CONVOCATION

After 1 and a half years, finally I manage to complete my master course in UPM. Although there were some problems at the final stage of my project, but eventually they were solved successfully. Today is my convocation. All my family members attended my convo full with excitement, especially my parents. Because this year a bit special, yesterday was my younger brother's graduation and today it's my convocation. When I walked into the hall, the feeling when I attended for the oath taking ceremony 1 and a half years ago suddenly appear. That day, I told myself, after 3 semester, I will be back here for my convacation again. And today I make it. When walking up to the stage and get the scroll from the Pro Canselor, I really feel proud of myself. To further my studies if without my wife's moral support, I don't think that I will quit from my previous company and join UniKL-MFI. My initial planning was joining UniKL-MFI as a stepping stone for me, coz I will have more time to concentrate on my studies. After that, I might go back to industries or further for my phD. But today, I might change my initial planning, coz I found "something" that I can't find it from the industries.
Chase for your dream, no matter how difficult it is, just face it! Solve it! If you think you can, then you can. I can do it, you can do it too!

ACR0213 3RD ASSIGNMENT - PIPE SIZING

Size up the pipe size for the system shown.

Deadline for submission 17/9/07 before 12p.m

Why boring???

Almost every class, students will do their presentation on a particular topic. And nearly all the presentation having the same response from the students. What response? Sleepy and boring! When the presenter scroll down the next slide, they will hope to see the wording "Thank you!" or "The End". If they see not their wanted slide, then they will ask when the presentation will end and how many remaining slides to go? Not only the students' presentation will have this type of response, even my presentation also having the same problem. Is that my lecture so boring? How to improve it? Purely theory, students will feel boring, calculation pulak, students will afraid to do it coz they don't like figures. Then how?

ACR0213 2ND ASSIGNMENT - KITCHEN EXHAUST SYSTEM

Question:
Size up the ducting system which installed in the kitchen of a food court.

Deadline for submission: 30 August 2007, before 12p.m

PJ HALF MARATHON

This month really full with RUNS. Perhaps it is because of the Merdeka month, so everyone also busy with organising this type of run.

Last Sunday joined the PJ half marathon. It was my 3rd half marathon. For me, it was really tough and exhausting. Maybe, I used to run for 10km, and when run for 21 km, I felt like wanted to stop half way. It was not easy. It really challenge my will power and spirit. Along the way, there were so many vehicles passing by, creating not only the dust and toxic gases, but it really affected my running mood. Know why? Cause I wanted to 'tumpang' them especially for the last 4 km. But, finally I manage to complete the run. I won myself cause never stop from running, but I lost to the qualifying time. Basically, I enjoy the run! Feel self-satisfaction!

SOLUTION FOR ACR 3933 2ND ASSIGNMENT - PSYCHROMETRICS

Question 1:
The air vapor mixture leaving the saturator will be saturated at 76 F. The relative humidity at saturation is 100%.

Question 2:
Sensible Heat Removed = 1.1*1 000cfm * (80-50) = 33 000 Btu/hr
Latent Heat removed = 0.69 * 1 000cfm * (77-54) = 15 870 Btu/hr
The factor 1.1 used in the sensible heat formula is a combination figure which converts cfm to lbs. per hour, and also takes into account the amount of heat required to raise 1 lb. of air 1 degree in temperature.
The factor 0.69 used in the latent heat formula is also a combination figure which converts cfm into lbs. per hour and also takes into account the amount of heat removed to condense moisture.

Question 3:
The mixture will be closest to the larger air quantity. In this case, it is the 80 DB condition. The total cfm in the mixture is 2 000 cfm. 500 cfm of outdoor air represents 500/2,000 or 1/4 of the mixture. The temperature difference between point 1 (80 DB) and point 2 (96 DB) is 96-80 or 16 degrees. One-quarter of 16 is 4 degrees. Therefore, the mixture is at 80+4 or 84 DB or point 3.
By locating 84F on line 1-2, we can determine the other properties by using the psychrometric chart.
a. DB = 84 F
b. WB = 69.2 F
c. DP = 62 F
d. Spec. hum. = 83.75 grains.

