Fundamentals of Mathematical Modeling Solutions: Complete Assignment

This document contains solutions to a mathematical modeling assignment, addressing ten different problems. The solutions cover a range of topics including the cooling of water, modeling a simple pendulum, optimizing a cake, analyzing a speed hump, modeling steady-state conduction and convection, calculating heat transfer based on surface temperature and air temperature, determining heat loss through a window, calculating heat transfer in a circular pipe, calculating thermal resistance, and determining heat transfer based on temperature differences and material properties. Each solution is presented with the relevant formulas, calculations, and final answers, providing a comprehensive guide to solving these mathematical modeling problems.

FUNDAMENTALSOFMATHEMATICALMODE
Solutions
Question 1
Time taken, in seconds. for the water in the cup to cool to a drinkable temperature of 46 °C
T 1−T3 = ( T 2−T3 ) e− yt y= U value A
mc =
9.49 W
m2 K × 0.004 m2
0.3 kg × 4150 J
kgK
=3.048996× 10−5 T 1=46 ,T 2=87 T 3=25
∴ 46−25= ( 87−25 ) e−−3.048996 ×10−5 t
t=
ln ( 46−25
46−25 )
−3.048996 ×10−5 =35507.1619 s ¿ 35507.162 s
Question 2
Modelling a simple pendulum, investigating the relationship between period and length of a
simple pendulum.
mass , m=2.12 , L=1.06 m , g=9.81 ms−2 , π =3.142T =2 π √ L
g =2 π √ 1.06 m
9.81 m s−2 =2.065372 s
∴ T =2.065 s
Question 3
Optimizing of the cake

A=2 πrh+2 π r2 V =π r2 h ⟹h= V
π r2 = 3894
π r2 A ( r )=2 πr ( 3894
π r2 )+ 2 π r2
d
dr ( A ( r ) )= d
dr (7788
r +2 π r2
)=4 πr− 7788
r2
d
dr ( A ( r ) )=0 ⟹ 4 πr− 7788
r2 =0
4 πr =7788
r2 ⟹ r3= 7788
4 π r =8.525869∴ h=h= V
π r 2 = 3894
π × 8.5258692 =17.0517 cmh=17.052 cm
Question 4
Speed hump modelling
Case 1: Leaving the first hump vehicle accelerates up to maximum allowed velocity
s1=vo t1 + 1
2 a1 t1
2 v0=2.88 m
s , v1=40 km
hr =100
9
m
s , a1=1.84 m
s2 vi =v0 + a1 t1 ⟹ t1= v1−v0
a1
t1=
100
9 −2.88
1.84 = 926
207 s=4.47343 s ∴ s1=vo t1 + 1
2 a1 t1
2=s1 =2.88 ( 926
207 ) + 1
2 × 1.84 × ( 926
207 )
2
¿ 31.29412775 m
Case 2: After reaching maximum velocity, vehicle decelerates back
s2=v1 t2 + 1
2 a2 t2
2
t2= v0−v1
a2
=
2.88−100
9
−6.69 =1.2303604
s2= 100
9 ( 1.23036 )+ 1
2 × (−6.69 ) × ( 1.23036 )2=8.607054 m
Total distance , S=s1 + s2=31.29412775+ 8.607054=39.901182 ¿ 39.901 m
Question 5
The steady state conduction and convection model.

l=0.13 , k =0.6 W
mC , T 1=22.94 ℃ , T2 =31.22℃h0 =10 W
m2 C , hi =150 W
m2 C
Hrate= T 2−T1
l
k + 1
h0
+ 1
hi
¿ 31.22−22.94
0.13
0.6 + 1
10 + 1
150
=25.608247 ¿ 25.608 W /m2
Question 6
A=heat transfer area of the surface (m2)
hc=convective heat transfer coefficient of the process(W /(m2 ° C))
T s=Temperature surface
T a=Temperature air W =hc A ( T s −T a )934=8.59 ×36.55 ( TS−0 )T s= 934
36.55 ×8.59 =2.974858 ℃
¿ 2.975 ℃
Question 7
Window dimensions and made
A=0.95 ×2.04=1.938 m2
Thickness , d=0.01448 m , k=0.81 W
m℃ , ∆ T =32−24=8 ℃Q= kA ∆ T
d
¿ 0.81× 1.938× 8
0.01448 =867.2818 W ¿ 867.282 W
Question 8
Heat transfer per unit length for a uniform circular pipe

Q= 2 πkl ∆ T
ln ( r2
r1 )
¿ 2 π ×155 ×1 ( 51.1−27.18 )
ln
( 116.45
2
77.47
2 ) =94032.43929 W
m
¿ 94032.439 W /m
Question 9
Thermal resistance
R= 1
h0
+ Lb
Kb
+ Lc
Kc
+ 1
hi
¿ 1
11.98 + 0.25
1.14 + 0.0283
1.11 + 1
33.7 =0.3579398 m2 K
W ¿ 0.358 m2 K /W
Question 10
Q= T 2−T 1
A
Q
A = T 2−T 1
1
h1
+ L
k + 1
h2
¿ 297.9−270
1
21.77 + 0.1
0.8 + 1
10.03
=109.0906 ¿ 109.091W /m2