Psychrometrics. PV = N R u T (9.01) PV = N M R T (9.02) Pv = R T (9.03) PV = m R T (9.04)

Transcription

1 Pycrometric Abtract. Ti capter include baic coverage of pycrometric propertie and pycrometric procee. Empai i upon propertie and procee relative to te environment and to proceing of biological material. Te capter alo include a dicuion of team propertie and ue of team table. Keyword. Cooling, drying, dew point temperature, entalpy, eating, umidity ratio, pycrometric, team quality, relative umidity, wet bulb temperature, team, team-air mixture, team table. 9.1 Introduction Air i a mixture of many different gae (a mixture we call dry air) and water vapor. If preure do not greatly exceed atmoperic preure, te perfect ga relationip are valid for tee air-water vapor mixture. Tee ideal ga relationip are ued to evaluate termodynamic propertie of te mixture. Ti field of tudy i called pycrometric. An undertanding of pycrometric i important in te drying and torage of elected food product. Proper drying condition are neceary to control te drying rate and reulting quality of dried product. In addition, pecific pycrometric condition ave been identified for optimum torage of many fre and dried product. Tee are furter dicued in oter capter. Pycrometric propertie are alo important in aeing environmental condition relative to plant and animal ealt. 9.2 Relationip Derived from te Perfect Ga Law Te propertie of air/water vapor mixture are baed upon ideal ga relationip. Tu, we will review te baic relationip needed to define important pycrometric propertie. Te perfect ga law may be written in different form a: were: m = ma of te ga, kg M = molecular weigt of te ga, kg/kmol N = number of mole of te ga P = preure, kpa PV = N R u T (9.01) PV = N M R T (9.02) Pv = R T (9.03) PV = m R T (9.04)

2 214 Food & Proce Engineering Tecnology R = ga contant for a particular ga, R u /M, kpa m 3 /kg K R u = univeral ga contant, R u = kpa m 3 /kmol K (or kj/kmol K) T = abolute temperature, K v = pecific volume of te ga, m 3 /kg V = total volume of te ga, m 3 Unit for te ga contant R and R u can be expreed in everal different form. Te unit given above for R u are probably te mot decriptive; owever, uing unit factor to convert unit, we can alo obtain unit of kn m/kmol K witout affecting te numeric value of Similarly, R may be expreed in unit of kpa m 3 /kg K, kn m/kg K, or kj/kg K witout affecting te numeric value. Molecular weigt and ga contant for air and water vapor (uing te unit given above) are: M a = 28.96, R a = , M w = , and R w = For a mixture of everal gae, Equation 9.01 to 9.04 may be written eparately for eac ga. If Equation 9.01 i ued, te equation for te total mixture become: P = (N 1 + N 2 + N ) R u T/V (9.05) For air containing water vapor, te total preure i given by: P = P a + P w (9.06) were P a i te partial preure of te dry air (including all it component gae) and P w i te partial preure of te water vapor. Tee two partial preure may be written (uing Equation 9.01) a: P a = N a R u T/V (9.07) P w = N w R u T/V (9.08) Te partial preure of water vapor i a very important parameter in pycrometric. It i frequently ued to calculate oter pycrometric propertie. For example, relative umidity (φ) i defined a te ratio of te amount of water vapor in te air (N w ) to te amount te air will old wen aturated at te ame temperature (N w ). Tu: Uing Equation 9.08, we can rewrite Equation 9.09 a: N w φ = (9.09) N w P w RuT Pw φ = = (9. 10) P w V V R T Te partial preure of water vapor at aturation (P w ) i a function only of temperature. However, ti relationip i complex, involving multiple exponential and logaritmic term (Wilelm, 1976). u P w

3 Capter 9 Pycrometric 215 Anoter pycrometric parameter of interet i te umidity ratio (W). Te umidity ratio i te ratio of te ma of water vapor in te air to te ma of te dry air: m w W = (9.11) ma Referring to te ideal ga equation in te form of Equation 9.04, we can write: V P W 622 w mw RwT RaPw Pw Pw = = = = = 0. (9.12) m V a P RwPa Pa Pa a RaT Te partial preure of te dry air (P a ) i te difference between atmoperic preure and te partial preure of te water vapor, alo called vapor preure: Tu: W P = P (9.13) a P w =. 622 P w 0 (9.14) P Pw or, olving for P w : P P W = w + (9.15) W Te degree of aturation (µ) i anoter pycrometric parameter tat i ometime ued. It i te umidity ratio divided by te umidity ratio at aturation: W µ = (9.16) A final parameter tat can be determined from te perfect ga law i te pecific volume (ν) of te moit air. Te pecific volume i defined in term of a unit ma of dry air. Tu, uing Equation 9.04 and 9.13: Subtituting from Equation 9.15 for P w : v m R T P W a a V a Ra T Ra T = = = = (9.17) ma ma Pa P Pw R T Ra T v = a = ( W ) P W P P 608 (9.18) W

