1
2
3
1
2
1
1
1
1

0
0
0
2
0
1
1
1
1

0
0
0
0
1
0
1
0
1

E = G1 + G2E1 + G3E2

(60)

1.77 1
1.77
0.73E1 = X  2.32   1  ------ + 1 – X  2.32   1  -----------2.50
1
1

(61)

E2 = X  2.32   1


(62)

1.77 

1 - + 1 – X  2.32   1.77  ---------0.50
-----2  3.69

2
2

and G1, G2, and G3 are given in Table 7. In Equations (61) and (62),
the function X with argument  is defined as
X() = /(Bi1.34 + )

Source: Pham (1984)
Notes:
As = area through which heat is transferred
Bil = Biot number for unfrozen phase
Bis = Biot number for frozen phase
Q1, Q2, Q3 = heats of precooling, phase change, and subcooling, respectively
Tm1, Tm2, Tm3 = corresponding log-mean temperature driving forces
Tc = final thermal center temperature
Tfm = mean freezing point, assumed 1.5 K below initial freezing point
To = mean final temperature
V = volume of food

Using data collected from a large number of freezing experiments, Cleland and Earle (1982b) developed empirical correlations
for the equivalent heat transfer dimensionality applicable to rectangular bricks and finite cylinders. For rectangular brick shapes with
dimensions D by 1D by 2D, the equivalent heat transfer dimensionality was given as follows:
E = 1 + W1 + W2

(56)

 Bi  5
 2 
2
W1 =  --------------- -------- +  --------------- ------------------------ Bi + 2 8 3  Bi + 2  1   1 + 1 

(57)

where

1

and
(58)

2

For finite cylinders where the diameter is smaller than the height,
the equivalent heat transfer dimensionality was given as
E = 2.0 + W2

G1

Infinite slab
Infinite cylinder
Sphere
Finite cylinder (diameter > height)
Finite cylinder (height > diameter)
Infinite rod
Rectangular brick
Two-dimensional irregular shape
Three-dimensional irregular shape

where

i=3
k3 = 6
Q3 = Cs (Tf m – Tc)V
Bi3 = Bis
 T fm – T m  –  T o – T m 
= ------------------------------------------------------ T fm – T m
ln  ---------------------
 To – Tm 

 Bi  5
 2 
2
W2 =  --------------- -------- +  --------------- ------------------------3
Bi
+
2

 8
 Bi + 2  2   2 + 1 

Shape

Source: Cleland et al. (1987a)

i=2
k2 = 4
Q2 = Lf V
Bi2 = Bis
Tm2 = Tf m – Tm

Subcooling

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Table 7 Geometric Constants

(59)

In addition, Cleland et al. (1987a, 1987b) developed expressions
for determining the equivalent heat transfer dimensionality of infinite slabs, infinite and finite cylinders, rectangular bricks, spheres,
and two- and three-dimensional irregular shapes. Numerical methods were used to calculate the freezing or thawing times for these
shapes. A nonlinear regression analysis of the resulting numerical
data yielded the following form for the equivalent heat transfer
dimensionality:

(63)

Using the freezing time prediction methods for infinite slabs and
various multidimensional shapes developed by McNabb et al.
(1990), Hossain et al. (1992a) derived infinite series expressions for
E of infinite rectangular rods, finite cylinders, and rectangular
bricks. For most practical freezing situations, only the first term of
these series expressions is significant. The resulting expressions for
E are given in Table 8.
Hossain et al. (1992b) also presented a semianalytically derived
expression for the equivalent heat transfer dimensionality of twodimensional, irregularly shaped foods. An equivalent “pseudoelliptical” infinite cylinder was used to replace the actual two-dimensional,
irregular shape in the calculations. A pseudoellipse is a shape that
depends on the Biot number. As the Biot number approaches infinity,
the shape closely resembles an ellipse. As the Biot number approaches zero, the pseudoelliptical infinite cylinder approaches an
infinite rectangular rod. Hossain et al. (1992b) stated that, for practical Biot numbers, the pseudoellipse is very similar to a true ellipse.
This model pseudoelliptical infinite cylinder has the same volume
per unit length and characteristic dimension as the actual food. The
resulting expression for E is as follows:
21 + -----Bi
E = 1 + --------------------2 2 2
 2 + --------Bi

