Question

A large amount of boiling water is used to cook mandi rice in a kitchen, but now the room is very humid. Next to the kitchen is a dining area of 50 cubic metres, where the temperature is 30 degree celcius and the relative humidity 90.01 percent.For the guests to have a more pleasant experience, design a system that can dehumidify the dining area by 62 percent relative humidity in an hour, just before the arrival of the guests

Answer 1

Given:

• Volume of dining room = 50 m^3
• Temp = 30 deg C
• Present RH = 90.01 %
• Desired RH = 90.01 - 62 = 28.01 %

From psychrometry chart,

Moisture content @ 30 deg C & 90.01% RH is 24.48 gm / Kg of air.

Moisture content @ 30 deg C & 28.01% RH is 7.42 gm / Kg of air.

Amount of extra moisture to be removed in order to maintain desired condition = 24.48 - 7.42 = 17.016 gm / Kg of air.

We know density of air = 1.225 Kg/m^3

Density = mass / volume

So, mass of air in the room = density * volume = 1.225 * 50 = 61.25 kg

Total moisture to be removed in order to maintain desired condition = 61.25 * 17.016 = 1042.475 gm.

This amount of moisture has to be removed in one hour (or 60 minutes) = 1042.475 / 60 = 17.374 gm/min.

Hence, a dehumidifier of 17.374 gm/min dehumidification rate has to be used.