3 The heat transfer coefficient is constant and uniform over the entire surface.. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions or the transient temperature
Trang 14-50 A hot dog is dropped into boiling water, and temperature measurements are taken at certain time
intervals The thermal diffusivity and thermal conductivity of the hot dog and the convection heat transfer coefficient are to be determined
Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal
symmetry about the centerline 2 The thermal properties of the hot dog are constant 3 The heat transfer coefficient is constant and uniform over the entire surface 4 The Fourier number is τ > 0.2 so that the
one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)
Properties The properties of hot dog available are given to be ρ = 980 kg/m3 and c p = 3900 J/kg.°C
Analysis (a) From Fig 4-16b we have
15 0 1
1
17 0 94 59
94 88
⎪
⎪
⎭
⎪
⎬
⎫
=
=
=
−
−
=
−
−
∞
∞
o o
o o
hr
k Bi r
r r
r
T
T
T
T
Water 94°C
Hot dog The Fourier number is determined from Fig 4-16a to be
20 0 47
0 94 20
94 59
15 0 1
2 0
=
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
=
−
−
=
−
−
=
=
∞
i
o
r t
T
T
T
T
hr
k
The thermal diffusivity of the hot dog is determined to be
/s m 10 2.017× −7 2
=
=
=
⎯→
⎯
=
s 120
m) 011 0 )(
2 0 ( 2 0 20
0
2 2
r r
o
α α
(b) The thermal conductivity of the hot dog is determined from
C W/m.
=
°
×
=
= (2.017 10−7m2/s)(980kg/m3)(3900J/kg C)
p
c
(c) From part (a) we have 1 = =0.15
o
hr
k
m 0.00165 m)
011 0 )(
15 0 ( 15
= r o
h
k
Therefore, the heat transfer coefficient is
C W/m
=
°
=
⎯→
⎯
=
m 0.00165
C W/m
771 0 00165
h
k
Trang 2
4-51 Using the data and the answers given in Prob 4-50, the center and the surface temperatures of the hot
dog 4 min after the start of the cooking and the amount of heat transferred to the hot dog are to be
determined
Assumptions 1 Heat conduction in the hot dog is one-dimensional since it is long and it has thermal
symmetry about the center line 2 The thermal properties of the hot dog are constant 3 The heat transfer coefficient is constant and uniform over the entire surface 4 The Fourier number is τ > 0.2 so that the
one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)
Properties The properties of hot dog and the convection heat transfer coefficient are given or obtained in
P4-47 to be k = 0.771 W/m.°C, ρ = 980 kg/m3, cp = 3900 J/kg.°C, α = 2.017×10-7 m2/s, and h = 467
W/m2.°C
Analysis The Biot number is
Water 94°C
Hot dog
) C W/m
771 0 (
) m 011 0 )(
C W/m 467
°
°
=
=
k
hr
The constants λ1andA1corresponding to this
Biot number are, from Table 4-2,
5357 1 and 0785
λ
The Fourier number is
2 0 4001 0 m)
011 0 (
s/min) 60 min /s)(4 m 10 017 2 (
2
2 7
L
t
α
τ
Then the temperature at the center of the hot dog is determined to be
C 73.8°
=
⎯→
⎯
=
−
−
=
=
=
−
−
∞
∞
o
i cyl
T T
e e
A T T
T T
2727 0 94
20
94
2727 0 )
5357 1 (
0
) 4001 0 ( 0785 2 ( 1
0 ,
0
2
2 τ λ θ
From Table 4-3 we read J0=0.1789 corresponding to the constant λ =2.0785 Then the temperature at the 1 surface of the hot dog becomes
C 90.4°
=
⎯→
⎯
=
−
−
=
=
=
−
∞
∞
) , ( 04878
0 94 20
94 ) ,
(
04878 0 ) 1789 0 ( )
5357 1 ( ) / ( )
,
1 0 1
2 2
t r T t
r
T
e r
r J e A T
T
T t
r
T
o o
o o i
The maximum possible amount of heat transfer is
J 13,440 C
) 20 94 )(
C J/kg
3900 )(
kg 04657 0 ( ) (
kg 04657 0 m) 125 0 ( m) 011 0 ( ) kg/m 980 (
max
2.