(64)

In Equation (64), the Biot number is based on the shortest distance from the thermal center to the food’s surface, not twice that
distance. Using this expression for E, the freezing time shape of
two-dimensional, irregularly shaped foods can be calculated with
Equation (55).
Hossain et al. (1992c) extended this analysis to predicting freezing times of three-dimensional, irregularly shaped foods. In this
work, the irregularly shaped food was replaced with a model ellipsoid shape having the same volume, characteristic dimension, and
smallest cross-sectional area orthogonal to the characteristic dimension, as the actual food item. An expression was presented for E of
a pseudoellipsoid as follows:

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Cooling and Freezing Times of Foods

20.11

Table 8 Expressions for Equivalent Heat Transfer Dimensionality
Expressions for Equivalent Heat Transfer Dimensionality E

Shape


– 1



 sin z n 

2 
2
E = 1 + -----   1 + -----  – 4  ------------------------------------------------------------------------------------------------------------ 

Bi   
Bi 
2

sin z  z
n=1 3 

z n  1 + -------------n-  ----n- sinh  z n  1  + cosh  z n  1  




Bi  Bi


Infinite rectangular rod
(2L by 21L)

where zn are roots of Bi = zn tan(zn) and Bi = hL/k, where L is the shortest distance from
the center of the rectangular rod to the surface.


yn  
y
3
4 
2
E = 2 + ------  1 + ------  – 8  y n J 1  y n   1 + ------- cosh   1 y n  + ----n- sinh   1 y n 


2 
Bi  
Bi
Bi 

Bi 
n=1
2

Finite cylinder, height exceeds
diameter
(radius L and height 21L)

–1 – 1





where yn are roots of yn J1(yn) – BiJ0( yn) = 0; J0 and J1 are Bessel functions of the first kind,
order zero and one, respectively; and Bi = hL/k, where L is the radius of the cylinder.

– 1

sin z n

2-   1 + ---2-  – 4 -----------------------------------------------------------------------------------------------------------E = 1 + ---

 2

Bi   
Bi 
z

n
n=1 z n  z n + cos z n sin z n   I 0  z n  1  + -----I 1  z n  1 




Bi
where zn are roots of Bi = zn tan(zn); I0 and I1 are Bessel function of the second kind,
order zero and one, respectively; and Bi = hL/k, where L is the radius of the cylinder.

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Finite cylinder, diameter exceeds
height
(radius 1L and height 2L)

Rectangular brick
(2L by 21L by 22L)

sin z n
 -----------------------------------------------------------------------------------------------------------
2
2



2
E = 1 + -----  1 + ----- – 4  3 
sin z  z

Bi  
Bi 
z  1 + -------------n- ----n- sinh  z n  1  + cosh  z n  1 
n=1 n
Bi  Bi



– 8 2 
2





z nm
sin z n sin z m  cosh  z nm  + ----------sinh  z nm 


Bi
2

n=1 m=1
2
2
2
1
1
z n z m z nm  1 + -----sin z n  1 + -----------sin z m



Bi
Bi 1

–1

–1





where zn are roots of Bi = zn tan(zn); zm are the roots of Bi1 = zm tan(zm); Bi = hL/k, where L is the
shortest distance from the thermal center of the rectangular brick to the surface; and znm is given as
2
2
2 2
2  2 
z nm = z n  2 + z m  ------- 
 1 

Source: Hossain et al. (1992a)

Table 9

Summary of Methods for Determining Equivalent Heat Transfer Dimensionality

Slab

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Infinite cylinder

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Sphere

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Finite cylinder (diameter > height)

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Hossain et al. (1992a)
Table 8

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Hossain et al. (1992a)
Table 8

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Hossain et al. (1992a)
Table 8

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Hossain et al. (1992a)
Table 8

2-D irregular shape (infinite ellipse)

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Hossain et al. (1992b)
Equation (64)

3-D irregular shape (ellipsoid)

Cleland et al. (1987a, 1987b)
Equations (60) to (63)

Hossain et al. (1992b)
Equation (65)

Finite cylinder (height > diameter)

