# nanopdf.com march-14

```College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
March 14 Homework Solutions
4-38
Reconsider problem 4-37 using EES (or other software) investigate the effect of the final
center temperature of the egg on the time it will take for the center to reach this
temperature. Let the temperature vary from 50oC to 95oC. Plot the time versus the
temperature and discuss the results.
Problem 4-37, assigned on last week’s homework is copied here: An ordinary egg can be
approximated as a 5.5-cm-diameter sphere whose properties are roughly k = 0.6 W/m2oC
and  = 0.14x10-6 m2/s. The egg is initially at a uniform temperature of 8oC and is dropped
into boiling water at 97oC. Taking the convection heat transfer coefficient to be h = 1400
W/m2&middot;oC, determine how long it will take for the center of the egg to reach 70oC.
Last week’s solutions showed how the problem can be solved using the charts. To compute
several times, we will use the approximate solution approach. At the end we will check to make
sure that the dimensionless time,  = t/ro2 is always greater than 0.2.
In order to use the approximate solution we have to know the values of A1 and 1 corresponding
to the Biot number computed below, where the sphere radius = (5.5cm)/2 = 0.0275 m.
Bi sphere 
 mo C 
hro 1400 W
  64.17
 2 o 0.0275 m
k
m C
 0.6 W 
Interpolating in Table 4-2 for the sphere we find that A1 = 3.0863 and 1 = 1.9969 for Bi = 64.17.
We use equation (4-28) on page 230 of the text for the center temperature of the sphere.
0 
2
T0  T
 A1e 1
Ti  T
 
t
r02

 0 
r02   0 



ln

t


ln 
12  A1 
12  A1 
1
We want to set this equation so that we can solve for t as a function of T 0. We can substitute the
known data to get a working equation as follows.
2
 1
 0 
T0  97 o C 

0.0275 m 


t   2 ln 
ln 
3.0863 8o C  97 o C 
1  A1 
0.14 x10 6 m 2
2

1.9969
s
 274.68o C 
1 min  274.68o C 
1 min  274.68o C 









 1354.6 s 
ln 

1354
.
6
s
ln

22
.
577
min
ln
o
60 s  T0  97 o C 
60 s  T0  97 o C 
 T0  97 C 
r02


The final equation gives the time in minutes when the desired center temperature is entered in
degrees Celsius. This formula is evaluated for temperatures between 50oC and 95oC and the
results are plotted using an Excel spreadsheet. The results are shown below. The table shows
that all values of t are greater than 0.2 an ex post facto demonstration that the use of the oneterm approximate solution is valid.
As expected the time increases as the desired temperature increases, however the computed
times seem quite long for cooking an egg. The assumptions used in this problem are
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
March 14 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 2
undoubtedly invalid. Perhaps the property data for the egg or the heat transfer coefficient are
incorrect.
50
55
60
65
70
75
80
85
90
39.9
42.4
45.3
48.5
52.4
57.0
62.8
70.7
82.9

0.4427
0.4709
0.5027
0.5391
0.5817
0.6331
0.6978
0.7851
0.9203
Time Required to Heat Egg
90
80
70
.
t (min)
60
Time (min)
T0 (oC)
50
40
30
20
10
0
50
55
60
65
70
75
80
85
90
95
Center Temperature (oC)

4-43E Long cylindrical AISI stainless steel
rods (k = 7.74 Btu/h&middot;ft&middot;oF and  = 0.135
ft2/h) of 4-in diameter are heat treated
by drawing them at a velocity of 7
ft/min through a 21-ft-long oven
maintained at 1700oF. The heat
transfer coefficient in the oven is 20
Btu/h&middot;ft2&middot;oF. If the rods enter at 70oF,
determine their centerline temperature
when they leave.
Assuming that the heat transfer is in the
radial direction only, we can use equation
(4.27) on page 230 to find the centerline
temperature. First we find the Fourier
number to make sure that we can use the approximate solution. Since the rods proceed at a
velocity of 7 ft/min through the 21 ft oven, their transit time is 3 min = 0.05 h. With this time, the
data given for  and the radius = 2 in, the Fourier number, , can be found.
0.135 ft 2
0.05 h
t
h
 2 
 0.243
2
r0

1 ft 
2 in 

12 in 

Next we have to find the Biot number so that we can find the correct values of A1 and 1.