3 2
=
°
−
°
=
−
=
=
=
=
=
∞
T T mc Q
L r m
i p
ρπ ρV
From Table 4-3 we read J1= 0.5701 corresponding to the constant λ =2.0785 Then the actual heat 1 transfer becomes
kJ 11,430
=
=
=
−
=
−
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
) kJ 440 , 13 ( 8504 0
8504 0 0785 2
5701 0 ) 2727 0 ( 2 1 ) ( 2
1
1
1 1 , max
Q
J Q
Q
cyl o
λ θ
Trang 34-52E Whole chickens are to be cooled in the racks of a large refrigerator Heat transfer coefficient that
will enable to meet temperature constraints of the chickens while keeping the refrigeration time to a minimum is to be determined
Assumptions 1 The chicken is a homogeneous spherical object 2 Heat conduction in the chicken is
one-dimensional because of symmetry about the midpoint 3 The thermal properties of the chicken are constant
4 The heat transfer coefficient is constant and uniform over the entire surface 5 The Fourier number is τ >
0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)
Properties The properties of the chicken are given to be k = 0.26 Btu/h.ft.°F, ρ = 74.9 lbm/ft3, cp = 0.98
Btu/lbm.°F, and α = 0.0035 ft2/h
Analysis The radius of the chicken is determined to be
Refrigerator
T∞ = 5°F
Chicken
Ti = 65°F
ft 2517 0 4
) ft 06676 0 ( 3 4
3 3
4
ft 06676 0 lbm/ft 9 74
lbm 5
3
3 3
3
3 3
=
=
=
⎯→
⎯
=
=
=
=
⎯→
⎯
=
π π
π
ρ ρ
V V
V V
o
r
m m
From Fig 4-17b we have
2 1
1
75 0 5 45
5 35
⎪
⎪
⎭
⎪
⎪
⎬
⎫
=
=
=
−
−
=
−
−
∞
∞
o o
o o
hr
k Bi r
r r
x
T
T
T
T
Then the heat transfer coefficients becomes
F Btu/h.ft
=
°
=
=
ft) 2517 0 ( 2
F Btu/.ft
26 0
2r o
k
h
Trang 44-53 A person puts apples into the freezer to cool them quickly The center and surface temperatures of the
apples, and the amount of heat transfer from each apple in 1 h are to be determined
Assumptions 1 The apples are spherical in shape with a diameter of 9 cm 2 Heat conduction in the apples
is one-dimensional because of symmetry about the midpoint 3 The thermal properties of the apples are constant 4 The heat transfer coefficient is constant and uniform over the entire surface 5 The Fourier
number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)
Properties The properties of the apples are given to be k = 0.418 W/m.°C, ρ = 840 kg/m3, cp = 3.81
kJ/kg.°C, and α = 1.3×10-7 m2/s
Analysis The Biot number is
Apple
Ti = 20°C
Air
T∞ = -15°C 861
0 ) C W/m
418 0 (
) m 045 0 )(
C W/m 8
°
°
=
=
k
hr
The constants λ1andA1corresponding to this
Biot number are, from Table 4-2,
2390 1 and 476
λ
The Fourier number is
2 0 231 0 m)
045 0 (
s/h) 600 3 h /s)(1 m 10 3 1 (
2
2 7
o
r
t
α
τ
Then the temperature at the center of the apples becomes
C 11.2°
=
⎯→
⎯
=
=
−
−
−
−
⎯→
⎯
=
−
−
∞
∞
0 )
231 0 ( 476 1 ( 0
1
0 ,
) 15 ( 20
) 15
2
T e
T e
A T T
T T
i sph
τ λ θ
The temperature at the surface of the apples is
C 2.7°
=
⎯→
⎯
=
−
−
−
−
=
=
=
−
−
∞
∞
) , ( 505
0 )
15
(
20
) 15
(
)
,
(
505 0 476 1
) rad 476 1 sin(
) 239 1 ( /
) / sin(
) , ( )
,
1
1 1
2 2
t r T t
r
T
e r
r
r r e
A T
T
T t r T
t
r
o o
o o
o o i
o sph
λ
The maximum possible heat transfer is
[20 ( 15)]C 42.75kJ )
C kJ/kg
81 3 )(
kg 3206 0 ( ) (
kg 3206 0 m) 045 0 ( 3
4 ) kg/m 840 ( 3 4
max
3.
