Slide 1

Jawaharlal Nehru Technological University Kakinada

III Year B. Tech. I Sem.Mass Transfer Operation - I

HumidificationPresented byProf. K. V. RaoAcademic Advisor / Visiting ProfessorSchool of Petroleum Courses

HUMIDIFICATION

Humidification is an interphase transfer operation that involves vaporization of liquid form a pure liquid phase into a gas that is practically insoluble in that liquid.

If the transfer is from gas to the liquid, it is known as dehumidification.

As concentration gradient does not exist in the liquid phase (Pure, single component) there is no mass transfer resistance in this phase.

Moreover, the mass transfer is invariably accompanied by heat transfer in these operations necessitating consideration of both equilibrium and enthalpy characteristics of the system in understanding and designing of the humidification or dehumidification operations.

Most of these operations deal with air-water mixture.2Psychrometry is the study of the properties of air-water mixtures. Atmospheric air is a mixture of dry air and water in equilibrium.

This mixture behaves like an ideal gas since the temperature of the dry air is usually very high compared to its critical temperature and the partial pressure of water is relatively low compared to its critical pressure.

In the following definitions and subsequent derivations the total pressure Pt is taken as 105 Pa, which is close to the atmospheric pressure.

The molecular weights of water and air are taken as 18 and 29 respectively. The reference temperature is chosen at 00C to calculate enthalpies of both gas and liquid phases.Definitions of Terms:

Absolute (Specific) Humidity: Mass of water vapor per unit mass of dry air. (kg of water vapor per kg of dry air)

Y = 18p / [29 (Pt p)] 1

Saturation Humidity: Humidity of air in equilibrium with water at the temperature of the air.

YS = 18ps / [29 (Pt ps)] 2

Percentage absolute (Specific) Humidity: It is the ratio of absolute humidity to saturation humidity to saturation humidity multiplied with 100 YP = 100 Y / YS 34Can be expressed in partial pressure terms as

YP = 100 [p / (Pt p)] / [ps / (Pt ps)]

Molal Humidity: moles of water vapor per unit mole of dry air

Y = y/ (1 y) = p /(Pt p)

Relative Humidity: Ratio of the partial pressure of water vapor to the saturation pressure (Vapor pressure of water at the temperature or air) YR = p / Ps

The percentage relative humidity is expressed as 100 (YR)

Humid volume: Volume of air expressed as m3 / kg of dry air i.e. sum of volume of 1kg dry air and water it contains.

VH = (RT / 29Pt) + (RTY / 18Pt) = RT (1/29 +Y / 18)/ Pt54 6 7When R = 8314 J/ kg mole K and Pt = 105N /M2 VH = 0.08314 T (1/29 + Y /18)

Humid Heat: It is the heat capacity of air-water mixture per kg of dry air. The heat capacity of dry air in a temperature range, 0 1000C is approximately equal to 1 kJ / kg K and that of water vapor 1.88kJ/ kg K.

CH = 1.00 + 1.88 Y kJ / kg K

Enthalpy of Air: Enthalpy of 1 kg dry air + enthalpy of water vapor Taking 00C reference temperature, it can be calculated as H = CH t+ Y (HV) kJ / kg of dry air89106HV = enthalpy of vaporization of water at 00C, kJ /kgt = temperature of air 0C

The vapor pressure and enthalpy of vaporization can be calculated from the following equations

PS (Pa) = exp [23.7093 4111 / (237.7 + t)] for 0 < t < 57 0C PS (Pa) = exp [23.1863 3809.4 / (226.7 + t)] for 57< t < 135 0C (HV)t = 2501.6 2.275t 0.0018t2 kJ /kg111213Adiabatic saturation temperature: Fig.1 shows an adiabatic saturator in which air flows through a spray of water. The entering air (unsaturated at the inlet conditions) is cooled and becomes saturated as some of the water vaporizes.

The walls of saturator are insulated to make the process adiabatic. The enthalpy of inlet air is equal to enthalpy of exit air under perfect adiabatic conditions. It is necessary to provide makeup water to maintain continuous recirculation of water.

The temperature of the makeup water should have the same temperature of the water sump. When the equilibrium is reached at steady state conditions, the temperature of the exit air will attain the adiabatic saturation temperature (tas), which will be equal to the water temperature.

