ch05
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ch05


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the Run command, the integration is performed from 0 to 0.1 s with a time step of
1×10-6 s. Then, using the Specify|Generation command, the generation rate is set to zero and the
Run|Continue command is executed. The temperature history is shown below.
(c) Finite-difference method of solution, IHT. The nodal arrangement for the clutch pair is shown
below with \u2206x = 0.1 mm and \u2206t = 1 ms. Nodes 02-10, 13-16 and 18-21 are interior nodes, and their
finite-difference equations (FDE) can be called into the Workspace using Tools|Finite Difference
Equations|One-Dimenisonal|Transient. Nodes 01 and 22 represent the mid-planes for the reaction and
composite plates, respectively, with adiabatic boundaries. The FDE for node 17 is derived from an
energy balance on its control volume (CV) considering different properties in each half of the CV.
The FDE for node 11 and 12 are likewise derived using energy balances on their CVs. At the
interface, the following conditions must be satisfied
11 12 f rp cpT T q q q\u2032\u2032 \u2032\u2032 \u2032\u2032= = +
The frictional heat flux is represented by a Lookup Table, which along with the FDEs, are shown in
the IHT code listed in Comment 2.
Continued (3)...
PROBLEM 5.118 (Cont.)
The temperature and heat flux histories are plotted below. The steady-state temperature was found as
156.5° C, which is in reasonable agreement with the energy balance result from part (a).
COMMENTS: (1) The temperature histories resulting from the FEHT and IHT based solutions are in
agreement. The interface temperature peaks near 225°C after 75 ms, and begins dropping toward the
steady-state condition. The mid-plane of the reaction plate peaks around 100 ms, nearly reaching
200°C. The temperature of the mid-plane of the composite plate increases slowly toward the steady-
state condition.
(2) The calculated temperature-time histories for the clutch pair display similar features as expected
from our initial sketches on T vs. t coordinates, part a. The maximum temperature for the composite is
very high, subjecting the bonded frictional material to high thermal stresses as well as accelerating
deterioration. For the reaction steel plate, the temperatures are moderate, but there is a significant
gradient that could give rise to thermal stresses and hence, warping. Note that for the composite plate,
the steel section is nearly isothermal and is less likely to experience warping.
(2) The IHT code representing the FDE for the 22 nodes and the frictional heat flux relation is shown
below. Note use of the Lookup Table for representing the frictional heat flux vs. time boundary
condition for nodes 11 and 12.
// Nodal equations, reaction plate (steel)
/* Node 01: surface node (w-orientation); transient conditions; e labeled 02. */
rhos * cps * der(T01,t) = fd_1d_sur_w(T01,T02,ks,qdot,deltax,Tinf01,h01,q''a01)
q''a01 = 0 // Applied heat flux, W/m^2; zero flux shown
Tinf01 = 40 // Arbitrary value
h01 = 1e-5 // Causes boundary to behave as adiabatic
qdot = 0
/* Node 02: interior node; e and w labeled 03 and 01. */
rhos*cps*der(T02,t) = fd_1d_int(T02,T03,T01,ks,qdot,deltax)
\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.
/* Node 10: interior node; e and w labeled 11 and 09. */
rhos*cps*der(T10,t) = fd_1d_int(T10,T11,T09,ks,qdot,deltax)
/* Node 11: From an energy on the CV about node 11 */
ks * (T10 - T11) / deltax + q''rp = rhos * cps * deltax / 2 * der(T11,t)
Continued (4)...
Temperature history for clutch pair, 100 ms lock-up time
0 200 400 600 800 1000
Engagement time, t (ms)
0
50
100
150
200
250
Te
m
pe
ra
tu
re
, T
 
