# Determination of Air Humidity

The humidity of air can be measured by either the dewpoint of the air or its wet bulb temperature.

Dewpoint means the temperature of saturated water vapor that has the same vapor pressure as the humid air in question. When the total pressure is constant, the constant vapor pressure means the same as the humidity x. In other words, dewpoint is the temperature of saturated air that has the same humidity as the air being considered.

By cooling a certain surface so cold that water starts condensing on it and measuring that temperature, the dewpoint can be measured. Combining this with the measurement of the dry bulb temperature, the state of air can be defined.

 FIGURE 4.10a Mollier diagram, p = 0.875 bar.

Example 4

The dry bulb temperature of air is 20 °C and the dewpoint is 8 °C. What is the relative humidity?

H’p (20 °C) = 0.0234 bar h’p (8 °C) = 0.01072 bar phip= p’h (20 °C) = 0.01072/0.0234 = 0.458 = 45.8%.

 The diagram applies to atmospheric

 Pressure of 1000 mbar

 SYMBOLS

 H = specific enthalpy, kj/kg dry air

 " = Sp1 : = a D

 Solute humidity, kg water/kg dry air

 0 = drv bulb temperature of air, °C

 0..= wet bulb temperature of air, °C



 H = partial pressure of water vapor, mbar

 P, = quantity of dry air,

 Kg drv air/m moist air

 Аhfщx — process direction

 0.005

 0.01

 0.015

 X 0.020

 Al>/Ax

 -5000 -2000 -1000

 000

 10000 8000 “000 6500 6000 5^00

 « 50000 2000(1

 0 5 10 15 20 25 30 35 40 Dry bulb temp °C FIGURE 4.10c Mollier diagram.

Example 5

The air pressure in a room is 950 mbar, the temperature is 20 °C, and the relative humidity <p = 40%. Define the dewpoint of the room.

H^(20 °C) = 0.0234 bar ph = 0.4 • 0.0234 bar = 0.00936 bar

0, is found by using the tables; h’p (Oj) = 0.00936 bar, and thus 0, = 6.0 °C.

The total pressure thus has no importance. If this result is sought from a Mollier diagram by finding the intersection of the humidity line (,y = humidity of air = constant) and the saturation curve, which gives the dewpoint temperature, a diagram constructed for a pressure of 950 mbar should be used. A decent ap­proximation can be found from a diagram constructed for pressure p — 1 bar.

When a damp cloth is laid in an air flow, it settles after a certain time to an equilibrium temperature, the so-called wet bulb temperature (0M), which is determined through heat and mass transfer. Negotiating the heat flow ob­tained by radiation and conduction, the heat balance of the wet cloth in a sta­tionary situation can be expressed as

A(0-HM) = m’, l(0M), {4.1 It)

Where 0 is the temperature of the air flow (°C), m’/’ is the water flow vaporizing from the damp cloth (kg/m2 s), /(0M) is the vaporization heat of water at temper­ature 0M (J/kg), and a is the convective heat transfer coefficient (W/m2 °C).

Air D (m2/s), and the heat conductivity of humid air A (W/m °C). In Table 4.7 the thermodynamic properties of saturated air, including the diffusion factor and the heat conductivity are presented.

Substituting Eqs. (4.112) and (4.113) into Eq. (4.111), we obtain

0- 0M = —^Le /(flM)ln……………. (/;………… , (4.116)

Pc-pRr ‘ P-h’p (0MY [

From which we observe that the heat transfer factor a has been reduced. The only factor in Eq. (4.116) depending on the airflow conditions of the measure­ment is the power n in the Lewis number. Because the value of the Lewis num­ber is very close to 1, the effect of n is very small.

The wet bulb temperature 0M can be solved for from Eq. (4.116) when the state of the air, the temperature t, and the partial pressure of water vapor ph are known. Inversely, if the temperature t and the wet bulb temperature f)Vl are known, the partial pressure and consequently the humidity of air can be found from Eq. (4.116).

Given the temperature of air 0 = °C and the wet bulb temperature 6M = 10 °C, calculate the humidity x, when the pressure of air is (a) p = I bar and (b) p = 0.90 bar.