March 14 homework solutions
Bi 
ME 375, L. S. Caretto, Spring 2007
Page 3
hro
20 Btu
1 ft  h  fto F 

  0.4307



2
in
k
12 in  7.74 Btu 
h  ft 2 o F
For this value of Biot number, we find the value of A1 = 1.0996 and 1 = 0.8790 from the data for
the infinite cylinder in Table 4-2. Now we can apply the approximate solution in equation (4-27)
0 
2
2
T0  T
 A1e 1  1.0995e 0.8790 0.243  0.911
Ti  T
We can then find the centerline temperature.
0 
T0  T
 0.911  T0  T  Ti  T 0.911  1700 o F  70 o F  1700 o F 0.911
Ti  T


T0 = 215oF
4-45
A long cylindrical wood log (k = 0.17 W/m&middot;oC and  = 1.28x10-7 m2/s) is 10 cm in diameter
and is initially at a uniform temperature of 15oC. It is exposed to hot gases at 550oC in a
fireplace with a heat transfer coefficient of 13.6 W/m2&middot;oC on the surface. If the ignition
temperature of the wood is 420oC, determine how long it will be before the log ignites.
Assuming that the heat transfer is in the radial direction only, we can use equation (4.24) on page
230 to find the surface temperature. Because we are trying to find the time, we cannot start by
finding the Fourier number to make sure that we can use the approximate solution. However, we
can find this as part of the solution and then make sure that our use of the approximate solution is
justified.
We first find the Biot number to determine the correct values of A1 and 1.
Bi cylinder
 mo C 
hro 13.6 W
  4.00

 2 o 0.05 m
k
m C
 0.17 W 
For this value of Biot number, we find the value of A1 = 1.4698 and 1 = 1.9081 from the data for
the infinite cylinder in Table 4-2. Now we can apply the approximate solution in equation (4-24)
where we find the value of the Bessel function by interpolation in Table 4-3 on page 231. (This
interpolation gives almost the same result as using the besselj(x,n) function in Excel since there is
only a small change from the tabulated values due to the interpolation.) For the surface of the
cylinder we set r/r0 = 1.

2
2
2
 r
T (r , t )  T
 A1e 1 J 0  1   1.4698e1.9081  J 0 1.90811  1.4698e1.9081  0.27711
Ti  T
 r0 
Solving this equation for the Fourier number,, when T(r = r0) = 420oC gives.
2
T (r , t )  T 420o C  550o C

 0.243  1.4698e 1.9081  0.27711
o
o
Ti  T
15 C  550 C
 


1
0.243
ln 
 0.142
2
1.9081 1.46980.27711
This means that all our calculations have been in vain! The chart for finding temperatures at
locations other than the center of the body is based on the same approximate solution process
March 14 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 4
that we tried to use here. Since we do have a Fourier number, we can see what time it is
equivalent to, even though we know that this will not be an accurate result.
r02 0.05 m  0.142
 2770 s
 0.128 x10 6 m 2
s
2
t
Converting to minutes gives t = 46.2 min .
Solving the exact differential equation for this case (which we have not covered in class) gives a
solution of  = 0.152; so applying the approximate solution at this point causes an error of about
7%.
4-72
A thick wood slab (k = 0.17 W/m&middot;oC and  = 1.28x10-7 m2/s) that is initially at a uniform
temperature of 25oC. It is exposed to hot gases at 550oC for a period of 5 minutes. The
heat transfer coefficient between the gases and the wood slab is 35 W/m2&middot;oC on the
surface. If the ignition temperature of the wood is 450oC, determine if the wood will ignite.
We are not given the dimension of the wood slab, but we are told that it is thick. Let’s assume
that we may model the surface of this “thick” slab as a semi-infinite surface. For such a surface
the following equation describes the temperature at any time and x location when the wood is
introduced into a convection environment at t = 0.
T  Ti
 x
 erfc
T  Ti
 4t
 hx h 2t 


k 2 
  k 
e

 x
h t 

erfc


k
4

t


In this case we want to find the surface temperature so we have x = 0, and the problem is
simplified to
 h 2t 

k 2 


T  Ti
 erfc0  e
T  Ti
 h 2t 

k 2 

 h t 

  1  e
erfc

 k 
 h t 

erfc

 k 
In the above equation we use the value of erfc(0) = 1. The terms in this equation are evaluated
below.
h t 35 W mo C 1.28 x107 m2
5 min  60 s  1.276
 2o
k
m  C 0.17 W
s
min
h2t
 1.2762  1.628
k2
With these values and Table 4-4 to find erfc(1.276), we can now find the dimensionless
temperature.
T  Ti
 1  e1.628erfc1.276  1  5.0940.0712  0.637
T  Ti
We can now find the surface temperature at the end of 5 minutes.
