2
$\begingroup$

I'm searching for an approach on how to calculate the adiabatic temperature loss of air when increasing its humidity by a humidifier.

Assumed we have the following setup:

Air measuring position 1 (before humidifier):

$T = 30{ °C}$

$x = 10\text{ g/kg (absolute humidity)}$

Air measuring position 2 (after humidifier):

$T = \text{? °C}$

$x = 20\text{ g/kg (absolute humidity)}$


The question now is: How can I calculate the air temperature at measuring position 2? With a Mollier-diagram it would be easy to figure out, but I want to solve it the analytical way.

I guess there is something possible with the enthalpy of air:

$$h_1=h_2$$

$$Q_1\cdot x_1\cdot T_1=Q_2\cdot x_2\cdot T_2$$

$$\frac{Q_1\cdot x_1\cdot T_1}{Q_2\cdot x_2}=T_2$$

What is missing? :-)

$\endgroup$
5
  • $\begingroup$ there are air conditioning handbooks in which the answers to questions like this are tabulated in charts and graphs. Do you have access to one of these? $\endgroup$ Mar 12 '19 at 0:50
  • $\begingroup$ Will the specific heat capacity of water be needed? $\endgroup$
    – Solar Mike
    Mar 12 '19 at 6:59
  • $\begingroup$ @nielsnielsen: Yes, I have access to fitting tables and graphs (e.g. Mollier-diagrams). However, I would prefer to calculate it by hand to paste it into an excel-file. $\endgroup$
    – Dave
    Mar 12 '19 at 12:47
  • $\begingroup$ @SolarMike: No, the specific heat capacity of water won't be needed. The water will have $7\text °C$ when getting injected in the humidifier. $\endgroup$
    – Dave
    Mar 12 '19 at 12:47
  • $\begingroup$ So the water won't absorb any heat from the air that is at 30 Deg C? Doesn't everything tend to the same temperature over time? $\endgroup$
    – Solar Mike
    Mar 12 '19 at 12:54
1
$\begingroup$

You are missing an equation of state that can be solved analytical or preprogram SRK or PR in a TI calculator when make it solve the enthalpy for you (enthalpy 1=enthalpy 2 but the phases are different and so is the volume), because what you want will have changes in the molar Volume and it has to be solved, that is why you solve it with a Mollier Diagram or Steam tables; Someone already solved/measured those properties so any engineer (without access to a simulator) can solve it with a pen and a napkin.

You can try solving it yourself (and it is quite fun) but please by all means use a computer because you will be wasting time for a simple problem otherwise.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.