EXPROPIACIONES

1. The temperature of the moist air is in contact with the spray of cold water is less than the dew point of the entering air.

2. Humidity of the air can be reduced by compressing it. During the compression, partial pressure of the water vapour increases. As soon as the water vapour reaches the saturation, condensation takes place. Due to the condensation, water vapour gets liquefied and drained off.

3. When moist air is passed over a solid adsorbent surface or it is passed through a liquid adsorbent spray, dehumidification bounds to take place. In this case, the dehumidification takes place because of lowering the water vapour pressure. The solid adsorbents used here are silica gel or activated alumina. The liquid adsorbents are the solution of inorganic salt, such as brine, lithium chloride and the organic compound, such as ethylene glycol. A typical approaches to dehumidification process is shown in Figures 2.12 and 2.13. The adsorbents are reactivated time to time and reused.

FIGURE 2.12 Spraying of water on moist air.

FIGURE 2.13 Adsorption of moist air.

EXAMPLE 2.47 An air (B)–water (A) sample has a dry bulb temperature 55°C and an absolute

humidity 0.030 kg water/kg dry air at 1 std. atm pressure. Calculate:

(i) The percentage humidity (ii) The molal absolute humidity

(iii) The partial pressure of water vapour in the sample (iv) The relative humidity

(v) Dew point (vi) Humid volume (vii) Humid heat (viii) Enthalpy.

Solution:

(i) The percentage humidity: Draw a vertical line at 55°C on humidity chart and extend the same till it meet the 100% saturation humidity curve. From the meet point draw a horizontal line towards the absolute humidity axis, which gives the saturation humidity. Then, percentage humidity can be

calculated as

Percentage humidity (HP) = where

Actual humidity = 0.030

Saturated humidity (Fig. 2.7) = 0.115

(ii) Molal absolute humidity: Molal absolute humidity =

= = =

(iii) Partial pressure of water vapour in the sample: The partial pressure of water vapour in the sample may be calculated using the equation

pA =

where

pA = Partial pressure of water vapour H = Molal absolute humidity

Pt = Total pressure

Now,

(iv) Relative humidity: The relative humidity (RH) may be calculated as

where

pA = 4667.35 N/m2

Therefore,

(v) Dew point: Draw a vertical line upward at 55 °C on humidity chart (Fig. 2.7). Also draw a horizontal line from the point of absolute humidity 0.030. This will meet the vertical line at a particular point. Extend the horizontal line through the meet point towards the 100% saturation humidity curve. It will meet the curve at a particular point. From this point, draw a line vertically downwards till it meet x-axis. Thus, we will get a temperature 31.5°C. This temperature is called dew point.

(vi) Humid volume: Draw a vertical line upward at 55°C on humidity chart Fig. 2.7. Extend the line upward till it meet the line of specific volume of dry air. From this meet point draw a horizontal line towards the y-axis which represents the specific volume of dry air. From the humidity chart, at 55°C, the specific volume of dry air is 0.93 m3/kg. The humid volume of saturated air is 1.10 m3/kg. Interpolating for 26.08 % humidity.

VH = 0.93 + (1.10 – 0.93) (0.2608)

= 0.974 m3/kg

(vii) Humid heat: Humid heat (HH) may be calculated using the equation

HH = Cg + H CV

where

Cg = 1005 kJ/kg °C (From the data book) H = 0.03 kg of water/kg of dry air

CV = 1884 kJ/kg °C (From the data book)

Note: Data is from Chemical Engineer’s Perry Handbook.

Therefore, HH = 1005 + (0.03)(1884)

= 1061.52 J (for wet air)/kg of dry air . K

(viii) Enthalpy: The enthalpy may be calculated using the equation

Enthalpy = Enthalpy of dry air + (Enthalpy of saturated air – Enthalpy of dry air)

× humidity where

Enthalpy of dry air = 56,000 J/kg (From data book)

Enthalpy of saturated air = 3,52,000 J/kg (From data book) Humidity = 0.2608

Therefore, Enthalpy = 56,000 + (352000 – 56000) × 0.2608 = 133.197 kJ/kg of dry air.

Note: Data is from Chemical Engineer’s Perry Handbook.

EXAMPLE 2.48 Air at a temperature of 25°C and pressure 750 mmHg has a relative humidity of

75%. Calculate:

(i) Humidity of air

(ii) Molal humidity of air

(iii) Weight of water condensed from 100 m3 of original wet air, if the temperature of the air is reduced to 15°C and the pressure is increased to 2 bar. At 25°C the vapour pressure of the water is 2.5 kN/m2.

