Unit operation (I) 1 Department of Chemical and
Description: Unit operation (I) 1 Department of Chemical and Petroleum Industries Engineering Fourth Year AL-Mustaqbal University Collage Lecture (4) Humidification (1) 2 Humidification and Water Cooling Humidification is the increase of the amount of
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slide1. Unit operation (I) 1 Department of Chemical and Petroleum Industries Engineering
Fourth Year
AL-Mustaqbal University Collage
Lecture (4)
Humidification (1)<br>
slide2. 2<br>
slide3. Humidification and Water Cooling Humidification is the increase of the amount of vapor present in a gas stream by passing the gas over a liquid which then evaporates into the gas. Dehumidification is the reduce of the vapor present in the gas. Humidity (H ): mass of vapor associated with unit mass of dry gas, (kg liquid/kg dry gas) Definitions Where:
Pw : is the partial pressure of vapor (water) in the gas (air).
Mw : is the molecular mass of vapor (water).
MA : is the molecular mass of gas (air).
P : is the total pressure.
H : is the humidity 3<br>
slide4. Where:
Pwo : is the partial pressure of vapor (water) in the saturated gas (air).
H o : is the humidity of saturated gas Partial pressure of vapour gas
Relative humidity (RH) = -------------------------------------------------- 100
Partial pressure of vapour in saturated gas Percentage humidity: (H / H o) 100 Humidity of saturated gas (H o): humidity of the gas when it is saturated with vapor at a given temperature 4<br>
slide5. Humid heat (s): heat required to raise unit mass of dry gas and its associated vapor through unit temperature difference at constant pressure, or: Where:
Ca : is the specific heat of gas (air)
Cw : is the specific heat of vapor (water).
For air-water system s = 1.0 +1.9 H [approximately]
Humid volume: is the volume occupied by unit mass of dry gas and its associated vapor.
Saturated volume: is the humid volume of saturated gas.
Dew point: is the temperature at which condensation will first occur. s = Ca + H Cw 5<br>
slide6. Ex:
In a process benzene is used as solvent, benzene is evaporated into dry nitrogen at 297 K and 101.3 KN/m2, the resulting mixture has a relative humidity of 60. It is required to recover 80 per cent of the benzene present by cooling to 283 K and compressing to a suitable pressure. What should this pressure be? The vapor pressure of benzene is 12.2 KN/m2 at 297 K and 6.0 KN/m2 at 283 K. Solution:
From the definition of percentage relative humidity (RH): In the benzene -nitrogen mixture: 6 At 297 K Pw = 12.2 × 60/100 = 7.32 KN/m2 Hence the humidity is : [with increasing pressure, N2 will saturated with benzene]
and hence at 283 K: H o = 0.217× 0.2 = 0.0433 kg/kg = 0.217 The humidity of the saturated gas<br>
slide7. Wet-bulb temperature
When a stream of unsaturated gas is passed over the surface of a liquid, the humidity of the gas is increased due to evaporation of the liquid. The temperature of the liquid falls below that of the gas and heat is transferred from the gas to the liquid. At equilibrium the rate of heat transfer from the gas just balances that required to vaporize the liquid and the liquid is said to be at the wet-bulb temperature. The rate of heat transfer from gas to the liquid is 7 The rate of evaporation is The rate of heat transfer in terms of mass transfer is From eqns. 1 and 3 Q = h A (θ - θw) ……..(1) kJ/s w = hD A ρA(H W - H ) ……..(2) kg/s Q = hD A ρA(H W - H )λ ……..(3)<br>
slide8. Where:
h : is heat transfer coefficient KW/m2K
hD : is mass transfer coefficient m/s
A : is area of heat transfer m2
θ : is the temp of gas (air) K
θw : is the temp of liquid (water) K
ρA : is the density of gas (air) kg/m3
Hw : is the humidity of air with saturated vapour at wet bulb temp kg water/ kg air
λ : is the latent heat of vaporization kJ /kg
For the air-water system, the ratio (h/ hD ρA) is about 1.0 kJ /kg K 8<br>
slide9. Temperature-humidity chart
The following quantities are plotted against temperature:
(i) The humidity H kg/kg
(ii) The specific volume of dry gas m 3 /kg
(iii) The saturated volume m 3 /kg
(iv) The latent heat of vaporization λ kJ/kg
(v) The relative humidity, RH %
(vi) The humid heat, s kJ/kg 9<br>
slide10. Figure (13.4) Humidity-Temperature chart 10<br>
slide11. Ex:
Air containing 0.005 kg water vapor per kg of dry air is heated to 325 K in a dryer and passed to the lower shelves. It leaves these shelves at 60 per cent humidity and is reheated to 325 K and passed over another set of shelves, again leaving at 60 per cent humidity. This is again repeated for the third and fourth sets of shelves, after which the air leaves the dryer. On the assumption that the material on each shelf has reached the wet-bulb temperature and that heat losses from the dryer may be neglected, determine:
(a) The temperature of the material (wet bulb temp) on each tray;
(b) The amount of water removed in kg/s, if 5 m3/s moist air leaves the dryer;
(c) The temperature to which the inlet air would have to be raised to carry out the drying in a single stage. 11 Solution:
(a) For each of the four sets of shelves, the condition of the air is changed to 60 per cent humidity along an adiabatic cooling line.