By solving for the steam pressure p/, from Eq. (4.116), we obtain

PcpRT 1 1

 (4.117)

P ~ph = (p — hp (0M) )exp (0-0M)

Mh p Le1’” K0M)

The diffusion factor D, which is contained in the Lewis number, is in­versely related to the total pressure:

D = jf(T). (4.118)

From Table 4.7 we obtain the diffusion factor at the temperature of 10 °C and pressure p = 1.0 bar: D(10 °C, p = 1.0 bar) = 23.3 ■ 10~6m2/s. The diffu­sion factor at the same temperature but a pressure p — 0.9 bar is, according to Eq. (4.118), D(10 °C, p(0.9 bar)) = 25.9 • 10’6m2/s.

For heat conductance there is a dependency on pressure almost like that of Eq. (4.118), so for good accuracy it is a valid approximation that D/A = g(T). From Table 4.7 we have A(10 °C, /;(1,0 bar)) = 0.02466 W/m K, and so D/A — 9.45 ■ 10-4m3 K/J. Vaporization heat and the pressure of saturated steam are, from Table 4.7;

L(0M) = 1(10 °C) = 2477 ■ 103 J/kg,

Pb(0M) = Ph№ °c) = 0.012271 bar.

Temperature T is taken as a mean boundary layer temperature:

T = (20 + 10)/2 + 273.15 = 2,88.15 K

Heat capacity,

P^p PjCpi ^ PhCph Pii^pj + X Ј i>h) i (^-119)

TABLE 4.7 Thermodynamic Characteristics of Saturated Air for a Total Pressure of 100 kN/m2

 Temperature (°С) Humidity (kg HjO/kg dry air) Water Vapor Partial Pressure (kPa) Water Vapor Partial Density (kg/m3) Water Vaporization Heat (kj/kg) Mixture enthalpy (kj/kg dry air) Dry air partial density (kB Kinematic viscosity (I0< mVs) Specific Heat (ki/Kkg) Heat conductivity (W/m K) Diffusion Factor Water-air (IO"mVs) Temperature CP 0 0.003 821 0.6108 0.004 846 2 500.8 9.55 1.285 13.25 1.0108 0.023 80 22.2 0 2 0.004 418 0.705 4 0.005 557 2 495.9 13.06 1.275 13.43 1.012 0 0.024 13 22.4 2 4 0.005 100 0.812 9 0.006 358 2 491.3 16.39 1.264 13.61 1.013 4 0.024 27 22.6 4 6 0.005 868 0.934 6 0.007 257 2 486.6 20.77 1.254 13.79 1.014 9 0.024 40 22.8 6 8 0.006 749 1.072 1 0.008 267 2 481.9 25.00 1.243 13.97 1.016 7 0.024 54 23.1 8 10 0.007 733 1.227 1 0.009 396 2 477.2 29.52 1.232 14.15 1.018 6 0.024 66 23.3 10 12 0.008 849 1.401 5 0.010 66 2 472.5 34.37 1.221 14.34 1.020 8 0.024 78 23.6 12 14 0.010 105 1.597 4 0.012 06 2 467.8 39.57 1.211 14.52 1.023 3 0.024 90 23.9 14 16 0.011 513 1.816 8 0.013 63 2 463.1 45.18 1.199 14.71 1.026 0 0.025 00 24.2 16 18 0.013 108 2.062 0.015 36 2 458.4 51.29 1.188 14.89 1.029 1 0.025 11 24.5 18 20 0.014 895 2.337 0.017 29 2 453.1 57.86 1.177 15.08 1.032 5 0.025 20 24.8 20 22 0.016 892 2.642 0.019 42 2 449.0 65.02 1.175 15.27 1.036 4 0.025 29 25.2 2? 24 0.019 131 2.982 0.021 77 2 442.0 72.60 1.154 15.46 1.040 7 0.025 37 25.5 24 26 0.021 635 3.360 0.024 37 2 439.5 81.22 1.141 15.65 1.045 5 0.025 44 25.9 26 28 0.024 435 3.775 0.027 23 2 434.8 90.48 1.129 15.84 1.050 9 0.025 08 26,3 28 .10 0.027 558 4.241 0.030 36 2 430.0 100.57 1.116 16.03 1.056 9 0.025 56 26.6 30 52 0.031 050 4.753 0.033 80 2 425.3 111..58 1.103 16.22 ! .063 ,5 0.025 61 27.0 32 .14 0.034 950 5.3 IS 0.037 58 2 420.5 І 23.72 1.090 16.41 1,071 0 0.025 65 27.4 .J’-t 36 0.039 289 5.940 0-041 71 2 415.8 136.99 ! .076 16.61 1.079 3 0.025 67 27.8 І Cl 38 0.044 136 6.624 0.046 22 2 411.0 151.60 1.061 16.80 1.088 5 0.025 69 28.3 38 40 0.049 532 7,375 0.051 14 2 406.2 167.64 1,046 17.00 1.098 9 0.025 69 29.7 40
 Ю