Cleland and Earle (1982a, 1982b)
Equations (58) and (59)

Infinite rod
Rectangular brick

Cleland and Earle (1982a, 1982b)
Equations (56) to (58)

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20.12

2010 ASHRAE Handbook—Refrigeration (SI)
221 + ---1 + ---Bi
Bi
E = 1 + --------------------- + --------------------2 2
2 2
 1 + --------1- 2 + --------2Bi
Bi

(65)

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In Equation (65), the Biot number is based on the shortest distance
from the thermal center to the surface of the food, not twice that
distance. With this expression for E, freezing times shape of threedimensional, irregularly shaped foods may be calculated using
Equation (55).
Table 9 summarizes the methods that have been discussed for
determining the equivalent heat transfer dimensionality of various
geometries. These methods can be used with Equation (55) to calculate freezing times.
Mean Conducting Path. Pham’s freezing time formulas, given
in Equations (50) and (51), require knowledge of the Biot number.
To calculate the Biot number of a food, its characteristic dimension
must be known. Because it is difficult to determine the characteristic
dimension of an irregularly shaped food, Pham (1985) introduced
the concept of the mean conducting path, which is the mean heat
transfer length from the surface of the food to its thermal center, or
Dm /2. Thus, the Biot number becomes
Bi = hDm/k

(66)

where Dm is twice the mean conducting path.
For rectangular blocks of food, Pham (1985) found that the mean
conducting path was proportional to the geometric mean of the
block’s two shorter dimensions. Based on this result, Pham (1985)
presented an equation to calculate the Biot number for rectangular
blocks of food:
Bi- = 1 +  1.5  – 1
-------
1
Bi o


–4

1
1  1 + ------4-
+  ----+ ----



Bi
 1
2
o

– 4 – 0.25





(67)

where Bio is the Biot number based on the shortest dimension of the
block D1, or Bio = hD1/k. The Biot number can then be substituted
into a freezing time estimation method to calculate the freezing time
for rectangular blocks.
Pham (1985) noted that, for squat-shaped foods, the mean conducting path Dm/2 could be reasonably estimated as the arithmetic
mean of the longest and shortest distances from the surface of the
food to its thermal center.
Equivalent Sphere Diameter. Ilicali and Engez (1990) and
Ilicali and Hocalar (1990) introduced the equivalent sphere diameter concept to calculate the freezing time of irregularly shaped
foods. In this method, a sphere diameter is calculated based on the
volume and the volume-to-surface-area ratio of the irregularly
shaped food. This equivalent sphere is then used to calculate the
freezing time of the food item.
Considering an irregularly shaped food item where the shortest
and longest distances from the surface to the thermal center were
designated as D1 and D2, respectively, Ilicali and Engez (1990) and
Ilicali and Hocalar (1990) defined the volume-surface diameter Dvs
as the diameter of a sphere having the same volume-to-surface-area
ratio as the irregular shape:
Dvs = 6V/As

(68)

where V is the volume of the irregular shape and As is its surface
area. In addition, the volume diameter Dv is defined as the diameter
of a sphere having the same volume as the irregular shape:
Dv = (6V/)1/3

(69)

Because a sphere is the solid geometry with minimum surface
area per unit volume, the equivalent sphere diameter Deq,s must be
greater than Dvs and smaller than Dv. In addition, the contribution of
the volume diameter Dv has to decrease as the ratio of the longest to

Table 10 Estimation Methods of Freezing Time of Regularly
and Irregularly Shaped Foods
Shape

Methods

Infinite slab

Cleland and Earle (1977), Hung and Thompson
(1983), Pham (1984, 1986a)

Infinite cylinder

Cleland and Earle (1979a), Lacroix and Castaigne
(1987a), Pham (1986a)

Short cylinder

Cleland et al. (1987a, 1987b), Hossain et al. (1992a),
equivalent sphere diameter technique

Rectangular brick

Cleland and Earle (1982b), Cleland et al. (1987a,
1987b), Hossain et al. (1992a)

Two-dimensional
irregular shape

Hossain et al. (1992b)

Three-dimensional
irregular shape

Hossain et al. (1992c), equivalent sphere diameter
technique

the shortest dimensions D2/D1 increases, because the object will be
essentially two-dimensional if D2/D1 » 1. Therefore, the equivalent
sphere diameter Deq,s is defined as follows:
2
1
Deq,s = --------------- Dv + --------------- Dvs
2 + 1
2 + 1

(70)

Thus, predicting the freezing time of the irregularly shaped food is
reduced to predicting the freezing time of a spherical food with diameter Deq,s. Any of the previously discussed freezing time methods
for spheres may then be used to calculate this freezing time.

