Deprtment of Mechnicl Engineering ME 3 Mechnicl Engineering hermodynmics Lecture 33 sychrometric roperties of Moist Air
Air-Wter Vpor Mixtures Atmospheric ir A binry mixture of dry ir () + ter vpor () he ir in the mixture is treted s pure substnce even though it is relly mixture itself Applictions Heting, ventilting, nd ir-conditioning (HVAC) Anlysis HVAC pressures re lys lo ~ tm Idel gs l cn be used for both ir nd ter vpor
roperties of Moist Air Both ir nd ter vpor re treted s idel gses tht obey Dlton s L of rtil ressures. Universl Gs Constnt: Btu R 1.986 lbmol-r M R Dry Air Wter Vpor 8.97 lbm/lbmol M 18.016 lbm/lbmol 0.06855 Btu/lbm-R R 0.110 Btu/lbm-R Since moist ir is binry mixture, y y 1 3
roperties of Moist Air he field of psychrometrics (ir-ter vpor properties) hs dopted other properties to represent the composition of the mixture rther thn the mole frction. Mole Frction Rtio y y Humidity Rtio m m hese properties re relted, n / n m M 8.97 m 1.608 n / n M m 18.016 m 4
roperties of Moist Air Reltive Humidity st rtil pressure of the ter vpor in sturted mixture rtil pressure of the ter vpor in the mixture Stte of the ter vpor in the mixture dp -s digrm of ter s rtil pressure of the dry ir otl pressure of the mixture De oint emperture y y, st st st 5
roperties of Moist Air All of these properties re relted. For exmple, m V / R m V / R R R / M R R / M M 18.016 lbm/lbmol M 8.97 lbm/lbmol 0.6 0.6 st st 0.6 0.6 6
Exmple Given: Moist ir t the folloing stte 70F 14 psi 60% 0.60 Find: Vrious psychrometric properties of the moist ir Solution: rtil pressure of the vpor 0.60 he prtil pressure of the ter in sturted mixture cn be found from ble C.1, st st 0.600.363 psi 0.179 psi st 7
Exmple De oint emperture he de point temperture is the sturtion temperture of the ter vpor t its prtil pressure. Using ble C.1, st 0.363 psi 0.179 psi 70F 55.5F dp 0.179 psi Interpolting... 55.5F dp If the mixture drops belo this temperture, the ter vpor ill strt condensing. s 8
Exmple Humidity Rtio m 0.179 psi 0.6 0.6 m 14 0.179 psi lbm 0.009834 0.009834 lbm Grins A ne unit! A grin is n ncient Egyptin mesure of the mss of one grin of brley (7000 grins/lbm). Since the humidity rtios re typiclly very smll, the HVAC industry hs dopted the use of grins to represent humidity rtio... lbm 7000 grins grins 0.009834 68.8 lbm lbm lbm 9
Exmple Mole Frction Rtio y 1.608 1.608 0.009834 0.015813 y Mole Frction of ech Component y y 0.015813 y 0.016 y 1 y 1 1 0.015813 y 1 y 1 1 y 0.984 y y 1 10.015813 Notice tht: y y 1 v 10
Intensive Moist Air roperties Consider the enthlpy of the mixture H H H H m h m h Question: Ho cn the specific enthlpy of the ir-ter vpor mixture be specified? Anser: he totl enthlpy must be divided by mss vlue. Which mss vlue should be used? 11
Intensive Moist Air roperties In ir conditioning pplictions, the ter vpor mss cn vry due to condenstion or evportion (dehumidifiction or humidifiction). hus, specific properties of the mixture re bsed on the dry ir, H m h m h H h h h m m h h h m Units: Btu/lbm or kj/kg 1
Intensive Moist Air roperties Using idel gs mixing for the components of moist ir, the internl energy, enthlpy, het cpcities, nd entropy of the mixture cn be clculted by, u u u,, h h h c c c p p p c c c v v v s s s 13
Deprtment of Mechnicl Engineering ME 3 Mechnicl Engineering hermodynmics Exmple Heting of Moist Air Strem 14
Exmple Given: Moist ir floing t 300 cfm enters heting unit t 65 F, 14 psi ith reltive humidity of 50%. he moist ir leves the heting unit t 110 F, 14 psi. Find: () he het trnsfer rte required (Btu/hr) (b) he reltive humidity of the ir leving the heter Q 1 14 psi 1 65F 1 0.50 V 300 cfm 14 psi 110F 15
Exmple 1 14 psi 1 65F 1 0.50 V 300 cfm Q 14 psi 110F he First L pplied to the heting system is, 1 1 1 Q m h m h m h m h he mss flo rte of the dry ir does not chnge (in this cse the ter vpor mss flo does not chnge either hy?). herefore, m m m 1 16
Exmple Rerrnging the First L, 1 1 1 Q m h m h m h m h 1 14 psi 1 65F 1 0.50 V 300 cfm Q m m 1 h h h 1 h 1 m m m h h h1 1h 1 Q 14 psi 110F he mss flo rte of the dry ir is, m V v 1 1 17
Exmple he specific volume of the dry ir t stte 1 is found using the idel gs EOS ith the prtil pressure of the dry ir, 1 14 psi 1 65F 1 0.50 V 300 cfm v 1 R 1 1 Q 14 psi 110F he prtil pressure of the dry ir t stte 1 is found knoing the reltive humidity, 1 1 st1 st1 st 1 1 1 1 18
Exmple he component enthlpy vlues cn be found using the idel gs model for ech component, 1 14 psi 1 65F 1 0.50 V 300 cfm Q 14 psi 110F he humidity rtio t stte 1 cn be found, 1 1 0.6 1 19
Exmple No ter vpor is dded to or tken from the moist ir from stte 1 to. herefore, 1 14 psi 1 65F 1 0.50 V 300 cfm Q 14 psi 110F 1 At this point, the problem cn be solved for the het trnsfer rte. We re lso interested in the reltive humidity t the exit of the heter. his cn be found from the humidity rtio t, st 0.6 st st 0
Exmple Solution (Key Vribles): 1 14 psi 1 65F 1 0.50 V 300 cfm Q 14 psi 110F Even though the humidity rtio stys constnt in this process, the moist ir leving the heter ill feel uncomfortbly dry. his is common problem encountered in heting processes. he dryness cn be llevited by injecting ter vpor into the moist ir strem leving the heter (humidifiction). 1
Exmple Wht ould hppen if the moisture content is neglected nd the mixture is treted s dry ir? 1 14 psi 1 65F 1 0.50 V 300 cfm Q 14 psi 110F Since no ter vpor is dded or removed from the moist ir in this process, neglecting the moisture results in smll error. Hoever, neglecting the moisture does not revel the reltive humidity t the exit!