# HW4_sols

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```							4-142 A watermelon is placed into a lake to cool it. The heat transfer coefficient at the surface of the watermelon
and the temperature of the outer surface of the watermelon are to be determined.
Assumptions 1 The watermelon is a homogeneous spherical object. 2 Heat conduction in the watermelon is one-
dimensional because of symmetry about the midpoint. 3 The thermal properties of the watermelon are constant. 4
The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is  > 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 watermelon are given to be k = 0.618 W/m.C,  = 0.1510-6 m2/s,  = 995 kg/m3
and      cp = 4.18 kJ/kg.C.
Analysis The Fourier number is
t        (0.15 10 6 m 2 /s) (4  60  40 min)  60 s/min                                 Lake
                                                                       0.252
ro2                              (0.10 m)   2                                                 15C
Water
which is greater than 0.2. Then the one-term solution can be written                                                              melon
in the form                                                                                                                      Ti = 35C
T0  T                20  15
 A1e 1             0.25  A1e 1 (0.252)
2                             2
 0,sph                        
Ti  T                35  15
It is determined from Table 4-2 by trial and error that this equation is satisfied when Bi = 10, which corresponds to
1  2.8363 and A1  1.9249 . Then the heat transfer coefficient can be determined from
hro      kBi (0.618 W/m.C)(10)
Bi        
 h                        61.8 W/m 2 .C
k        ro      (0.10 m)
The temperature at the surface of the watermelon is
T (ro , t )  T              sin( 1 ro / ro )                                sin( 2.8363 rad)
 A1 e 1                     (1.9249)e ( 2.8363) (0.252)
2                                          2
 (ro , t ) sph 
Ti  T                       1 ro / ro                                         2.8363
T (ro , t )  15

 0.0269  T (ro , t )  15.5C
35  15

4-150 Copper balls ( = 8933 kg/m3, k = 401 W/mC, cp = 385 J/kgC,  = 1.16610-4 m2/s) initially at 180°C are
allowed to cool in air at 30C for a period of 2 minutes. If the balls have a diameter of 2 cm and the heat transfer
coefficient is 80 W/m2C, the center temperature of the balls at the end of cooling is
(a) 78C               (b) 95C                (c) 118C            (d) 134C             (e) 151C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.

D=0.02 [m]
Cp=385 [J/kg-K]
rho= 8933 [kg/m^3]
k=401 [W/m-K]
V=pi*D^3/6
A=pi*D^2
m=rho*V
h=80 [W/m^2-C]
Ti=180 [C]
Tinf=30 [C]
b=h*A/(rho*V*Cp)
time=2*60 [s]
Bi=h*(V/A)/k

"Lumped system analysis is applicable. Applying the lumped system analysis equation:"
(T-Tinf)/(Ti-Tinf)=exp(-b*time)

“Some Wrong Solutions with Common Mistakes:”
(W1_T-0)/(Ti-0)=exp(-b*time) “Tinf is ignored”
(-W2_T+Tinf)/(Ti-Tinf)=exp(-b*time) “Sign error”
(W3_T-Ti)/(Tinf-Ti)=exp(-b*time) “Switching Ti and Tinf”
(W4_T-Tinf)/(Ti-Tinf)=exp(-b*time/60) “Using minutes instead of seconds”
\

4-162 Polyvinylchloride automotive body panels (k = 0.092 W/mK, cp = 1.05 kJ/kgK,  = 1714 kg/m3), 1-mm
thick, emerge from an injection molder at 120 oC. They need to be cooled to 40oC by exposing both sides of the
panels to 20oC air before they can be handled. If the convective heat transfer coefficient is 15 W/m2K and radiation
is not considered, the time that the panels must be exposed to air before they can be handled is
(a) 0.8 min        (b) 1.6 min        (c) 2.4 min      (d) 3.1 min        (e) 5.6 min

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.

T=40 [C]
Ti=120 [C]
Ta=20 [C]
r=1714 [kg/m^3]
k=0.092 [W/m-K]
c=1050 [J/kg-K]
h=15 [W/m^2-K]
L=0.001 [m]
Lc=L/2
b=h/(r*c*Lc)
(T-Ta)/(Ti-Ta)=exp(-b*time)

4-162 Polyvinylchloride automotive body panels (k = 0.092 W/mK, cp = 1.05 kJ/kgK,  = 1714 kg/m3), 1-mm
thick, emerge from an injection molder at 120 oC. They need to be cooled to 40oC by exposing both sides of the
panels to 20oC air before they can be handled. If the convective heat transfer coefficient is 15 W/m2K and radiation
is not considered, the time that the panels must be exposed to air before they can be handled is
(a) 0.8 min        (b) 1.6 min        (c) 2.4 min      (d) 3.1 min        (e) 5.6 min

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.

