```11-42
11-57 The center temperature of potatoes is to be lowered to 6qC during cooling. The cooling time and if
any part of the potatoes will suffer chilling injury during this cooling process are to be determined.
Assumptions 1 The potatoes are spherical in shape with a radius of r0 = 3 cm. 2 Heat conduction in the
potato is one-dimensional in the radial direction because of the symmetry about the midpoint. 3 The
thermal properties of the potato are constant. 4 The heat transfer coefficient is constant and uniform over
the entire surface. 5 The Fourier number is W > 0.2 so that the one-term approximate solutions (or the
transient temperature charts) are applicable (this assumption will be verified).
Properties The thermal conductivity and thermal diffusivity of potatoes are given to be k = 0.50 W/mqC
and D = 0.13u10-6 m2/s.
Air
Analysis First we find the Biot number:
2qC
hro (19 W/m 2 .qC)(0.03 m)
4 m/s
1.14
Bi
0.5 W/m.qC
k
Potato
From Table 11-2 we read, for a sphere, O1 = 1.635
Ti = 25qC
and A1 = 1.302. Substituting these values into the
one-term solution gives
T0 T f
Ti Tf
T0
A1e O1W o
2
62
25 2
1.302e (1.635)
2
W
o W = 0.753
which is greater than 0.2 and thus the one-term solution is applicable. Then the cooling time becomes
W
Dt
Wro2
D

o t
ro2
(0.753)(0.03 m) 2
0.13 u 10 -6 m 2 / s
5213 s 1.45 h
The lowest temperature during cooling will occur on the surface (r/r0 = 1), and is determined to be
T ( r ) Tf
Ti Tf
A1e O1W
2
sin(O1 r / ro ) T ( ro ) Tf
o
O1 r / ro
Ti Tf
T (ro ) 2
25 2
Substituting,
T0
sin(O1 ro / ro ) To Tf sin(O1 ro / ro )
=
O1 ro / ro
Ti Tf
O1 ro / ro
§ 6 2 · sin(1.635 rad)

o T (ro ) = 4.44qC
¸
¨
1.635
which is above the temperature range of 3 to 4 qC for chilling injury for potatoes. Therefore, no part of
the potatoes will experience chilling injury during this cooling process.
Alternative solution We could also solve this problem using transient temperature charts as follows:
1
Bi
0.50 W/m.o C
k
hro
(19W/m 2 . o C)(0.03m)
T0 Tf
Ti Tf
Therefore,
t
62
25 2
W ro2
D
0.174
(0.75)(0.03) 2
0.13 u 10 6 m 2 / s
½
0.877°
Dt
°
¾W = 2
ro
°
°
¿
0.75
(Fig. 11 - 17a)
5192 s 1.44 h
The surface temperature is determined from
1
Bi
k
hro
r
ro
1
which gives Tsurface
½
0.877°
° T ( r ) Tf
¾
° To Tf
°¿
Tf 0.6(To Tf )
0.6
(Fig.11 17b)
2 0.6(6 2)
4.4qC
The slight difference between the two results is due to the reading error of the charts.
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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