 0.055 560 8.198 0.056 50 2 401,4 0.062 278 ‘».010 0.062 33 2 396.6 0.069 778 10.085 0.068 67 2 391.8 0.078 146 Il.161 0.075 53 2 387.0 0.087 516 12.335 0.082 98 2 382.1
 0.098 018 13.613 0.091 03 2 377.3 0.109 76 15.002 0.099 74 2 372.4 0.122 97 16.509 0.109 1 2 367.6 0.137 90 18.146 0.119 3 2 362.7 0.154 72 19.92 0.130 2 2 357.9
 0.173 80 21.84 0.141 9 2 353.0 0.195 41 23.91 0.154 5 2 348.1 0.220 21 26.14 0.168 0 2 343.1 0.248 66 28.55 0.182 6 2 338.2 0.281 54 31.16 0.198 1 2 333.3
 0.319 66 33.96 0.214 6 2 328.3 0.364 68 36.96 0.232 4 2 ,323.3 0.417 90 40.19 0.251 4 2 318.3 0.480 48 43.65 0.271 7 2 313.3 0.559 31 47.36 0.293 3 2 308.3

 1.030 17.20 I. 110 3 0.025 68 29.1 42 ! ,014 17.39 1.123 2 0.026 66 29.6 44 0.9979 16.59 1.1375 0.025 63 30,0 46 0.9791 17.79 1,153 4 0.025 58 30.5 48 0,9606 ! 7.99 1.171 3 0.025 52 30.9 50
 0.9411 18.19 1.191 3 0.025 45 31.4 52 0.9207 18.39 1.213 7 0.025 36 31.9 54 0.8999 18.59 1.238 9 0.025 26 32.4 56 0.8768 18.79 1.267 3 0.025 14 32.9 58 0.8532 18.99 1.299 4 0.025 01 33.4 60
 0.8283 19.19 1.335 7 0.024 87 34 62 0.8021 1.9.38 1.377 0 0.024 71 34.5 64 0.7746 19.57 1.424 1 0.024 55 35.1 66 0.7456 19,76 1.478 2 0.024 37 35.7 68 0,7150 19.94 1.541 8 0.024 18 36.3 70
 0.6829 20.01 1.613 2 0.023 99 36.9 72 0.6489 20.28 1.698 6 0.023 79 37.6 74 0.6132 20.44 1,799 4 0.023 60 38,3 76 0.5755 20.58 1.919 9 0.023 41 39.0 78 0.5358 20.71 2.066 4 0.023 23 39.8 80

(continues)

 Temperature ("Q Humidity (kg H20/kg dry air) Water Vapor Partial Pressure (kPa) Water Vapor Partial Density (kg/mJ) Water Vaporization Heat (4/kg) Mixture enthalpy (kj/kg dry air) Dry air partial density ОЧлгчЛ"3) Kinematic Viscosity (I04m3/s) Specific heat (kJ/K kg) Heat conductivity (W/m K) Diffusion factor water-air (104m3/s) Temperature (”C) 82 0.655 73 51.33 0.316 2 2 303.2 1 820.46 0.4939 20.81 2.247 7 0.023 07 40.7 82 84 0.777 81 55.57 0.340 6 2 298.1 2 148.92 0.4497 20.90 2.476 7 0.022 94 41.5 84 86 0.937 68 60.50 0.366 6 2 293.0 2 578.73 0.4031 20.96 2.773 9 0.022 85 42.5 86 88 1.152 44 64.95 0.394 2 2 287.9 3 155.67 0.3542 20.99 3.170 8 0.022 81 43.6 88 90 1.458 73 70.11 0.423 5 2 282.8 3 978.42 0.3026 20.99 3.730 4 0.02283 44.7 90 92 1.927 18 75.61 0.454 5 2 277.6 5 236.61 0.2482 20.94 4.574 0.022 95 46.0 92 94 2.731 70 81.46 0.487 3 2 272.4 7 395.49 0.1909 20.84 5.987 0.023 18 47.4 94 96 4.426 70 87.69 0.522 1 2 267.1 11 944.39 0.1305 20.69 8.820 0.023 55 49.0 96 98 10.303 06 94.30 0.558 8 2 261.9 27 711.34 0.06694 20.47 17.338 0.024 09 50.8 98 100 • 101.325 0.597 7 2 256.7 • 0 20.08 . 0.024 86 52.8 100