Remarks:
The temperature of the mixture can also be determined by calculation as follows:
tm = [(cfm1*t1)+(cfm2*t2)]/(cfm1 + cfm2) = [(1500*80)+(500*96)]/(1500+500) = 84 F

SOLUTION FOR ACR3422 PSYCHROMETRICS 1ST ASSIGNMENT

Question 1:
Locate points for recirculated and outside air on the chart. connect with a straight line. Read specific volumes of air at each point from chart. Convert cfm of air to pounds of air and find total weight of mixture.
7500/13.22 = 567.3 lbs./min
2500/14.37 = 173.9 lbs./min
Total weight = 741.2 lbs/min
Dry bulb of mixture =
(567.2/741.2)*57=43.6
(173.9/741.2)*96=22.5
Therefore, DB=66.1 deg F
From intersection of 66.1 deg F DB line with line between two points, read WB=62.1F, h=27.83 Btu/lb. dry air, humidity ratio = 77.1 gr. of moisture/lb. of dry air.
Approximate solution does not convert cfm to pounds per minute.
(7500/10 000)*57 = 42.75
(2500/10 000)*96 = 24.00
Mixture DB = 66.75 deg F.
At intersection of this line with line between two points, read WB = 62.5 F, h = 28.1 Btu/lb. dry air, humidity ratio = 78.0 grains of moisture/lb.of dry air.

Question 2:
Locate the initial condition on the chart. Draw a horizontal line through the initial point to the 95 F DB line. Read 7.3% RH, WB = 59.2 F and DP = 23.5 F.
Read enthalpy of initial air as 11.1 Btu/lb=h1
Read enthalpy of final air as 25.6 Btu/lb. = h2.
Heat added equals difference 14.5 Btu/lb.

Question 3:
Locate the initial and final conditions on the chart.
For the initial condition read W1 = 81.7; h1 = 32.48
For the final condition read W2 = 56.2; h2 = 21.4
Heat removed: h2 - h1 = 21.4 - 32.48 = -11.08 Btu/lb.dry air.
Moisture removed: W2 - W1 = 56.2 - 81.7 = -25.5 grains.
To determine the SHR, draw a straight line between the initial and final conditions. Draw a line parallel to this one from the reference point at 78 F DB and 50% RH to the sensible heat ratio scale. Read SHR = 63.8

CONSTRUCTION OF CONTROLLED SPACES

1. any vapor barrier must be continuous, without breaks or tears.
2. all lap joining must be tightly closed. (this is particularly critical when mechanical or caulked joints are used).
3. insulation between vapor barriers can be a potential problem. if construction occurs in humid weather, moisture can be sealed in between the two vapor barriers. Sealed-in water vapor will travel into the controlled space and impose an extra drying load on the drying equipment. this extra load lasts only until the insulation dries out, but meanwhile humidity control is difficult. if a heat source is present (even heat from the sun), serious damage can be caused by the expanding trapped vapor. there have been cases when so-called "non-permeable" materials have split open at a joint because of vapor pressure. examples include a floor or tiled wall that has literally lifted from its mounting surface because the surface was wet during application.
4. final inside vapor barriers should be applied only after the enclosed area has been dried. drying equipment should be used to withdraw as much moisture as possible before the final barrier is applied. of course, without a barrier in place, equipment cannot dry the air to design specifications, but a significant amount can and should be removed before all the vapor barrier material is in place. (Although this strategy is not included in most industrial planing suggestions, the concept of drying the structure before applying the final vapor barrier is a precaution that is often overlooked and can help prevent problems associated with trapped water vapor).
5. all doors, service or personnel, should be weather-stripped or air locked through vestibules if the desired conditions warrant it. any crack or opeingng around a door will admit water vapor.
6. when a conveyor opening or similar elements are used, a drop curtain, shroud, or tunnel will reduce the migration of water vapor.

Whose fault? Engineering control or Administrative control?

This afternoon while I was having lunch, there was an accident occurred in front of the restaurant. A Vios bang with a Harrier. Quite serious because the front wheel of the Vios seems like came out from the car body already. The 2 drivers were arguing on whose fault. Everybody was blaming on the opposite party. The cross road without stop-look-go sign. So, whose fault was that? Just think about this???

THE NATURE OF WATER VAPOR

Consider two closed rooms adjacent to one another. If the partial pressure of the water vapor in room #1 is greater than the partial pressure of the water vapor in room #2, then the water vapor will travel through the wall into room #2 regardless of the composition of the wall.
Let's take this hypothetical example a step further. If the absolute humidity of the air in room #1 is greater than that of the air in room #2, then the water vapor pressure will be higher in room #1. Therefore, when drying room #2, the problem of water vapor coming through the wall from room #1 must be considered.
A vapor barrier can slow down the passage of vapor from wet to drier areas, but it cannot keep water vapor out; it can only slow the rate of penetration.
The choice of a vapor barrier is based on the degree of dryness required in the controlled space; the efficiency of the equipment being used for drying; and the cost of construction.
Commercial vapor barriers, moisture resistant construction material,p paints an other coatings, offer a variety of design alternatives. Manufacturers of vapor barrier materials can supply specific information on their products.