4 216 Food & Proce Engineering Tecnology 9.3 Oter Important Parameter Te dew point temperature (t dp ) i te temperature at wic moiture begin to condene if air i cooled at contant preure. Te dew point temperature i directly related to partial preure of te water vapor (P w ); owever, tat relationip i complex, involving everal logaritmic term (ASHRAE, 1997). Since P w i alo related to te umidity ratio W, ti mean tat pecifying any one of te tree parameter t dp, P w, and W pecifie all tree. Te wet-bulb temperature (t wb ) i te temperature meaured by a enor (originally te bulb of a termometer) tat a been wetted wit water and expoed to air movement tat remove te evaporating moiture. Te evaporating water create a cooling effect. Wen equilibrium i reaced, te wet-bulb temperature will be lower tan te ambient temperature. Te difference between te two (te wet bulb depreion) depend upon te rate at wic moiture evaporate from te wet bulb. Te evaporation rate, in turn, depend upon te moiture content of te air. Te evaporation rate decreae a te air moiture content increae. Tu, a mall wet bulb depreion indicate ig relative umidity, wile a large wet bulb depreion i indicative of low relative umidity. Te entalpy () of moit air i one of te mot frequently ued pycrometric parameter. It i a meaure of te energy content of te air and depend upon bot te temperature and te moiture content of te air. It i determined by adding te entalpy of te moiture in te air (W w ) to te entalpy of te dry air ( a ): or: ( c t) = a + W w = c pa t + W fg + pw (9.19) ( t) = 1.006t + W (9.20) Equation 9.20 i baed upon a pecific eat for air of kj/kg K and a zero value of at t = 0 C. Te entalpy of water i baed upon: a zero value of at 0 C (liquid tate); fg = 2501 kj/kg at 0 C; and an average pecific eat for water vapor of kj/kg K. Equation 9.20 provide a good approximation for te entalpy of moit air over a wide range of temperature; owever, te error increae rapidly at temperature above 100 C. Empirical relationip, cart, or table mut ten be ued to determine. 9.4 Pycrometric Cart If two independent pycrometric parameter are known, all te pycrometric parameter previouly dicued can be computed numerically. However, numerical calculation i practical only if te calculation metod i computerized. Te fatet and eaiet metod of evaluating pycrometric propertie i often te ue of a pycrometric cart. Te pycrometric cart ow elected pycrometric propertie a a function of eac oter. Propertie own on mot pycrometric cart are drybulb, wet-bulb, and dew point temperature; relative umidity; umidity ratio; entalpy; and pecific volume. Figure 9.01 i a typical pycrometric cart. It cover te

5 Capter 9 Pycrometric 217 Figure Pycrometric cart for normal temperature (SI unit). (Courtey of te American Society of Heating, Refrigerating and Air-Conditioning Engineer.)

6 218 Food & Proce Engineering Tecnology normal temperature range at tandard atmoperic preure. Oter cart are available for iger and lower temperature range and for preure above and below one atmopere. In addition, different cart verion are available for tee common pycrometric range. Te general arrangement of all cart i te ame; owever, difference in appearance are obviou. Te procedure for reading te pycrometric parameter differ little among te cart except for te entalpy and (occaionally) te wet-bulb temperature. We will ue Figure 9.01 to gain an undertanding of pycrometric cart. Before reading furter, you ould examine ti figure cloely. Identify te line repreenting parameter uc a dew point, wet-bulb and dry-bulb temperature, umidity ratio, etc. Eac poition on a pycrometric cart repreent a unique tate point. Any two independent pycrometric propertie define uc a tate point. We will now ue two propertie to define a tate point and ten determine te value of te oter pycrometric propertie. Let u aume tat we ued a pycrometer, a device tat meaure wet-bulb and dry-bulb temperature, and obtained te following meaurement: t wb = 17.1 C t = 24 C To find te tate point repreented by tee value, we follow a vertical line upward from t = 24 C at te bottom of te cart. We ten locate te point were ti line interect wit an imaginary line jut above, and parallel to, te daed line loping downward to te rigt from a wet-bulb temperature of 17.0 C. Te interection of te line (circled on Figure 9.01) repreent te deired tate point. Note tat te temperature on te curved line at te upper left of te cart may repreent eiter dew point (aturation temperature), wet-bulb, or dry-bulb temperature. A contant wet-bulb temperature i repreented by a line loping downward to te rigt; a contant dew point temperature i repreented by a orizontal line; and a contant dry bulb line i repreented by a vertical (or near vertical) line. At aturation (100% relative umidity), te dry bulb, wet-bulb and dew point temperature are equal. At any oter condition, tey are not equal and follow te order t > t wb > t dp. We ave now located te tate point (indicated by te center of te circle on Figure 9.01). Moving left orizontally from ti point, we find tat te dew point i 13.0 C. Alo, moving rigt along ti orizontal line we find tat te umidity ratio i 9.3 g/kg of dry air. Relative umidity i repreented by line curving upward to te rigt. Our tate point appear to be exactly on te 50% line. Tu, φ = 50%. Similarly, te pecific volume i found by interpolating between te 0.85 and 0.86 contant pecific volume line wic lope arply downward to te rigt. We find tat v = m 3 /kg. Te partial preure of te water vapor can be calculated from Equation 9.15, noting tat P = kpa and W mut be dimenionle ( kg/kg of dry air, intead of te 9.3 g/kg noted above): P w = = kpa