Solution: (i) Humidity of air:

Therefore,

Now, total pressure (Pt) = 750 mmHg =

= 99.967 kN/m2 Humidity (H) =

(ii) Molal humidity:

Hm =

= 0.01186 × = 0.0191

where

H = Humidity at given temperature Mw = 18

MA = 29

pA = 2.5 kN/m2 from steam table at 15°C Pt = 197.43 kN/m2 (2 bar)

Therefore,

Water condensed = Initial humidity – Final humidity = 0.01186 – 0.00796

= 0.0039

Humid volume =

=

= 0.66 m3 of wet air/kg of dry air 0.66 m3 of wet air = 1 kg of dry air

100 m3 of wet air = kg of dry air Amount of dry air = 151.52 kg

Water condensed from 100 m3 of air = 151.52 × 0.0039 = 0.591 kg of water.

EXAMPLE 2.49 Conditioned air at 760 mmHg total pressure, 50°C and at a humidity of 0.01 kg

water per kg of bone dry air enters the drier. It leaves the drier at 760 mmHg total pressure and 50°C, with RH 83%. Vapour pressure of water at 50°C is 92.5 mmHg. If 50 kg of water enters into the air

stream per hour, calculate the rate of bone dry air flowing through the dryer. Solution: Here, … … where RH = 83% pw = ? ps = 92.5 mmHg Therefore, 83 = pw = 76.775 mmHg Now, where MA = 18 MB = 29 pA = 76.775 mmHg Pt = 760 mmHg Therefore,

Water removed = Humidity at removal – Humidity at entry level = 0.0697 – 0.01

= 0.0597

0.0597 kg of water = 1 kg of dry air 50 kg of water = kg of dry air

= 837.52 kg

EXAMPLE 2.50 0.6 m3/s of gas is to be dried from a dew point of 294 K to a dew point 277.5

K.How much water must be removed and what will be the volume of the gas after drying? The vapour pressure of the gas at 294 K is 0.85 kN/m2.

Solution:

Mole fraction of water in the inlet = = 0.0246

Mole fraction of water in the outlet = = 0.00839

Mole fraction of air in the inlet = 0.975 Mole fraction of air in the outlet = 0.9916 22.414 m3 1 kgmol

Amount of moist air = = 0.0248 kgmol/s

Dry air balance over the dryer:

0.0248 × 0.975 = G × 0.9916 G = 0.0243 kgmol/s

Now, 1 kgmol contain 22.414 m3

Therefore, in 0.0243 kgmol the amount of dry air leaving = = 0.554 m3

Amount of water removed = (0.0246)(0.0248) – (0.00839)(0.0243) = 0.000406 kgmol/s

= 0.00731 kg/s Also, kgmol =

Mass of vapour = Molecular weight × kgmol Now, 22.414 m3 contains 1 kgmol

Therefore, 1 m3 will contain

Amount of vapour (in kg) per m3 of vapour at the inlet =

= 0.0184

kg of vapour per m3 of gas at the outlet =

= 0.0066

Water removed/s = (0.0184 – 0.0066) × 0.6 = 0.00708

EXAMPLE 2.51 Brick material containing 70% moisture is to be dried at the rate of 0.15 kg/s in a

counter current drier to give a product containing 5% moisture (wet basis). The drying medium consists of air heated to 373 K and containing water vapour equivalent to a partial pressure of 1 kN/m2. The air leaves the drier at 313 K and 70% RH. Calculate how much air is required to remove the moisture. The vapour pressure of water at 313 K is 7.4 kN/m2.

Solution:

Feed water (in Brick) = 0.15 × 0.7 = 0.105 kg/s Weight of dry solids = 0.15 × 0.3 = 0.045 kg/s

Weight of water in the product = 0.05 kg/s (wet basis)

Weight fraction of water present in the product stream of bricks

where

Mw = Mass of water in kg

Ms = Mass of solids in kg (0.045 kg/s)

Mw = 0.00237 kg/s

Water to be removed from the brick = 0.105 – 0.00237 = 0.10263 kg/s Humidity of entering air =

=

= 0.006 For exit air,

RH = 70 = pw =

pw = 5.18 kN/m2

Humidity at exit air =

=

= 0.0335

Increase in humidity = 0.0335 – 0.006 = 0.0275

Let ‘m’ be the mass flow rate of dry air in kg/s then,

m × 0.0275 = 0.10263

m = 3.732 kg of dry air/s

Weight of (wet) moist air entering = 3.732 + 0.02345 = 3.7554 kg/s

Exercises

2.1 Calculate the kilogram atoms of carbon which weigh 36 kg.

In document CONCESIÓN CONEXIÓN VIAL MELIPILLA CAMINO DE LA FRUTA (página 51-55)