Has to be iterated because steam pressure ph and therefore humidity x are at this stage unknown. When humidity is low, as in this example, an approxima­tion is pcp = PiCpj. The calculation can be repeated when needed using the steam pressure from Eq. (4.117). In this example, we don’t do that. For calcu­lating the density and heat capacity, we will use an approximation pcp s p„ so /? s pj and cp = Cpj. From this approximation it follows that

^ (4.1201

Mh p Mh p Mh P”

Whose value is

^ct,= 1617J/kg ”C.

From Eq. (4.76) we get the partial pressure of dry air:

TOC o "1-5" h z

 La) Pi

_PjMi_pMi_ 105.0,029 _1211 k, j RT ~ RT 8.314 -288.15 B ’

/b) 0=P№isЈMi = 0-9 • 105 • 0.029 = 089 . /m3

(o) Pi rt ~ RT 8.314-288.15 g ’

The Lewis number in cases a and b is as follows:

(a) Le = f^- = PiCbЈ = 1.211 1006 • 9.45 ■ 10-4 = 1.151

‘ A ; A

It is assumed that n = 0.5, so Le1-” = Le1 °"’ = 1.073

(b) Le = s PtcD = 1.089 — 1006 • 9.45 • 10 4 = 1.035

A ‘ A

Thus Le1 = 1.017. Substituting all values into Eq. (4.117), we obtain

P-pi, = (1.0-0.012271)

Bar = 0.99376 bar

 Exp

(20 — 10) • 1617 • — rTni.——————- 5

1.073 7477 ? . 1fl

2477.2 — 10J

ph = (1.0 — 0.99376) bar = 0.00624 bar = 624 Pa

X = 0.6220 = 0.6220 • — = 0.00391

P~Ph 10-624

(b) p — ph = (0.9-0.012271)

 Bar = 0.89345 bar

Exp (20 — 10) ■ 1617 • ——— 5

L i-Ol7 2477.2-103

Ph = (0.9 — 0.89345) bar = 0.00655 bar = 655 Pa

X = 0.6220 = 0.6220 • 655— = 0.00456

P^Pb 10 — 655

Comparing the results for (a) and (b), we see that pressure has a consider­able effect on the result. This is important to remember, especially in industrial ventilation and process measurements with notable underpressures and over­pressures.

We will now derive an approximation for Eq. (4.116) that can be used when the partial pressure of water vapor in air is low compared with the total pressure.

 Ph(8M)-PhPb (Om) Pb P~Ph (8M) P-Ph (&m) P~Ph

 + Pk (8m) Pb P-Ph (0m)

First we note that with fairly good accuracy it is valid that

|n-r-^ = ln

P ~Pb ( 8m)

Where using Eq. (4.83) leads to the approximation

Where x'(6M) is the humidity of saturated air at temperature 6M when the total pressure of air is p.

Substituting approximations (4,120) and (4.121) to Eq. (4.116), we obtain

 (4.122)

X'(0M)-x _ cp 1 8m KOm) Le1"”

The Lewis number for air is approximately 1 (see Example 6), so with good accuracy Le1_" h 1 , and we get an approximation from Eq. (4.122):

 (4.123)

X'(8M)-x

T-8M 1(0M)

Above we considered the question of which temperature the damp cloth settles to when it is thermally insulated against all surroundings but the air­flow, and when it can be assumed that there is no radiation heat transfer be­tween the cloth and the airflow. In this consideration the state of the air has been constant.