Initial condition of air: θ = 325 K, H = 0.005 kg/kg
On humidifying to 60 per cent humidity:
At the end of the 1st pass: θ = 301 K, H = 0.015 kg/kg, and θw = 296 K
At the end of the 2nd pass: θ = 308 K, H = 0.022 kg/kg and θw = 301 K
At the end of the 3rd pass: θ = 312 K, H = 0.027 kg/kg and θw = 305 K
At the end of the 4th pass: θ = 315 K, H = 0.032 kg/kg and θw = 307 K<br>
slide12. Total increase in humidity = (0.032 - 0.005) = 0.027 kg water/kg dry air
The air leaving the system is at 315 K and 60 per cent humidity.
(b) From Figure 13.4, specific volume of dry air = 0.893 m3/kg (At leaving air from dryer, 315 K)
Specific volume of saturated air (saturated volume) = 0.968 m3/kg 12 The humid volume of air of 60 per cent humidity = Vdry air + Vvapor
= 0.893 + (0.968 – 0.893)×0.6 = 0.938 m3/kg
Mass of air passing through the dryer = 5/0.938 = 5.34 kg/s
Mass of water evaporated = (5.34 x 0.027) = 0.144 kg/s
(c) If the material is to be dried by air in a single pass, the air must be heated before entering the dryer such that its wet-bulb temperature is 307 K.
For air with a humidity of 0.005 kg/kg, this corresponds to a dry bulb temperature of 370 K.<br>
slide13. The various steps in this calculation are shown in Figure 13<br>
slide14. Enthalpy Humidity Chart Figure (13.5) Humidity-Enthalpy diagram for air-water vapor system 14<br>
slide15. Mixing of two streams of humid gas The mixing of two gases of humidity H 1 and H 2, at temperatures θ1 and θ2, and with enthalpies H1 and H2 to give a mixed gas of temperature θ, enthalpy H, and humidity , H . If the masses of dry gas concerned are m1, m2, and m respectively, then taking a balance on the dry gas, vapor, and enthalpy: Ex:
In an air-conditioning system, 1 kg/s air at 350 K and 10% relative humidity is mixed with 5 kg/s air at 300 K and 30% relative humidity. What is the enthalpy, humidity, and temperature of the resultant stream? 15 m1 H 1 + m2 H 2 = mH m1 + m2 = m m1H1 + m2H2 = mH Solution:
From Figure 13.4: At θ1 = 350 K and relative humidity = 10%; H 1 = 0.043 kg/kg
At θ2 = 300 K and relative humidity = 30%; H 2 = 0.0065 kg/kg m = m1+m2 = 6 kg/s<br>
slide16. The enthalpy of mixed air is calculated from the Eq.