7 Capter 9 Pycrometric 219 Te pycrometric cart own in Figure 9.01 i only one of everal verion of pycrometric cart available. Two oter cart are own in Figure 9.02 and Figure 9.02 cover lower temperature range not covered by Figure Figure 9.03 i a greatly condened cart tat extend to iger temperature. All tee cart preent te data in SI unit. Similar cart are available uing US unit. Altoug tere are ome difference in cart arrangement, te ame interpretation procedure apply for all pycrometric cart. In reading te pycrometric cart, you ould keep te following point in mind: 1. Entalpy, umidity ratio, and pecific volume are referenced to a unit ma of dry air. 2. On cart uing US unit, te umidity ratio may be given in pound of moiture per pound of dry air or in grain of moiture per pound of dry air. Since one pound contain 7000 grain, te two meaurement can be converted a needed. In SI unit, te umidity ratio i expreed in eiter gram of water or kilogram of water per kilogram of dry air. 3. Contant entalpy and contant wet bulb line are not parallel, altoug tere i only a ligt difference in lope. Becaue of ti, mot cart do not ow bot contant wet-bulb and contant entalpy line. Only one parameter i uually plotted, altoug te oter may be plotted at wide interval. In te example ued earlier, te entalpy i 48 kj/kg dry air. Note te ort loping line along te rigt edge and bottom of te cart. Tee repreent contant entalpy line correponding to imilar line along te entalpy cale at te upper left. By uing a traigt edge to align te point in te circle on Figure 9.01 wit equal value of entalpy along te bottom and upper left of te cart, we find tat te entalpy value i 48 kj/kg dry air. Some cart ue an entalpy correction curve. Tee curve ow te adjutment to correct for deviation between contant entalpy and contant wet-bulb line. On tee cart, te contant wet-bulb line i followed to find a value for entalpy from te entalpy cale at upper left. Tat value i ten adjuted by ubtracting te value of te entalpy correction Pycrometric Table Tabulated value of pycrometric propertie (ee Table 9.01) can alo be ued for analyi of pycrometric data. However, ue of te table i more time conuming tan ue of a cart. Ue of te table alo a limitation. For example, te wet bulb temperature cannot be directly related to any tabulated propertie. However, we can ue te table to determine oter pycrometric propertie at t = 24 C and 50% relative umidity. Ti i te ame point ued in te cart analyi; owever, we can expect te value of ome propertie to differ ligtly from toe determined by te cart. Reult obtained uing te table would normally be conidered more accurate tan toe from a pycrometric cart. To find te partial preure (or vapor preure) of water vapor in te air, we note tat relative umidity i defined a φ = P w /P w (Equation 9.10). At t = 24 C, te table give P w = Solving for P w = φ P w = 0.5(2.9852) = kpa. Ti i eentially te ame value found from te pycrometric cart. We can now ue Equation 9.14 to olve for te umidity ratio (W) at 24 C and 50% relative umidity:

8 220 Food & Proce Engineering Tecnology Figure Pycrometric cart for low temperature (SI unit). (Courtey of te American Society of Heating, Refrigerating and Air-Conditioning Engineer.)

9 Capter 9 Pycrometric 221 Figure Pycrometric cart for ig temperature (SI unit). (Courtey of te American Society of Heating, Refrigerating and Air-Conditioning Engineer.)

10 222 Food & Proce Engineering Tecnology W Pw = P P w kg H 2O = = kg DA Since te dew point temperature i defined a te temperature at wic 100% relative umidity i reaced a te air i cooled, ti would be te temperature at wic W = kg H 2 O/kg DA. Searcing te table for te temperature correponding to W = kg H 2 O/kg DA, we find tat t dp = 13 C at W = kg H 2 O/kg DA. Ti i ligtly above but very cloe to te look-up value of tat we are uing. If neceary, we could interpolate to find te temperature correponding to W = Note tat ti ould alo be te temperature correponding to P w = kpa. Note alo tat W = at 24 C; but ince we ave only 50% relative umidity, tat moiture level i never reaced. Te remaining two propertie, pecific volume (v) and entalpy () can be determined from te table by noting tat eac are made up of an air component and a water component. For example, te energy content () of te wet air i te um of energy from te dry air and energy from te moiture preent. Te entalpy column of te table give value of a (entalpy of te dry air), (entalpy of te aturated air), and a (entalpy added by te moiture to produce aturated air and equal to a ). We can determine at any moiture content by uing tee value and te equation: W = a + µ a, were µ = (9.21) Rewriting te equation, ubtituting for µ, and inerting value for all te propertie, we find: W = a + a W W = = kj kg DA Ti compare to a value of 48 found from te cart. Te olution for pecific volume follow exactly te ame procedure: W v = va + va W = m = kg DA Again, te reulting value i very cloe to te value determined from te pycrometric cart. Typically, ue of te tabulated value, wile more time conuming, will give more accurate reult tan ue of te pycrometric cart.

11 Capter 9 Pycrometric 223 Temp ( C) t Table Selected termodynamic propertie of moit air. [a] Humidity Ratio (kg w /kg da ) Volume Entalpy Condened Water Vapor Entalpy Preure (kj/kg) (kpa) w p w (m 3 /kg da ) (kj/kg da ) W v a v a v a a [a] Abridged data taken from ASHRAE Handbook of Fundamental, Capter 6. Courtey of te American Society of Heating, Refrigerating and Air-Conditioning Engineer.

12 224 Food & Proce Engineering Tecnology 9.5 Procee on te Pycrometric Cart Te pycrometric cart i ued in many application bot witin and outide te food indutry. It i ued extenively to analyze product drying and air cooling, primarily becaue it i a imple and quick alternative to more cotly metod. Drying wit air i an extremely cot-effective metod to reduce te moiture of a biological material, and te addition of a mall amount of eat ignificantly improve te air drying potential. Air at a particular temperature and preure a a limit to it drying capability and o it i ueful to undertand te cange in it drying potential a moiture i added or removed by te food material, people, animal, etc. Hot, dry environment for animal, including uman, can be made more comfortable by evaporative cooling, a low-cot metod tat lower te air temperature a a trade-off for increaing umidity. Te addition of moiture to te air i valuable in certain application a een in ummertime greenoue evaporative cooling. Fruit and vegetable arveted in ot weater often can be cooled wit evaporatively cooled air to reduce repiration. Large indutrial plant, particularly food proceing plant, generating wate eat and needing cooling often utilize cooling tower to reduce teir refrigeration requirement. Cooling tower, operating on te evaporative cooling principle, provide a way to cool water by rejecting te unwanted eat to te atmopere. In all tee application, we are concerned wit following a proce tat can be decribed on te pycrometric cart. By a proce we mean moving from one tate point to anoter tate point on te cart. We will look at a few imple procee, and diplay te pat of tee procee on mall pycrometric cart in Figure Selected relevant propertie are alo noted on eac cart (Cart I, II, III, IV, V, VI). Tee are ideal procee auming no eat tranfer from te urrounding. In actual procee, tere will alway be ome eat gain or lo. a. Heating or cooling. Tee procee follow a contant moiture line (contant umidity ratio). Tu, temperature increae or decreae but moiture content and dew point are uncanged. Proce A B (Cart I) repreent a eating proce, wile proce A B (Cart II) repreent cooling. If cooled below te dew point (Cart III), te contant umidity ratio line i followed until te dew point (Point D) i reaced. Furter cooling follow te aturation (100% relative umidity) line until te final temperature (Point B) i reaced. Moiture i condened during te part of te proce tat follow te aturation line. b. Moiture addition (i.e., evaporation) wile uing only energy from te air (Cart IV). Bot drying and evaporative cooling follow ti proce. It i repreented by a contant wet-bulb line (Proce A B.) Temperature, moiture content, etc., cange but te wet-bulb temperature remain contant. Ti can be verified by an energy balance analyi. Note tat entalpy increae ligtly in ti proce. Ti i due to energy preent in te water before it i evaporated. c. Heating and drying (Cart V). Ti combination of te procee own in Cart I and IV i common in drying application. Air i eated (Proce A B) and paed over te material to be dried (Proce B C). A econd tage of eating (C D) and drying (D E) i ometime included.