The enthalpy balance on air will give the relationship as obtained below.8

Make UpAir outAir inFig. 1 Adiabatic Saturatort1, Y1, H19Taking Tas as reference temperature, the enthalpies of air at inlet and exit condition can be equated as

(1.00 + 1.88 Y 1) (t1 tas) + Y1 (HV)tas = (1.00 + 1.88 Y2)(tas tas) + Y2 (HV)tas

As Y2 = Yas, the above equation becomes, after rearrangement (Y1 = Yas) / (t1 tas) = (1.00 + 1.88 Y 1) / (HV)tas = CH / (HV)tas

.14 1510Wet bulb temperature: It is the non-equilibrium steady state temperature attained at the liquid (water) when small liquid surface comes into contact with a large amount of gas (air) under adiabatic conditions.

Air in y,tAir out y, tFig. 2 Experimental Determination of Wet blub Temperature11Fig. 2 shows dry bulb and wet bulb thermometers, placed in an insulated duct through which air flows continuously under turbulent conditions. The quantity of air is large such that the humidity and temperature of air remain constant.

A small quantity of water from the wick evaporates and cools the wick which in turn cools the bulb of the wet bulb thermometer.

At steady state, the latent heat of vaporization is exactly balanced by the convective heat transfer from the air to wick. It can be expressed asMA NA A (HV)twb = hc A (t twb)MA=Molecular weight of waterNA=Molar flux of water k mol / m2 sA=Surface area of wick, m2HV=Enthalpy of vaporization at twb J / kgHc=Convective heat transfer coefficient, w / m2 K16Where12The expression for molar flux of A (water) through a stagnant B (air) can be written as NA = ky (y yi) / (1 y)1m

Since y is small; yi = ywb (The interphase composition of water vapor is equal to the composition of water calculated at the wet bulb temperature); y = YMB / MA; ywb = Ywb MB/MA. Then the equation 17 becomes NA = ky (MB / MA) (Ywb Y)

Then MA ky (MB / MA) (Ywb Y) (HV)twb = hc (t twb)

rearranging, (Y Ywb) / (t twb) = (hc / ky MB) / (HV)twb17181920The numerator, hc / ky MB is termed as psychrometric ratio. For air water mixture, the value is equal to CH, which is approximately equal to 1 kJ / kg K.

This is known as Lewis relation or Lewis number. Even though it is true that the wet bulb temperature is equal to adiabatic saturations, it may be realized that the former is a non equilibrium temperature and the latter is an equilibrium temperature at steady state conditions.

In case of other gas mixture these temperatures will not be equal.

Dew Point Temperature: It is the temperature at which a given mixture of air and water vapor becomes saturated as it is cooled at constant pressure and humidity.14Psychrometric (or Humidity) Chart:

The psychrometric chart is a useful tool for calculations involving air-water system. Most of the psychrometric properties defined and described above are shown graphically in the chart which help solve a wide range of problems relating to heating, cooling, humidification and dehumidification. A typical psychrometric chart is shown in Fig. 3.

The X-axis shown dry bulb temperature, measured as the temperature of the humid air. The Y-axis shows the absolute humidity. The 100% curve gives the humidity of saturated air as a function of air temperature.

There are other curves, right side of the 100% curve, each represents % saturation of air ranging from 80 10%. In some charts relative humidity curves are used instead those of % saturation.

The adiabatic or wet bulb temperature lines are plotted using equation 15 or 20 for a given constant value of temperature.

Fig. 3:Psychrometric Chart

Specific volume line run almost vertically from the upper left to the lower right of the chart. The enthalpy curve runs parallel to the saturation curve.

On some psychrometric charts, the enthalpy lines coincide with the wet bulb or adiabatic saturation lines.

A standard psychrometric chart is plotted for a sea-level barometric pressure of 101.3 kpa.

The chart is used with little error for barometric pressures ranging from 97 to 104 kPa.

High altitude analyses require a special chart because the barometric pressure affects denisity and therby, thermodynamic properties of air.17Applying the Chart:

To determine the humidity and wet bulb temperature when the dry bulb temperature, t and % saturation humidity of humid air are given.Fig. 4a. Illustrates the determination of humidity.

Locate the dry bulb temperatures on X axis

Draw a vertical line from t to the % saturation curve to intersect it at A.

From A, read humidity horizontally on Y Axis.

Fig. 4b. Illustrates the determination of wet bulb temperature.

From point A, follow the adiabatic cooling or saturation line to the 100% saturation curve to intersect it at B.

From B, read the wet bulb temperature vertically on X axis.

