EM220 BACHELOR OF ENGINEERING (HONS) MECHANICAL
MEC500 NUMERICAL METHOD WITH APPLICATIONS
CASE STUDY OF INDOOR AIR POLLUTIONBY USING LINEAR ALGEBRAIC AND
INVERSE MATRICES
LECTURER:EN ARMANSYAHNAME:NURULHANNAH BINTI KAMARUDDINSTUDENT
ID:2012293588
BACKGROUNDIndoor air pollution deals with air contamination in
enclosed spaces such as homes, offices and work areas. Indoor
pollution sources that release gases or particles into the air are
the primary cause of indoor air quality problems in homes.
Inadequate ventilation can increase indoor pollutant levels by not
bringing in enough outdoor air to dilute emissions from indoor
sources and by not carrying indoor air pollutant out of the homes.
High temperature and humidity levels can also increase
concentrations of some pollutants.
PROBLEM STATEMENTThe serving area consists of two rooms for
smokers and kids and one elongated room which divided into two
sections. Room 1 and section 3 have sources of carbon monoxide from
smokers and grill works respectively. In addition, room 1 and 2
gain carbon monoxide from air intakes alongside the freeway.
Figure 1: Top view of rooms in a restaurant
Figure 1 above show the top view of rooms in a restaurant. The
one-way arrows represent volumetric airflows while the two-way
arrows represent the diffusive mixing between the rooms.
OBJECTIVES1. Determine how the various sources contribution
affects the kids room.2. Determine the concentration of carbon
monoxide in each room3. Determine the percentage of the carbon
monoxide in kids room due to smokers, grill and air intakes
alongside of the road.4. Solving mathematical modelling and problem
using MATLAB
ENGINEERING SOLVING
a) By analytical methodThe steady-state mass balance equation
is;(Load) + (flow in) (flow out) + (Mixing) = 0For room 1;1000 +
(2)(200) - 200 + 25( = 01000 + 400 - 200 + 25 - 251400 - 225 + 25 =
0----------------------- (1)
For room 2;(50(2) + (200-100) (150) + 25(100 + 100 - 150 + 25 -
25100 + 125 - 175--------------------- (2)
For room 3;2000 + (200) (200) + 25( + 50(2000 + 200 - 200 + 25 -
25 + 50 - 502000 + 225 - 275 + 50------------------ (3)For room
4;(200) (200) + 50( + 25(25 250 - 275 = 0----------------- (4)
From equation (1); 0.111 6.222
Substitute into (3) we get;225(0.111 + 6.222) - 275 + 5024.975
1399.95 - 275 + 50-2500.025+ 50 ------------------ (5)
From equation (2); 0.571 + 0.714
Substitute into (4) we get;25(0.571 + 0.714) + 250 - 27514.275 +
17.85 + 250 - 27514.275 + 250 257.15 = 0 -------------------
(6)
From equation (5); -67.999 + 5.0005
Substitute into (6)14.275 + 250 257.15(-67.999 + 5.0005 +
25014.275 + 17485.943 1285.879 + 25017500.218 =
1035.87916.894--------- Ans
Therefore; -67.999 + 5.0005 = -67.999 + 5.0005(16.894) = 16.479
--------- Ans
0.571 + 0.714 = 0.571 + 0.714(16.479) = 12.337 ---------- Ans
0.111 6.222 = 0.111(16.894) + 6.222 = 8.0972---------- Ans
b) By using numerical method
Why use inverse matrix
For matrices, there is no such as division process. Matrices can
be add, subtract and multiply but cannot be divided. In this case
study, to determine the coefficient of c with given A and B matrix,
suppose that it will be divided with matrix A. However, division
cannot be done. That is why inverse matrix is implemented. Inverse
matrix would allow matrix A to be cancel off from the matrix
equation and solve for the coefficient of c.
Solution
Forming the linear algebraic equation;
225 - 25 = 1400----------------- (1)-125 + 175-----------------
(2)-225 + 275 - 50----------------- (3)-25250 + 275 =
0----------------- (4)
Transforming all the equation into matrix form [Ac = B];
=
Hence solve it using MATLAB command. Figure 2: The command for
solving the linear algebraic. Figure 2 show the coding of the
command to determine the concentration of carbon monoxide in each
room. Based on the result obtained, it is surprising that the
smoking section has the lowest carbon monoxide levels content. The
highest concentration occurs in room 3 and 4 with section 2 having
and intermediate level. Besides that, the element of the matrix
inverse can be used to determine the percentage of the carbon
monoxide in the kids room section due to each source. For
example;Percentage of carbon dioxide due to smokers: = 0.0034483
(1000) = 3.4483
Figure 3 Figure 4Figure 3 and 4 show the distribution of carbon
monoxide in kids room in percentage due to each source. The faulty
grill is clearly the most significant source.
RESULT & DATA TABULATION
True valueApproximation value Et, %
C18.09728.09660.0074
C212.33712.3450.0648
C316.89416.8970.0178
C416.47916.4830.0243
Table 1: The comparison between analytical method and numerical
method results of carbon monoxide content in each room.
Figure 5: Graph of carbon monoxide content result by using
analytical method and numerical method.
PERCENTAGE OF CARBON MONOXIDE IN KIDS ROOM
Figure 6: The distribution of carbon monoxide due to each
source.
CONCLUSIONBased on the result obtained, it shows that the
approximate error between the true value which is by analytical
method and approximate value by numerical method is slightly
different. It can be said that the error is almost approaching
zero. However, by using numerical method, the value is more
accurate due to the decimal places.
REFERENCE1.Applied Numerical Methods with MATLAB for Engineers
and Scientists, Steven C.
Chapra2.http://www.purplemath.com/modules/mtrxinvr.htm3.http://www.mathsisfun.com/algebra/matrix-inverse.html4.
http://www.epa.gov/iaq/ia-intro.html