If, instead, the air is damped adiabatically with the wet cloth, so that the state of the air varies, the cloth will settle to a slightly different temperature. Each state of air (6, x) is represented by a certain wet bulb temperature 9M, which can be calculated from Eq. (4.116) or its approximation (4.123), when the partial pressures of water vapor are low compared with the total pressure. When the state of air reaches the saturation curve, we have an interesting spe­cial case. Now the temperatures of the airflow and the cloth are identical. This equilibrium temperature is called the adiabatic cooling border or the thermo­dynamic wet bulb temperature (0aj).

When air is humidified with water flow mv and when the incoming and outgoing humid airflows are denoted m, and m2, the energy balance of the conditioning chamber can be written as

Equation (4.124) is illustrated in Fig. 4.11.

 ■ mvh v 9 = 0 > M h j :> W2^2 > > ■i? /
 FIGURE 4.1 i Energy balance for an adiabatic conditioning chamber.

In Eq. (4.124) the incoming enthalpy flow of humid air is

Tn1h i = mnhll + mblhhl (4.12.5a)

And the outgoing enthalpy flow is

M2h2 = mi2bt2 + • (4.125b)

While the dry air flow stays constant, it can be written that

Riij = thj i = m, 2 (4.126)

And using Eq. (4.126) the enthalpy flows (4.125a-b) can be written as

Fhybi = ntjhki (4.127a)

Tn2h2 = mibk2, (4.127b)

Where according to Eq. (4.87)

Hk | = h, j +x[hhi (4.128a)

H kl = ^i2 + (4.128b)

When all the water fed to the conditioning chamber vaporizes, the follow­ing humidity balance is valid:

= ™b2 — f^hl ~ — *i) • (4.129)

Substituting Eqs. (4.127a-b) and (4.129) into the energy balance (4.124), we obtain

4^ = H.„ (4.130)

Ax 1

Where Abk = bk2~bkl and Ax = x2-xx. When air is humidified adiabati — cally with water with temperature the enthalpy of water is now

Hv = cpv6A d (4.131)

If, in addition, the air is humidified so that it reaches the saturation point, with the corresponding temperature 0arf, we will now use the notations

Bkl = hkad, „ , ,

(4.132)

 FIGURE 4.12 The state change of air in adiabatic humidification. A is the initial state and B is the sat­urated final I state.

Using the notations in (4.132) we can without danger of mix-up leave out the subscript 1 of the incoming point and write = x and hkl = hk. Using these notations and Eq. (4.131), Eq. (4.130) can be written as

H. zhtel — Cpv0sd. (4.133)

So the state point of air in the Mollier diagram is shifted in a direction where the dependency between the enthalpy and the humidity change according to the Eq. (4.133) is valid. This result is illustrated in Fig. 4.12.

Equation (4.133) can be formally written in a form resembling Eq. (4.123). To demonstrate this, we first write Eq. (4.133) as

On the other hand,

Hk = cpi9 + x(cphe+ lho) = 0

And

Substituting these into the equation above, we obtain by grouping terms appropriately

On the other hand, the vaporization heat of water at temperature #ad is

And the specific heat of humid air per kilogram of dry air is

Substituting Eqs. (4.135) and (4.136) in Eq. (4.134), we get

— = Stih (4 1 37)

Which is equivalent to Eq. (4.133).

Equation (4.137) is almost exactly the same as the approximation equa­tion (4.123) derived for wet bulb temperature. When the partial pressure of water vapor is low compared with the total pressure—in other words when the humidity x is low—the specific heat of humid air per kilogram of humid air, cp, and the specific heat of humid air per kilogram of dry air, cpk, are al­most the same: cp Ј cpk. Therefore, in a situation where the humidity is low and Les 1, the thermodynamic wet bulb temperature is very nearly rhe same as the technical wet bulb temperature 6M.

Example 7

(Table 4.7)Draw in the Mollier diagram at the 14 °C point of the satura­tion curve (a) the state change line of the adiabatic humidification and (b) an

Auxiliary line, associated with the wet bulb temperature measurement, by means of which the state can be defined. The pressure of air is p = 1 bar.