Fig. 4c. Illustrates the determination of dew point.

Draw horizontal line from point A and extend it to the 100% saturation curve to point C.

From C, read the dew point temperature vertically on X axis.

19To find out the humidity and the % saturation humidity when the dry bulb and wet bulb temperatures are given.Fig 5. Illustrates the determination of humidity

Draw a vertical line to 100% saturation line form the wet bulb temperature twb. It intersects at S.

Locate K the intersection point of adiabatic cooling line from S and the vertical line drawn from X axis at the point of dry bulb temperature, t.

Read the value of humidity horizontally on Y axis.

Point K gives the % saturation humidity. If it is not falling on any one of the % saturation curves, proper interpolation may be necessary.

Cooling Towers

Basically, a cooling tower is a direct contact heat exchanger for removing waste heat from process water.

The water to be cooled is distributed through the tower packing material called fill, while air is blown across or counter-currently to the flow or water.

Cooling is accomplished mainly through evaporation of some quantity of water (up to 1% of the circulated water for every 100F cooling). As the evaporation process takes place, latent heat exchange occurs (about 1000 Btu are removed per 1b of water evaporated) reducing the water temperature.

The heat is rejected into the atmosphere through humid air and the cooled water is returned to the process system.Theory

Fig. 6. Shows a schematic diagram of a cooling tower giving the inlet and outlet conditions of air and water.

As a small portion of water evaporates, it can be reasonably assumed that the rate of water remains constant. ie. L1 = L2 = L. The heat capacity of water, cw, is constant over the temperange of the cooling tower.

An enthalpy balance over differential volume, SdZ, of the column gives

GdH = d [LCw (TW T0)]2122

Fig. 6: Schematic Diagram of a Cooling Tower 2When the reference temperature is taken as zero and differentiating the RHS of equation 21,

G x dH = L x Cw x dtwIntegration over the tower givesH2 H1=L x Cw x dtw / G

The above equation is linear and is the operating line for the cooling tower.The rate of sensible heat transferred from the interface to the bulk of the gas-vapour mixture in the differential volume is given by the expression.G x CH x dtG=hG x aH x (tGi tG)222324WheretGi is the temperature at the interface Incorporating the Mass transfer coefficient for the heat transfer coefficient by means of the Lewis relation, the above equation can be written as G x CH x dtG=Ky x aH x MB x CH x (tGi tG) x SdZ

The rate of mass transfer in gas phase from the interface is G x dY=Ky x aM x MA x (YAi YA) x SdZSince YAi YAi MB / MA and YA YA MB / MA, the equation becomes in terms as humiditiesG x dY=Ky x aM x MB x (YAi YA) x SdZ25262725The equation 27 is multiplied both sides by the latent heat of aporization, H and combining it with equation 25 gives the total heat transfer rate to the gas phase from the interface. Assuming aH = aM = a, the rate of equation becomes,

G x [CH dTG + dY HV]=G dH= RHSRHS= Ky x a x MB x CH x (tGi tG) + HV x (YAi YA) x S dZ

The saturation gas enthalpy at the interface and bulk gas enthalpy interface are defined as

Hi=CH tGi + HV x YAiH=CH tG + HV x YA282930Subtracting (30) from (29)

Hi H = CH (tGi tG) + HV x (YAi YA)Substuition of equation (31) for the square braketted term in the RHS of the equation results in

G x dH = Ky x a x MB (Hi H) S dZRearranging the equation (32) and integrating for the height of the cooling tower, Z, assuming that G / Ky am is constant, it becomes 3132

33Where (NTU)G is known as number of individual gas enthalpy transfers units based on gas phase and (HTU)G is the corresponding height of an individual unit .

(NTU)G=dH / (Hi H) (HTU)G=G / Ky a MB STo determine the (NTU)G, the values of interfacial gas phase enthalpies Hi and the corresponding enthalpy of the bulk gas phase at any given cross section in the cooling tower are necessary.

A relation between the gas phase enthalpies and the rate of heat transfer from the gas phase to the bulk liquid phase to the interface establishes the calculation of enthalphies at the liquid (water) temperatures. 353428Thus,L Cw dtw=hL a (tw twi) S dZ

Combining equations (22), (32) and (36),

hL a (tw twi) S dZ=Ky x a x MB (Hi H) S dZRearranging,

- hL a / Ky a MB=(Hi H) / (twi tw)

At the interface, twi = tGi, because the resistance to the heat and mass transfer is negligible here. Then the equation (38) becomes

hL a / Ky a MB=(Hi H) / (tGi tw)

Graphical Procedure for (NTU)G36373839The determination of (NTU)G involves a two step graphical procedure.

