Psychrometric Applictions The reminder of this presenttion centers on systems involving moist ir. A condensed wter phse my lso be present in such systems. The term moist irrefers to mixture of dry ir nd wter vpor in which the dry ir is treted s pure component. The Dlton model pplies to moist ir. By identifying gs 1with dry ir nd gs 2 with wter vpor, Tble 12.2gives moist ir property reltions on mss bsis. The study of systems involving moist ir is known s psychrometrics.
Moist Air Consider closed system consisting of moist ir occupying volume Vt mixture pressure p nd mixture temperture T. In moist ir the mount of wter vpor present is much lessthn the mount of dry ir: m v << m n v << n. The Dlton model pplies to the mixture of dry ir nd wter vpor:
Moist Air 1.The overll mixture nd ech component, dry ir nd wter vpor, obey the idel gs eqution of stte. 2. Dry ir nd wter vpor within the mixture re considered s if they ech exist lone in volume Vt the mixture temperture Twhile ech exerts prt of the mixture pressure..the prtil pressuresp nd p v of dry ir nd wter vpor re, respectively p = y p p v = y v p (Eq. 12.41b) wherey nd y v re the mole frctions of the dry ir nd wter vpor, respectively. These moist ir expressions conform toeqs. (c)of Tble 12.2.
Moist Air 4.The mixture pressure is the sum of the prtil pressures of the dry ir nd the wter vpor: p= p + p v 5.A typicl stte of wter vpor in moist ir is fixed using prtil pressure p v nd the mixture temperture T. The wter vpor is superhetedt this stte. Mixture pressure, p T Typicl stte of the wter vpor in moist ir,
6.When p v corresponds to p g t temperture T, the mixture is sid to be sturted. 7.The rtio of p v nd p g is clled the reltive humidity, φ: Moist Air p v φ = p (Eq. 12.44) g T, p Reltive humidity is usully expressed s percent nd rnges s dry ir only 0 φ 100% (p v = 0) sturted ir (p v = p g )
Humidity Rtio The humidity rtioωof moist ir smple is the rtio of the mss of the wter vpor to the mss of the dry ir. v m m = ω (Eq. 12.42) v v 0.622 p p p = ω (Eq. 12.4) Since m v << m, the vlue of ωis typiclly << 1. Using the idel gs eqution of stte nd the reltionship p = p p v v m m ω= RT V p M RT V p M / / v v = v v p M p M = = v v v p p p M M 18.02/28.97 = 0.622
Mixture Enthlpy Vlues for U, H, nd Sfor moist ir cn be found by dding contributions of ech component. For exmple, the enthlpy H is H = H + (Eq. 12.45) + H v = mh mvhv which conforms to Eq. (d)in Tble 12.2. Dividing by m nd introducing ω, the mixture enthlpy per unit mss of dry iris H mv = h + hv = h +ωhv m m (Eq. 12.46) For moist ir, the enthlpy h v is very closely given by the sturted vpor vlue corresponding to the given temperture. hv hg ( T) (Eq. 12.47)
Heting Moist Air in Duct Exmple: Moist ir enters duct t 10 o C, 80% reltive humidity, is heted s it flows through the duct, nd exits t 0 o C. No moisture is dded or removed nd the mixture pressure remins constnt t 1 br. For stedy-stte opertion nd ignoring kinetic nd potentil energy chnges, determine () the humidity rtio, ω 2, nd (b) the rte of het trnsfer, in kj per kg of dry ir.
