Idea Transcript
Chapter O ne
Psychrom etry
Chapter One Psychrometry 1.1- Introduction:The psychrometry is that branch of engineering science, which deals with the study of moist air i.e. dry air mixed with water vapour or humidity. It is also includes the study of behavior of dry air and water vapour mixture under various sets of conditions. Though the earth's atmosphere is a mixture of gases including nitrogen (N2), oxygen (O2), argon (Ar) and carbon dioxide (CO2), yet for the purpose of psychrometry, it is considered to be a mixture of dry air and water vapour only.
1.2- Psychrometric Terms:Though there are many Psychrometric terms, yet the following are important from the subject point view: 1. Dry air: The pure dry air is a mixture of a number of gases such as nitrogen, oxygen, carbon dioxide, hydrogen, argon, helium etc. but the nitrogen and oxygen have the major portion of the combination. The dry air is considered to have the composition as given in the following table:
Table (1.1) Composition of dry air No.
Constituent
By volume
By mass
Molecular mass
1.
Nitrogen(N2)
78.03%
57.47%
28
2.
Oxygen(O2)
20.99%
23.19%
32
3.
Argon(Ar)
0.94%
1.29%
40
4.
Carbon-dioxide (CO2)
0.03%
0.05%
44
5.
Hydrogen(H2)
0.01%
--
2
The molecular mass of dry air is taken as 28.966 and the gas constant of air (Ra) is equal to 0.287 kJ/kg K or 287 J/kg K. The molecular mass of water vapour is taken as 18.016 and the gas constant for water vapour (Rv) is equal to 0.461 kJ/kg K or 461 J/kg K. Notes: (a) The pure dry air does not ordinarily exist in nature because it always contains some water vapour. (b) The term air, wherever used in this text, means dry air containing moisture in the vapour form. (c) Both dry air and water vapour can be considered as perfect gases because both exist in the atmosphere at low pressure. Thus all the perfect gas terms can be applied to them individually. (d) The density of dry air is taken as 1.293 kg/m3 at pressure 1.01325 bar or 101325 kN/m2 and temperature 0ºC (273 K).
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2.Moist air: it is a mixture of dry air and water vapour. The amount of water vapour depends upon the absolute pressure and temperature of the mixture.
3.Saturated air: It is a mixture of dry air and water vapour. When the air has diffused the maximum amount of water vapour into it. The water vapour, usually, occurs in the form of superheated steam as an invisible gas. However, when the saturated air is cooled, the water vapour in the air starts condensing, and the same may be visible in the form of moist, fog or condensation on cold surfaces. 4.Degree of saturation: It is the ratio of actual mass of the water vapour in unit mass of dry air to the mass of water vapour in the same mass of dry air when it is saturated at the same temperature.
5.Humidity: It is the mass of water vapour present in 1 kg of dry air, and generally is expressed in terms of gram per kg of dry air (g/kg dry air). It is also called specific humidity or humidity ratio. 6.Absolute humidity: It is the mass of water vapour present in 1 m3 of dry air, and is generally expressed in terms of gram per cubic-meter of dry air (g/m3 of dry air). It is also expressed in terms of grains per cubic meter of dry air. Mathematically, one kg of water vapour is equal to 15430 grains.
7.Relative humidity: It is the ratio of actual mass of the water vapour in a given volume of moist air to the mass of water vapour in the same volume of saturated air at the same temperature and pressure. It is briefly written as RH.
8.Dry bulb temperature: it is the temperature of air recorded by thermometer, when it is not affected by the moisture present in the air. The dry bulb temperature (briefly written as DBT) is generally denoted by td or tdb. 9.Wet bulb temperature: it is the temperature of air recorded by thermometer, when its bulb is surrounded by a wet cloth exposed to the air. Such a thermometer is called *wet bulb thermometer. The wet bulb temperature (briefly written as WBT) is generally denoted by tw or twb.
