During heat moisture treatment procedure the moisture level of sample is adjusted to 20, 25 and 30% by adding appropriate volumes of water. Is there any other instrument require for treatment or else can i use by hot air oven itself
Hi Suriya,
To understand your question better, the following essential information should be provided:
1. Is it a gas, liquid or a solid sample?
2. Is the target moisture content on mass or volume basis?
Different methods should be used to add moisture to the sample. The following serve as examples but not the only solution.
1. To add moisture to a gas sample, one probably should use a humidifier (which is essentially a thermostat with good temperature control and a large gas-liquid contact area to achieve VLE). First, calculate the saturation temperature of water vapour at the pressure of the gas stream that correspond to the 20, 25 and 30% (I assume vol.% for this example). The humidifier will do the rest for you.
2. To add moisture to a liquid sample, first weigh your initial sample mass (W0) and its initial moisture content (M0, in wt.%). Then calculate the dry mass and water mass in the original sample, which are:
Wd = W0*(100-M0)/100 and Wm0 = W0 - Wd,
respectively. With the target moisture content M1 = 20, 25 and 30% (assuming wt.% for this example), you know the total humidified sample mas should be
W1 = Wd*100/(100-M1)
Thus the amount of water to be added is
Wwat = W1 - W0.
Since the density of water is almost exactly 1 g/cm3 at room temperature, you can either weigh the amount or water or measure its volume. Mechanical agitation may be required during moisture addition.
3. To add moisture to a solid sample, the math is the same as for a liquid. However, the challenge is usually how to ensure uniform moisture distribution in the sample. This depends on whether your solid sample is a fine powder, a porous medium, or something else. It could take anywhere from a few minutes to a few days to reach uniform moisture distribution in your sample.
Hope this helps.
since the density of distilled water is equal 1g/cm3 so the volume and weight of water is equal. You need to calculate the weight of water related to the moisture level of sample that you wish.
Hi Suriya,
To understand your question better, the following essential information should be provided:
1. Is it a gas, liquid or a solid sample?
2. Is the target moisture content on mass or volume basis?
Different methods should be used to add moisture to the sample. The following serve as examples but not the only solution.
1. To add moisture to a gas sample, one probably should use a humidifier (which is essentially a thermostat with good temperature control and a large gas-liquid contact area to achieve VLE). First, calculate the saturation temperature of water vapour at the pressure of the gas stream that correspond to the 20, 25 and 30% (I assume vol.% for this example). The humidifier will do the rest for you.
2. To add moisture to a liquid sample, first weigh your initial sample mass (W0) and its initial moisture content (M0, in wt.%). Then calculate the dry mass and water mass in the original sample, which are:
Wd = W0*(100-M0)/100 and Wm0 = W0 - Wd,
respectively. With the target moisture content M1 = 20, 25 and 30% (assuming wt.% for this example), you know the total humidified sample mas should be
W1 = Wd*100/(100-M1)
Thus the amount of water to be added is
Wwat = W1 - W0.
Since the density of water is almost exactly 1 g/cm3 at room temperature, you can either weigh the amount or water or measure its volume. Mechanical agitation may be required during moisture addition.
3. To add moisture to a solid sample, the math is the same as for a liquid. However, the challenge is usually how to ensure uniform moisture distribution in the sample. This depends on whether your solid sample is a fine powder, a porous medium, or something else. It could take anywhere from a few minutes to a few days to reach uniform moisture distribution in your sample.
Hope this helps.
Mr. Li
you have good enough explained Mrs. Suriya's question
Thanks
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