APPLIED THERMODYNAMICS TUTORIAL 6 AIR-VAPOUR MIXTURES

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1 APPLIED THERMODYNAMICS TUTORIAL 6 AIR-APOUR MIXTURES In thi tutoril you will do the following. Revie the UNIERSAL GAS LAW Lern DALTONS LAW OF PARTIAL PRESSURES Aly thee lw to mixture of wter vour nd ir. Solve roblem involving ir conditioning lnt. Solve roblem involving cooling tower. Solve roblem involving tem condener. Let' trt by reviing the Univerl g lw nd the lw of rtil reure.

2 1.1 UNIERSAL GAS LAW mrot = N ~ where Ro i the univerl contnt J/kmol K Ñ i the reltive moleculr m which i 18 for wter vour treted g nd for dry ir treted ingle g. 1.2 PARTIAL PRESSURES The reure exerted by g on the urfce of continment i due to the bombrdment of the urfce by the molecule. The reltive ditnce between molecule i very lrge o if two or more ge exit in the me ce, their behviour i unffected by the other nd o ech g roduce reure on the urfce ccording to the g lw bove. Ech g occuie the totl volume nd h the me temerture T. If two ge A nd B re conidered, the reure due to ech i : m = N ~RoT b = m b N ~RoT b The totl reure on the urfce of continment i = + b Thi i Dlton Lw of rtil reure. Now let ee how thee lw re lied to mixture of vour nd ir. 2

3 2. AIR - APOUR MIXTURES In the following work, wter vour i treted g. Conider mixture of dry ir nd vour. If the temerture of the mixture i cooled until the vour trt to condene, the temerture mut be the turtion temerture (dew oint) nd the rtil reure of the vour mut be the vlue of in the fluid tble t the mixture temerture. If the mixture i wrmed u t contnt reure o tht the temerture rie, the vour mut become uerheted. It cn be hown tht the rtil reure of the vour nd the dry ir remin the me t the turtion temerture. Let condition (1) be t the turtion condition nd condition (2) be t the higher temerture. i contnt o it follow tht : 1 = T1 T 2 2 The initil rtil reure of the vour i: = 1 m 1 N ~RoT 1 The finl reure of the vour i : = 2 m 2 N ~RoT 2 Since 1 = 2 then 1 = 2 T 1 T 2 By the me roce it cn be hown tht 1 = 2 If i contnt then the rtil reure re contnt nd the rtil reure of the vour my eily be found by looking u the turtion reure t the dew oint if it i known. When the ir i contct with wter, it will evorte the wter nd the wter will cool down until it i t the turtion temerture or dew oint. Thi ide i ued in wet bulb thermometer for exmle, which meure the dew oint. When tble condition re reched, the ir become turted nd equl to the temerture of the wter nd o it temerture i the dew oint (t ) in fluid tble. 3

4 WORKED EXAMPLE No.1 Moit ir t 1 br nd 25oC e over wter nd emerge t 1 br nd 18oC. Clculte the rtil reure of the ir nd vour before cooling. SOLUTION When cooled 18oC mut be the turtion temerture o the rtil reure of the vour i in the fluid tble nd i br. The rtil reure of the vour w the me before cooling o the rtil reure of the ir mut be = br. Now let' look t the definition nd ue of humidity. 3. HUMIDITY There re two wy to exre humidity SPECIFIC AND RELATIE SPECIFIC HUMIDITY ω ω = m of wter vour/m of dry ir Strting with the g lw N ~ m = RoT RoTN ~ ω = RoTN ~ ω = N ~ = N ~ = x = Thi derivtion i often requeted in Exmintion. 4

5 3.2. RELATIE HUMIDITY φ φ = m of vour/mximum oible m of vour The mximum oible m of wter vour which cn be held by ir i when the vour i turted nd the temerture of the mixture i the turtion temerture. m = olume/ecific volume = /v When turted, v = vg t the mixture temerture. m m φ = = = g v v v v Alterntively v = /m N v = RoT φ = g nd v g = N g g RoT = rtil reure of the ctul vour g = rtil reure when turted. ω = ( ) ω φ = g nd φ = g 5

6 WORKED EXAMPLE No.2 Moit ir t 1 br nd 25oC i cooled to 18oC by ing it over wter t 18oC. It emerge t 18oC nd 1 br with reltive humidity of 1.0. Auming tht there i no net wter borbed nor lot, clculte the reltive nd ecific humidity before cooling. SOLUTION Thi i the me the reviou roblem o the dew oint mut be 18oC nd the rtil reure of the vour i t 18oC nd i br. = = br g t 25oC = br It follow tht if no net wter i gined nor lot then the ecific humidity mut be the me before nd fter nd i : ω = 0.622(/) = φ = /g =0.651 Thee re the humidity vlue which will reult in no evortion nor condention. If φ<0.651 then there will hve been evortion. If φ>0.651 then condention will hve tken lce on contct with the wter nd cooling lo. 6