* A wet bulb thermometer has its bulb covered with a piece of soft cloth (or silk wick) which is exposed to the air. The lower part of this cloth is dipped in a small basin of water. The water from the basin rises up in the cloth by capillary action, and then gets evaporated. It may be noted that if relative humidity of air is high (i.e. the air contains more water vapour), there will be little evaporation and thus there will be a small cooling effect. On the other hand, if relative humidity of air is low (i.e. the air contains less water vapour), there will be more evaporation, and thus there will be more cooling effect.
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10. Dew point temperature: It is the temperature of air recorded by thermometer, when the moisture (water vapour) present in it begins to condense. In other words, the dew point temperature is the saturation temperature (tsat) corresponding to the partial pressure of water vapour (pv). it is, usually, denoted by tdp. Since pv is very small, therefore saturation temperature by water vapour (pv) is also low (less than the atmospheric or dry bulb temperature). The water vapour in air exists in the superheated state and the moist air containing moisture in such a form (i.e. superheated state) is said to be unsaturated air. When the partial pressure of water vapour (pv) is equal to the saturation pressure (ps), the water vapour is in dry condition and the air will be saturated air. Notes: a) The dew point temperature is the temperature at which the water vapour begins to condense. b)
For saturated air, the dry bulb temperature, wet bulb temperature and dew point
temperature is same.
11. Dew point depression: It is the difference between the dry bulb temperature and dew point temperature of air. 12. Wet bulb depression: It is the difference between the dry bulb temperature and wet bulb temperature at any point. The wet bulb temperature indicates relative humidity of air.
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13. Psychrometer: There are many types of psychrometers, but the sling psychrometer as shown in Fig.(1.1) is widely used. It consists of a dry bulb thermometer and a wet bulb thermometer mounted side by side in a protective case that is attached to handle by a swivel connection so that the case can be easily rotated. The dry bulb thermometer is directly exposed to air and measures the actual temperature of the air. The bulb of the wet bulb thermometer is covered by a wick thoroughly wetted by distilled water. The temperature measured by this wick covered bulb of a thermometer is the temperature of liquid water in the wick and is called wet bulb temperature. The sling psychrometer is rotated in the air for approximately one minute after which the readings from both the thermometers are taken. This process is repeated several times to assure that the lowest possible wet bulb temperature is recorded.
(a)
Fig.(1.1) Sling and aspiration psychrometers (a) Sling psychrometer; (b) Aspiration psychrometers
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1.3-Dalton's law of partial pressure: It state, "The total pressure exerted by the mixture of air and water vapour is equal to the sum of the pressures, which each constituent would exert, if it occupied the same space by itself". It is possible to show that if Dalton's law holds, each component of the mixture obeys the general gas law. As a consequence, it is sometimes more convenient to re-express the law in two parts: (i) The pressure exerted by each gas in a mixture of gases is independent of the presence of the other gases, and (ii) The total pressure exerted by a mixture of gases equals the sum of the partial pressures. Figure (1.2) illustrates this. It shows an air-tight container under three different conditions, from which it can be seen that the partial pressures exerted by the water vapour in (b) and (c) are equal, as are those exerted by the dry air in (a) and (c) and, that in (a), (b) and (c), the total pressure equals the sum of the partial pressures. As in the two gas laws already considered, Dalton's law agrees with the results achieved by the kinetic theory of gases and, to some extent, finds substantiation in experiment.
Fig.(1.2) Dalton's law of partial pressure referred to a mixture of dry air and water vapour.
In other words, the total pressure exerted by air and water vapour mixture is equal to the barometric pressure. Mathematically, barometric pressure of the mixture, pb=pa+pv where
(1.1)
pa=Partial pressure of dry air, and pv= Partial pressure of water vapour.