m1H1 + m2H2 = mH
1×192 + 5×42 = 6×H → H = 67 kJ /kg 16 using the Eq. m1 H 1 + m2 H 2 = mH The humidity of mixed air is H = 0.0125 kg/ kg From fig. 13.5 :
at θ 1 = 350 K and H 1 = 0.043 kg/kg , H 1 = 192 kJ /kg
at θ 2 = 300 K and H 2 = 0.0065 kg/kg , H 2 = 42 kJ /kg From Figure 13.5: at H = 67 and H = 0.0125 kg/kg; θ = 309 K The data used in the example is<br>
Fourth Year
AL-Mustaqbal University Collage
Lecture (4)
Humidification (1)<br>
slide2. 2<br>
slide3. Humidification and Water Cooling Humidification is the increase of the amount of vapor present in a gas stream by passing the gas over a liquid which then evaporates into the gas. Dehumidification is the reduce of the vapor present in the gas. Humidity (H ): mass of vapor associated with unit mass of dry gas, (kg liquid/kg dry gas) Definitions Where:
Pw : is the partial pressure of vapor (water) in the gas (air).
Mw : is the molecular mass of vapor (water).
MA : is the molecular mass of gas (air).
P : is the total pressure.
H : is the humidity 3<br>
slide4. Where:
Pwo : is the partial pressure of vapor (water) in the saturated gas (air).
H o : is the humidity of saturated gas Partial pressure of vapour gas
Relative humidity (RH) = -------------------------------------------------- 100
Partial pressure of vapour in saturated gas Percentage humidity: (H / H o) 100 Humidity of saturated gas (H o): humidity of the gas when it is saturated with vapor at a given temperature 4<br>
slide5. Humid heat (s): heat required to raise unit mass of dry gas and its associated vapor through unit temperature difference at constant pressure, or: Where:
Ca : is the specific heat of gas (air)
Cw : is the specific heat of vapor (water).
For air-water system s = 1.0 +1.9 H [approximately]
Humid volume: is the volume occupied by unit mass of dry gas and its associated vapor.
Saturated volume: is the humid volume of saturated gas.
Dew point: is the temperature at which condensation will first occur. s = Ca + H Cw 5<br>
slide6. Ex:
In a process benzene is used as solvent, benzene is evaporated into dry nitrogen at 297 K and 101.3 KN/m2, the resulting mixture has a relative humidity of 60. It is required to recover 80 per cent of the benzene present by cooling to 283 K and compressing to a suitable pressure. What should this pressure be? The vapor pressure of benzene is 12.2 KN/m2 at 297 K and 6.0 KN/m2 at 283 K. Solution:
From the definition of percentage relative humidity (RH): In the benzene -nitrogen mixture: 6 At 297 K Pw = 12.2 × 60/100 = 7.32 KN/m2 Hence the humidity is : [with increasing pressure, N2 will saturated with benzene]
and hence at 283 K: H o = 0.217× 0.2 = 0.0433 kg/kg = 0.217 The humidity of the saturated gas<br>
slide7. Wet-bulb temperature
When a stream of unsaturated gas is passed over the surface of a liquid, the humidity of the gas is increased due to evaporation of the liquid. The temperature of the liquid falls below that of the gas and heat is transferred from the gas to the liquid. At equilibrium the rate of heat transfer from the gas just balances that required to vaporize the liquid and the liquid is said to be at the wet-bulb temperature. The rate of heat transfer from gas to the liquid is 7 The rate of evaporation is The rate of heat transfer in terms of mass transfer is From eqns. 1 and 3 Q = h A (θ - θw) ……..(1) kJ/s w = hD A ρA(H W - H ) ……..(2) kg/s Q = hD A ρA(H W - H )λ ……..(3)<br>
slide8. Where:
h : is heat transfer coefficient KW/m2K
hD : is mass transfer coefficient m/s
A : is area of heat transfer m2
θ : is the temp of gas (air) K
θw : is the temp of liquid (water) K
ρA : is the density of gas (air) kg/m3
Hw : is the humidity of air with saturated vapour at wet bulb temp kg water/ kg air
λ : is the latent heat of vaporization kJ /kg
For the air-water system, the ratio (h/ hD ρA) is about 1.0 kJ /kg K 8<br>
slide9. Temperature-humidity chart
The following quantities are plotted against temperature:
(i) The humidity H kg/kg
(ii) The specific volume of dry gas m 3 /kg
(iii) The saturated volume m 3 /kg
(iv) The latent heat of vaporization λ kJ/kg
(v) The relative humidity, RH %
(vi) The humid heat, s kJ/kg 9<br>
slide10. Figure (13.4) Humidity-Temperature chart 10<br>
slide11. Ex:
Air containing 0.005 kg water vapor per kg of dry air is heated to 325 K in a dryer and passed to the lower shelves. It leaves these shelves at 60 per cent humidity and is reheated to 325 K and passed over another set of shelves, again leaving at 60 per cent humidity. This is again repeated for the third and fourth sets of shelves, after which the air leaves the dryer. On the assumption that the material on each shelf has reached the wet-bulb temperature and that heat losses from the dryer may be neglected, determine:
(a) The temperature of the material (wet bulb temp) on each tray;
(b) The amount of water removed in kg/s, if 5 m3/s moist air leaves the dryer;
(c) The temperature to which the inlet air would have to be raised to carry out the drying in a single stage. 11 Solution:
(a) For each of the four sets of shelves, the condition of the air is changed to 60 per cent humidity along an adiabatic cooling line.