T=40 [C]
Ti=120 [C]
Ta=20 [C]
r=1714 [kg/m^3]
k=0.092 [W/m-K]
c=1050 [J/kg-K]
h=15 [W/m^2-K]
L=0.001 [m]
Lc=L/2
b=h/(r*c*Lc)
(T-Ta)/(Ti-Ta)=exp(-b*time)

5-19 A plane wall with no heat generation is subjected to specified temperature at the left (node 0) and heat flux at
the right boundary (node 8). The finite difference formulation of the boundary nodes and the finite difference
formulation for the rate of heat transfer at the left boundary are to be determined.
Assumptions 1 Heat transfer through the wall is given to be steady, and the thermal conductivity to be constant. 2
Heat transfer is one-dimensional since the plate is large relative to its thickness. 3 There is no heat generation in the
medium.
Analysis Using the energy balance approach and taking the direction of
all heat transfers to be towards the node under consideration, the finite
difference formulations become
Left boundary node:         T0  40                                                           No heat generation

T  T8                T  T8                        40C                                 3000 W/m2
Right boundary node:         kA 7      q0 A  0 or k 7
                    3000  0                           x
x                    x
Heat transfer at left surface:                                                                    
T T                                         0 1 2 3 4 5 6 7 8

Qleft surface  kA 1 0  0
x

5-24 A uranium plate is subjected to insulation on one side and convection on the other. The finite difference
formulation of this problem is to be obtained, and the nodal temperatures under steady conditions are to be
determined.
Assumptions 1 Heat transfer through the wall is steady since there is no indication of change with time. 2 Heat
transfer is one-dimensional since the plate is large relative to its thickness. 3 Thermal conductivity is constant. 4
Properties The thermal conductivity is given to be k = 34 W/m°C.
Analysis The number of nodes is specified to be M = 6. Then the nodal spacing x becomes
L      0.05 m
x                     0.01 m
M 1        6 -1
This problem involves 6 unknown nodal temperatures, and                                                        
e
thus we need to have 6 equations to determine them uniquely.
Node 0 is on insulated boundary, and thus we can treat it as                      Insulated
x
an interior note by using the mirror image concept. Nodes 1,                                                                 h, T
2, 3, and 4 are interior nodes, and thus for them we can use                                                        
0 1         2
the general finite difference relation expressed as                                                                3   4 5
Tm1  2Tm  Tm1 em  
     0 , for m = 0, 1, 2, 3, and 4
x 2
k
Finally, the finite difference equation for node 5 on the right surface subjected to convection is obtained by applying
an energy balance on the half volume element about node 5 and taking the direction of all heat transfers to be
towards the node under consideration:
T1  2T0  T1        
e
Node 0 (Left surface - insulated) :                            0
x  2           k
T0  2T1  T2
e
Node 1 (interior):                                    0

x     2    k
T1  2T2  T3 e   
Node 2 (interior):                               0
x  2        k
T2  2T3  T4 e    
Node 3 (interior):                                0
x  2        k
T3  2T4  T5 e    
Node 4 (interior):                                0
x 2        k
T  T5
Node 5 (right surface - convection : h(T  T5 )  k 4
)                           e(x / 2)  0

x
where
x  0.01 m, e  6 105 W/m 3 , k  34 W/m  C, h  60 W/m 2  C, and T  30C.

This system of 6 equations with six unknown temperatures constitute the finite difference formulation of the
problem.
(b) The 6 nodal temperatures under steady conditions are determined by solving the 6 equations above
simultaneously with an equation solver to be
T0 = 552.1C,       T1 = 551.2C, T2 = 548.5C, T3 = 544.1C, T4 = 537.9C, and T5 = 530.0C
Discussion This problem can be solved analytically by solving the differential equation as described in Chap. 2, and
the analytical (exact) solution can be used to check the accuracy of the numerical solution above.

5-140 A hot brass plate is having its upper surface cooled by impinging jet while its lower surface is insulated. The
explicit finite difference equations, the maximum allowable value of the time step, and the temperature at the center
plane of the brass plate after 1 minute of cooling are to be determined.
Assumptions 1 Transient heat conduction is one-dimensional. 2 Thermal properties are constant. 3 Convection heat
transfer coefficient is uniform. 4 Heat transfer by radiation is negligible. 5 There is no heat generation.
Properties The properties of the brass plate are given as  = 8530 kg/m3, cp = 380 J/kg∙K, k = 110 W/m∙K, and  =
33.9  10−6 m2/s.
Analysis The nodal spacing is given to be x = 2.5 cm. Then the
number of nodes becomes M = L/x +1 = 10/2.5 + 1 = 5. This
problem involves 5 unknown nodal temperatures, and thus we
need to have 5 equations. The finite difference equation for node
0 on the top surface subjected to convection is obtained by
applying an energy balance on the half volume element about
node 0 and taking the direction of all heat transfers to be towards
the node under consideration:
T1i  T0i    x T0i 1  T0i
h(T  T0i )  k                cp
x        2       t
            hx  i      i       hx 
or       T0i 1  1  2  2     T0    2T1  2    T 
             k                   k    
Node 4 is on insulated boundary, and thus we can treat it as an interior node by using the mirror image concept.
Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation
expressed as
Tm1  Tm
i      i
Tm1  2Tm  Tm1 
i       i    i