1.3.1- STANDARD ATMOSPHERE The temperature and barometric pressure of atmospheric air vary considerably with altitude as well as with local geographic and weather conditions. The standard atmosphere gives a standard of reference for estimating properties at various altitudes. At sea level, standard temperature is 15°C; standard barometric pressure is 101.325 kPa. Temperature is assumed to decrease linearly 5
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with increasing altitude throughout the troposphere (lower atmosphere), and to be constant in the lower reaches of the stratosphere. The lower atmosphere is assumed to consist of dry air that behaves as a perfect gas. Gravity is also assumed constant at the standard value, 9.806 65 m/s2. Table 1.2 summarizes property data for altitudes to 10 000 m.
Pressure values in Table 1.2 may be calculated from
(
p = 101.325 1 − 2.25577 ×10 −5 Z
)
5.2559
…………………….. (1.2)
The equation for temperature as a function of altitude is t = 15 – 0.0065Z ………………………………………. (1.3) where Z = altitude, m p = barometric pressure, kPa t = temperature, °C Equations (1.2) and (1.3) are accurate from −5000 m to 11 000 m. For higher altitudes, comprehensive tables of barometric pressure
Table(1.2)
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1.4-Psychrometric Relations We have already discussed some psychrometric terms in Art. 1.2. These terms have some relations between one another. The following psychrometric relations are important from the subject point of view: 1. Specific humidity, humidity ratio or moisture content: It is the mass of water vapour present in 1 kg of dry air (in the air-vapour mixture) and is generally expressed in g/kg of dry air. It may also be defined as the ratio of mass of water vapour to the mass of dry air in a given volume of the airvapour mixture. Let
pa , va , Ta , ma and Ra = Pressure, volume, absolute temperature, mass and gas constant
respectively for dry air, and pv , vv , Tv , mv and Rv = Corresponding values for water vapour. Assuming that the dry air and water vapour behave as perfect gases, we have for dry air, pa va = ma Ra Ta ………….………. (1.4) pv vv = mv Rv Tv …………..……….(1.5)
and for water vapour va = vv
Also and
Ta = Tv = Td
….(where Td is dry bulb temperature)
From equations (1.4) and (1.5), we have pv mv Rv = pa ma Ra ∴ Humidity ratio, W =
mv Ra pv = ma Rv pa
Substituting Ra =0.287 kJ/kg K for dry air and Rv =0.461 kJ/kg K for water vapour above equation, we have
W=
0.287 × pv p pv = 0.622 × v = 0.622 × 0.461× pa pa pb − pv
W = 0.622 ×
…(Q pb = pa + pv )
pv …………………………. (1.6) pb − pv
For saturated air (i.e. when the air holding maximum amount of water vapour), the humidity ratio or maximum specific humidity, Ws = Wmax = 0.622 ×
ps …………………….…….. (1.7) pb − p s
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ps = partial pressure of air corresponding to saturation
temperature (i.e. dry bulb
temperature t d ). 2.Degree of saturation or percentage humidity: We have already discussed that the degree of saturation is the ratio of actual mass of water vapour in unit mass of dry air to the mass of water vapour in the same mass of dry air when it is saturated at the same temperature (dry bulb temperature). In other words, it may be defined as the ratio of actual of specific humidity to the specific humidity of saturated air at the same dry bulb temperature. It is, usually, defined by µ . Mathematically, degree of saturation, 0.622 × pv p 1− s p − pv p p − p s pv pb W = µ = = b = v b 0.622 × ps Ws t p s pb − pv p s 1 − pv d pb − p s pb 1− pv µ= p s 1 −
ps pb pv pb
………………………………….. (1.8)
Notes :(a) The partial pressure of saturated air ( ps ) is obtain from the steam tables corresponding to dry bulb temperature t d . (b) If the relative humidity, φ = pv ps is equal to zero, then the humidity ratio, W=0, i.e. for dry air, µ = 0 . (c) If the relative humidity, φ = pv ps is equal to 1, then W = Ws and µ = 1 . Thus µ varies between 0 and 1. 3.Relative humidity: We have already discussed that the relative humidity is the ratio of actual mass of water vapour (mv ) in a given volume of moist air to the mass of water vapour (ms ) in the same volume of saturated air at the same temperature and pressure. it is usually denoted by φ . Mathematically, relative humidity,
φ= Let
mv ms
…………………………….... (1.9)
pv , vv , Tv , mv and Rv =Pressure, volume, absolute temperature, mass and gas constant
respectively for water vapour in actual conditions, and ps , vs , Ts , ms and Rs = Corresponding values for water vapour in saturated air.