(C
)
Midplane, reaction plate, T01
Interface, T11 or T12
Midplane, composite plate, T22
Heat flux histories for clutch pair during engagement
0 20 40 60 80 100
Engagement time, t (ms)
-0.5
0
0.5
1
1.5
2
H
e
a
t f
lu
x,
 
q'
' 
(W
/m
^2
 
*
 
10
^7
)
Frictional heat flux, q''f
Reaction plate, q''rp
composite plate, q''cp
PROBLEM 5.118 (Cont.)
// Friction-surface interface conditions
T11 = T12
q''f = LOOKUPVAL(HFVST16,1,t,2) // Applied heat flux, W/m^2; specified by Lookup Table
/* See HELP (Solver, Lookup Tables). The look-up table, file name "HFVST16' contains
 0 16e6
 0.1 0
 100 0 */
q''rp + q''cp = q''f // Frictional heat flux 
// Nodal equations - composite plate
// Frictional material, nodes 12-16
/* Node 12: From an energy on the CV about node 12 */
 kfm * (T13 - T12) / deltax + q''cp = rhofm * cpfm * deltax / 2 * der(T12,t)
/* Node 13: interior node; e and w labeled 08 and 06. */
rhofm*cpfm*der(T13,t) = fd_1d_int(T13,T14,T12,kfm,qdot,deltax)
\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026
/* Node 16: interior node; e and w labeled 11 and 09. */
rhofm*cpfm*der(T16,t) = fd_1d_int(T16,T17,T15,kfm,qdot,deltax)
// Interface between friction material and steel, node 17
/* Node 17: From an energy on the CV about node 17 */
kfm * (T16 - T17) / deltax + ks * (T18 - T17) / deltax = RHS
RHS = ( (rhofm * cpfm * deltax /2) + (rhos * cps * deltax /2) ) * der(T17,t)
// Steel, nodes 18-22
/* Node 18: interior node; e and w labeled 03 and 01. */
rhos*cps*der(T18,t) = fd_1d_int(T18,T19,T17,ks,qdot,deltax)
\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.
/* Node 22: interior node; e and w labeled 21 and 21. Symmetry condition. */
rhos*cps*der(T22,t) = fd_1d_int(T22,T21,T21,ks,qdot,deltax)
// qdot = 0
// Input variables
// Ti = 40 // Initial temperature; entered during Solve
deltax = 0.0001
rhos = 7800 // Steel properties
cps = 500
ks = 40
rhofm = 1150 //Friction material properties
cpfm = 1650
kfm = 4
// Conversions, to facilitate graphing
t_ms = t * 1000
qf_7 = q''f / 1e7
qrp_7 = q''rp / 1e7
qcp_7 = q''cp / 1e7
PROBLEM 5.119
KNOWN: Hamburger patties of thickness 2L = 10, 20 and 30 mm, initially at a uniform temperature
Ti = 20°C, are grilled on both sides by a convection process characterized by T\u221e = 100°C and h =
5000 W/m2\u22c5K.
FIND: (a) Determine the relationship between time-to-doneness, td, and patty thickness. Doneness
criteria is 60°C at the center. Use FEHT and the IHT Models|Transient Conduction|Plane Wall. (b)
Using the results from part (a), estimate the time-to-doneness if the initial temperature is 5 °C rather
than 20°C. Calculate values using the IHT model, and determine the relationship between time-to-
doneness and initial temperature.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, and (2) Constant properties are approximated as
those of water at 300 K.
PROPERTIES: Table A-6, Water (300K), \u3c1 = 1000 kg/m3, c = 4179 J/kg\u22c5K, k = 0.613 W/m\u22c5K.
ANALYSIS: (a) To determine T(0, td), the center point temperature at the time-to-doneness time, td,
a one-dimensional shape as shown in the FEHT screen below is drawn, and the material properties,
boundary conditions, and initial temperature are specified. With the Run|Calculate command, the early
integration steps are made very fine to accommodate the large temperature-time changes occurring
near x = L. Use the Run\uf8e6Continue command (see FEHT HELP) for the second and subsequent steps
of the integration. This sequence of Start-(Step)-Stop values was used: 0 (0.001) 0.1 (0.01 ) 1 (0.1)
120 (1.0) 840 s.
Continued \u2026..
PROBLEM 5.119 (Cont.)
Using the View|Temperature vs. Time command, the temperature-time histories for the x/L = 0
(center), 0.5, and 1.0 (grill side) are plotted and shown below for the 2L = 10 mm thick patty.
Using the View|Temperatures command, the time slider can be adjusted to read td, when the center
point, x = 0, reaches 60°C. See the summary table below.
The IHT ready-to-solve model in Models|Transient Conduction|Plane Wall is based upon Eq. 5.40 and
permits direct calculation of td when T(0,td) = 60°C for patty thickness 2L = 10, 20 and 30 mm and
initial temperatures of 20 and