13 Capter 9 Pycrometric 225 d. Adiabatic mixing (no eat tranfer) of air (Cart VI). Moit air from two ource and at different tate point i mixed to produce air at a tird tate point. Relationip among te propertie at te tree tate point are etablied from ma and energy balance for te air and water component: m + m = m air ma balance (9.22) aa aa A ab ab ac m W + m W = m W water ma balance (9.23) aa A ab B B ac ac C C m + m = m energy balance (9.24) Tee equation are uually olved to relate pecific propertie to te pecific mae involved. For example, we can ue Equation 9.22 to 9.24 to obtain te following relationip: m m W W aa C B C B = = (9.25) ac A B WA WB Subcript A and B in te above equation repreent entering air tream, wile ubcript C repreent condition after mixing. Note tat ti proce doe not follow a contant property line on te pycrometric cart. Te poition of Point C along te line between Point A and B i determined by te proportion of air at eac tate point included in te mix. If equal amount of air are provided from eac ource, Point C will be at approximately te midpoint of te line. If 90% of te air i from tate Point B, Point C will be approximately 90% of te ditance from Point A and 10% of te ditance from Point B. Ti caled approac i ueful a a quick etimate of te reulting tate point. It i alo ueful a a quick verification of numerical calculation. However, ue of te above equation to calculate te tate point of mixed air i te preferred approac. e. Adiabatic Saturation. Te drying proce wa identified earlier a a contant wetbulb proce. Wile ti i te generally accepted approac, a review of te adiabatic aturation proce i provided ere for added clarification. An adiabatic aturation proce occur wen te umidity of air i increaed a it flow troug an inulated camber a own below. Water evaporate into te air a it pae troug te camber. If te camber i long enoug for equilibrium to occur, ten te exit air will be aturated at an equilibrium temperature t*. Wit no eat tranfer, te energy balance ten become: + ) = 1 ( W W1 If te water temperature inide te camber i equal to te exit temperature, te above equation for adiabatic aturation become: * * * + W W = 1 ( 1) f f

14 226 Food & Proce Engineering Tecnology Cart I Heating Coil B A B t dp A A B W A = W B Senible Heating t A t B Cooling Coil Cart II A A B t dp B B A W A = W B Senible Cooling t B t A Cooling Coil wit Condenation Cart III B d A A W A A B B W B Senible Cooling followed by condenation formation t B t d t A Figure Example of pycrometric procee (continued next page).

15 Capter 9 Pycrometric 227 Evaporative Pad Cart IV B A B W B A B A W A Evaporative Cooling t B t A Evaporative Pad Cart V t wbc = t wbb t wbd E W E A B C D E A B C D W C = W D A B W A = W B Preeater Reeater Senible Heating- Evaporative Cooling t A t C t B t D Cart VI Adiabatic Mixing B B WB C A B C A A C W C W A Cooling and Deumidification t A t C t B Figure 9.04 (continued). Example of pycrometric procee.

16 228 Food & Proce Engineering Tecnology were: 1 = entalpy of entering air * f = entalpy of aturated liquid at t * * = entalpy of aturated air at t * t W t W 1 1 W 1 = umidity ratio of entering air Water t * 2 2 * W = umidity ratio of aturated air at t * (ti i alo W 2, ince t 2 = t * ) Te econd term of te equation i te amount of energy added to te air a it pae troug te camber. Ti i energy already preent in te liquid water tat i evaporated into te air. Te temperature, t *, tat atifie te above equation i called te termodynamic wet-bulb temperature. Te termodynamic wet-bulb temperature i not quite te ame a te wet-bulb temperature. Te wet-bulb temperature i obtained by cooling in a combined eat and ma tranfer proce governed by an equation ligtly different from te adiabatic aturation equation. Comparion of te two equation introduce a dimenionle Lewi number Le: Le = kc p were: = urface eat tranfer coefficient at te wet-bulb temperature (W/m 2 K) c p = pecific eat of mot air (J/kg K) k = air-film coefficient of ma tranfer (kg/m 2 ) If ti Lewi number i exactly equal to one, ten te wet-bulb and termodynamic wet-bulb temperature are equal. Fortunately, for air-water vapor mixture te Lewi number i very cloe to one. Tu, te wet-bulb temperature i normally conidered te ame a te adiabatic aturation, or termodynamic wet-bulb temperature. Ti relationip i valid only for air and water vapor mixture. It i not valid wit oter material (e.g., air and alcool) were te Lewi number may ubtantially deviate from one. Mot real procee do not occur along a contant property line. A number of factor influence eac proce uc tat, for example, a contant wet-bulb proce i very difficult to attain. In uc cae, we may not be able to follow te proce on te cart. Intead, we identify te initial and final tate point and determine te cange in propertie between te two. Te following example will ow ow pycrometric cart may be ued to evaluate procee. Air Example 9.1 Air at 30 C and 10% relative umidity enter an evaporative cooler were moiture i evaporated until te temperature reace 21 C. Wat i te relative umidity of te air a it leave?