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We know that for water vapour in actual conditions, pv vv = mv Rv Tv ………………….. (1.10) Similarly, for water vapour in saturated air, p s vs = ms Rs Ts ………………….. (1.11) According to the definitions, vv = v s and
Tv = Ts
Also
Rv = Rs = 0.416 kJ / kg K
∴ from equations (1.10) and (1.11), relative humidity,
φ=
mv pv = ………………………………. (1.12) ms p s
Thus, the relative humidity may also be defined as the ratio of actual partial pressure of water vapour in moist air at a given temperature (dry bulb temperature) to the saturation pressure of water vapour (or partial pressure of water vapour in saturated air) at the same temperature. The relative humidity may also be obtained as discussed below: We know that degree of saturation, 1− pv µ= p s 1 −
∴
φ=
ps pb pv pb
ps 1− p b =φ 1 − φ × p s pb
µ p 1 − (1 − µ ) s pb
p …… Qφ = v ps
………………………. (1.13)
Note: For saturated air, the relative humidity is 100% 4.pressure of water vapour: The partial pressure of water vapour in the air, when the air is un saturated we can be calculated according to the experimental equation,
pv = psw − pb A(td − t w ) …………………….. (1.14) where
pv =Partial pressure of water vapour in wet air psw =Saturation pressure corresponding to wet bulb temperature (from steam table(table 1.4)) pb =Barometric pressure, td =Dry bulb temperature, t w =Wet bulb temperature, 9
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A= Constant equal to
6.66 × 10 −4 C −1
or
5.94 × 10 −4 C −1
, t w ≥ 0° C , t w < 0° C
5.Vapour density or absolute humidity: We have already discussed that the vapour density or absolute humidity is the mass of water vapour present in 1 m3 of dry air.
vv =Volume of water vapour in m3/kg of dry air at its partial pressure,
Let
va = Volume of dry air in m3/kg of dry air at its partial pressure
ρv =Density of water vapour in kg/m3 corresponding to its partial pressure and dry bulb temperature td , and
ρ a =Density of dry air in kg/m3 of dry air. We know that mass of water vapour,
and mass of dry air,
mv = vv ρ v
………………………………… (1.15)
ma = va ρ a
………………………………… (1.16)
Dividing equation (1.15) by equation (1.16), mv vv ρ v = ma va ρ a
Since va = vv , therefore humidity ratio, W=
mv ρ v = ma ρ a
We know that Since va = pa
1
1
ρa
= Ra Td
ρa
or ρ v = W ρ a ……………………..(1.17)
pa va = ma Ra Ta and ma = 1kg , therefore substituting these values in above expression we get or
ρa =
pa …………………….. (1.18) Ra Td
Substituting the value of ρ a equ. (1.18) in equation (1.17), we have
ρv =
W pa W ( pb − pv ) = Ra Td Ra Td
ρv =
W ( pb − p v ) ……………..……………….. (1.19) Ra Td
where
………(Q pb = pa + pv )
pa =Pressure of air in kN/m2, Ra =Gas constant for air=0.287 kJ/kg K, and Td =Dry bulb temperature in K
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6.Enthalpy (Total heat) of moist air: The enthalpy of moist air is numerically equal to the enthalpy of dry air plus the enthalpy of water vapour associated with dry air.
