CONTINOUS HUMIDIFICATION PROCESSES latent heat of vaporisation given off cools water water can be cooled only to wet bulb temperature only small amount of water is lost evaporation of warm water when in contact with unsaturated airWater-cooling tower packed tower

Water-cooling tower

Water-cooling tower packed tower air flows upward warm water cascades downNatural draft water cooler or forced draft water cooler adiabatic warm water flows counter-currently to an air streamwater can be discharged to the river or recirculated and reused

Natural draft water cooler

Natural draft water cooler

CONTINOUS HUMIDIFICATION PROCESSESWater vapour is transferred from the vapour state to the liquid stateTransfer of water from the liquid phase into a gaseous mixture of air & water vapourHumidificationDehumidificationwaterAir &water vapour

Humidity ChartProperties of air-water vapour mixtures at 1 atm abs pressureHumidity vs dry bulb temperature0.00850.0115Dew point =10oC, humidity = ?Wet bulb temp. =20oC, dry bulb temp. = 30oC,humidity = ?

Humidity chartFigure 9.3-2 Humidity Chart

where:L= water flowrate (kg/s.m2)TL = water temperature (oC)G = dry air flowrate (kg/s.m2)H = humidity of air (kg water/kg dry air)Hy = entahlpy of air-vapour mixture(J/kg dry air)kGa = gas-phase volumetric coefficient (kmol/s.m3.Pa)P = atmospheric pressure (Pa)MB = molecular weight of airTheory and calculations for water-cooling towersz= height of tower (m)Using film mass-transfer coefficientUsing overall mass transfer coefficient

Procedure:2. Determine Hy1 from Hy1 = (1.005 + 1.88H1)103(TG1-0) + 2.501 x 106H13. Calculate Hy2 from G(Hy2-Hy1) = LcL(TL2-TL1) where cL = 4.187 x 103 J/kg.K4. Plot the equilibrium line (table 10.5-1) on the Hy vs TL graph5. Draw the operating-line with the point (TL1,Hy1) and (TL2,Hy2) on the graphDesign of water-cooling tower using film mass-transfer coefficient1. Determine H1 from the humidity chart given TG1 & Tw

7. Obtain several values of Hyi with the corresponding Hy. Tabulate values of Hyi, Hy, (Hyi-Hy) and 1/(Hyi-Hy) Design of water-cooling tower using film mass-transfer coefficient

Procedure:2. Determine Hy1 from Hy1 = (1.005 + 1.88H1)103(TG1-0) + 2.501 x 106H13. Calculate Hy2 from G(Hy2-Hy1) = LcL(TL2-TL1) where cL = 4.187 x 103 J/kg.K4. Plot the equilibrium line (table 10.5-1) on the Hy vs TL graph5. Draw the operating-line with the point (TL1,Hy1) and (TL2,Hy2) on the graphDesign of water-cooling tower using overall mass-transfer coefficient1. Determine H1 from the humidity chart given TG1 & Tw

7. Obtain several values of Hy* with the corresponding Hy. Tabulate values of Hy*, Hy, (Hy*-Hy) and 1/(Hy*-Hy) 6. Plot several vertical linesDesign of water-cooling tower using overall mass-transfer coefficient

Example 10.5-1:Tower Design Using Film CoefficientGiven: kGa = 1.207 x 10-7 kgmol/s.m3.Pa & hLa/kGaMBP = 4.187x104 J/kg.K1. Determine H1 from the humidity chart given TG1 & Tw

0.0165At TG1=29.4oC & Tw = 23.9oC, H1 = ?2. Determine Hy1 from Hy1 = (1.005 + 1.88H1)103(TG1-0) + 2.501 x 106H1Hy1 = (1.005 + 1.88[0.0165])103(29.4-0) + 2.501 x 106(0.0165) = 7.17 x 103 J/kgExample 10.5-1:

3. Calculate Hy2 from G(Hy2-Hy1) = LcL(TL2-TL1) where cL = 4.187 x 103 J/kg.KExample 10.5-1: 1.356 (Hy2-7.17 x 103) = 1.356 (4.187 x 103)(43.3-29.4) Hy2 = 129.9 x 103 J/kg dry air