Step1: Evaluating a set of values of (Hi H) , in the specified limits of operation on a plot of (H - Hw) coordinates.

Curve AB shown in fig is the saturated enthalpy curve for air at various water temperatures in the given range. The following equation may be used to calculate saturation enthalpy of air: (t in oC)

40The operating line CD is drawn with a slope LCw/ G, initially locating point C, which has the coordinates (H1, tw1. The wet bulb temperature of the inlet air and the temperature of water, leaving the tower may be used to locate the point C as shown in figure.

A line with a slope - hL a / Ky a MB is plotted from an intermediate point (H,tw) on the operating line, to intersect the equilibrium curve. The coordinates of the intersection represent the interface values of Hi and twi. Thus a number of values of Hi for the corresponding values of H are obtained in between H1 and H2.

Step 2: A plot of 1/ (Hi-H) vs H results in a curve as in fig B. the area under the curve gives the number of transfer units (NTU)G.Overall Transfer Units:

Very often, it is assumed that the liquid film resistance is negligible, making tw, twi and tGi identical. According to the equation, - hL a / Ky a MB equals to infinity. It enables to estimate Hi as H* which is the intersection point of vertical line drawn from the point (H,tw) and the equilibrium curve. Then, (H* - H) represents the overall driving force. The overall gas enthalpy number of the transfer units (NTU)OG can be evaluated by substituting (H* - H) for individual driving force (Hi H). In that case, the overall coefficient Ky replaces Ky in the equation (35) for the height of overall gas enthalpy transfer unit.

(NTU)OG=dH / (H* H)(HTU)OG=G / Ky a MBThen,

Z =(NTU)OG (HTU)OG = (NTU)G (HTU)G414243The transfer unit method, described above, for the tower height, Z, is theoretically sound provided appropriate values of HTU and NTU are used. In the absence of data on hL / ky a, the assumption of negligible temperature gradient in the liquid phase can be invoked to estimate (NTU)OG, and is made equal to (NTU)G in the design calculations.

Fig. 7: Determination of Transfer Units, NTUG INDUSTRIAL COOLING TOWERS35COOLING TOWERSWHAT IS COOLING TOWER?A cooling tower is an equipment used to reduce the temperature of a water stream by extracting heat from water and emitting it to the atmosphere. Cooling towers make use of evaporation whereby some of the water is evaporated into a moving air stream and subsequently discharged into the atmosphere. As a result, the remainder of the water is cooled down significantly.

36

37Components of a cooling towerThe basic components of a cooling tower include the frame and casing, fill, cold-water basin, drift eliminators, air inlet, louvers, nozzles and fans.

Frame and casing. Most towers have structural frames that support the exterior enclosures (casings), motors, fans, and other components. With some smaller designs, such as some glass fiber units, the casing may essentially be the frame.

Fill. Most towers employ fills (made of plastic or wood) to facilitate heat transfer by maximizing water and air contact. There are two types of fill:

Splash fill: water falls over successive layers of horizontal splash bars, continuously breaking into smaller droplets, while also wetting the fill surface. Plastic splash fills promote better heat transfer than wood splash fills.38Film fill: consists of thin, closely spaced plastic surfaces over which the water spreads, forming a thin film in contact with the air. These surfaces may be flat, corrugated, honeycombed, or other patterns. The film type of fill is the more efficient and provides same heat transfer in a smaller volume than the splash fill.

Cold-water basin. The cold-water basin is located at or near the bottom of the tower, and it receives the cooled water that flows down through the tower and fill. The basin usually has a sump or low point for the cold-water discharge connection.

In many tower designs, the coldwater basin is beneath the entire fill. In some forced draft counter flow design, however, the water at the bottom of the fill is channeled to a perimeter trough that functions as the coldwater basin. Propeller fans are mounted beneath the fill to blow the air up through the tower.

39With this design, the tower is mounted on legs, providing easy access to the fans and their motors.

Drift eliminators. These capture water droplets entrapped in the air stream that otherwise would be lost to the atmosphere.

Air inlet. This is the point of entry for the air entering a tower. The inlet may take up an entire side of a tower (cross-flow design) or be located low on the side or the bottom of the tower (counter-flow design).