Evaluation of Freezing Time Estimation Methods
As noted previously, selecting an appropriate estimation method
from the plethora of available methods can be challenging for the
designer. Thus, Becker and Fricke (1999a, 1999b, 1999c, 2000a,
2000b) quantitatively evaluated selected semianalytical/empirical
food freezing time estimation methods for regularly and irregularly
shaped foods. Each method’s performance was quantified by comparing its numerical results to a comprehensive experimental freezing time data set compiled from the literature. The best-performing
methods for each shape are listed in Table 10.

Algorithms for Freezing Time Estimation
The following suggested algorithm for estimating the freezing
time of foods and beverages is based on the modified Plank equation
presented by Cleland and Earle (1977, 1979a, 1979b). This algorithm is applicable to simple food geometries, including infinite
slabs, infinite cylinders, spheres, and three-dimensional rectangular
bricks.
1. Determine thermal properties of food (see Chapter 19).
2. Determine surface heat transfer coefficient for the freezing process (see Chapter 19).
3. Determine characteristic dimension D and dimensional ratios 1
and 2 using Equations (16) and (17).
4. Calculate Biot, Plank, and Stefan numbers using Equations (28),
(29), and (30), respectively.
5. Determine geometric parameters P and R given in Table 5.
6. Calculate freezing time using Equation (31) or (32), depending
on the final temperature of the frozen food.
The following algorithm for estimating freezing times of foods
and beverages is based on the method of equivalent heat transfer
dimensionality. It is applicable to many food geometries, including
infinite rectangular rods, finite cylinders, three-dimensional rectangular bricks, and two- and three-dimensional irregular shapes.
1. Determine thermal properties of the food (see Chapter 19).
2. Determine surface heat transfer coefficient for the freezing process (see Chapter 19).

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Cooling and Freezing Times of Foods

20.13

3. Determine characteristic dimension D and dimensional ratios 1
and 2 using Equations (16) and (17).
4. Calculate Biot, Plank, and Stefan numbers using Equations (28),
(29), and (30), respectively.
5. Calculate freezing time of an infinite slab using a suitable
method. Suitable methods include
(a) Equation (31) or (32) in conjunction with the geometric
parameters P and R given in Table 5.
(b) Equation (36) or (37) in conjunction with Equations (33),
(34), and (35).
6. Calculate the food’s equivalent heat transfer dimensionality.
Refer to Table 9 to determine which equivalent heat transfer
dimensionality method is applicable to the particular food geometry.
7. Calculate the freezing time of the food using Equation (55).

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SAMPLE PROBLEMS FOR ESTIMATING
FREEZING TIME
Example 3. A rectangular brick-shaped package of beef (lean sirloin)
measuring 0.04 by 0.12 by 0.16 m is to be frozen in a blast freezer.
The beef’s initial temperature is 10°C, and the freezer air temperature is –30°C. The surface heat transfer coefficient is estimated to be
40 W/(m2 ·K). Calculate the time required for the thermal center of
the beef to reach –10°C.
Solution: Because the food is a rectangular brick, the algorithm based
on the modified Plank equation by Cleland and Earle (1977, 1979a,
1979b) is used.
Step 1: Determine the thermal properties of lean sirloin.
As described in Chapter 19, the thermal properties can be calculated as follows:
At –1.7°C
At –40°C At –10°C (Initial At 10°C
(Fully
(Final
Freezing (Initial
Frozen)
Temp.)
Temp.)
Point)

Property

s = 1018 s = 1018 l = 1075 l = 1075


Hs = 83.4 Hl = 274.2

cl = 3.52
Specific heat, kJ/(kg ·K) cs = 2.11


Density, kg/m3
Enthalpy, kJ/kg

Thermal conductivity,
W/(m ·K)

ks = 1.66







Volumetric enthalpy difference between the initial freezing point
and –10°C:
H10 = l Hl – s Hs
H10= (1075)(274.2) – (1018)(83.4) = 210 