Solution: Heting Moist Air in Duct ()At stedy stte, mss rte blnces for the dry ir nd wter vpor red: 1 v1 = = 2 v2 (dry ir) (wter vpor) Since the mss flow rtes of the dry ir nd wter vpor do not chnge from inlet to exit, they re denoted for simplicity s m nd m v. Moreover, since no moisture is dded or removed, the humidity rtio does not chnge from inlet to exit: ω 1 = ω 2. The common humidity rtio is denoted by ω. ω = v
Heting Moist Air in Duct The humidity rtiois evluted using dt t the inlet: The prtil pressure of the wter vpor t the inlet, p v1, cn be evluted from the given inlet reltive humidity φ 1 nd the sturted pressure p g1 t 10 o C from Tble A-2: p v1 = φ 1 p g1 = 0.8(0.01228 br) = 0.0098 br The humidity rtio cn be found from: ω = 0.622 p p v p v = 0.0098 0.622 1 0.0098 kg (vpor) = 0.00616 kg (dry ir)
Heting Moist Air in Duct (b) The stedy-stte form of the energy rte blnce reduces to: 0 0= Q & cv W & cv+ ( h1+ vhv1) ( h2 + vhv2) Solving for Q cv Q& cv = ( h2 h1) + v ( hv2 hv1) Noting tht m v = ωm, we get Q& cv = ( h2 h1) + ω( hv2 hv1)
Heting Moist Air in Duct Q& cv = ( h2 h1) + ω( hv2 hv1) For the dry ir, h 1 nd h 2 re obtined from idel gs tble Tble A-22 t 10 o C nd 0 o C, respectively. For the wter vpor, h v1 nd h v2 re obtined from stem tble Tble A-2 t 10 o Cnd 0 o C, respectively, using h v h g Q& cv Q& cv kj = (0.2 28.1) kg (dry ir) 0.00616 kg (vpor) kg (dry ir) = (20.1+ 0.22) + (2556. 2519.8) kj kg (dry ir) = 20.2 kj kg (vpor) kj kg (dry ir) The contribution of the wter vpor to the het trnsfer mgnitude is reltively minor.
Dew Point Temperture When moist ir is cooled, prtil condenstionof the wter vpor initilly present cn occur. This is observed in condenstion of vpor on window pnes, pipes crrying cold wter, nd formtion of dew on grss. An importnt specil cseis cooling of moist ir t constnt mixture pressure, p. The figure shows smple of moist ir, initilly t Stte1, where the wter vpor is superheted. The ccompnying T-v digrm loctes sttes of wter. Let s study this system s it is cooled in stges from its initil temperture.
Dew Point Temperture In the first prt of the cooling process, the mixture pressure nd wter vpor mole frction remin constnt. Since p v = y v p, the prtil pressure of the wter vpor remins constnt. Accordingly, the wter vpor cools t constnt p v from stte 1 to stte d, clled the dew point. The temperture t stte d is clled the dew point temperture. As the system cools below the dew point temperture, some of the wter vporinitilly present condenses. The rest remins vpor.
Dew Point Temperture At the finl temperture, the system consists of the dry ir initilly present plus sturted wter vpor nd sturted liquid. Since some of the wter vpor initilly present hs condensed, the prtil pressure of the wter vpor t the finl stte, p g2, is less thn the prtil pressure initilly, p v1. The mount of wter tht condenses, m w, equls the difference in the initil nd finl mounts of wter vpor: m w = m v1 m v2
Dew Point Temperture Using m v = ωm nd the fct tht the mount of dry ir remins constnt, the mount of wter condensed per unit mss of dry ir is where m m w =ω ω 1 2 ω1 = 0. 622 p p v1 p v1 ω 2 = 0. 622 p p g2 p g2 ndpdenotes the mixture pressure, which remins constnt while cooling occurs.
Dry-bulb Temperture nd Wet-bulb Temperture In engineering pplictions involving moist ir, two redily-mesured tempertures re commonly used: the dry-bulb nd wet-bulb tempertures. The dry-bulb temperture, T db, is simply the temperture mesured by n ordinry thermometer plced in contct with the moist ir. The wet-bulb temperture, T wb, is the temperture mesured by thermometer whose bulb is enclosed by wick moistened with wter.
