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.
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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%.
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The diagram applies to atmospheric |
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Pressure of 1000 mbar |
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SYMBOLS |
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H = specific enthalpy, kj/kg dry air |
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" = Sp1 : = a D |
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Solute humidity, kg water/kg dry air |
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0 = drv bulb temperature of air, °C |
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0..= wet bulb temperature of air, °C |
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<p = relative humidity, % |
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H = partial pressure of water vapor, mbar |
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P, = quantity of dry air, |
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Kg drv air/m moist air |
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Аhfщx — process direction |
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0.005 |
|
0.010 |
|
0.015 |
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X 0.020 |
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Al>/Ax |
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-5000 -2000 -1000 |
|
000 |
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10000 8000 “000 6500 6000 5^00 |
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« 50000 2000(1 |
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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)
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TABLE 4.7 Thermodynamic Characteristics of Saturated Air for a Total Pressure of 100 kN/m2
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|
Ю |
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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 |
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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 |
34.0 |
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 |
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0.6489 |
20.28 |
1.698 6 |
0.023 79 |
37.6 |
74 |
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0.6132 |
20.44 |
1,799 4 |
0.023 60 |
38,3 |
76 |
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0.5755 |
20.58 |
1.919 9 |
0.023 41 |
39.0 |
78 |
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0.5358 |
20.71 |
2.066 4 |
0.023 23 |
39.8 |
80 |
(continues)
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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 |
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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
From Eq. (4.76) we get the partial pressure of dry air:
TOC o "1-5" h z
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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
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
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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)
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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.
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Ph(8M)-PhPb (Om) Pb P~Ph (8M) P-Ph (&m) P~Ph |
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+ 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.
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■ mvh v |
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9 = 0 |
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> |
|||
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M h j |
:> |
W2^2 |
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> |
|||
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> |
■i? / |
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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)
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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)
-*■ -^ad
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
^k ~ ^ad — *ad) ^
On the other hand,
Hk = cpi9 + x(cphe+ lho) = 0
And
^kad Xad(cpfo 0ad + hio)
Substituting these into the equation above, we obtain by grouping terms appropriately
Cpi(8 — 0ad) XCphiO — ^ad) + cph@ad ~ cpz/^ad) ^ j ^4 J
~xad(^ho cph@ad ~ cpt/^ad) — ®
On the other hand, the vaporization heat of water at temperature #ad is
^(^ad) ^bo Cph^ad ~~ ad (4.135)
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)
O-flad Mad)’
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.
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