Fundamentals of Mathematical Mode Solutions

This document provides solutions for various questions related to Fundamentals of Mathematical Mode. It includes topics such as cooling time, simple pendulum, cake optimization, speed hump modeling, steady state conduction and convection, heat transfer, thermal resistance, and more.

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