Dry-bulb Temperture nd Wet-bulb Temperture The figure shows wet-bulb nd dry-bulb thermometers mounted on n instrument clled psychrometer. Flow of moist ir over the two thermometers is induced by bttery-operted fn. Owing to evportion from the wet wick to the moist ir, the wet-bulb temperture reding is less thn the dry-bulb temperture: T wb < T db. Ech temperture is esily red from its respective thermometer. Moist Air in
Psychrometric Chrt Grphicl representtions of moist-ir dt re provided by psychrometric chrts. Psychrometric chrts re given in Fig. A-9. These chrts re constructed for moist ir mixture pressure of 1 tm. Severl importnt fetures of the psychrometric chrt re discussed in Sec. 12.7, including
Psychrometric Chrt
Dry-bulb temperture, T db. Psychrometric Chrt Moist ir stte T db
Humidity rtio, ω. Psychrometric Chrt Moist ir stte ω
Psychrometric Chrt Dew point temperture, T dp. Since the dew point is the stte where moist ir becomes sturted when cooled t constnt pressure, the dew point for given stte is determined from the chrt by following line of constnt ω(constnt p v ) to the sturtion line where φ = 100%. Moist ir stte T dp
Reltive humidity, φ. Psychrometric Chrt Moist ir stte
Psychrometric Chrt Mixture enthlpy per unit mss of dry ir, (h + ωh v ). The vlue of (h + ωh v ) is clculted using h = c p T Fig. 12.9: Tin o C,c p = 1.005 kj/kg-k (h + ωh v ) Moist ir stte
Psychrometric Chrt Wet-bulb temperture, T wb. Lines of constnt wet-bulb temperture re pproximtely lines of constnt mixture enthlpy. T wb Moist ir stte
Psychrometric Chrt Volume per unit mss of dry ir, V/m. Lines giving V/m cn be interpreted s the volume of dry ir or of wter vpor (ech per unit mss of dry ir) becuse in keeping with the Dlton model ech component is considered to fill the entire volume. Moist ir stte V/m
Psychrometric Chrt Exmple: Using Fig. A-9, determine reltive humidity, humidity rtio, nd mixture enthlpy, in kj/kg (dry ir) corresponding to dry-bulb nd wet-bulb tempertures of 0 o C nd 25 o C, respectively.
Solution: Psychrometric Chrt (h + ωh v ) = 76 kj/kg dry ir 25 o C φ= 67% ω= 0.0181 kg wter/kg dry ir
Anlyzing Air-Conditioning Systems The next series of slides demonstrtes the ppliction of mss nd energy rte blnces together with property dt to typicl ir-conditioning systemsusing the psychrometric principles introduced thus fr. Fetured pplictions include Dehumidifiction Humidifiction Mixing of two moist ir strems An ppliction of psychrometric principles to cooling tower is lso considered.
Dehumidifiction The im of dehumidifier is to remove some of the wter vpor in the moist ir pssing through the unit. This is chieved by llowing the moist irto flow cross cooling coilcrrying refrigernt t temperture low enough tht some wter vpor condenses.
Dehumidifiction The figure shows control volume enclosing dehumidifier operting t stedy stte. Moist ir enters t stte 1. As the moist ir flows over the cooling coil, some wter vpor condenses. Sturted moist ir exits t stte 2(T 2 < T 1 ). Condenste exits s sturted liquid t stte. Here, we tke T = T 2. m, T 1, ω 1 1 m w T = T 2 2 φ 2 = 100%, T 2 < T 1, ω 2 < ω 1
Dehumidifiction For the control volume, let us evlute The mount of condenste exiting per unit mss of dry ir: m w /m nd The rte of het trnsfer between the moist ir nd cooling coil, per unit mss of dry ir: Q cv /m. 1 m, T 1, ω 1 φ 2 = 100%, T 2 < T 1, ω 2 < ω 1 m w 2 T = T 2
Dehumidifiction Mss rte blnces. At stedy stte, mss rte blnces for the dry ir nd wter re, respectively 1 v1 & = m2 (dry ir) = w + v2 (wter) 1 2 Solving for the mss flow rte of the condenste m & = w Then, with m v1 = ω 1 m nd m v2 = ω 2 m, where m denotes the common mss flow rte of the dry ir, we get the following expression for the mount of wter condensed per unit mss of dry ir w =ω ω (1) v1 v2 1 2 m, T 1, ω 1 φ 2 = 100%, T 2 < T 1, ω 2 < ω 1 m w T = T 2
Dehumidifiction Energy rte blnce. With W cv = 0nd no significnt kinetic nd potentil energy chnges, the energy rte blnce for the control volume reduces t stedy stte to 0= Q & cv+ ( h1+ v1hv1) whw ( h2 + v2hv2) (2) With m v1 = ω 1 m, m v2 = ω 2 m, nd Eq. (1), Eq.(2)becomes Q& cv = ( h + ωh ) ( h + ωh ) + ( ω ω ) h v 2 v 1 1 2 w () Since het trnsfer occurs from the moist ir to the cooling coil, Q cv /m will be negtive in vlue.
