Nalareksa ResearchIntro to Nalareksa
Nalareksa is a method of air analysis by using water as a mediator for the research. To investigate out the elements in air, we could use water as a mediator for the container of the elements from the air if we know the balance condition between air and water. Nalareksa was founded by Prof. Ir. Hardjoso Prodjopangarso as his intention for his research could be useful by small people in Indonesia.
In “old Javanese”, the word nala means beloved ones and the word reksa means nourishing. Literally, nalareksa represent the world as a beloved one that needs nourishing from humans. In this case, taking precautions to pollutants that may cause destruction to the world. Every equipments, sciences and technologies forNalareksa were created by the simplest approach so it could be reached by common. That’s why all the instruments in Nalareksa are made using easy-to-find materials in Indonesia. As already known, people became realizes of climate changes as the result of the significant increase in anthropogenic greenhouse gases (gas emission caused by fossil fuels usage for transportations) and deforestation in concentrations via enhanced greenhouse effects. CO2/CO in large concentrations are created reducing benefit elements such as O2/O3 concentration for humans.
The instruments for direct analyzing elements in the air are still rare. As for water laboratory are more common to the society and have good analyzing capabilities. To analyze CO2 and O2 inside water are pretty easy to do and the chemical elements are easy enough to get. The amount time to do this analyze are also very short. It only took 5 to 15 minutes for analyzing elements inside of water.
Using simple lists and formulas, elements parts per million (ppm) inside of water could be converted into elements contains by air in % volume. By comparing with the elements limitation in certain air condition, we could analyze air condition in our environments.
Instrument to do Nalareksa are:
a. 1 aerator to take air sample
b. 1 tube filled with aquadest. Aquadest will be aerated for 60 minutes with air that will be analyzed.
c. The tube will be closed and left with two holes. One hole will be connected with a device to control pressure (aeration) and another hole with adjustable output volume of the air.
d. Temperature above the surface of the water and the pressure could be identified and it will be used by a conversion method using tables and formulas.
Conversion examples from air elements to water elements and vice versa
1. O2 in the air are 20.8 % volume, for example in temperature 30° C the Oxygen concentration in the water is:
37.2 x 0.208 ppm = 7.73 ppm
2. For example CO2 in water is 0.7 ppm, the percentage of CO2 in the air could be found by this calculations:
On 15°C then 2000 x X = 0.7 ppm, CO2. X in the calculation of the process is 0.7/2000 = 0.035%
Air substitutions (luchtvesing = fresh) in the rooms is created by:
a. Temperature difference
b. Air flow
Elements that is dangerous for human (elements from humans and lamps):
a. Acid substance (koolzur)
b. Water damp
c. Other disadvantaging materials (kitchen)
There is also heat inside the room affected by internal (human’s warm afgtile) and external (sun) factors that need cooling down by air motion.
Numerous elements produced:
a. Carbon Dioxide (CO2): human produces 20 liters of Carbon Dioxide (CO2) every hour, petrol lamp produce 220 liter CO2 every hour/50NK 0.1 electric lamp, free air contain 0.04% CO2
b. Water Steam: Human produce 40-80 gr/hour, petrol lamp produce 6 gr/NK, 1 hour of outside air with 660 humidity (vachtigheid cells)
c. Burning elements: Inside a house normally there is 205 (zuur staf). Humans needs 8% minimum
Amount of acid substance allowed for humans (maximum):
– Hospital = 0.07% acid
– General buildings = 0.01% acid
– Housing and schools = 0.15% acid
Air movements are:
1. Horizontal movement caused by wind
2. Vertical movements caused by inside and outside temperature difference
Controlling air movements:
a. Creating holes near the plafond (upside) opposite side with the door and windows. The method could be replace or combine with plafond holes and roofs.
b. Windows’ down section should be designed as low as possible. (zolaag megelijk).
c. Windows’ upper section (bovenlicht) designed highest as possible to avoid air disturbance (tocht). For:
– Ventilation holes lower than 2 m, air maximum velocity is 0.3-0.5 m/s.