C s  Tf – T m   2148   –1.7 –  – 30  
Ste = ---------------------------- = ---------------------------------------------------= 0.289
3
H 10
210  10
Step 5: Determine the geometric parameters P and R for the rectangular
brick.
Determine P from Table 5.
34
P1 = ------------------------------------------- = 0.316
234 + 3 + 4

P 2 = 0.316 1.026 +  0.5808   0.211 


+ 0.289  0.2296   0.211  + 0.0182
---------------- + 0.1050 
0.964

= 0.379
P = 0.379 + 0.316{0.1136 + 0.289[(5.766)(0.316) – 1.242]}
= 0.468
Determine R from Table 5.
1- = 4   3 – 4   3 – 1  +  4 – 1  2  1  2 = 10.6
--Q
2
1
r = ---  3 + 4 + 1 +  3 – 4   3 – 1  +  4 – 1 
3

12


 = 3.55


2
1
s = ---  3 + 4 + 1 –  3 – 4   3 – 1  +  4 – 1 
3

12


 = 1.78


3.55 
1 -  3.55 – 1   3 – 3.55   4 – 3.55  ln  ------------------
R 1 = ---------------------
 10.6   2 
 3.55 – 1 
 1.78 
–  1.78 – 1   3 – 1.78   4 – 1.78  ln  ------------------- 
 1.78 – 1 
1
+ ------   2   3  +  2   4  – 1  = 0.0885
72
R2 = 0.0885{1.202 + 0.289[(3.410)(0.211) + 0.7336]} = 0.144

103

kJ/m3

Volumetric specific heats:
Cs = s cs = (1018)(2.11) = 2148 kJ/(m3·K)
Cl = l cl = (1075)(3.52) = 3784 kJ/(m3·K)
Step 2: Determine the surface heat transfer coefficient.
The surface heat transfer coefficient is estimated to be 40 W/(m2 ·K).
Step 3: Determine the characteristic dimension D and the dimensional
ratios 1 and 2.
For freezing time problems, the characteristic dimension D is twice
the shortest distance from the thermal center of the food to its surface.
For this example, D = 0.04 m
Using Equations (16) and (17), the dimensional ratios then become
1 = 0.12/0.04= 3
1 = 0.16/0.04= 4
Step 4: Using Equations (28) to (30), calculate the Biot, Plank, and
Stefan numbers.
hD  40.0   0.04 
Bi = ------- = ------------------------------- = 0.964
1.66
ks

C l  T i – T f   3784   10 –  – 1.7  
- = 0.211
Pk = -------------------------- = -----------------------------------------------3
H 10
210  10

R = 0.144 + 0.0885{0.7344 + 0.289[(49.89)(0.0885) – 2.900]}
= 0.248
Step 6: Calculate the beef’s freezing time.
Because the final temperature at the thermal center of the beef is
given to be –10°C, use Equation (31) to calculate the freezing time:
8

2

2.10  10
0.468   0.04   0.248   0.04  = 5250 s = 1.46 h
 = ------------------------------ --------------------------------- + -----------------------------------–1.7 –  – 30 
40.0
1.66
Example 4. Orange juice in a cylindrical container, 0.30 m diameter by
0.45 m tall, is to be frozen in a blast freezer. The initial temperature of
the juice is 5°C and the freezer air temperature is –35°C. The surface
heat transfer coefficient is estimated to be 30 W/(m2 ·K). Calculate the
time required for the thermal center of the juice to reach –18°C.
Solution: Because the food is a finite cylinder, the algorithm based on
the method of equivalent heat transfer dimensionality (Cleland et al.
1987a, 1987b) is used. This method requires calculation of the freezing
time of an infinite slab, which is determined using the method of Hung
and Thompson (1983).
Step 1: Determine the thermal properties of orange juice.

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20.14

2010 ASHRAE Handbook—Refrigeration (SI)

Using the methods described in Chapter 19, the thermal properties
of orange juice are calculated as follows:

Property

At –40°C
(Fully
Frozen)

At –18°C
(Final
Temp.)