Q& cv Dehumidifiction = ( h + ωh ) ( h + ωh ) + ( ω ω ) h v 2 For the condenste, h w = h f (T 2 ), where h f is obtined from Tble A-2. Options for evluting the underlined terms of Eq. () include ω 1 nd ω 2 re known. Since T 1 nd T 2 re lso known, h 1 nd h 2 cn be obtined from idel gs tble Tble A-22, while h v1 nd h v2 cn be (h + ωh v ) 1 obtined from stem tble Tble A-2 using h v = h g. (h + ωh v ) 2 Alterntively, using the respective temperture nd humidity rtio vlues to fix the sttes, (h + ωh v ) t sttes 1 nd 2 cn be red from psychrometric chrt. T 2 T 1 v 1 1 2 w ()
Humidifiction The im of humidifieris to increse the mount of wter vpor in the moist ir pssing through the unit. This is chieved by injecting stem or liquid wter.
Humidifiction The figure shows control volume enclosing humidifier operting t stedy stte. Moist ir enters t stte 1. Stem or liquid wter is injected. Moist ir exits t stte 2with greter humidity rtio, ω 2 > ω 1. W cv = 0, Q cv = 0 m 1 h, m
Humidifiction For dibticopertion, the ccompnying psychrometric chrts show sttes 1nd 2for ech cse. With reltively high-temperture stem injection, the temperture of the moist ir increses. With liquid injectionthe temperture of the moist ir my decresebecuse the liquid is vporized by the moist ir into which it is injected. W cv = 0, Q cv = 0 m 1 h, m
Humidifiction For the control volume, let us evlute The humidity rtio, ω 2, nd The temperture, T 2.
Humidifiction Mss rte blnces. At stedy stte, mss rte blnces for the dry ir nd wter re, respectively 1 v1 & = m2 (dry ir) + = v2 (wter) Then, since m v1 = ω 1 m nd m v2 = ω 2 m, where m denotes the common mss flow rte of the dry ir, we get ω 2 =ω1+ (1) W cv = 0, Q cv = 0 Since ω 1, m, nd m re specified, the humidity rtio ω 2 cn be clculted from Eq. (1) m 1 h, m
Humidifiction Energy rte blnce. With no significnt kinetic nd potentil energy chnges, the energy rte blnce for the control volume reduces to 0= Q & cv W & cv+ ( h1+ v1hv1) + h ( h2 + v2hv2) Since W cv nd Q cv re ech zero in this cse 0= ( m & h1+ v1hv1) + h ( h2+ v2hv2) With m v1 = ω 1 m nd m v2 = ω 2 m, Eq.(2) becomes 0= ( h1+ ω 1hv1) + ( ) h ( h2+ ω2hv2) Solving Eq. () ( h ω h (4) 2 + 2hv2) = ( h1+ ω1hv1) + ( ) (2) ()
Humidifiction ( h ω h (4) 2 + 2hv2) = ( h1+ ω1hv1) + ( ) Options for determining T 2 from Eq. (4)include Use the psychrometric chrt: The first term on the right side of Eq. (4)cn be red from the chrt using T 1 nd ω 1 to fix the stte. Since the second term on the right is known, the vlue of (h 2 + ω 2 h v2 ) cn be clculted. This vlue together with ω 2 fixes the exit stte, which llows T 2 to be determined by inspection. (h 1 + ω 1 h v1 ) (h 2 + ω 2 h v2 ) 1 T 1 T 2 2 ω 2 ω ω 1
Humidifiction ( h ω h (4) 2 + 2hv2) = ( h1+ ω1hv1) + ( ) Options for determining T 2 from Eq. (4)include An itertive solution using dt from Tble A-22:h (T)for the dry ir nd Tble A-2:h v = h g (T) for the wter vpor: The vlue of the right side of Eq. (4)is known becuse the dt re either known or cn be obtined from the indicted tbles using T 1. On the left side of Eq. (4), ω 2 is known from the mss rte blnce. Accordingly, the only unknown is T 2, which cn be found itertively: For ech ssumed vlue of T 2, Tble A-22 gives h 2 nd Tble A-2 gives h v2. This llows the left side to be clculted. Itertion with T 2 continues until the clculted vlue on the left grees with the known vlue on the right.
