Adiabatic Saturation

Author: Bruce Wernick
Last updated: 11 November 2015

The definition of the thermodynamic wet bulb temperature can be used to calculate the entering air humidity.

We start with the ideal arrangement known as an adiabatic saturator shown in figure 1. The walls are insulated so there is no heat flow, the outlet air is saturated and the make-up water is arranged to be equal to the leaving air wet bulb temperature.

Figure 1. Theoretical adiabatic saturator

Unsaturated air enters and gains moisture from the sump. We solve the system by applying a energy and mass balance.

Energy IN = Energy OUT
and
mass IN = mass OUT.

m_a h_ai + m_w h_w = m_a h_ao .......Energy conservation

m_w = m_a (w_ao - w_ai) .......Mass conservation

Substitute m_w into the energy equation

m_a h_ai + m_a h_w (w_ao - w_ai) = m_a h_ao

and cancel the common air mass flow

h_ai + h_w (w_ao - w_ai) = h_ao

Rearrange (and notice the new property)

h_ai - w_ai h_w = h_ao - w_ao h_w ....Sigma Heat equation

Before going further, it is worth looking at this equation. The term (h_a - w_a h_w) is property that we call the Sigma heat. Notice that it is the same at inlet and outlet. In other words, the adiabatic saturation process is also a constant sigma heat process.

za = ha - wa hw

za = ha - wa cpw wb

But the inlet enthalpy can be defined as

h_ai = C_pa db_i + w_ai h_wvi

and similarly, the leaving air enthalpy is

h_ao = C_pa db_o + w_ao h_wvo

Substitute for h_ai and h_ao

(C_pa db_i + w_ai h_wvi) - w_ai h_w = (C_pa db_o + w_ao h_wvo) - w_ao h_w

Rearrange...

C_pa db_i + w_ai (h_wvi - h_w) = C_pa db_o + w_ao (h_wvo - h_w)

Now, we can solve for w_ai...

w_ai = [w_ao (h_wvo - h_w) + C_pa (db_o - db_i)] / (h_wvi - h_w)

All of the unknown values can be read directly from the saturation tables.

w_ao = w_s(wb) ...Saturation humidity at the wet bulb

db_o = wb ...Saturated at outlet (leaving dry bulb = web bulb)

h_wvo = h_g(wb) ...Saturated water vapor at wet bulb temperature

h_wvi = h_g(db) ...Saturated water vapor at dry bulb temperature

Finally...

w_ai = [w_s(wb) (h_g(wb) - h_w) - C_pa (db - wb)] / (h_g(db) - h_w)

This is the equation that links the moist air state properties. So, given (Pa,db,wb), we can now calculate w_a and therefore all of the other properties explicitly.

Example...

Given db=35°C and wb=25°C at Pa=101.325 kPa

Table 1. Moist Air Properties
t	w_s
25	0.020170
35	-

Table 2. Saturated Water and Steam properties
t	h_f	h_g
25	104.81	2546.47
35	-	2564.53

w_a = [0.020170 (2546.47-104.81)-1.006 (35-25)]/(2564.53-104.81)

= 15.932 g/kg

Sigma Heat - a useful arrangement.

Our original equation allows for the direct calculation of w_a from db and wb. Now, we need to use db and w_a to get the wet bulb temperature.

Here, Sigma heat gives the ideal choice of property. Sigma heat is a property that is constant for a given wet bulb. So all we need to do is find wb such that the Sigma heat is equal to that at saturation.

Sg_i = Cp_a db + wa_i (h_g,db - h_w)
Sg_o = Cp_a wb + w_sw (h_g,wb - h_w)

We know that wb <= db so a valid starting guess would be wb=db. We also know that wb >= dew point so this is a good 2nd guess. From there, we will interpolate between the last two values.

Example...

Given db=35.0C and w_a=15.946 g/kg, calculate wb.

step	wet bulb	Sg_o-Sg_i
1	35.000	50.385
2	21.209	-14.289
3	24.256	-2.974
4	24.855	-0.585
5	24.971	-0.113
6	24.994	-0.022
7	24.998	-0.004

Calculate dry bulb

In our derivation of the adiabatic cooler, we calculated the entering air humidity W_ai.

Another useful arrangement would be to calculate the entering dry bulb, given the humidity.

C_pa db + w_ai (h_g(db) - h_w) = C_pa wb + w_s(wb) (h_g(wb) - h_w)

Can you see the problem here?

The entering water vapor enthalpy is a function of dry bulb so it can't be extracted.

As a reasonable simplification, we can use a linear fit of enthalpy vs temperature

h_g = H_go + C_pv db

C_pa db + w_ai (H_go + C_pv db - h_w) = C_pa wb + w_s(wb) (h_g(wb) - h_w)

That's better, now we can rearrange for db to get...

db = [C_pa wb + w_s(wb)(h_g(wb) - h_w) - w_ai (H_go - h_w)]/(C_pa + w_ai C_pv)

Let's test our assumption...

Given wb = 25.0°C and W_ai = 15.000 g/kg

db = (1.006×25.0 + 0.020170×(2546.47 - 104.81) - 0.015×(2500.77 - 104.81)) / (1.006 + 0.015×1.82)

= 37.21947°C

Using db = 37.21947°C and wb = 25.0°C gives W_a = 15.002 g/kg

I would say that a 0.013% error is acceptable.