17 Capter 9 Pycrometric 229 Solution: Te initial condition (at t 1 = 30 C and φ 1 = 10%) are (uing Figure 9.01): t wb = 13.3 C 1 = 37 kj/kg dry air w 1 = 2.6 g/kg dry air Ti metod of moiture addition i a contant wet-bulb proce (Cart IV of Figure 9.04.). By following a contant wet-bulb line to t 2 = 21 C, we find: φ 2 = 41% ANSWER Example 9.2 Steam i added to air at 30 C and 10% relative umidity until te temperature reace 38 C and te umidity ratio (W) i 18 g water/kg dry air. How muc water i added to one kilogram of air? Solution: Ti proce doe not follow a contant property line. Intead, we define tate point one (Example 9.1) and tate point two from te given information. Tu, tate point two a te following propertie: t 2 = 38 C (given) W 2 = 18 g/kg dry air (given) φ 2 = 43% t wb = 26.9 C t dp = 23.2 C 2 = 84.5 kj/kg dry air Te moiture added i W 2 W 1 = = 15.4g/kg dry air ANSWER Example 9.3 Air at 20 C, 50% relative umidity (State 1) i eated to 55 C (State 2). Te eated air pae troug a dryer and exit at a relative umidity of 90% (State 3). How muc water i removed for eac cubic meter of air entering te eater? How muc eat i added during te eating proce? Solution: Te procee followed for te eating and drying are: (1) contant umidity ra-

18 230 Food & Proce Engineering Tecnology tio (or dew point) for eating, and (2) contant wet-bulb temperature for drying. Ti i repreented by procee A B and B C own on Cart V of Figure From Figure 9.01, te propertie at eac tate point are: Property State 1 State 2 State 3 t ( C) 20 [a] 55 [a] 26.1 t wb ( C) [b] t dp ( C) [b] 24.3 φ (%) 50 [a] 7 90 [a] (kj/kg) W (g H 2 O/kg dry air) v (m 3 /kg dry air) [a] Propertie given in te problem tatement. [b] Propertie uncanged from previou tate point From te table: W rem = W 3 W 2 = = 12.2 g H 2 O/kg DA Q = 2 1 = = 36.3 kj/kg DA We mut now convert to volumetric value baed upon te entering air: g H 2O kg DA g H 2O W rem = = ANSWER 3 3 m m kg DA kj kg DA kj Q = = m m kg DA ANSWER Example 9.4 Air at State Point 3 of Example 9.3 i cooled to 20 C. (1) How muc water i removed from te air? (2) How muc eat i removed? Solution: From Example 9.3, Point 3 i defined a: t = 26.1 C; t dp = 24.3 C; φ = 90%; and = 75.6 kj/kg. (Tee value are given in te table for te previou example.) Uing Figure 9.01, we find W=19.5 g H 2 O/kg DA.

19 Capter 9 Pycrometric 231 From ti point, te cooling proce follow a orizontal line (contant W) to 100% relative umidity. Air i now at te dew point temperature (24.3 C). Te proce ten follow te aturation (φ =100%) line to t=t wb =t dp =20 C. Te entalpy and umidity ratio at ti point are 57.2 kj/kg DA and g H 2 O/kg DA. Tu te anwer are: W = = 4.75 g H 2 O/kg DA ANSWER 1 Q = = = 17.9 kj/kg DA ANSWER 2 Example 9.5 If te exit flow rate for te air of Example 9.4 i 800 L/, wat are te rate of water and eat removal? Solution: At te exit condition, v = 0.85 m 3 /kg DA. Tu, te ma flow rate i: L 800 a = = 3 m L kg DA m m kg DA g H 2O kg DA g H 2O W = = ANSWER kg DA kj kg DA kj Q = = = kw kg DA ANSWER Example 9.6 Air at 80 C and φ = 4% (Point A) i mixed wit air at 40 C and φ = 50% (Point B). Te ratio i 60% of air at 80 C to 40% at 40 C. Wat i te tate point of te reulting mix (Point C)? Solution: We can ue Equation 9.24 to find propertie of te mixed air, but we mut firt ue pycrometric cart to find te entalpy and umidity ratio for Point A and B.

20 232 Food & Proce Engineering Tecnology We mut ue te ig temperature cart (Figure 9.03) for Point A. At t = 80 C and φ = 4%, we find: W A = 12 g/kg DA A = 115 kj/kg DA For Point B, we may ue eiter Figure 9.03 or 9.01; owever, it i eaier to read te property value from Figure Uing tat figure, we find at t = 40 C and φ = 50%: W B = 23.5 g/kg DA B = 101 kj/kg DA maa Air at propertie of Point A repreent 60% of te mix. Tu = m aa Uing Equation 9.24, we can olve for c (or W c ) a = + ( ) or: C = ( ) = = kj/kg DA W C = ( ) = = 16.6 g/kg DA C B m m ac A ac B ANSWER ANSWER Example 9.7 Air at 30 C and 60% relative umidity i cooled to 18 C. (1) How muc energy i removed from te air? (2) I any water removed? If o, ow muc? Solution: Ti proce follow te pat A d B own on Cart III of Figure Initial condition at point A from Figure 9.01 are: t = 30 C W = 16 g/kg DA φ = 60% = 70.8 KJ/kg DA A te air i cooled, te temperature drop until te air i aturated. At ti point (Point d): t = t wb = t dp = 21.3 C and φ = 100% Continued cooling follow te aturation line to Point B were te propertie are: t = t wb = t dp = 18 C W = 13 g/kg DA φ = 100% = 51 KJ/kg DA Te energy removed from te air i: Q = A B = = 19.8 KJ/kg DA Similarly, te water removed i: W = W A W B = = 3 g/kg DA ANSWERS