At 5°C
(Initial
Temp.)

Density, kg/m3
Enthalpy, kJ/kg

s = 970


s = 970
Hs = 40.8

l = 1038
Hl = 381.5

Specific heat, kJ/(kg ·K)

cs = 1.76



cl = 3.89

Thermal cond., W/(m ·K)

ks = 2.19





Use the method presented by Cleland et al. (1987a, 1987b), Equations (60) to (63), to calculate the equivalent heat transfer dimensionality. From Table 7, the geometric constants for a cylinder are
G1 = 2

X(1.132) = 1.132/(4.111.34 + 1.132) = 0.146
E2 = 0.146/1.5 + (1 – 0.146)(0.50/1.53.69) = 0.193
Thus, the equivalent heat transfer dimensionality E becomes
E = G1 + G2E1 + G3E2

H18= l Hl – s Hs
H18= (1038)(381.5) – (970)(40.8) = 356 

103

kJ/m3

Volumetric specific heats:

E = 2 + (0)(E1) + (1)(0.193) = 2.193
Step 7: Calculate freezing time of the orange juice using Equation (55):

Cs = s cs = (970)(1.76) = 1707 kJ/(m3·K)

shape = slab/E = 135 000/2.193 = 61 600 s = 17.1 h

Cl = l cl = (1038)(3.89) = 4038 kJ/(m3·K)

Licensed for single user. © 2010 ASHRAE, Inc.

Step 2: Determine the surface heat transfer coefficient.
The surface heat transfer coefficient is estimated to be 30 W/(m2 ·K).
Step 3: Determine the characteristic dimension D and the dimensional
ratios 1 and 2.
For freezing time problems, the characteristic dimension is twice
the shortest distance from the thermal center of the food item to its surface. For the cylindrical sample of orange juice, the characteristic
dimension is equal to the diameter of the cylinder:
D = 0.30 m
Using Equations (16) and (17), the dimensional ratios then become
1 = 2 = 0.45 m/0.30 m= 1.5
Step 4: Using Equations (28) to (30), calculate the Biot, Plank, and Stefan numbers.
Bi = hD/ks = (30.0)(0.30)/2.19= 4.11
Cl  T i – T f   4038   5 –  – 0.4  
Pk = -------------------------- = ---------------------------------------------= 0.0613
3
H 18
356  10
C s  T f – T m  1707   – 0.4 –  – 35  
- = 0.166
Ste = ----------------------------- = ---------------------------------------------------3
H 18
356  10
Step 5: Calculate the freezing time of an infinite slab.
Use the method of Hung and Thompson (1983). First, find the
weighted average temperature difference given by Equation (33).
T = [–0.4 – (–35)] +
2

G3 = 1

 = 2.32/1.77
= 2.32/1.51.77 = 1.132
2

Initial freezing temperature: Tf = –0.4°C
Volumetric enthalpy difference between Ti = 5°C, and –18°C:

G2 = 0

Calculate E2:

2

 5 –  –0.4    4038  2  –  –0.4 –  – 18    1707  2 
----------------------------------------------------------------------------------------------------------------------------- = 34.0 K
3
356  10
Determine the parameter U:
U = 34.0/[–0.4 – (–35)] = 0.983
Determine the geometric parameters P and R for an infinite slab
using Equations (34) and (35):
P = 0.7306 – (1.083)(0.0613) + (0.166)[(15.40)(0.984 0.983)
– 15.43 + (0.01329)(0.166)/4.11] = 0.616
R = 0.2079 – (0.2656)(0.983)(0.166) = 0.165
Determine the freezing time of the slab using Equation (36):
8

3.56  10 0.616   0.30   0.165   0.30  2
 = ------------------------- --------------------------------- + ------------------------------------ = 135 000 s = 37.5 h
34.0
2.19
30.0
Step 6: Calculate the equivalent heat transfer dimensionality for a finite
cylinder.