T  Ti
T  25o C

 0.637 T  25o C  525o C 0.637   360oC
o
o
T  Ti 550 C  25 C


This is less than the ignition temperature so the wood will not ignite .
March 14 homework solutions
4-83
ME 375, L. S. Caretto, Spring 2007
Page 5
A 5-cm-high rectangular ice block (k = 2.22 W/m&middot;oC
and  = 0.124 x 10-7 m2/s) initially at -20oC is placed
on table on its square base 4 cm by 4 cm in size in
a room at 18oC. The heat transfer coefficient on
the exposed surfaces of the ice block is 12
W/m2&middot;oC. Disregarding any heat transfer from the
base to the table, determine how long it will be
before the ice block starts melting. Where on the
ice block will the first liquid droplets appear?
We can find the solution to this problem as the solution
to three separate one-dimensional infinite slabs. The
first slab, which is parallel to the table, will have a half
length, L = 5 cm, the height of the block. This is valid
because we are told to assume that there is no heat flow through the table. A condition of no
heat flow is the same as a condition of zero temperature gradient, which is the condition at the
centerline of our infinite slab. The other two solutions will be for the sides of the ice block. Both
of these solutions will be for the same dimensions – a slab with a half-length of 2 cm. For the
three-dimensional solution, the dimensionless temperature at any point (x,y,z) in the slab, at any
time t, is given by the product of three one-dimensional solutions.
x, y, z, t   1x, t 1 y, t 1z, t 
We will assume that the Fourier number, , is greater than 0.2 so that we can use the
approximate solution and check this assumption after we solve for . If we denote the coordinate
distance as xi and the half-length as Li for each of the coordinate directions, the one-dimensional
solution for an infinite slab is.
1 xi , t   A1e  1 cos 1  A1e
2

21t
L2i
 x 
cos 1 i 
 Li 
We expect that the first drops of liquid to form will be at the upper corners of the ice block where
the exposure to air is the greatest and the overall distance from the center of the cold ice is the
greatest. For this corner, the dimensionless coordinates, xi/Li, in the above equation, are 1 for all
three directions.
We first have to find the Biot number to determine the values of A1 and 1. In general we would
have three such calculations for a three-dimensional problem, but here we have only two
calculations since two of the dimensions are the same. For the vertical distance of 5 cm = 0.05
m, the Biot number is.
Bi slab 
 mo C
hL 12 W
 2 o 0.05 m 
 2.22 W
k
m  C


  0.2703


Interpolating in Table 4-2 we find that A1 = 1.0408 and 1 = 0.4951 for this Biot number. Both of
the horizontal solutions will have the same Biot number for their half thickness of 2 cm = 0.02 m.
Bi slab 
 mo C
hL 12 W
 2 o 0.02 m 
 2.22 W
k
m  C


  0.1091


Interpolating in Table 4-2 we find that A1 = 1.0713 and 1 = 0.3208 for this Biot number.
Since two of our one-dimensional solutions are the same, we can write our three-dimensional
product solution as follows.
March 14 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 6
x, y, z, t   12 x, t 1z, t 
The value of  = t/Li2, will be different for each dimension because the length parameter will be
different for each dimension. However, the value of time, t, will be the same. So we have to write
the equation in terms of the time, t. Substituting the one dimensional solutions, with the
appropriate values of A1, 1, , and the length parameter for each dimension and setting the
dimensionless distance in the cosine term to 1 for each solution gives.
2
T  T
Ti  T
0.3208 2 0.124 x10  7 m 2 t





0 C  18 C
s

0.02 m 2




0
.
4737

1
.
0713
e
cos
0
.
3208

 
 20o C  18o C




2
7 2
0.4951 0.124 x10 m t





s

0.02 m 2
cos0.4951
1.0408e




o
o
We can not solve this equation explicitly for t, but we can solve it by an iteration algorithm or
using a calculator or a program such as Matlab or Excel. Doing this we find that t = 77,500 s =
21.5 h. We have to check the value of t to make sure that it is greater than 0.2. We only have to
check the value for a length parameter of 0.05 m. If t &gt; 2 for this length, then it will also be &gt; 0.2
for the length parameter of 0.02 m.

t 0.124 x10 7 m 2 77,500 s 

 0.384  0.2
s
L2
0.05 m 2
So the criterion that t &gt; 0.2 is satisfied and our approach is justified; the answer is t = 21.5 h .