3 3
=
°
−
−
°
=
−
=
=
⎥⎦
⎤
⎢⎣
⎡
=
=
=
∞
T T mc Q
r m
i p
π ρ ρV
Then the actual amount of heat transfer becomes
kJ 17.2
=
=
=
=
−
−
=
−
−
=
kJ) 75 42 )(
402 0 ( 402
0
402 0 )
476 1 (
) rad 476 1 cos(
) 476 1 ( ) rad 476 1 sin(
) 749 0 ( 3 1 ) cos(
) sin(
3
1
max
3 3
1
1 1 1 ,
max
Q
Q
Q
Q
sph o
λ
λ λ λ θ
Trang 54-54 EES Prob 4-53 is reconsidered The effect of the initial temperature of the apples on the final center
and surface temperatures and the amount of heat transfer is to be investigated
Analysis The problem is solved using EES, and the solution is given below
"GIVEN"
T_infinity=-15 [C]
T_i=20 [C]
h=8 [W/m^2-C]
r_o=0.09/2 [m]
time=1*3600 [s]
"PROPERTIES"
k=0.513 [W/m-C]
rho=840 [kg/m^3]
C_p=3.6 [kJ/kg-C]
alpha=1.3E-7 [m^2/s]
"ANALYSIS"
Bi=(h*r_o)/k
"From Table 4-2 corresponding to this Bi number, we read"
lambda_1=1.3525
A_1=1.1978
tau=(alpha*time)/r_o^2
(T_o-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau)
(T_r-T_infinity)/(T_i-T_infinity)=A_1*exp(-lambda_1^2*tau)*Sin(lambda_1*r_o/r_o)/(lambda_1*r_o/r_o)
V=4/3*pi*r_o^3
m=rho*V
Q_max=m*C_p*(T_i-T_infinity)
Q/Q_max=1-3*(T_o-T_infinity)/(T_i-T_infinity)*(Sin(lambda_1)-lambda_1*Cos(lambda_1))/lambda_1^3
8 3.051 -1.97 9.283
28 18.75 9.36 17.35
Trang 60 5 10 15 20 25 30
-10
-5
0
5
10
15
20
25
T
i [C]
T o
T 0
T r
6
8
10
12
14
16
18
20
T
i [C ]
Trang 74-55 An orange is exposed to very cold ambient air It is to be determined whether the orange will freeze in
4 h in subfreezing temperatures
Assumptions 1 The orange is spherical in shape with a diameter of 8 cm 2 Heat conduction in the orange
is one-dimensional because of symmetry about the midpoint 3 The thermal properties of the orange are constant, and are those of water 4 The heat transfer coefficient is constant and uniform over the entire surface 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient
temperature charts) are applicable (this assumption will be verified)
Properties The properties of the orange are approximated by those of water at the average temperature of
about 5°C, k = 0.571 W/m.°C and α =k/ρc p =0.571/(999.9×4205)=0.136×10−6m2/s (Table A-9)
Analysis The Biot number is
0 1 051 1 ) C W/m
571 0 (
) m 04 0 )(
C W/m 15
≈
=
°
°
=
=
k
hr
Orange
Ti = 15°C
Air
T∞ = -15°C The constants λ1andA1corresponding to this
Biot number are, from Table 4-2,
2732 1 and 5708
λ
The Fourier number is
2 0 224 1 m)
04 0 (
s/h) 600 3 h /s)(4 m 10 136 0 (
2
2 6
o
r
t
α
τ
Therefore, the one-term approximate solution (or the transient temperature charts) is applicable Then the temperature at the surface of the oranges becomes
C 5.2
=
⎯→
⎯
=
−
−
−
−
=
=
=
−
−
∞
∞
) , ( 0396
0 )
6
(
15
)
6
(
)
,
(
0396 0 5708 1
) rad 5708 1 sin(
) 2732 1 ( /
) / sin(
) , ( )
,
1
1 1
2 2
t r T t
r
T
e r
r
r r e
A T
T
T t r T t
r
o o
o o
o o i
o sph
λ
which is less than 0°C Therefore, the oranges will freeze
Trang 84-56 A hot baked potato is taken out of the oven and wrapped so that no heat is lost from it The time the