h = ha + W hv …………………………………….(1.20) ha= 1.005 Tdb ha=1.007 Tdb - 0.026
0ºC >Tdb>-10 ºC ……………………..(1.21a ) 60ºC >Tdb>0 ºC ……………………..(1.21b )
Where
h: The enthalpy moist air kJ/kg of dry air ha: The enthalpy of dry air kJ/kg of dry air hv: The enthalpy of water vapour kJ/kg of dry air Assuming that the water vapour in the rang of temperatures from 0 ºC to 60ºC is generate from water at 0 ºC temperature and the specific heat for the superheated water vapour is constant. We can use the following equation to calculate enthalpy of water vapour:
hv = 2501 + 1.84t ………………..(1.22 ) Now we can substituting the equations (1.21 ) and (1.22 ) in equation (1.20 ) to get on the approximating equation accepting accuracy to calculating enthalpy of moist air at rang of temperature from 0 ºC to 60ºC :
h = (1.007t − 0.026) + W (2501 + 1.84t ) ………………..(1.23 ) Example 1.1: The readings from a sling psychrometer are as follows: dry bulb temperature = 30º C;
wet bulb temperature = 20º C; barometer reading =740 mm of Hg. Using steam tables, determine: 1.Dew point temperature, 2.Relative humidity, 3.Specific humidity, 4. Degree of saturation, 5. Vapour density, 6. Enthalpy of mixture per kg of dry air. Solution: Given: td=30º C , tw = 20º C , pb = 740 mm Hg 1. Dew point temperature
First of all, let us find the partial pressure of vapour (pv). From steam tables (table 1.4), we find that the saturation pressure corresponding to wet bulb temperature of 20º C is psw=0.023388 bar We know that barometric pressure, pb = 740 mm of Hg
……(Given)
=740×133.4=98 716 N/m2
… (Q 1mm of Hg=133.4 N/m2)
= 0.98716 bar
… (Q 1bar=105 N/m2)
Q Partial pressure of vapour
pv = psw − pb A(td − t w ) =0.023388- 0.98716×6.66×10-4×(30-20)= 0.0168 bar 11
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Since the dew point temperature is the saturation temperature corresponding to the partial pressure of water vapour (pv), therefore from steam tables(table 1.4), we find that corresponding to a pressure of 0.01703 bar, the dew point temperature is tdp≈15º C
Ans.
2. Relative humidity
From steam tables (table 1.4), we find that the saturation pressure of vapour corresponding to dry bulb temperature of 30º C is ps=0.04246 bar We know that relative humidity,
φ=
pv 0.0168 = = 0.3959 ps 0.04246
Ans.
or 39.59 %
3. Specific humidity
We know that Specific humidity, W=
0.622 pv 0.622 × 0.0168 0.01044 = = 0.010768 kg / kg of dry air = pb − p v 0.98716 − 0.0168 0.97036
= 10.768 g / kg of dry air
4. Degree of saturation
We know that Specific humidity of saturated air, Ws =
0.622 ps 0.622 × 0.04246 0.02641 = = = 0.027955 kg / kg of dry air pb − p s 0.98716 − 0.04246 0.944
We know that degree of saturation,
µ=
W Ws
0.010768 = = 0.385 or 38.5% td 0.027955
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Note: The degree of saturation (µ) may also be calculated from the following relation:
µ=
pv ps
pb − ps 0.0168 0.98716 − 0.04246 = = 0.385 or 38.5% pb − pv 0.04246 0.98716 − 0.0168
Ans.
4. Vapour density
We know that vapour density,
ρv =
W ( pb − pv ) 0.010768 (0.98716 − 0.0168) = = 0.012015 kg / m 3 of dry air Ra Td 287 (273 + 30 )
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5. Enthalpy of mixture per kg of dry air
We know that enthalpy of moist air,
h = (1.007t d − 0.026) + W (2501 + 1.84t d ) =(1.007×30-0.026)+0.010768×(2501+1.84×30)=57.7 kJ/kg of dry air
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Example 1.2: The humidity ration of atmospheric air at 28º C dry bulb temperature and 760 mm of
mercury is 0.016 kg/kg of dry air. Determine: 1. partial pressure of water vapour; 2. Relative humidity; 3. Dew point temperature; 4. Specific enthalpy; and 5. Vapour density. Solution. Given: td=28º C , pb = 760 mm Hg , W=0.016 kg/kg of dry air 1. Partial pressure of water vapour
Let
pv= Partial pressure of water vapour.