PU 14 °C) = 0.01597 bar (Table 4.7)

*ad = 0.6220 ■ ^Ol^Z— = 0.01009 = x’(0M)

/(14 °C) = 24678 kj/kg (Table 4.7)

Pi = 1.211 kg/m3 (Table 4.7)

Pb = 0.01206 kg/m3 (Table 4.7)

P = Pi + Ph = 1.223 kg/m3 pc, p = + phcph = 1.211 ■ 1.006 + 0.01206 ■ 1.85 = 1.241 kj/m3 °C

Cp = 1.015 kj/kg °C

D = 23.9 ■ 10~6 m2/s (Table 4.7)

A = 0.0249 Wm/°C

I p = D()C > = 23.9 ■ 10~6 • 1241 = i 191 A 0.0249

N = 0.5

(a) We choose an auxiliary point on the isotherm 9 = 25 °C. According to Eq.

(4.137) , ‘

*ad-* = ‘ (25 “ M) = 0-00457 ’

So it follows that the place of the auxiliary point is * = 0.00552.

(b) For the auxiliary point on the isotherm 0 = 25 °C, according to Eq. (4.122),

X (0m) ~x ~ 2467.8 191°’^ ~ = 0-00415 ?

And it follows that the place of the auxiliary point is x = 0.00594. The results are illustrated in Fig. 4.13.

 50000 20000 10000 8000 7000 6500 6000 5500

 For wet bulb temperature. Ј = constant enthalpy. A adiabatic humfdification line.

The enthalpy of humid air responding to the point 14 °C on the satura­tion curve is

Hklld = 1.0016- 14+ 0.01009 (1.85 — 14+ 2501) = 39.58 kj/kg.

The humidity at the temperature of 25 °C corresponding to this enchalpy is determined from the equation

39.58 = 1.006 ■ 25 + *( 1.85 -25 + 2501),

And so a — = 0.00567.

Comparing this value with the {a) and (b) point results of Example 7, we discover that the line of constant enthalpy lies between the determination line of wet bulb temperature and the adiabatic humidification line. The nearer the Lewis number is to 1, the nearer the wet bulb temperature is to the adiabatic humidification temperature.

In practical calculations the iVlollier diagram’s constant enthalpy line can be used as the auxiliary line for the wet bulb temperature line to a satisfactory accuracy. The intersection of the constant enthalpy line with the isotherm re­sponding to the temperature of air gives the humidity of air. For more accurate calculations Eq. (4.116) should be used, or its approximation (4.122) when the steam pressures (ph and pj.’,) are low compared to the total pressure of air (p).

As an example of using a Mollier diagram in defining the state of air, we can take a typical measurement from the local exhaust hood of a paper machine. The temperature of the exhaust air is 82 °C and its wet bulb temperature 60 °C. In Fig. 4.10d we move from the saturation curve at the point 60 °C straight up along the constant enthalpy line ( = 460 kj/kg d. a.) until we reach the isotherm

0 = 82 °C. The intersection represents the state of air, and from Fig. 4.10<i we see that to the accuracy of the diagram x = 0.14 and the corresponding humidity relation f = x/x'(S2 °C) = 0.20. Based on the values x = 0.14 and p = 1.0 bar the relative vapor pressure <p can be calculated. From Eq. (4.84) we have pi, = 0.183 bar and from Table 4.7 p’h (82 °C) = 0.5133 bar; then on the ba­sis of definition (4.110) <p = (ph/p’h ) = 0.356 = 35.6%. We see that the values of f and <p clearly differ from each other.

According to Eq. (4.122) when Le= l, 1(0M) = 2450 kj/kg and CpBt.0 kj/kg °C:

Ph = Ph (««)- 6.6 • 10 4 ■ p ■ (0- 0M)i (4.138)

And by means of this the state of air can be approximately calculated. Often we call the temperature of air the dry bulb temperature to distinguish it from the wet bulb temperature.

It is important to emphasize that, especially in process measurements, radiation can have an essential influence on the wet bulb temperature, and therefore generally the wet bulb temperature is dependent on the measurement device and the method of measurement. If the airflow is very low, the radiation can have a remarkable contribution in addition to the convective heat transfer. Basically, an equation analogous to Eq.

(4.138) can be empirically determined for each wet bulb temperature and method of measurement.