MATLAB Solutions for Mechanical Engineering Problems | Desklib

2018
Computer Application for Mechanical Engineering
United States
Institutional Affiliation
Student name
Student ID number
Date of Submission

Project 4.2
Solution
Use MATLAB to fill in the following table,
X(min) \ T(C) 15 20 25 30
0.1 620.6897 416.6667 382.9787 375.0000
0.2 641.6639 430.7466 395.9203 387.6720
0.3 660.1080 443.1281 407.3007 398.8153
0.4 673.7973 452.3176 415.7473 407.0859
0.5 681.0810 457.2072 420.2415 411.4865
0.6 681.0810 457.2072 420.2415 411.4865
0.7 673.7973 452.3176 415.7473 407.0859
0.8 660.1080 443.1281 407.3007 398.8153
0.9 641.6639 430.7466 395.9203 387.6720
1.0 620.6897 416.6667 382.9787 375.0000
Using the search method, the function must be written in the form,
f ( x )=0
Perform a search to obtain the intervals in which the real roots lie. Sub-divide the x domain into
N equal subdivisions, giving,
x1 , x2 , x3 , ... , x N+1 ∧xi +1=xi+ ∆ x
One can now locate at which point the function changes the sign. It is set to occur when,
f ( xi ) f ( xi +1 ) < 0
The sign change usually indicates that a real root has been passed and could also indicate a
discontinuity in the function. When the intervals are established using the search methods, other
methods can be used to determine the real roots.
There is convection at the boundary in cartesian coordinates where the geometry is the same but
the boundary conditions vary. The characteristic length of the slab is given as

V
As
= WH 2 L
2WH =L
The initial heat equation is given as,
∂2 T ( x , t )
∂ x2 = 1
α
∂ T ( x ,t )
∂ t
Boundary conditions,
at x=0 , x=L;−k ∂ T
∂ x +h ( T −T ∞ )=0
X = x
L , τ= αt
L2 θ ( X , τ )= T ( x ,t )−T ∞
T i−T ∞
Bi= hL
k ... Biot number
The heat conduction problem is given as,
∂2 θ
∂ X2 = ∂θ
∂ τ {at X =0
at X =1
∂ θ
∂ X =0
∂ θ
∂ X + Biθ=0
∧¿ 0 ≤ X ≤ 1 for τ =0
θ=1
(T ¿¿ H−T s )/(T s−T ∞ )=(L/k . A)/( 1
h. A )¿
¿ H Tinternalresistance
H T externalresistance
= hL
k =Bi