We know that humidity ratio (W), 0.016 =
0.622 pv 0.622 pv = pb − p v 760 − pv
12.16-0.016 pv =0.622 pv
or
0.638 pv=12.16
pv =19.06 mm of Hg =19.06×133.4=2542.6 N/m2
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2. Relative humidity
From steam tables (table 1.4), we find that the saturation pressure of vapour corresponding to dry bulb temperature of 28º C is ps=3.7822 kPa =3782.2 N/m2 Relative humidity, ∴
φ=
pv 2542.6 = = 0.672 p s 3782.2
Ans.
or 67.2%
3. Dew point temperature
Since the dew point temperature is the saturation temperature corresponding to the partial pressure of water vapour(pv), therefore from steam tables (table 1.4), we find that corresponding to a pressure of 2542.6 N/m2(0.025426 bar), the dew point temperature is, tdp=21.1º C
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4. Specific enthalpy
From steam tables(table 1.4), latent heat of vaporization of water corresponding to a dew point temperature of 21.1º C, We know that enthalpy of moist air,
h = (1.007t d − 0.026) + W (2501 + 1.84t d ) =(1.007×28-0.026)+0.016×(2501+1.84×28) =69.01 kJ/kg of dry air
Ans.
5. Vapour density
We know that vapour density,
ρv =
W ( pb − pv ) 0.016(760 − 19.06 ) = = 0.0183 kg / m 3 of dry air Ra Td 287(273 + 28)
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1.5- Thermodynamic Wet Bulb Temperature or Adiabatic Saturation Temperature
The thermodynamic wet bulb temperature or adiabatic saturation temperature is the temperature at which the air can be brought to saturation state, adiabatically, by the evaporation of water into the following air. The equipment used for the adiabatic saturation of air, in its simplest form, consists of an insulated chamber containing adequate quantity of water. There is also an arrangement for extra water (know as make-up water) to flow into the chamber from its top, as shown in fig. (1.3). Let the unsaturated air enters the chamber at section 1. As the air passes through the chamber over a long sheet of water, the water evaporates which is carried with the flowing stream of air, and specific humidity of air increases. The make up water is added to the chamber at this temperature to make the water level constant. Both the air and water are cooled as the evaporation takes place. This process continues until the energy transferred from the air to the water is equal to the energy required to vaporize the water. When steady conditions are reached, the air flowing at section 2 is saturated with water vapour. The temperature of the saturated air at section 2 is known as thermodynamic wet bulb temperature or adiabatic saturation temperature. During the adiabatic saturation process, the partial pressure of vapour increases, although the total pressure of the air-vapour mixture remains constant. The unsaturated air initially at dry bulb temperature td1 is cooled adiabatically to dry bulb temperature td2 which is equal to the adiabatic saturation temperature tw. It may be noted that the adiabatic saturation temperature is taken equal to the wet bulb temperature for all practical purposes.
1.3
Let
h1=Enthalpy of unsaturated air at section 1, W1= Specific humidity of air at section 1, h2, W2=Corresponding values of saturated air at section 2, and hfw =Sensible heat of water at adiabatic saturation temperature. 14
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Balancing the enthalpies of air at inlet and outlet (i.e. at section 1 and 2), h1 + (W2 − W1 )h fw = h2 …………………..………….(1.24) h1 − W1h fw = h2 − W2 h fw …………………………….(1.25)
or
The term (h2 − W2 h fw ) is known as sigma heat and remains constant during the adiabatic process. We know that
h1 = ha1 + W1 hv1 h2 = ha 2 + W2 hv 2
and
ha1 =Enthalpy of 1 kg of dry air at dry bulb temperature td 1
where
*
hv1 =Enthalpy of superheated vapour at td 1 per kg of vapour,
ha 2 =Enthalpy of 1 kg of dry air at wet bulb temperature t w , and hv 2 =Enthalpy of superheated vapour at wet bulb temperature t w per kg of vapour.