4-86
Consider a cubic block whose sides are 5 cm long and a cylindrical block whose height
and diameter are also 5 cm. Both blocks are initially at 20oC and are made of granite (k =
2.5 W/m&middot;oC and  = 1.15x10-6 m2/s). Now both blocks are exposed to hot gases at 500oC in
a furnace on all of their surfaces with a heat transfer coefficient of 40 W/m2&middot;oC. Determine
the center temperature of each geometry after 10, 20, and 60 min.
For the cube we have three product solutions, but they are all identical because the lengths are
identical and the location at which we want to evaluate the temperature is identical for each
coordinate. Thus we can write
x, y, z, t  
T  T
 1 x, t 1  y, t 1 z, t   1 xi , t 3
Ti  T
For the minimum time of 10 min = 600 s, we can find the Fourier number, , = t/L2, where L is
the half thickness. For this problem, L = (5 cm)/2 = 2.5 cm = 0.025 m, and the Fourier number is

t 1.15 x10 6 m 2 600 s 

1.104  0.2
s
L2
0.025 m 2
Since  &gt; 0.2, we can apply the approximate solution
1xi , t   A1e
 21
cos 1  A1e

21t
L2i
 x 
cos 1 i 
 Li 
March 14 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 7
We can find the values of A1 and 1 from the Biot number.
Bi slab 
 mo C
hL 40 W
 2 o 0.025 m 
 2.5 W
k
m  C


  0.4


For Bi = 0.4 we find that A1 = 1.0580 and 1 = 0.5932 from Table 4-2. We can now compute the
center temperature of the cube where x = y = z = 0. (Recall that cos(0) = 1.)
3
  21t

  21t 
2
2 
 x 

T  T 
x, y, z, t  
  A1e Li cos 1 i    A1e Li 
Ti  T
 Li 







3
The argument of the exponential is
21t

2
L
0.5932 2
1.15 x10 6 m 2 600 s 
0.3885
s
0.025 m 2
Substituting this value and the value for A1 into our equation for  gives the temperature as
3
  21t 
2 
3

T  T  Ti  T  A1e Li   500o C  25o C  500o C 1.050e0.3885  323o C







At t = 20 minutes = 1200 s and 60 minutes = 3600 s, -12t/L2 = 0.7770 and 2.331, respectively.
Applying the same equation gives T = 445oC and 500oC at these times.
The cylinder can be solved as the product of the solution for the infinite cylinder and that of the
infinite slab. Since the cylinder radius is the same as the half width of the cube solved above, the
Fourier number will be the same and all times will have a Fourier number greater than 0.2
allowing the use of the approximate solution. The temperature will be the product of two
solutions.
x, r , t  
2

 x 
T  T
 slab r , t cyl r , t    A1e1 cos 1 i 
Ti  T
 Li 

 2   r 
 A1e 1 J 0  1 
 r0  c
s
The subscripts s for slab and c for cylinder indicate that the values of A1 and 1 will be different for
the different solutions. The Biot numbers will be the same for both solutions because the halflength of the cylinder height is the same as its radius.
Bi slab 
 mo C
hL
hr
40 W
 Bicyl  o  2 o 0.025 m 
 2.5 W
k
k
m  C


  0.4


For Bi = 0.4 Table 4-2 shows that that A1 = 1.0580 and 1 = 0.5932 for the slab and A1 = 1.0931
and 1 = 0.8516 for the cylinder. We can now compute the center temperature of the cylinder
where x = r = 0. (Note that cos(0) = J0(0) = 1.)
2
2
T  T  21  0   21  0 
  A1e
cos 1   A1e
J 0  1    A1e1   A1e1 
 s 
 c
Ti  T 
 L  s 
 r0  c 
Substituting numerical values for t = 10 min = 600 s, where we have previously computed  =
1.104 gives.
March 14 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 8
2
2
T  T   21    21 
 A1e
A1e
 1.0580e0.5932  1.104   1.0931e 0.8516  1.104    0.352
 

 s 
 c 
Ti  T 
We can then find the temperature.



T  T  Ti  T   500o C  25o C  500o C 1.050e0.3885  331o C
3
At t = 20 minutes = 1200 s and 60 minutes = 3600 s,  = 2.208 and 6.624, respectively. Applying
the same equations give T = 449oC and 500oC at these times.
The answers to this problem are summarized in the table below.
Geometry
Cube
Cylinder
Temperatures at times shown below
10 min
20 min
60 min
323oC
445oC
500oC
o
o
331 C
449 C
500oC
At the large values of Fourier number in this problem, t = 1.1, 2.2, and 6.6, the temperature
approaches and then equals the value of T.
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