7 MASS BALANCE Conider the worked exmle 2 gin only thi time uoe the reltive humidity t inlet i 0.5. Thi men tht wter i evorted. Conider 1 kg of dry ir ing through from inlet to outlet. At outlet φ = 1 nd 2 = g t 18oC = br. 2 = = br ω 2 = 0.622/g= = m/m hence for 1 kg of dry ir there mut be kg of turted vour. At inlet φ 1 = 0.5 Thi time the m of the vour t inlet nd outlet re not the me o the ecific humidity i different t inlet. φ 1 = 0.5 = ω 1 (-)/(0.622g) t 25oC. Remember i the turtion reure t the dew oint (18oC) nd g i the turtion reure t the ctul temerture (25oC). 0.5 = ω 1 ( )/(0.622 x ) = ω 1 Hence ω 1 = 0.01 = m 1 /m Since the ir m i 1 kg throughout, then the m of vour t inlet i 0.01 kg. It follow tht the m of wter evorted i = kg 7

8 SELF ASSESSMENT EXERCISE No.1 1. Reet the worked exmle 2 but thi time the reltive humidity 0.8 t inlet. I wter condened or evorted? (m 1 = kg o wter i condened) 2. Define ecific humidity ω nd rove tht ω = Ñ/Ñ( - ) Humid ir t 1 br flow through n inulted veel over ool of wter nd emerge turted. The temerture re 25oC nd 18oC t inlet nd outlet reectively. The m of wter i mintined contnt t 18oC ll the time. Clculte the reltive humidity t inlet uming contnt reure throughout. (An ) 8

9 4. ENERGY BALANCE Conider imle ir conditioner. Moit ir i drwn in nd cooled o tht wter condene out. The ir t thi oint mut be t the dew oint. Figure 1 Alying the lw of energy conervtion we get : M cθ 1 + m 2 h 1 = m w c w θ 2 + m c θ 2 + m 2 h 2 + energy removed The uffixe, nd w refer to ir, vour nd wter reectively. Treting vour g, the cceted vlue of the ecific het ccity i kj/kg K. The enthly of tem reltive to 0oC i then h = hg θ - t) t i the turtion temerture of the vour. It i robbly bet to ue the thermodynmic tble or the h - chrt ulied in the exm whenever oible to find the enthly of vour t low reure nd temerture. WORKED EXAMPLE No.3 Moit ir enter conditioning unit t 25oC nd 1 br with reltive humidity of 0.7. It i ed through cooler cuing the temerture to fll to 18oC nd condente i formed. Clculte the m of condente formed er kg of dry ir nd the energy removed er kg. SOLUTION Following the me method in the reviou exmle, the m of vour t exit i m 2 = kg er kg of dry ir. The vour reure t inlet i 1.0 br. At inlet the reltive humidity i 0.7 = φ 1 = ω 1 ( - )/(0.622g) t 25oC. ω 1 =0.622 x x 0.7/( ) = The m of vour t inlet i then m 1 = kg/kg 9

10 The condente formed mw = = kg Conducting n energy blnce: 1 x x h 1 = 1x x h 2 + mwθw + Φ h 1 = hg (θ - t) t br. h 1 = (25-18)(1.864) = kJ/kg h 2 = hg t 18oC ince it i turted. h 2 = kj/kg The blnce become 1 x x (2546.9) = 1x x (2533.9) x4.168x18 + Φ Φ = 9.87 kj er kg of dry ir. In ir conditioning it i norml to het the ir to the required temerture before it leve the unit. Figure 2 Suoe the ir i heted to 22oC before leving. Wht i the het trnfer required in the heter? The me re unchnged o we only need n energy blnce between 2 nd 3. 1x x h 2 + Φ = 1 x x h 3 h 3 = hg (θ - t) t br. t = 18oC h 3 = ) = kj/kg h 2 = kj/kg 1 x x x Φ = 1 x x x Φ = Φ = 4.11 kj 10

11 SELF ASSESSMENT EXERCISE No. 2 Air hving reure, temerture nd reltive humidity of 1 br, 26oC nd 0.65 reectively, flow into n ir conditioner t tedy rte nd i dehumidified by cooling nd removing wter from it. The ir i then heted to roduce n outlet temerture nd reltive humidity of 24oC nd reectively. The reure i contnt throughout. Determine the het trnfer in the cooler nd heter er kg of conditioned ir t exit. Drw u comlete m blnce. (36.22 kj/kg nd kj/kg) 11