potato is baked in the oven and the final equilibrium temperature of the potato after it is wrapped are to be determined
Assumptions 1 The potato is spherical in shape with a diameter of 9 cm 2 Heat conduction in the potato is one-dimensional because of 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 τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified)
Properties The properties of the potato are given to be k = 0.6
W/m.°C, ρ = 1100 kg/m3, cp = 3.9 kJ/kg.°C, and α = 1.4×10-7
m2/s
Oven
T∞ = 170°C
Potato
T0 = 70°C
Analysis ( a) The Biot number is
3 )
C W/m
6 0 (
) m 045 0 )(
C W/m 40
°
°
=
=
k
hr
The constants λ1andA1corresponding to this
Biot number are, from Table 4-2,
6227 1 and 2889
λ
Then the Fourier number and the time period become
163 0 )
6227 1 ( 69 0 170 25
170
1
0 ,
−
−
⎯→
⎯
=
−
−
∞
T T
T T
i sph
which is not greater than 0.2 but it is close We may use one-term approximation knowing that the result may be somewhat in error Then the baking time of the potatoes is determined to be
min 39.3
=
=
×
=
/s m 10 4 1
m) 045 0 )(
163 0 (
2 7
2 2
α
τr o
t
(b) The maximum amount of heat transfer is
kJ 237 C ) 25 170 )(
C kJ/kg
900 3 )(
kg 420 0 ( ) (
kg 420 0 m) 045 0 ( 3
4 ) kg/m 1100 ( 3 4
max
3.
3 3
=
°
−
°
=
−
=
=
⎥⎦
⎤
⎢⎣
⎡
=
=
=
p
o
T T mc Q
r
Then the actual amount of heat transfer becomes
kJ 145
=
=
=
=
−
−
=
−
−
=
kJ) 237 )(
610 0 ( 610
0
610 0 )
2889 2 (
) 2889 2 cos(
) 2889 2 ( ) 2889 2 sin(
) 69 0 ( 3 1 ) cos(
) sin(
3 1
max
3 3
1
1 1 1 ,
max
Q Q
Q
Q
sph o
λ
λ λ λ θ
The final equilibrium temperature of the potato after it is wrapped is
C 114°
=
° +
°
= +
=
⎯→
⎯
−
=
) C kJ/kg
9 3 )(
kg 420 0 (
kJ 145 C
25 )
(
p i eqv i
eqv p
mc
Q T T T
T mc
Q
Trang 94-57 The center temperature of potatoes is to be lowered to 6°C 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 τ > 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⋅°C and α = 0.13×10-6 m2/s
Analysis First we find the Biot number:
C W/m 0.5
) m 03 0 ( C) W/m 19
(
Bi
2
=
°
°
=
=
k
hr o
From Table 4-2 we read, for a sphere, λ1 = 1.635
and A1 = 1.302 Substituting these values into the
one-term solution gives
2 25
2 6
1 0
−
−
→
=
−
−
∞
T
T
T
T
i
Potato
Ti = 25°C
Air 2°C
4 m/s
which is greater than 0.2 and thus the one-term solution is applicable Then the cooling time becomes
h 1.45
=
=
×
=
=
⎯→
⎯
s / m 10 0.13
m) 03 0 )(
753 0 (
2
τ α
o
r t r
t
The lowest temperature during cooling will occur on the surface (r/r 0 = 1), and is determined to be
o o
o o i
o o o
o o i
o o
o
r r T
T
T T r r
r r T
T
T r T r
r
r r e
A
T
T
T
r
T
/
) / sin(
= /
) / sin(
) ( /
) / sin(
)
(
1
1 1
1 0 1
1 1
2
λ
λ λ
λ θ λ
λ τ λ
∞
∞
∞
∞
−
∞
∞
−
−
=
−
−
→
=
−
−
1.635
rad) 635 1 sin(
2 25
2 6 2 25
2 ) (
°
⎯→
⎯
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
−
−
o
r T
which is above the temperature range of 3 to 4 °C 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:
17a) -4 (Fig