21 Capter 9 Pycrometric 233 Example 9.8 Air at 50 C and a wet-bulb temperature of 30 C (Point A) i mixed wit air at 24 C and 50% relative umidity (Point B) in proportion of 70% ot air and 30% cooler air. Wat are te propertie of te air after mixing? Solution: Te propertie at Point C, te mixed air tate point, can be computed uing Equation Relevant propertie at Point A and B are: W A = 18.7 g/kg DA W B = 9.3 g/kg DA A = 98.9 KJ/kg DA B =47.8 KJ/kg DA Solving Equation 9.24, noting tat m aa i 70% of te total mixture, m ac, we ave: = C WC and 0.7 = and: C = 0.7( ) W C = 0.7( ) C = 83.6 KJ/kg DA W C = 15.9 g/kg DA Locating Point C from te value of and W give oter property value a: φ C = 30% t = 42.4 C ANSWER Example 9.9 Water i often cooled by evaporation into te air, uing a cooling tower. Incoming ot water i prayed into te air were a portion i evaporated, lowering te temperature of te remaining water by 10 to15 C. (Evaporation of 1 kg of water provide ufficient cooling to lower te temperature of approximately 56 kg of liquid water by 10 C.) Conider water entering a cooling tower at 40 C and leaving at 30 C. Air enter at 25 C wit a relative umidity of 40% and exit at 34 C wit 90% r. For a water flow rate of 50 kg/, and ufficient evaporation to meet te above condition, determine te required air flow rate (kg/) troug te cooling tower. Solution: Te olution of cooling tower problem conit of a combination of te evaporative cooling proce on te pycrometric cart and an energy balance between air and water tream. Te proce decribing te air umidification along

22 234 Food & Proce Engineering Tecnology a contant wet-bulb line i needed to determine te entalpy at te two tate point. Read from pycrometric cart at 25 C dry bulb temperature and 40% r, te entalpy i 45.5 kj/kg, and at 34 C and 90% r te entalpy i kj/kg. Te energy lot by te cooling of water mut equal energy gained by air. Tu: a = 50 m kg w c pw m m t w = m a = mw c pw 4.18 kj kg C a a t w a ( 40 30) C kj ( ) kg kg m a = 30.7 of air ANSWER Note: Ti olution aumed a contant m w, ignoring te approximately 2% of water tat wa evaporated. Ti approac greatly implifie te olution and a little effect upon te computed air flow. 9.6 Ice-Water-Steam Water in it tree form (olid, liquid, ga) i ued extenively in food, agricultural, and oter application. Te propertie of water, in all tree form, make it well uited for many proceing application. In many application we are concerned wit te ue of water in one or more of it form for eating or cooling. To undertand ti ue, we mut firt undertand certain water propertie tat are relevant to tee application. If ice at 20 C (or any oter ub-freezing temperature) and atmoperic preure i eated very lowly, it temperature would increae until te melting point (0 C, 32 F) i reaced (Figure 9.05). It would ten melt at tat temperature. If additional eat i added after te ice i melted, te water temperature will increae until it reace 100 C (212 F). At 100 C, te water will evaporate (boil) into ga (team). Any furter addition of eat will caue an increae in team temperature. Ti proce of eating and cange of tate i own in Figure Te comparative energy requirement for eating (average pecific eat) and cange of tate (latent eat) are alo own.

23 Capter 9 Pycrometric fg = 2257 kj kg f = 335 kj kg. c p = 1.97 kj kg K c p = kj kg K Temperature ( C) Figure Heat content (entalpy) of water at a preure of one atmopere a a function of temperature and tate (olid, liquid, or ga).

24 236 Food & Proce Engineering Tecnology C kpa Abolute Preure (kpa) Figure Boiling point temperature a a function of preure for water. Te above example aume tat te water i eld at atmoperic preure. However, team i uually produced and ued at preure iger tan atmoperic. Tu, temperature, preure, and energy content of te team are all important factor to be conidered. A key conideration ere i te effect of preure upon te boiling point temperature of water. Ti cange in te boiling point of water wit preure i own in Figure Te energy required to melt te ice (latent eat of fuion) and to evaporate te water (latent eat of vaporization) are important conideration in te ue of water for proceing and environmental control application. However, in ti unit, we will be concerned excluively wit te propertie of team eiter a pure team ( dry team ) or a mixture of team and liquid water droplet ( wet team ). Figure 9.07 i a temperature entalpy diagram for team. By following eating procee on te cart we can get a better picture of te water-team ytem. If we begin wit liquid water at 25 C and eat it lowly (at atmoperic preure) to 100 C, te entalpy of te water would follow line A B. At 100 C, te water would begin to evaporate (at contant preure) until all te liquid i converted to team (point C). Furter eating will upereat te team (line C D). Similarly, eating water at a iger preure of 600 kpa would produce line A B H I J. Te curved dome containing point A, B, H, X, I and C i commonly called te vapor dome. To te left of te dome, liquid water i preent. To te rigt of te dome i upereated (dry) team. Te curved line forming te left alf of te dome repreent aturated liquid; tat on te rigt repreent aturated vapor. Between tee two line i wet team progreing from all liquid on te left