Initial condition of air: θ = 325 K, H = 0.005 kg/kg
On humidifying to 60 per cent humidity:
At the end of the 1st pass: θ = 301 K, H = 0.015 kg/kg, and θw = 296 K
At the end of the 2nd pass: θ = 308 K, H = 0.022 kg/kg and θw = 301 K
At the end of the 3rd pass: θ = 312 K, H = 0.027 kg/kg and θw = 305 K
At the end of the 4th pass: θ = 315 K, H = 0.032 kg/kg and θw = 307 K<br>
slide12. Total increase in humidity = (0.032 - 0.005) = 0.027 kg water/kg dry air
The air leaving the system is at 315 K and 60 per cent humidity.
(b) From Figure 13.4, specific volume of dry air = 0.893 m3/kg (At leaving air from dryer, 315 K)
Specific volume of saturated air (saturated volume) = 0.968 m3/kg 12 The humid volume of air of 60 per cent humidity = Vdry air + Vvapor
= 0.893 + (0.968 – 0.893)×0.6 = 0.938 m3/kg
Mass of air passing through the dryer = 5/0.938 = 5.34 kg/s
Mass of water evaporated = (5.34 x 0.027) = 0.144 kg/s
(c) If the material is to be dried by air in a single pass, the air must be heated before entering the dryer such that its wet-bulb temperature is 307 K.
For air with a humidity of 0.005 kg/kg, this corresponds to a dry bulb temperature of 370 K.<br>
slide13. The various steps in this calculation are shown in Figure 13<br>
slide14. Enthalpy Humidity Chart Figure (13.5) Humidity-Enthalpy diagram for air-water vapor system 14<br>
slide15. Mixing of two streams of humid gas The mixing of two gases of humidity H 1 and H 2, at temperatures θ1 and θ2, and with enthalpies H1 and H2 to give a mixed gas of temperature θ, enthalpy H, and humidity , H . If the masses of dry gas concerned are m1, m2, and m respectively, then taking a balance on the dry gas, vapor, and enthalpy: Ex:
In an air-conditioning system, 1 kg/s air at 350 K and 10% relative humidity is mixed with 5 kg/s air at 300 K and 30% relative humidity. What is the enthalpy, humidity, and temperature of the resultant stream? 15 m1 H 1 + m2 H 2 = mH m1 + m2 = m m1H1 + m2H2 = mH Solution:
From Figure 13.4: At θ1 = 350 K and relative humidity = 10%; H 1 = 0.043 kg/kg
At θ2 = 300 K and relative humidity = 30%; H 2 = 0.0065 kg/kg m = m1+m2 = 6 kg/s<br>
slide16. The enthalpy of mixed air is calculated from the Eq.
m1H1 + m2H2 = mH
1×192 + 5×42 = 6×H → H = 67 kJ /kg 16 using the Eq. m1 H 1 + m2 H 2 = mH The humidity of mixed air is H = 0.0125 kg/ kg From fig. 13.5 :
at θ 1 = 350 K and H 1 = 0.043 kg/kg , H 1 = 192 kJ /kg
at θ 2 = 300 K and H 2 = 0.0065 kg/kg , H 2 = 42 kJ /kg From Figure 13.5: at H = 67 and H = 0.0125 kg/kg; θ = 309 K The data used in the example is<br>