SYMBOLS
A1
A2
As
B1
B2
Bi
Bi1
Bi2
Bi3
Bic
Bil
Bio
Bis
c
Cl
Cs
D
D1
D2
Deq,s
Dm
Dv
Dvs
E
E0
E1
E2
E
f
f1
f3
fcomp
G
G1
G2
G3
h
I0(x)
I1(x)
j
j1
j3
jc
jcomp
jm
js
J0(x)
J1(x)
j
k

=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=

cross-sectional area in Equation (8), m2
cross-sectional area in Equation (8), m2
surface area of food, m2
parameter in Equation (7)
parameter in Equation (7)
Biot number
Biot number for precooling = (Bil + Bis)/2
Biot number for phase change = Bis
Biot number for subcooling = Bis
Biot number evaluated at kc = hD/kc
Biot number for unfrozen food = hD/kl
Biot number based on shortest dimension = hD1/k
Biot number for fully frozen food = hD/ks
specific heat of food, J/(kg·K)
volumetric specific heat of unfrozen food, J/(m3 ·K)
volumetric specific heat of fully frozen food, J/(m3 ·K)
slab thickness or cylinder/sphere diameter, m
shortest dimension, m
longest dimension, m
equivalent sphere diameter, m
twice the mean conducting path, m
volume diameter, m
volume-surface diameter, m
equivalent heat transfer dimensionality
equivalent heat transfer dimensionality at Bi = 0
parameter given by Equation (61)
parameter given by Equation (62)
equivalent heat transfer dimensionality at Bi  
cooling time parameter
cooling time parameter for precooling
cooling time parameter for subcooling
cooling parameter for a composite shape
geometry index
geometric constant in Equation (60)
geometric constant in Equation (60)
geometric constant in Equation (60)
heat transfer coefficient, W/(m2 ·K)
Bessel function of second kind, order zero
Bessel function of second kind, order one
cooling time parameter
cooling time parameter for precooling
cooling time parameter for subcooling
cooling time parameter applicable to thermal center
cooling time parameter for a composite shape
cooling time parameter applicable to mass average
cooling time parameter applicable to surface temperature
Bessel function of first kind, order zero
Bessel function of first kind, order one
lag factor parameter given by Equation (24)
thermal conductivity of food, W/(m·K)

This file is licensed to Abdual Hadi Nema (ahaddi58@yahoo.com). License Date: 6/1/2010

Licensed for single user. © 2010 ASHRAE, Inc.

Cooling and Freezing Times of Foods
kc = thermal conductivity of food evaluated at (Tf + Tm )/2,
W/(m·K)
kl = thermal conductivity of unfrozen food, W/(m·K)
ks = thermal conductivity of fully frozen food, W/(m·K)
L = half thickness of slab or radius of cylinder/sphere, m
Lf = volumetric latent heat of fusion, J/m3
m = inverse of Biot number
M 12 = characteristic value of Smith et al. (1968)
N = number of dimensions
p1 = geometric parameter given in Table 4
p2 = geometric parameter given in Table 4
p3 = geometric parameter given in Table 4
P = Plank’s geometry factor
P = geometric factor for rectangular bricks calculated using method
in Table 5
P1 = intermediate value of Plank’s geometric factor
P2 = intermediate value of Plank’s geometric factor
Pk = Plank number = Cl (Ti – Tf )/H
Q = parameter in Table 5
Q1 = volumetric heat of precooling, J/m3
Q2 = volumetric heat of phase change, J/m3
Q3 = volumetric heat of subcooling, J/m3
r = parameter given in Table 5
R = Plank’s geometry factor
R = geometric factor for rectangular bricks calculated using method
in Table 5
R1 = intermediate value of Plank’s geometric factor
R2 = intermediate value of Plank’s geometric factor
s = parameter given in Table 5
Ste = Stefan number = Cs (Tf – Tm )/H
T = product temperature, °C
Tc = final center temperature of food, °C
Tf = initial freezing temperature of food, °C
Tfm = mean freezing temperature, °C
Ti = initial temperature of food, °C
Tm = cooling or freezing medium temperature, °C
To = mean final temperature, °C
Tref = reference temperature for freezing time correction factor, °C
u = parameter given in Table 1
U = parameter in Equations (34) and (35) = T/(Tf – Tm)
v = parameter given in Table 2
V = volume of food, m3
w = parameter given in Table 3
W1 = parameter given by Equation (57)
W2 = parameter given by Equation (58)
x = coordinate direction
X() = function given by Equation (63)
Xb = parameter in Equation (12)
Xg = parameter in Equations (11) and (12)
y = coordinate direction
Y = fractional unaccomplished temperature difference
Yc = fractional unaccomplished temperature difference based on final
center temperature
Ym = fractional unaccomplished temperature difference based on final
mass average temperature
yn = roots of transcendental equation; yn J1( yn ) – Bi J0( yn ) = 0
z = coordinate direction
zm = roots of transcendental equation; Bi1 = zm tan(zm)
zn = roots of transcendental equation; Bi = zn tan(zn)
znm = parameter given in Table 8