75 0 t
= 174
0 2 25
2 6
877 0 m) C)(0.03
W/m (19
C W/m
50 0 Bi
1
2 0
o 2 o
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
=
−
−
=
−
−
=
=
=
∞
i
o
r T
T
T
T
hr
k
α τ
×
=
/ m 10 13 0
) 03 0 )(
75 0 (
2 6
2 2
s
r
α τ
The surface temperature is determined from
17b) 4 (Fig
0.6 )
( 1
877 0 1
−
=
−
−
⎪
⎪
⎭
⎪⎪
⎬
⎫
=
=
=
∞
∞
T T
T r T
r
r
r
h
k
Bi
o o
o
which gives T surface=T∞+0.6(T o−T∞)=2+0.6(6−2)=4.4°C
The slight difference between the two results is due to the reading error of the charts
Trang 104-58E The center temperature of oranges is to be lowered to 40°F during cooling The cooling time and if
any part of the oranges will freeze during this cooling process are to be determined
Assumptions 1 The oranges are spherical in shape with a radius of ro =1.25 in = 0.1042 ft 2 Heat
conduction in the orange is one-dimensional in the radial direction because of the symmetry about the
midpoint 3 The thermal properties of the orange are constant 4 The heat transfer coefficient is constant and uniform over the entire surface 5 The Fourier number is τ > 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 oranges are given to be k = 0.26 Btu/h⋅ft⋅°F and α = 1.4×10-6 ft2/s
Analysis First we find the Biot number:
Orange
D = 2.5 in
85% water
Ti = 78°F
Air 25°F
1 ft/s
C Btu/h.ft
0.26
) ft 12 / 25 1 ( F) Btu/h.ft 6 4 ( Bi
2
=
°
°
=
=
k
hr o
From Table 4-2 we read, for a sphere, λ1 = 1.9569 and A1 =
1.447 Substituting these values into the one-term solution gives
426 0
= 447
1 25 78
25 40
1
0
−
−
→
=
−
−
∞
T
T
T
T
i
which is greater than 0.2 and thus the one-term solution is applicable
Then the cooling time becomes
×
=
=
→
s / ft 10 1.4
ft) 12 / 25 1 )(
426 0 (
2 2
τ α
o
r t r
t
The lowest temperature during cooling will occur on the surface (r/r 0 = 1), and is determined to be
o o
o o i
o o o
o o i
o o
o
r r T
T
T T r r
r r T
T
T r T r
r
r r e
A
T
T
T
r
T
/
) / sin(
= /
) / sin(
) ( /
) / sin(
)
(
1
1 1
1 0 1
1 1
2
λ
λ λ
λ θ λ
λ τ λ
∞
∞
∞
∞
−
∞
∞
−
−
=
−
−
→
=
−
−
1.9569
rad) 9569 1 sin(
25 78
25 40 25 78
25 ) (
°
⎯→
⎯
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
−
−
o o
r T r
T
which is above the freezing temperature of 31°C for oranges Therefore, no part of the oranges will freeze during this cooling process
Alternative solution We could also solve this problem using transient temperature charts as follows:
17a) -4 (Fig
43 0 283
0 25 78
25 40
543 0 ft) F)(1.25/12 º
Btu/h.ft (4.6
F Btu/h.ft.º 0.26
1
2 0
2
=
=
⎪
⎪
⎭
⎪
⎪
⎬
⎫
=
−
−
=
−
−
=
=
=
∞
i
o
r t
T
T
T
T
r
h
k
/s ft 10 1.4
5/12ft) (0.43)(1.2
2 6
2 2
=
=
×
=
α
τr o t
The lowest temperature during cooling will occur on the surface (r/r o =1) of the oranges is determined to
be
17b) 4 (Fig
0.45 )
( 1
543 0 1
0
−
=
−
−
⎪
⎪
⎭
⎪⎪
⎬
⎫
=
=
=
∞
∞
T T
T r T
r
r
r
h
k
Bi
o
o
which gives T surface=T∞+0.45(T0−T∞)=25+0.45(40−25)=31.8°F
The slight difference between the two results is due to the reading error of the charts