Louvers. Generally, cross-flow towers have inlet louvers. The purpose of louvers is to equalize air flow into the fill and retain the water within the tower. Many counter flow tower designs do not require louvers.

40Nozzles. These spray water to wet the fill. Uniform water distribution at the top of the fill is essential to achieve proper wetting of the entire fill surface. Nozzles can either be fixed and spray in a round or square patterns, or they can be part of a rotating assembly as found in some circular cross-section towers.

Fans. Both axial (propeller type) and centrifugal fans are used in towers. Generally, propeller fans are used in induced draft towers and both propeller and centrifugal fans are found in forced draft towers. Depending upon their size, the type of propeller fans used is either fixed or variable pitch.

A fan with non-automatic adjustable pitch blades can be used over a wide kW range because the fan can be adjusted to deliver the desired air flow at the lowest power consumption. Automatic variable pitch blades can vary air flow in response to changing load conditions.41TYPES OF COOLING TOWERS

Figure 2. Cross flow natural draft cooling tower Figure 3. Counter flow natural draft cooling tower 42MECHANICAL DRAFT COOLING TOWERS

Figure 4. Forced Draft Cooling Tower43

44

45ASSESSMENT OF COOLING TOWERSThis section describes how the performance of cooling powers can be assessed.3 The performance of cooling towers is evaluated to assess present levels of approach and range against their design values, identify areas of energy wastage and to suggest improvements.

During the performance evaluation, portable monitoring instruments are used to measure the following parameters:

Wet bulb temperature of airDry bulb temperature of airCooling tower inlet water temperatureCooling tower outlet water temperatureExhaust air temperatureElectrical readings of pump and fan motorsWater flow rateAir flow rate 46

47a) Range (see Figure 7). This is the difference between the cooling tower water inlet and outlet temperature. A high CT Range means that the cooling tower has been able to reduce the water temperature effectively, and is thus performing well. The formula is:

CT Range (C) = [CW inlet temp (C) CW outlet temp (C)]

b) Approach (see Figure 7). This is the difference between the cooling tower outlet coldwater temperature and ambient wet bulb temperature. The lower the approach the better the cooling tower performance. Although, both range and approach should be monitored, the `Approach is a better indicator of cooling tower performance.

CT Approach (C) = [CW outlet temp (C) Wet bulb temp (C)]48Effectiveness. This is the ratio between the range and the ideal range (in percentage), i.e. difference between cooling water inlet temperature and ambient wet bulb temperature, or in other words it is = Range / (Range + Approach). The higher this ratio, the higher the cooling tower effectiveness.

CT Effectiveness (%) = 100 x (CW temp CW out temp) / (CW in temp WB temp)

d) Cooling capacity. This is the heat rejected in kCal/hr or TR, given as product of mass flow rate of water, specific heat and temperature difference.

49e) Evaporation loss. This is the water quantity evaporated for cooling duty. Theoretically the evaporation quantity works out to 1.8 m3 for every 1,000,000 kCal heat rejected. The following formula can be used (Perry):

Evaporation loss (m3/hr) = 0.00085 x 1.8 x circulation rate (m3/hr) x (T1-T2)

T1 - T2 = temperature difference between inlet and outlet water

f) Cycles of concentration (C.O.C). This is the ratio of dissolved solids in circulating water to the dissolved solids in make up water.

g) Blow down losses depend upon cycles of concentration and the evaporation losses and is given by formula:

Blow down = Evaporation loss / (C.O.C. 1)50h) Liquid/Gas (L/G) ratio. The L/G ratio of a cooling tower is the ratio between the water and the air mass flow rates. Cooling towers have certain design values, but seasonal variations require adjustment and tuning of water and air flow rates to get the best cooling tower effectiveness.

Adjustments can be made by water box loading changes or blade angle adjustments. Thermodynamic rules also dictate that the heat removed from the water must be equal to the heat absorbed by the surrounding air. Therefore the following formulae can be used:

L(T1 T2) = G(h2 h1)L/G = (h2 h1) / (T1 T2)

Where:

L/G = liquid to gas mass flow ratio (kg/kg)T1 = hot water temperature (OC)T2 = cold-water temperature (OC)h2 = enthalpy of air-water vapor mixture at exhaust wet-bulb temperature (same units as above)h1 = enthalpy of air-water vapor mixture at inlet wet-bulb temperature (same units as above)51

Thank You52