25 Capter 9 Pycrometric 237 to all vapor on te rigt. Note tat above point X, tere i no ditinct cange from liquid to ga. Point X i te critical point ( kpa, C). Above te critical point, te cange from liquid to ga i gradual and cannot be identified a occurring at a pecific point. Note: Strictly peaking, we do not follow te aturated liquid line during te eating proce decribed above. Te aturation line repreent value of entalpy at aturated temperature and preure. Te preure along tat aturation line and below Point B and H are le tan kpa and 600 kpa, repectively; and we are applying eat at a contant preure of or 600 kpa, repectively. However, te effect of preure upon entalpy i mall. On te grap own, we would be unable to detect any deviation from te aturated liquid line. Beneat te vapor dome, a line of contant temperature i alo a line of contant preure. Tu, if eiter property i pecified, te oter i alo known. In te upereat area, ti i not true. Bot temperature and preure mut be pecified to identify a pecific tate point in te upereated region. Figure 9.07 preent a good viual image of te preure, temperature, entalpy relationip. However, accurate value of tee propertie cannot be readily obtained from ti figure. Tabulation of tee propertie, commonly called team table, are ued to obtain accurate property value. 400 X Critical Point J D 300 STEAM 200 LIQUID H 600 kpa I 100 B kpa C A Heat Content (Entalpy) kj kg Figure Temperature-Entalpy diagram for water. Te curve A B H X I C i called a vapor dome. Te region beneat te vapor dome repreent a condition of wet team.

26 238 Food & Proce Engineering Tecnology 9.7 Steam Table Table 9.02, 9.03, and 9.04 are te tree type of team table normally ued. Te firt two table ow propertie a a function of temperature (Table 9.02) and preure (Table 9.03). Bot table give te ame type of information; tu, te table to be ued depend upon te known property (temperature or preure). Tee table repreent condition witin te vapor dome. Table 9.04 give propertie for upereated team. Te coice of wic table to ue depend upon te condition given. Any given combination of temperature and preure may repreent eiter wet team (Table 9.02 or 9.03) or upereated team (Table 9.04). It may be neceary to look at eac table before you find te correct tate point. However, a notation tat te team i aturated or tatement of a value for team quality (ee below) reveal tat te wet team table (Table 9.02 or 9.03) ould be ued. For all table, te propertie of interet are temperature (t), preure (P), entalpy (), and pecific volume (v). Entropy () i an additional property often ued in team calculation. Te commonly ued ubcript are f for fluid propertie, g for gaeou propertie, and fg for te difference between fluid and gaeou propertie. Ue of te upereated team table i imply a matter of finding te appropriate tate point given any two of te propertie p, t, v,, or. Temperature and preure are mot commonly ued. To ue Table 9.02 and 9.03, it i often neceary to know te quality of te team. Quality i an indication of ow muc of te wet team i vapor. It i defined a te ma of te vapor divided by te total ma (liquid plu vapor). Quality i uually identified by te ymbol x. At x = 0, we ave aturated liquid. At x = 1, we ave aturated vapor. For x between 0 and 1, we ave a mixture of liquid and vapor (wet team). If we know te entalpy or pecific volume for a given tate point under te vapor dome, we can determine te quality, x: v v f f x = or x = (9.26) v fg Converely, te quality (x) can be ued to determine te entalpy or pecific volume of wet team. We all make all calculation baed upon a unit ma of water. Noting tat te entalpy of evaporated water (team) i g and te entalpy of liquid water i f, we find te entalpy a follow: 1. Saturated team (x = 1) (ince we ave only aturated vapor, te entalpy i g ): = x g = 1 g = g 2. Saturated liquid (x = 0): = (1 x) f = (1 0) f = f 3. Mixture (ere we mut take te amount of liquid and vapor into account): = liquid + vapor = x g + (1 x) f = x ( g f ) + f = x fg + f = f + x fg (9.27) Te above example wa for entalpy. For pecific volume, te ame rule applie, e.g.: v = x (v g v f ) + v f = x v fg + v f = v f + x v fg (9.28) fg

27 Capter 9 Pycrometric 239 Example 9.10 Find te entalpy, pecific volume, and preure for team at 175 C and a quality of 0.5. Locate te approximate poition of ti tate point on Figure Solution: We know te preure and, wit a quality of 0.5, we know tat te team i wet. Tu, we cooe Table 9.02, te table for wet team propertie a a function of temperature. From te table, at 175 C, P = kpa ANSWER v f = m 3 /kg v fg = m 3 /kg f = kj/kg fg = kj/kg Solving for v and : v = v f + x v fg = ( ) = m 3 /kg ANSWER = f + x fg = (2030.7) = kj/kg ANSWER At 175 C and an entalpy of kj/kg, te approximate tate point location on Figure 9.07 i near te letter P of 600 kpa along line H I. ANSWER Example 9.11 Find te temperature and entalpy of team at 300 kpa and m 3 /kg. Locate te approximate poition of te tate point on Figure Solution: From Table 9.03 at 300 kpa, we ee tat v g = m 3 /kg. v g i te volume of aturated vapor at te given preure. Ti value of v g = i le tan te pecific volume given (0.6506). Since te pecific volume of te team i greater tan tat of aturated vapor, te team mut be upereated. Uing Table 9.04, at 300 kpa and te pecific volume given, we find: t = 160 C = kj/kg ANSWER For 160 C, 300 kpa, and kj/kg, te tate point location on Figure 9.07 i below, and to te rigt of Point I, between line I J and C D. ANSWER Note: (1) If te value of v ad not been given exactly in Table 9.04, it would ave been neceary for u to interpolate between value in te table. (2) If te value of v ad been le tan v g = , ti would indicate tat we ave wet team and we would ave ued Table 9.03 to find t (at 300 kpa), olve for te quality (x), and compute te entalpy ().

