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# ch05

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```5°C. The IHT code is provided in Comment 3, and the results are
tabulated below.
Solution method Time-to-doneness, t (s) Ti ( C)
Patty thickness, 2L (mm)
10 20 30
FEHT 66.2 264.5 591 20
IHT 67.7
80.2
264.5
312.2
590.4
699.1
20
5
Eq. 5.40 (see
Comment 4)
x x
x x
5
20
Considering the IHT results for Ti = 20°C, note that when the thickness is doubled from 10 to 20 mm,
td is (264.5/67.7=) 3.9 times larger. When the thickness is trebled, from 10 to 30 mm, td is
(590.4/67.7=) 8.7 times larger. We conclude that, td is nearly proportional to L2, rather than linearly
proportional to thickness.
Continued \u2026..
PROBLEM 5.119 (Cont.)
(b) The temperature span for the cooking process ranges from T\u221e = 100 to Ti = 20 or 5°C. The
differences are (100-20 =) 80 or (100-5 =) 95°C. If td is proportional to the overall temperature span,
then we expect td for the cases with Ti = 5°C to be a factor of (95/80 =) 1.19 higher (approximately
20% ) than with Ti = 20°C. From the tabulated results above, for the thickness 2L = 10, 20 and 30
mm, the td with Ti = 5°C are (80.2/67.7 = ) 1.18, (312 / 264.5 =) 1.18, and (699.1/590.4 =) 1.18,
respectively, higher than with Ti = 20°C. We conclude that td is nearly proportional to the temperature
span (T\u221e - Ti).
COMMENTS: (1) The results from the FEHT and IHT calculations are in excellent agreement. For
this analysis, the FEHT model is more convenient to use as it provides direct calculations of the time-
to-doneness. The FEHT tool allows the user to watch the cooking process. Use the
View\uf8e6Temperature Contours command, click on the from start-to-stop button, and observe how color
band changes represent the temperature distribution as a function of time.
(2) It is good practice to check software tool analyses against hand calculations. Besides providing
experience with the basic equations, you can check whether the tool was used or functioned properly.
Using the one-term series solution, Eq. 5.40:
( ) ( )2do 1
i
T 0, t T
C exp Fo
T T
\u3b8 \u3b6\u2217 \u221e
\u221e
\u2212
= = \u2212
\u2212
( )2d 1Fo t / L C , Bi , Table 5.1\u3b1 \u3b6= =
Ti (°C) 2L (mm) o\u3b8\u2217 Bi C1 1\u3b6 Fo td (s)
20 10 0.5000 24.47 1.2707 1.5068 0.4108 70.0
5 30 0.4211 73.41 1.2729 1.5471 0.4622 709
The results are slightly higher than those from the IHT model, which is based upon a multiple- rather
than single-term series solution.
(3) The IHT code used to obtain the tabulated results is shown below. Note that T_xt_trans is an
intrinsic heat transfer function dropped into the Workspace from the Models window (see IHT
Help|Solver|Intrinsic Functions|Heat Transfer Functions).
// Models | Transient Conduction | Plane Wall
/* Model: Plane wall of thickness 2L, initially with a uniform temperature T(x,0) = Ti, suddenly subjected
to convection conditions (Tinf,h). */
// The temperature distribution is
T_xt = T_xt_trans(&quot;Plane Wall&quot;,xstar,Fo,Bi,Ti,Tinf) // Eq 5.39
// The dimensionless parameters are
xstar = x / L
Bi = h * L / k // Eq 5.9
Fo= alpha * t / L^2 // Eq 5.33
alpha = k/ (rho * cp)
// Input parameters
x = 0 // Center point of meat
L = 0.005 // Meat half-thickness, m
//L = 0.010
//L = 0.015
T_xt = 60 // Doneness temperature requirement at center, x = 0; C
Ti = 20 // Initial uniform temperature
//Ti = 5
rho = 1000 // Water properties at 300 K
cp = 4179
k = 0.613
h = 5000 // Convection boundary conditions
Tinf = 100
PROBLEM 5.120
KNOWN: A process mixture at 200°C flows at a rate of 207 kg/min onto a 1-m wide conveyor belt
traveling with a velocity of 36 m/min. The underside of the belt is cooled by a water spray.
FIND: The surface temperature of the mixture at the end of the conveyor belt, Te,s, using (a) IHT for
writing and solving the FDEs, and (b) FEHT. Validate your numerical codes against an appropriate
analytical method of solution.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction at any z-location, (2)
Negligible heat transfer from mixture upper surface to ambient air, and (3) Constant properties.