12 5. COOLING TOWERS Cooling tower fll into two tye, dry nd wet. Dry cooling tower re no more thn very lrge ir conditioner nd the theory i the me lredy outlined. Wet cooling tower work on the rincile of rying wrm wter downwrd o tht het nd vour i ed to the ir which rie nd crrie wy ltent het leving the wter t lower temerture to collect in ool t the bottom of the tower. The wter i then recycled from the ool. The moit ir leve the to of the tower lume. The tower h venturi he to it the roce by cuing light reure reduction in the ry re followed by reurgence the to widen. Thi cue condention to form nd mke the lume viible. Some of the condente rin down into the ool. We cn y with certinty tht the ir leve the tower with 100% humidity. The bet wy to undertnd the roblem i to do worked exmle follow. WORKED EXAMPLE No.4 A cooling tower mut cool 340 kg of wter er minute. The wter i ulied t 42oC nd it i ryed down into the column of ir which enter the bottom of the tower t rte of 540m3/min with temerture of 18oC nd reltive humidity of 60%. The moit ir leve the to of the tower turted t 27oC. The whole roce occur t contnt reure of br. Determine the temerture of the cooled wter in the ool nd the rte t which mke u wter mut be ulied to relce tht evorted. Figure 3 12

13 SOLUTION R = 287 J/kg K for ir nd 462 J/kg K for vour. INLET AIR g1 = br t 18oC φ 1 = 0.6 = 1 / g1 1 = br hence 1 = = 1 br m = /RT = 1 x 105 x 540/(287 x 291) = kg/min m 1 = /RT = x 105 x 540/(462 x 291 ) = kg/min OUTLET AIR φ2 = 1 2 = g2 = br hence 2 = br φ2 = x / = m 2 = x = kg/min Wter evorted = = kg/min Mke u wter = kg/min m w2 = = ENERGY BALANCE In thi exmle enthly vlue from the tem tble nd chrt will be ued. h w1 = 42oC = kj/kg h w2 i unknown h 1 = x 27 kj/kg h 1 = br & 18oC = 2530 kj/kg (from h- chrt) h 2 = 27oC = kj/kg Blncing energy we get (340 x175.8) + {646.6 x x (18-27)} + {4.971 x 2530) (14.66 x ) = h w2 h w2 = kj/kg nd from the tble the temerture mut be 29.5 oc. The temerture of the cooled wter i 29.5 oc. 13

14 SELF ASSESSMENT EXERCISE No Derive the exreion for ecific humidity ω= 0.622(/) Wter flow t kg/h nd 40oC into cooling tower nd i cooled to 26oC. The unturted ir enter the tower t 20oC with reltive humidity of 0.4. It leve turted ir t 30oC. The reure i contnt t 1 br throughout. Clculte i. the m flow of ir er hour. (4 636 kg/h) ii. the m of wter evorted er hour. (100.5 kg/h) 2. The cooling wter for mll condener i ent to mll cooling tower. 7 m3/ of ir enter the tower with reure, temerture nd reltive humidity of br, 15oC nd 0.55 reectively. It leve turted t 32oC. The wter flow out of the tower t 7.5 kg/ t 13oC. Uing m nd energy blnce, determine the temerture of the wter entering the tower. (Anwer 33.9oC) 3. A fn ulie 600 dm3/ of ir with reltive humidity of 0.85, temerture 30oC nd reure 1.04 br into n ir conditioner. Moiture i removed from the ir by cooling nd both the ir nd condente leve t the me temerture. The ir i then heted to 20oC nd h reltive humidity of 0.6. Determine the following. i. The m of dry ir nd wter t entrnce to the conditioner. ( kg/ nd kg/) ii. The m of wter vour delivered t exit. ( kg/) iii. The m of wter extrcted from the cooler. ( kg/) iv. The temerture t exit from the cooler. (12oC) v. The het trnfer in the cooler. 14

15 6. CONDENSERS It i inevitble tht ir will be drwn into tem condener oerting with vcuum. The effect of thi i to reduce the turtion reure nd temerture of the tem reulting in colder condente tht would otherwie be obtined. Thi in turn men more het required to turn it bck into tem in the boiler nd reduced therml efficiency for the ower lnt. The ir mut be removed from the condener in order to kee the rtil reure mll oible. Thi i done with n extrctor um. Some vour will be removed with the ir but thi lo i tolerble becue of the energy ved. Figure 4 The olution to roblem on condener i imilr to tht for cooling tower nd require m nd energy blnce. It i norml to neglect the rtil reure of the ir t inlet it mke little difference to the nwer. SELF ASSESSMENT EXERCISE No. 4. Dicu the reon why ir mixed with tem in condener i not deirble. b. Wet tem with dryne frction of 0.9 enter condener t br reure t rte of kg/h. The condente leve t 25oC. Air lo enter with the tem t rte of 40 kg/h. The ir i extrcted nd cooled to 20oC. The rtil reure of the ir t inlet i negligible nd the roce i t contnt reure. The cooling wter i t 10oC t inlet nd 21oC t outlet. i. Determine the m of vour extrcted with the ir. (50 kg/h) ii. Clculte the flow rte of the cooling wter required. ( kg/h) 15

Mon 2:04:28 PM. Introduction

Mon 2:04:28 PM. Introduction Mon 2:04:28 PM Slide Nr. 0 of 18 Slide Introduction Air in the tmohere normlly contin ome wter vor (or moiture) nd i referred to tmoheric ir. By contrt, ir tht contin no wter vor i clled dry ir. It i often