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# ch05

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```planes and considered as adiabatic. We
have chosen to use the finite-element software FEHT as the solution tool.
Using FEHT, an outline of the symmetrical wall section is drawn, and the material properties are
specified. To determine the initial conditions for the cool-down process, we will first find the
temperature distribution for steady-state operation. As such, specify the boundary condition for the
inner surface as a constant temperature of 900°C; the other boundaries are as earlier described. In the
This distribution represents the initial temperature distribution, Ti (x, y, 0), for the wall at the onset of
the cool-down process.
Next, in the Setup menu, click on Transient; for the nodes on the inner surface, in the Specify |
Boundary Conditions menu, deselect the Temperature box (900°C) and set the Flux box to zero for the
adiabatic condition (part (a)); and, in the Run command, click on Continue (not Calculate). Be sure to
change the integration time scale from seconds to hours.
Because of the high ratio of wall section width (nearly 8.5 m) to the thickness (1 m), the conduction
heat transfer through the section is nearly one-dimensional. We chose the x,y-section 1 m to the right
of the centerline (1 m, y) as the location for examining the temperature-time history, and determining
the cool-down time for the inner surface to reach the safe working temperature of 35°C.
Continued \u2026..
PROBLEM 5.128 (Cont.)
Time-to-cool, Part (a), Adiabatic inner surface. From the above temperature history, the cool-down
time, ta, corresponds to the condition when Ta (1 m, 0, ta) = 35°C. As seen from the history, this
location is the last to cool. From the View | Tabular Output, find that
at 1306 h 54 days= = <
Continued \u2026..
PROBLEM 5.128 (Cont.)
Time-to-cool, Part (b), Cooled inner surface. From the above temperature history, note that the center
portion of the wall, and not the inner surface, is the last to cool. The inner surface cools to 35°C in
approximately 175 h or 7 days. However, if the cooling process on the inner surface were
discontinued, its temperature would increase and eventually exceed the desired safe working
temperature. To assure the safe condition will be met, estimate the cool down time as, tb,
corresponding to the condition when Tb (1 m, 0.75 m, tb) = 35°C. From the View | Tabular Output,
find that
bt 311 h 13 days= = <
COMMENTS: (1) Assuming the furnace can be approximated by a two-dimensional symmetrical
section greatly simplifies our analysis by not having to deal with three-dimensional corner effects. We
justify this assumption on the basis that the corners represent a much shorter heat path than the straight
wall section. Considering corner effects would reduce the cool-down time estimates; hence, our
analysis provides a conservative estimate.
(2) For background information on the Continue option, see the Run menu in the FEHT Help section.
Using the Run | Calculate command, the steady-state temperature distribution was determined for the
normal operating condition of the furnace. Using the Run | Continue command (after clicking on
Setup | Transient), this steady-state distribution automatically becomes the initial temperature
distribution for the cool-down transient process. This feature allows for conveniently prescribing a
non-uniform initial temperature distribution for a transient analysis (rather than specifying values on a
node-by-node basis.
PROBLEM 5.129
KNOWN: Door panel with ribbed cross-section, initially at a uniform temperature of 275°C, is
ejected from the hot extrusion press and experiences convection cooling with ambient air at 25°C and
a convection coefficient of 10 W/m2\u22c5K.
FIND: (a) Using the FEHT View|Temperature vs. Time command, create a graph with temperature-
time histories of selected locations on the panel surface, T(x,0,t). Comment on whether you see
noticeable differential cooling in the region above the rib that might explain the appearance defect; and
Using the View|Temperature Contours command with the shaded-band option for the isotherm
contours, select the From start to stop time option, and view the temperature contours as the panel
cools. Describe the major features of the cooling process you have seen. Use other options of this
command to create a 10-isotherm temperature distribution at some time that illustrates important
features. How would you re-design the ribbed panel in order to reduce this thermally induced paint
defect situation, yet retain the stiffening function required of the ribs?
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional conduction in the panel, (2) Uniform convection coefficient
over the upper and lower surfaces of the panel, (3) Constant properties.
PROPERTIES: Door panel material (given): \u3c1

= 1050 kg/m3, c = 800 J/kg\u22c5K, k = 0.5 W/m\u22c5K.
ANALYSIS:
(a) Using the Draw command, the shape of the symmetrical element of the panel (darkened region in
schematic) was generated and elements formed as shown below. The symmetry lines represent
adiabatic surfaces, while the boundary conditions for the exposed web and rib surfaces are
characterized by (T\u221e, h).
Continued \u2026..
PROBLEM 5.129 (Cont.)
After running the calculation for the time period 0 to 400 s with a 1-second time step, the temperature-
time histories for three locations were obtained and the graph is shown below.
As expected, the region directly over the rib (0,0) cooled the slowest, while the extreme portion of the
web (0, 13 mm) cooled the fastest. The largest temperature differences between these two locations
occur during the time period 50 to 150 s. The maximum difference does not exceed 25°C.
Continued \u2026..
PROBLEM 5.129 (Cont.)
(b) It is possible that the temperature gradients within the web-rib regions \u2013 rather than just the upper
surface temperature differentials \u2013 might be important for understanding the panel\u2019s response to
cooling. Using the Temperature Contours command (with the From start to stop option), we saw that
the center portion of the web and the end of the rib cooled quickly, but that the region on the rib
centerline (0, 3-5 mm), was the hottest region. The isotherms corresponding to t = 100 s are shown
below. For this condition, the temperature differential is about 21°C.
From our analyses, we have identified two possibilities to consider. First, there is a significant surface
temperature distribution across the panel during the cooling process. Second, the web and the
extended portion of the rib cool at about the same rate, and with only a modest normal temperature
gradient. The last region to cool is at the location where the rib is thickest (0, 3-5 mm). The large
temperature gradient along the centerline toward the surface may be the cause of microstructure
variations, which could influence the adherence of paint. An obvious re-design consideration is to
reduce the thickness of the rib at the web joint, thereby reducing the temperature gradients in that
region. This fix comes at the expense of decreasing the spacing between the ribs.
Problems 5.1-5.25
5.1
5.2
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5.5
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5.24
5.25
Problems 5.26-5.50
5.26
5.27
5.28
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5.50
Problems 5.51-5.75
5.51
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5.75
Problems 5.76-5.100
5.76
5.77
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5.79
5.80
5.81
5.82
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5.84
5.85
5.86
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5.90
5.91
5.92
5.93
5.94```