Greek
 = thermal diffusivity of food, m2/s
1 = ratio of second shortest dimension to shortest dimension,
Equation (16)
2 = ratio of longest dimension to shortest dimension, Equation (17)
1 = geometric parameter from Lin et al. (1996b)
2 = geometric parameter from Lin et al. (1996b)
H = volumetric enthalpy difference, J/m3
H1 = volumetric enthalpy difference = Cl (Ti – Tf m ), J/m3
H2 = volumetric enthalpy difference = L f + Cs (Tf m – Tc ), J/m3
H 10 = volumetric enthalpy difference between initial freezing
temperature Tf and –10°C, J/m3
H 18 = volumetric enthalpy difference between initial temperature Ti
and –18°C, J/m3

20.15
T
T1
T2
Tm1
Tm2
Tm3

1
2
3
shape
slab






=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=

weighted average temperature difference in Equation (33), K
temperature difference = (Ti + Tfm )/2 – Tm, K
temperature difference = Tfm – Tm, K
temperature difference for precooling, K
temperature difference for phase change, K
temperature difference for subcooling, K
cooling or freezing time, s
precooling time, s
phase change time, s
tempering time, s
freezing time of an irregularly shaped food, s
freezing time of an infinite slab-shaped food, s
geometric parameter from Lin et al. (1996b)
parameter given by Equation (26)
density of food, kg/m3
argument of function X, Equation (63)
first root of Equation (14)

REFERENCES
Becker, B.R. and B.A. Fricke. 1999a. Evaluation of semi-analytical/empirical
freezing time estimation methods, part I: Regularly shaped food items.
International Journal of HVAC&R Research (now HVAC&R Research)
5(2):151-169.
Becker, B.R. and B.A. Fricke. 1999b. Evaluation of semi-analytical/empirical
freezing time estimation methods, part II: Irregularly shaped food items.
International Journal of HVAC&R Research (now HVAC&R Research)
5(2):171-187.
Becker, B.R. and B.A. Fricke. 1999c. Freezing times of regularly shaped
food items. International Communications in Heat and Mass Transfer
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Becker, B.R. and B.A. Fricke. 2000a. Evaluation of semi-analytical/empirical
freezing time estimation methods, part I: Regularly shaped food items
(RP-888). Technical Paper 4352, presented at the ASHRAE Winter Meeting, Dallas.
Becker, B.R. and B.A. Fricke. 2000b. Evaluation of semi-analytical/empirical
freezing time estimation methods, part II: Irregularly shaped food items
(RP-888). Technical Paper 4353, presented at the ASHRAE Winter Meeting, Dallas.
Cleland, A.C. 1990. Food refrigeration processes: Analysis, design and simulation. Elsevier Science, London.
Cleland, A.C. and R.L. Earle. 1977. A comparison of analytical and numerical methods of predicting the freezing times of foods. Journal of Food
Science 42(5):1390-1395.
Cleland, A.C. and R.L. Earle. 1979a. A comparison of methods for predicting the freezing times of cylindrical and spherical foodstuffs. Journal of
Food Science 44(4):958-963, 970.
Cleland, A.C. and R.L. Earle. 1979b. Prediction of freezing times for foods
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Cleland, A.C. and R.L. Earle. 1982b. Freezing time prediction for foods—
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Cleland, A.C. and R.L. Earle. 1984. Freezing time predictions for different
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Cleland, D.J., A.C. Cleland, and R.L. Earle. 1987a. Prediction of freezing
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156-164.
Cleland, D.J., A.C. Cleland, and R.L. Earle. 1987b. Prediction of freezing
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