PROPERTIES: Process mixture (m), \u3c1m = 960 kg/m3, cm = 1700 J/kg\u22c5K, and km = 1.5 W/m\u22c5K;
Conveyor belt (b), \u3c1b = 8000 kg/ m3, cb = 460 J/kg\u22c5K, and kb = 15 W/m\u22c5K.
ANALYSIS: From the conservation of mass requirement, the thickness of the mixture on the
conveyor belt can be determined.
m c c mm A V where A W L\u3c1= =\ufffd
3
m207 kg / min 1 min/ 60s 960 kg / m 1m L 36 m / min 1min/ 60s× = × × × ×
mL 0.0060 m 6 mm= =
The time that the mixture is in contact with the steel conveyor belt, referred to as the residence time, is
( )res ct L / V 30 m / 36 m / min 1 min/ 60 s 50 s= = × =
The composite system comprised of the belt, Lb = 3 mm, and mixture, Lm = 6 mm, as represented in
the schematic above, is initially at a uniform temperature T(x,0) = Ti = 200°C while at location z = 0,
and suddenly is exposed to convection cooling (T\u221e, h). We will calculate the mixture upper surface
temperature after 50 s, T(0, tres) = Te,s .
(a) The nodal arrangement for the composite system is shown in the schematic below. The IHT model
builder Tools|Finite-Difference Equations|Transient can be used to obtain the FDEs for nodes 01-12
and 14-19.
Continued \u2026..
PROBLEM 5.120 (Cont.)
For the mixture-belt interface node 13, the FDE for the implicit method is derived from an energy
balance on the control volume about the node as shown above.
in out stE E E\u2032\u2032 \u2032\u2032 \u2032\u2032\u2212 =\ufffd \ufffd \ufffd
a b st,m st,bq q E E\u2032\u2032 \u2032\u2032 \u2032\u2032 \u2032\u2032+ = +\ufffd \ufffd
( )( )
p 1 p 1 p 1 p 1 p 1 p
12 13 14 13 13 13
m b m m b b
T T T T T T
k k c c x / 2
x x t
\u3c1 \u3c1
+ + + + +
\u2212 \u2212 \u2212
+ = + \u2206
\u2206 \u2206 \u2206
IHT code representing selected FDEs, nodes 01, 02, 13 and 19, is shown in Comment 4 below (\u2206x =
0.5 mm, \u2206t = 0.1 s). Note how the FDE for node 13 derived above is written in the Workspace. From
the analysis, find
Te,s = T(0, 50s) = 54.8°C <
(b) Using FEHT, the composite system is drawn and the material properties, boundary conditions, and
initial temperature are specified. The screen representing the system is shown below in Comment 5
with annotations on key features. From the analysis, find
Te,s = T(0, 50s) = 54.7°C <
COMMENTS: (1) Both numerical methods, IHT and FEHT, yielded the same result, 55°C. For the
safety of plant personnel working in the area of the conveyor exit, the mixture exit temperature should
be lower, like 43°C.
(2) By giving both regions of the composite the same properties, the analytical solution for the plane
wall with convection, Section 5.5, Eq. 5.40, can be used to validate the IHT and FEHT codes. Using
the IHT Models|Transient Conduction|Plane Wall for a 9-mm thickness wall with mixture
thermophysical properties, we calculated the temperatures after 50 s for three locations: T(0, 50s) =
91.4°C; T(6 mm, 50s) = 63.6°C; and T(3 mm, 50s) = 91.4°C. The results from the IHT and FEHT
codes agreed exactly.
(3) In view of the high heat removal rate on the belt lower surface, it is reasonable to assume that
negligible heat loss is occurring by convection on the top surface of the mixture.
Continued \u2026..
PROBLEM 5.120 (Cont.)
(4) The IHT code representing selected FDEs, nodes 01, 02, 13 and 19, is shown below. The FDE for
node 13 was derived from an energy balance, while the others are written from the Tools pad.
// Finite difference equations from Tools, Nodes 01 -12 (mixture) and 14-19 (belt)
/* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */
rhom * cm * der(T01,t) = fd_1d_sur_w(T01,T02,km,qdot,deltax,Tinf01,h01,q''a01)
q''a01 = 0 // Applied heat flux, W/m^2; zero flux shown
qdot = 0
Tinf01 = 20 // Arbitrary value
h01 = 1e-6 // Causes boundary to behave as adiabatic
/* Node 02: interior node; e and w labeled 03 and 01. */
rhom*cm*der(T02,t)```