Now the equation (1.25) may be written as:
(ha1 + W1 hv1 ) − W1h fw = (ha 2 + W2 hv 2 ) − W2h fw W1 (hv1 − h fw ) = W2 (hv 2 − h fw ) + ha 2 − ha1
∴
W1 =
W2 (hv 2 − h fw ) + ha 2 − ha1 ………………………………….(1.25) hv1 − h fw
* In psychrometry, the enthalpy of superheated vapour at dry bulb temperature saturated vapour corresponding to dry bulb temperature
td 1 15
td 1 is taken equal to the enthalpy of
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Example 1.3: Atmospheric air at 0.965 bar enters the adiabatic saturator. The wet bulb temperature
is 20ºC and dry bulb temperature is 31ºC during adiabatic saturation process. Determine: 1. humidity ratio of the entering air; 2. Vapour pressure and relative humidity at 31ºC; and 3. dew
point temperature.
Solution. Given: pb = 0.965 bar ; tw =20º C ; td =31º C 1. Humidity ratio of the entering air
W1 = Humidity ratio of the entering air, and
Let
W2 = Humidity ratio of the saturation air. First, let us find the value of W2. From psychrometric or steam tables(table 1.4), we find that saturation pressure of vapour at 20ºC,
pv2 =0.023388 bar Enthalpy of saturated vapour at 20ºC,
hs2 = hg2=2537.38 kJ/kg Sensible heat of water at 20ºC,
hfw= 83.9 kJ/kg and enthalpy of saturated vapour at 31ºC,
hs1 = hg1=2557.32 kJ/kg We know that enthalpy of unsaturated air corresponding to dry bulb temperature of 31ºC,
ha1=m cp td =1×1.005×31=31.155 kJ/kg
… (Taking cp for air =1.005 kJ/kg K)
Enthalpy of 1 kg of saturated air corresponding to wet bulb temperature of 20ºC,
ha2=m cp tw =1×1.005×20=20.1 kJ/kg We know that W2 =
W1 = =
0.622 pv 2 0.622 × 0.023388 = = 0.0154 kg / kg of dry air pb − pv 2 0.965 − 0.023388
W2 (hv 2 − h fw ) + ha 2 − ha1 hv1 − h fw
0.0154(2537.38 − 83.9 ) + 20.1 − 31.155 = 0.0108 kg / kg of dry air 2557.32 − 83.9
2. Vapour pressure and relative humidity at 31ºC
Let
pv1 = Vapour pressure at 31ºC.
We know that humidity ratio of the entering air (W1 ),
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0.622 pv1 pb − pv1 0.622 pv1 0.965 − pv1
⇒ pv1 = 0.0164 bar
From psychrometric or steam tables (table 1.4), we find that the saturation pressure corresponding to 31ºC is
ps = 0.04495 bar ∴ Relative humidity,
φ=
pv1 0.0164 = = 0.365 ps 0.04495
or 36.5%
3. Dew point temperature
Since the dew point temperature (tdp) is the saturation temperature corresponding to the partial pressure of water vapour (pv1), therefore from psychrometric or steam tables (table 1.4), we find that corresponding to pressure of 0.0164 bar, the dew point temperature is 1.7055 − 1.598 1.64 − 1.598 = 15 − 14 X − 14 ⇒ X =
1.64 − 1.598 + 14 = 14.39 1.7055 − 1.598
P bar
X=tdp =14.39ºC 1.7055 1.64 1.598
T ºC 14
17
X
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Table(1.3)
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Table(1.3)
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Psychrom etry Table(1.4)
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Table(1.4)
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Table(1.4)
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