28 240 Food & Proce Engineering Tecnology 9.8 Steam -Air Mixture In ti unit, we ave examined ytem involving team (100% water vapor) and typical air and water vapor mixture (low level of water vapor.) We will now examine team-air mixture involving ig water vapor concentration. Te iue of team air mixture in food proceing i important becaue te temperature of a mixture of air and team at any preure i lower tan te temperature of pure team at te ame preure. Tu, te firt tep in batc proceing wit a retort, ome canner, or preure cooker i to vent until a teady tream of team i produced. Failure to vent and purge te container of air will reult in a mixture of air and water vapor, cauing lower proceing temperature and, poibly, inadequate terilization. Our examination will make te implifying aumption tat ideal ga relationip apply. Wile not necearily valid for all condition, ti aumption i a good approximation; and it will allow u to view te effect of increaing amount of air in a team environment. Te trend we ee will be valid even toug te numeric value we calculate may differ ligtly from te correct value. Equation 9.05, 9.07, and 9.08 relate total and partial preure, repectively, to te number of mole of ga preent, te temperature, and te total volume. Uing ti relationip, we can olve for te partial preure of water vapor in a team-air mixture. We note tat ti mixture will be at a uniform temperature and tat it occupie a fixed volume. Applying te above referenced equation, we find: Pa N a RuT / V N a = = = ya a P NR T / V N u ( y = mole fraction of air) (9.29) or Pa = yap (9.30) Uing te above information we can olve for te partial preure of water vapor (team) at any ma ratio of water vapor. Firt, we note tat te molecular weigt of air and water are and , repectively. We need tee value to calculate te mole of air and water preent in a given mixture. If we aume a mixture of unit ma, at 200 kpa, containing a ma ratio of 0.2 air (20% air, by ma), we can compute P w by te following tep. Te mole of air and water are: ma 0.2 mw 0.8 N a = = = and N w = = = M M a Ti reult in a total of mole ( ) in te ytem. Te mole fraction of air and water in te ytem are ten: N = w N y = = and = a w ya = = Nt Nt Te partial preure of water vapor (team) for ti mixture i ten: P w = yw t P = = kpa w

29 Capter 9 Pycrometric Temp Pre Temperature (C) Steam Preure (kpa) Ma Fraction of Air Figure Effect of air upon team partial preure and temperature in team-air mixture at 200 kpa total preure. From te team table, te boiling point at 200 kpa and kpa are C and C, repectively. Tu, te reult of mixing air and team in a ratio of 2 : 8 by ma reult in almot a 6 C drop in te boiling point temperature at a total preure of 200 kpa. Ti temperature drop could ignificantly affect a termal proceing operation. Wile a team-air mixture containing 20% air i probably extreme, ti doe illutrate te effect of air in te mixture. Figure 9.08 ow te effect of differing air-team mixture (at 200 kpa) upon te team partial preure and boiling point temperature. 0 Example 9.12 We will conider te effect of failure to vent by examining te ue (or miue) of a ome canner. A typical ome canner i approximately 320 mm in diameter and 270 mm deep wit a ligtly domed lid. Figure 9.09 ow a cro-ection repreenting a typical ome canner containing everal can. Te typical operating procedure i to add ot water to a dept of about 25 mm, inert te can to be proceed, intall te lid, eat, vent until continuou pure team i oberved, eal, and proce for te required time. We will aume te following condition for our example:

30 242 Food & Proce Engineering Tecnology STEAM WATER Figure Sketc of ome canner owing approximate proportion of water and team. canner volume (V t ) = 21.7 L = m 3 water volume (V w ) = 2.0 L = m 3 (for a dept of about 25 mm) tree can wit a total volume (V c ) = 2.8 L = m 3 initial temperature of all component (t i ) = 25 C proceing preure (P r ) = 200 kpa ( retort proceing preure) canner immediately ealed witout venting and eated until te preure reace 200 kpa (Except for te initial temperature and te mall number of can, all value above are realitic. In actual condition, te water and can temperature would probably be well above 25 C. However, we will make ti aumption to implify te problem.) If te canner ad been properly vented to inure pure team inide, te inide temperature would be C at 200 kpa. Wat i te temperature witout te venting? Solution: We mut firt determine te ma and mole fraction of water and air in te container. Water: From Table 9.05 at 25 C, we find ρ w = kg/m 3. Tu: kg 3 mw = ρ w Vw = m = kg 3 m

31 Capter 9 Pycrometric 243 Air: From Table 9.01 at 25 C, v a = m 3 /kg, tu ρ a = kg/m 3 (We could alo determine v a, but wit le preciion, from Figure 9.01 at 25 C and W = 0.) Te volume of air in te canner (V a ) i tat pace not occupied by te can or te water: 3 V a = Vt Vw Vc = = 16.9 L = m kg 3 ma = ρ ava = m = kg 3 m We can now determine te number of mole of water and air preent: ma mw N a = = = and N w = = = M a M w and N t = N a + N w = = Mole Te mole fraction of water vapor and te reulting partial preure of water vapor are: N w y w = = = and Pw = ywpw = = kpa N t Interpolating from Table 9.03, te temperature correponding to a preure of kpa i C. Ti i approximately 0.2 below te value for pure team at 200 kpa ( C). A difference of only 0.2 eem mall; owever, at te ig temperature ued for proceing, it can affect te required proceing time. Tu, venting i important to inure tat proceing condition are known.

32 244 Food & Proce Engineering Tecnology Table Propertie of aturated team and aturated water (temperature). Computed uing ASME oftware accompanying ASME Steam Table (ASME, 1993). Temperature Preure Volume (m 3 /kg) Entalpy (kj/kg) Entropy (kj/kg K) ( C) (K) (kpa) Water Evap. Steam Water Evap. Steam Water Evap. Steam t T p vf vfg vg f fg g f fg g