or       Tm1   (Tm1  Tm1 )  (1  2 )Tm
i         i      i                 i

Thus, the explicit finite difference equations are
            hx  i      i       hx 
Node 0: T0i 1  1  2  2     T0    2T1  2    T 
             k                   k    
Node 1: T1i 1   (T0i  T2i )  (1  2 )T1i

Node 2: T2i 1   (T1i  T3i )  (1  2 )T2i

Node 3: T3i 1   (T2i  T4i )  (1  2 )T3i

Node 4: T4i 1   (T3i  T3i )  (1  2 )T4i
where
x = 2.5 cm, k = 110 W/m∙K, h = 220 W/m2∙K, T∞ = 15°C,  = 33.9  10−6 m2/s, and hx/k = 0.05.
(b) The upper limit of the time step t is determined from the stability criterion that requires all primary coefficients
to be greater than or equal to zero. The coefficient of T0i is smaller in this case, and thus the stability criterion for
this problem can be expressed as
hx                           1                            x 2
1  2  2       0      →                           →   t 
k                      2(1  hx / k )              2 (1  hx / k )

Since   t / x 2 . Substituting the given quantities, the maximum allowable value of the time step is determined
to be
(0.025 m) 2
t                                                                              8.779 s
2(33.9  10 6 m 2 / s)[1  (220 W/m 2  K )(0.025 m) /(110 W/m  K )]
Therefore, any time step less than 8.779 s can be used to solve this problem. For convenience, let us choose the time
step to be ∆t = 6 s. Then the mesh Fourier number becomes
t           (33.9  10 6 m 2 / s)(6 s)
                                               0.32544           (for ∆t = 6 s)
x 2                 (0.025 m) 2
(c) With the initial nodal temperatures of 650°C, the results obtained from successive iterations are listed in the
following table:

Nodal temperature,°C
Time             Time,
step, i          s             T0i          T1i               T2i            T3i          T4i
0                0             650          650               650            650         650
1                6             629.3        650               650            650         650
2                12            622.8        643.3             650            650         650
3                18            616.3        638.8             647.8          650         650
4                24            611.4        634.4             645.6          649.3       650
5                30            607.0        630.6             643.2          648.3       649.5
6                36            603.1        627.0             640.7          647.0       648.7
7                42            599.5        623.7             638.3          645.5       647.6
8                48            596.2        620.6             635.9          643.9       646.3
9                54            593.2        617.6             633.5          642.1       644.7
10               60            590.3        614.8             631.1          640.1       643.0

The temperature at the center plane of the brass plate after 1 minute of cooling is
T2  T (0.05 m, 60 s)  631.1C
10

(d) From Chapter 4, the approximate analytical solution is given as
T ( x, t )  T
 A1e 1 cos(1 x / L)
2
 wall 
Ti  T
where
hL (220 W/m2  K)(0.10 m)
Bi                              0.2
k      110 W/m  K
t         (33.9  10 6 m 2 /s)(60 s)
                                            0.2034  0.2
L2                  (0.10 m) 2
1  0.4328            and       A1  1.0311 (from Table 4-2)
Hence,
T ( x, t )  (Ti  T ) A1e  1 cos(1 x / L)  T
2

T (0.05 m, 180 s)  (650 C  15 C)(1.0311)e (0.4328)                         cos(0.4328)(0.5)  15 C
2
( 0.2034)

 630.6C
Discussion The comparison between the approximate analytical and numerical solutions is within ±0.08%
agreement.
5-142 The unsteady forward-difference heat conduction for a constant area, A, pin fin with perimeter, p, exposed to
air whose temperature is T0 with a convection heat transfer coefficient of h is
k       *            hpx 2            2k        hp  *
Tm1 
*
Tm1  Tm1 
*
T0   1                  Tm
c p x 2                 A         c p x
 
2   c p A 

2k        hp
In order for this equation to produce a stable solution, the quantity                     must be
c p x 2   c p A
(a) Negative         (b) zero               (c) Positive          (d) Greater than 1    (e) Less than 1

```
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