– Ventilation holes near to plafond, air maximum velocity is 1.5-2 m/s, and smallest holes could be 4-6 m/s.
If gas encounters water, then some part of the gas will be absorbed to water. Eventually, there will be an equilibrium, which means the amount of elements released by the water to the air will be equal to the amount of elements that absorbed by air and water. Water is a very strong solvent, referred to as the universal solvent, dissolving many types of substances. Substances that will mix well and dissolve in water, e.g. salts, sugars, acids, alkalis, and some gases: especially oxygen, carbon dioxide (carbonation), are known as “hydrophilic” (water-loving) substances, while those that do not mix well with water (e.g. fats and oils), are known as “hydrophobic” (water-fearing) substances. All the major components in cells (proteins, DNA and polysaccharides) are also dissolved in water.
The amount of the gas elements that concentrated in water on this equilibrium called (concentration), equilibrium (kk) concentration. This equilibrium has a linear comparison with air pressure to water.
Equilibrium solution = c.s = a.p, with a = elements absorption coefficient. The coefficient is not constant, it depends on gas types and the lower the temperature, and the coefficient will be smaller.
Gas absorption (ppm) in pure water (on) every atmosphere pressure
(a = Elements coefficient)
On 10° C, air with condensed water vapor in open-air will contains percentage volume:
Nitrogen (N2) = 77.1 %, Oxygen (O2) = 20.8%, water vapor = 1.2 %, Argon (Ar) = 0.9% and Carbon Dioxide (CO2) = 0.03% (for industrial area is 0.1%)
Rain drops that falls through the air above mentioned will contains (condensed level)
N2 = 23.2 x 0.771 = 17.9 gram/m3
O2 = 54.3 x 0.208 = 11.3 gram/m3
CO2 = 2345 x 0.0003 = 0.7 gram/m3
54.3 = O2 coefficient absorbed by water (mg/lt)
0.208 = O2 part from all gas (volume percentage) from air
Equilibrium Concentration for O2 contacted with external air
Water salts will lower Cs (kk) in seawater with S salt 35000 PPm, O2 18 % reduced, CO2 12% reduced, if versproeid in the air (dresdener). If it’s in the river, it is really dependable with V, falls coefficient, the numbers deficits above mentioned are end limits (not relative limits).
Research objective and intention
To analyze the elements in the air or in water in order to identify air pollution level and air-condition in our environment.
1. Aerator (Used for gathering air samples)
2. Nalareksa instruments
3. 500 ml Glass Tube (To measure gas pumped by aerator)
4. Air pressure measuring tools (in the shape of connected bottles. To make both surface brighter, water inside the measuring tools could be in color)
5. Dry and Wet Thermometer
2. NaOH 0.1 N
3. PP indicator
Research application on the field
1. Put Aquades to 500 ml Glass Tube, placed the Glass Tube in certain place of research
2. Wait for 60 minutes to determine how much are the dry temperature and wet temperature (air sample)
3. After collecting data and samples, proceed to laboratory examination
Research Data and Results
– Put water sample in to O2 bottle until it’s full
– Add 10 drops of MnSO4 and 10 drops of H2S2, stir and let it sediment
– Add 20 drops of high concentrated H2SO4; stir until the residue became solution with the water until the solution color is yellow.
– Take 100ml water sample from the bottle and put in to Erlenmeyer
– Add 10 drops of amylum (C6H10O5)11 until the color turns dark blue
– Titration with Na2S2O3 1/40 N until the colpr turns light blue, the ammount Na2S2O31/40 N needed for titration (=tl)
|O2 = 1000/100 x (tl) x f x 0.2 = … Mg/lt||%O2 = O2/a x|
– Put 100 ml water sample to Erlenmeyer pumpkin
– Add 3 drops of PP indicator
– Titration with NaOH 0.1 N
– Write the amount of NaOH 0.1 needed (x)
– CO2 ammount:
|CO2 = 1000/100 x ( x/20 ) x 4,4 = ….. Mg/lt|
|% CO2 = CO2 / a x 100%|
|% Air = mg / a x 100 %|