Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot. isothermal metal substrate at T h = 300 ° C . The properties of the acrylic are ρ = 1990 kg/m 3 , c = 1470 J/kg ⋅ K, and k = 0.21 W/m ⋅ K . Neglecting the thermal contact resistance between the acrylic and the metal substrate, determine how long it will take for the insulated back side of the acrylic to reach its softening temperature, T soft = 90 ° C . The initial acrylic temperature is T i = 20 ° C .
Consider an acrylic sheet of thickness L = 5 mm that is used to coat a hot. isothermal metal substrate at T h = 300 ° C . The properties of the acrylic are ρ = 1990 kg/m 3 , c = 1470 J/kg ⋅ K, and k = 0.21 W/m ⋅ K . Neglecting the thermal contact resistance between the acrylic and the metal substrate, determine how long it will take for the insulated back side of the acrylic to reach its softening temperature, T soft = 90 ° C . The initial acrylic temperature is T i = 20 ° C .
Solution Summary: The author explains that in hot isothermal metal substrate, the dimensionless Biot number tends to infinity because of the constant temperature boundary conditions.
Consider an acrylic sheet of thickness
L
=
5
mm
that is used to coat a hot. isothermal metal substrate at
T
h
=
300
°
C
.
The properties of the acrylic are
ρ
=
1990
kg/m
3
,
c
=
1470
J/kg
⋅
K,
and
k
=
0.21
W/m
⋅
K
.
Neglecting the thermal contact resistance between the acrylic and the metal substrate, determine how long it will take for the insulated back side of the acrylic to reach its softening temperature,
T
soft
=
90
°
C
.
The initial acrylic temperature is
T
i
=
20
°
C
.
(I) [40 Points] Using centered finite difference approximations as done in class, solve the equation for O:
d20
dx²
+ 0.010+ Q=0
subject to the boundary conditions shown in the stencil below. Do this for two values of Q: (a) Q = 0.3,
and (b) Q= √(0.5 + 2x)e-sinx (cos(5x)+x-0.5√1.006-x| + e −43*|1+.001+x* | * sin (1.5 − x) +
(cosx+0.001 + ex-1250+ sin (1-0.9x)|) * x - 4.68x4. For Case (a) (that is, Q = 0.3), use the stencil in Fig.
1. For Case (b), calculate with both the stencils in Fig. 1 and Fig 2. For all the three cases, show a table as
well as a plot of O versus x. Discuss your results. Use MATLAB and hand in the MATLAB codes.
1
0=0
x=0
2
3
4
0=1
x=1
Fig 1
1 2 3 4 5 6 7 8 9 10
11
0=0
x=0
0=1
x=1
Fig 2
Fig 2
(II) [60 Points] Using centered finite difference approximation as done in class, solve the equation:
020 020
+
მx2 მy2
+0.0150+Q=0
subject to the boundary conditions shown in the stencils below. Do this for two values of Q: (a) Q = 0.3,
and (b) Q = 10.5x² + 1.26 * 1.5 x 0.002 0.008. For Case (a) (that is, Q = 0.3) use Fig 3. For Case (b),
use both Fig. 3 and Fig 4. For all the three cases, show a table as well as the contour plots of versus
(x, y), and the (x, y) heat flux values at all the nodes on the boundaries x = 1 and y = 1. Discuss your
results. Use MATLAB and hand in the MATLAB codes. (Note that the domain is (x, y)e[0,1] x [0,1].)
0=0
0=0
4
8
12
16
10
Ꮎ0
15
25
9
14
19
24
3
11
15
0=0
8-0
0=0
3
8
13
18
23
2
6
сл
5
0=0
10
14
6
12
17
22
1
6
11
16
21
13
e=0
Fig 3
Fig 4
Textbook: Numerical Methods for Engineers, Steven C. Chapra and Raymond P. Canale, McGraw-Hill, Eighth
Edition (2021).
Ship construction question. Sketch and describe the forward arrangements of a ship. Include componets of the structure and a explanation of each part/ term.
Ive attached a general fore end arrangement. Simplfy construction and give a brief describion of the terms.
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