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