Originally Posted by **charon010** Hello everyone!
I have a practical problem which I have to solve.
The problem is the following:
We want to maintain constantly a relative humidity of 65% at 10 celsius degrees in a room of 10 square meters territory.
To maintain this humidity we have an ultrasonic humidifier which is able to vaporate 250 ml/h water maximum. (pressure in the room is 1 atm of course)
The question is: to how big hourly evaporation should we set the humidifier to maintain the required humidity?
My attempt:
With the help of the psychrometric chart of humidity we can see that at 10 celsius and at 65% RH 1 kg of air contains about 7 g of water.
So we have to maintain 7g of vapor in the air constantly.
But I don't know at which rate the humidity decreases, and how the territory of the room affects it. So basicilly it is an equilibrium process and we need to find the equilibrium rate of evaporation of the humidifier.
Thank you for your help |

Aim: maintain a relative humidity of 0.65 with changes in room temperature by changing the mass of water vapour in the air

Assume:

Volume is constant = 10h m3

Pressure is constant = 1 atm = 101325 Pa

Water vapour is an ideal gas.

Absolute humidity is defined as the mass of water vapour per unit volume.

$\displaystyle H = \frac{m_v}{V}$

This can be defined in terms of the partial vapour pressure, $\displaystyle P_v$,

$\displaystyle P_v V = m_v R_v T$,

where $\displaystyle R_v = 4615 J kg^{-1} K^{-1}$.

Relative humidity, U, is defined as the ratio of partial vapour pressure to the total pressure,

$\displaystyle U = \frac{P_v}{P} = \frac{m_v R_v T}{PV}$

We can calculate the mass of water vapour in the room by solving for $\displaystyle m_v$ with known U = 0.65. We call this $\displaystyle m_0$, which will be at a room temperature $\displaystyle T = T_0 = 10$ deg Celsius = 283.15 K.

$\displaystyle m_0 = \frac{UPV}{R_v T_0}$

Then, you can work out the change in $\displaystyle m_v$ for a given change in temperature using

$\displaystyle \frac{m_0 R_v T_0}{PV} = \frac{m_1 R_v T_1}{PV}$

$\displaystyle m_0 T_0 = m_1 T_1$

So... the amount of mass of water to remove is

$\displaystyle \Delta m = m_0 - m_1$

$\displaystyle = m_0 - \frac{m_0 T_0}{T_1}$

$\displaystyle = m_0 \left(1 - \frac{T_0}{T_1}\right)$

For a typical 3m high room, I get

$\displaystyle m_0 = \frac{0.65 \times 101325 \times 30}{4615 \times 283.15} =$ 1.512 kg

Therefore, your equation for amount to remove is

$\displaystyle \Delta m = 1.512 \left(1 - \frac{283.13}{T_1}\right)$

Just make sure that your target temperature, $\displaystyle T_1$, is in Kelvin. If you wanted to, you could approximate the mass removal as a linear removal over a duration $\displaystyle \Delta t$ by replacing the $\displaystyle \Delta m$ with $\displaystyle \dot{m}\Delta t$ where $\displaystyle \dot{m}$ is the rate of change of mass removal of your vaporator.

In real rooms, the pressure and temperature are varying with time, so it's not as simple as this, but equations like this should give you a head-start for solving more sophisticated problems and afford you some ball-park figures at least.