4. Plot the equilibrium line (table 10.5-1) on the Hy vs TL graph5. Draw the operating-line with the point (TL1,Hy1) and (TL2,Hy2) on the graphExample 10.5-1:

Example 10.5-1:Taking a point ,for example, Hy2 = 129.9 x 103 J/kg dry air & TL2 = 43.3oCHy = 184.3 x 103 J/kg dry air184.3

Example 10.5-1:7. Obtain several values of Hyi with the corresponding Hy. Tabulate values of Hyi, Hy, (Hyi-Hy) and 1/(Hyi-Hy) 71.794.483.5108.494.9124.4106.5141.8162.1129.9184.7

Example 10.5-1:

HyiHyHyi-Hy1/(Hyi-Hy)94.4 x 103108.4 x 103124.4 x 103141.8 x 103162.1 x 103184.7 x 10371.7 x 10383.5 x 10394.9 x 103106.5 x 103118.4 x 103129.9 x 10322.7 x 10324.9 x 10329.5 x 10335.3 x 10343.7 x 10354.8 x 1034.41 x 10-54.02 x 10-53.39 x 10-52.83 x 10-52.29 x 10-51.82 x 10-5

Example 10.5-1:Given: kGa = 1.207 x 10-7 kgmol/s.m3.Pa & hLa/kGaMBP = 4.187x104 J/kg.K

Tower Design Using Overall Mass Transfer CoefficientProblem 10.5-2(b): It is planned to cool water from 110oF to 85oF in a packed countercurrent water cooling tower using entering air at 85oF with a wet bulb temperature of 75oF. The water flow is 2000lbm/h.ft2 and the air flow is 1400 lbm/h.ft2. The overall mass transfer coefficient is KGa=6.90 lbmol/h.ft3.atm. Calculate the tower height needed.

Procedure:2. From point M (TL1,Hy1), draw a straight line to intercept the equilibrium line at TL2, point N4. Calculate Gmin from Gmin(Hy2 max-Hy1) = LcL(TL2-TL1) where cL = 4.187 x 103 J/kg.K3. Obtain the value of Hy2 max.Minimum value of air flow, Gmin1. Plot the equilibrium line on the Hy vs TL graph5. G = 1.3-1.5GminHy2max

Problem 10.5-2(a): It is planned to cool water from 110oF to 85oF in a packed countercurrent water cooling tower using entering air at 85oF with a wet bulb temperature of 75oF. The water flow is 2000lbm/h.ft2 and the air flow is 1400 lbm/h.ft2. The overall mass transfer coefficient is KGa=6.90 lbmol/h.ft3.atm. Calculate the minimum air rate that can be used.

where HG = height of a gas enthalpy transfer unit (m) where HOG = height of an overall gas enthalpy transfer unit (m)Design of water-cooling tower using height of transfer unitz = HGNG = HOGNOG

Problem 10.5-1:A forced-draft countercurrent water-cooling tower is to cool water from 43.3oC to 26.7oC. The air enters the bottom of the tower at 23.9oC with a wet bulb temperature of 21.1oC.The value of HG for the flow conditions is HG=0.533 m. The heat transfer resistance in the liquid phase will be neglected; since hL is very large. Hence, values of Hy* should be used. Calculate the tower height needed if 1.5 times the minimum air rate is used.Ans: z=1.70 m

Dehumidification tower humidity & temperature of air reduced Operating line lies below the equilibrium lineHumidification/ cooling towerDehumidification tower Water is cooled & air humidified Operating line lies above the equilibrium line

GROUP 1 (Report and presentation):Question1.Explain about the theory of humidification?2.The applications of humidification process3.How to design the water cooling tower?4.Solve the given problem

Use power point for your presentation.Group 1:

PROBLEM-Group 1You are to design a cooling tower A cooling tower receives warm water at 43oC at a mass flow rate of 7000 kg/m2.hr. A cooling range of 13oC is to be achieved by countercurrent contact with air. The air enters at a rate of 4200 kg/m2.hr at a dry-bulb temperature of 31oC and a humidity of 0.01520kg/kg dry air. Given that MBKGaP = 2500 kg/h.m3, where KGa is the overall mass-transfer coefficient. Determine:(i)The number of overall gas-phase transfer units (NTU)(ii)The height of packing needed (Z)

*When a relatively warm liquid is brought into direct contact with gas that is unsaturated, some of the liquid is vaporized. The liquid temperature will drop mainly due to the latent heat of evaporation. E.g. humidifying air for control of moisture content of air in drying or air condition; dehumidifying air, where cold water condenses some water vapor from warm air; and water cooling tower, where evaporation of water to the air cools warm water. The most important application is on water cooling towers. Warm water discharged from HEX and condensers will be cooled down before it is reused or discarded. The water cannot be cooled below its wet bulb temperature since the driving force for the evaporation of the water is approximately the vapor pressure of the water vapor pressure it would have at wet bulb temperature. Only about 1.5% of water will be lost during evaporation. So in tower design, the water flow rate is assumed to be constant.*In the processing of materials it is often necessary either to increase (humidification) or decrease (dehumidification) the amount of vapor present in a gas stream. The most widespread application of humidification and dehumidification involves the air-water system. But the most important application of humidification principle is water cooling tower. In order that hot condenser water may be re-used in a plant, it is normally cooled by contact with an air stream. The equipment usually takes the form of a tower in which the hot water is run in at the top and allowed to flow downwards over a packing against a countercurrent flow of air which entera at the bottom of the cooling tower.*A cooling tower is a special type of heat exchanger where the warm water and the air are brought in direct contact for evaporative cooling. Industrial processes use a lot of water for cooling purposes. In the above diagram, cooling tower is used to remove heat from process stream. The largest users of cooling water are the power plants in which a huge quantity of low-pressure steam from the turbine is condensed in big condenser. Hot water can not be discharged directly into the river since it will cause thermal pollution and will affect all the aquatic creatures. If treated water is used, it is better if this water can be reused.*A natural draft cooling tower has a large reinforced concrete shell of hyperbolic shape. It is so called because natural flow of air occurs through the tower. Natural draft towers are used in big installations like steam power plants requiring a huge quantity of cooling water. While for forced draft cooling tower, one or more fans located at the tower bottom to push air into the tower.**Involves the transfer of material between a pure liquid phase and a fixed gas that is nearly insoluble in the liquid. The operation is simpler compared to absorption/stripping since the liquid contains only one component so there is no concentration gradients and no resistance to mass transfer in the liquid phase.*Dry-bulb temperature is the temperature of air measured by a thermometer whose bulb is dry. This is a true temperature of the air.Web-bulb temperature is the temperature attained by a small amount of evaporating water in a manner such that the sensible heat transferred from the air to the liquid is equal to the latent heat required for the evaporation. The wet-bulb temperature is measured by passing air over the bulb of a thermometer which is covered with a cloth-wick saturated with water.*Find Humidity if T wet bulb=20 C and T dry bulb=30 C. Clue: At 100% relative humidity both T wet bulb = T dry bulb.*For sizing a cooling tower, we need to determine the height of packing required to achieve the desired cooling of the water.Assumptions:The rate of vaporization of water is much less than the rate of water input to the tower (typically , 1% loss of water due to vaporization).Adiabatic cooling of water occurs in the tower (no heat loss or gain at the tower wall).L will remain constant.************Ans: H=0.0165, Hy1=30.8 Btu/lbm, Hy2=66.7 Btu/lbm, z=21.8 ft**In designing the cooling tower, often the air flow rate (G) is not fixed but must be set. For the actual tower, a value of G greater than G minimum (Gmin) must be used. Usually a value of G =(1.3 1.5) times G min is used.G min =935 lbm/hr.ft2**Another form of equations have been used here.*1.Explain about the theory of humidification?2.The applications of humidification process3.How to design the water cooling tower4.Solve the given problem*