Adibtic Mixing of Two Moist Air Strems In ir-conditioning systems, frequent component is one tht mixes moist ir strems s shown in the figure: For the cse of dibtic mixing, let us consider how the following quntities t the exit of the control volume, m, ω, nd T, cn be evluted knowing the respective quntities t the inlets.
Adibtic Mixing of Two Moist Air Strems Mss rte blnces. At stedy stte, mss rte blnces for the dry ir nd wter vpor re, respectively 1 v1 & + m2 = m (dry ir) + v2 & = v (wter vpor) With m v = ωm, these equtions combine to give ω & & & + 1m 1+ ω2m2 = ω( m1 m2) & Alterntively 1 2 = ω ω ω ω 1 2 (1) These equtions cn be solved for ω using known vlues of ω 1, ω 2, m 1, nd m 2.
Adibtic Mixing of Two Moist Air Strems Energy rte blnce. Ignoring the effects of kinetic nd potentil energy, the energy rte blnce for the control volume reduces t stedy stte to = Q & cv W & cv+ ( 1h1+ v1hv1) + ( 2h2+ v2hv2) ( h + Since W cv nd Q cv re ech zero in this cse 0 v v 1 ( h1+ ω 1hv1) + 2 ( h2+ ω2hv2) = ( h+ ωhv) h (Eq. 12.56c) The enthlpies of the wter vpor re evluted using h v = h g. With m = m 1 + m 2, Eq. 12.56ccn be solved to give n expression with the sme form s Eq. (1) 1 2 = ( h ( h 1 + ω h + ωh 1 g g1 ) ( h ) ( h 2 + ω h 2 + ω h Using known dt, this eqution cn be solved for (h + ωh g ), from which T cn be evluted. g2 g ) ) (2) )
Adibtic Mixing of Two Moist Air Strems From study of Eqs. (1)nd (2)we conclude tht on psychrometric chrt stte lies on stright line connecting sttes 1nd 2, s shown in the figure 1 2 = ω ω ω ω 1 2 (1) 1 2 = ( h ( h 1 + ω h + ωh 1 g g1 ) ( h ) ( h 2 + ω h 2 + ω h g2 g ) ) (2)
Adibtic Mixing of Two Moist Air Strems Exmple: For dibtic mixingof two moist ir strems with the dt provided in the tble below, use the psychrometric chrt to determine ()ω, in kg (vpor)/kg (dry ir), nd (b)t in o C. Stte 1 2 T ( o C) 24 5 ω (kg (dry ir)/kg (vpor)) 0.0094 0.002 m (kg(dry ir)/min) 497 180 (h + ωh g ) * (kj/kg (dry ir)) 48 10 * The vlues of (h + ωh g ) re red from Fig. A-9 using the respective temperture nd humidity rtio vlues.
Adibtic Mixing of Two Moist Air Strems Solution: () Inserting known vlues in Eq. (1), 497 180 ω 0.002 0.0094 ω we get ω = 0.0074 kg (vpor)/kg (dry ir). (b)then from Fig. A-9 = T = 19 o C
Adibtic Mixing of Two Moist Air Strems Solution: () Inserting known vlues in Eq. (1), 497 180 ω 0.002 0.0094 ω we get ω = 0.0074 kg (vpor)/kg (dry ir). (b)then from Fig. A-9 Alterntively, Eq. (2) cn be used to determine (h + ωh g ) = 8 kj/kg (dry ir). Then, from Fig. A-9 = T = 19 o C