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1 9. Drying 이광이광남남 9. Drying 1

2 정의 The removal of water from process materials and other substances The removal of other organic liquids from solids 증발과건조의차이 amount means evaporation relatively large boiling drying relatively small air flow 9. Drying 2

3 건조방법 열을가하고수증기로제거하는데사용되는물리적 조건에따른구분 상압에서가열된공기와의접촉에의한열의직접접촉. 생성된수증기는공기에의해제거 진공건조물의증발은저압에서보다빠르게일어나며열은 material wall이나복사와같은간접적인방법으로공급 Freeze drying 냉동된물질에서물은승화 9. Drying 3

4 건조장치 1. 회분식단건조기, 진공선반간접건조기 2. 연속식연속식터널건조기, 회전건조기, 원통건조기, 곡물의건조 9. Drying 4

5 9. Drying 5

6 1. Humidity ; H ( 습도 ) Humidity and Humid Chart kg H O p kgmol H O 18.02kg H O kgmol dry air = 2 A 2 2 H kg dry air P p A kgmol dry air kgmol H 2 O kg dry air pa H = (9.3 1) P p A 2. Saturation humidity : H s ( 포화습도 ) H pas s = (9.3 2) P p As 3. Percentage humidity : H p ( 비교습도 ) H p H = 100 (9.3 3) H s 9. Drying 6

7 4. Percentage relative humidity : H R ( 비교상대습도 ) H R pa = 100 (9.3 4) p As H H R P H P H pa pas = 100 = 100 / H P p P p S A As pa P pas = 100 (9.3 5) p P p As A 9. Drying 7

8 9. Drying 8

9 solution o 26.7 C물의증기압은 p = 3.5kPa As pa H = = = [ kg H2O / kg dry air] P p A pas HS = = = [ kg H2O / kg dry air] P p H H P R As H = 100 = 100 = 78.3% H S pa 2.76 = 100 = 100 = 78.9% p 3.5 As 9. Drying 9

10 5. Dew point of an air-water vapor mixture 공기와수증기의혼합물이포화가되는온도 6. Humid heat of an air-water mixture 1kg 의 dry air 와이에포함된수증기의온도를 1 K (or 1 ) 올리는데필요한열량 c S [kj/kg dry air]= H kj/kg water vapor K kj/kg dry air K 9. Drying 10

11 7. Humid volume of an air-water vapor mixture (v H ) 절대압 kPa 과주어진 gas temp. 에서 dry air 1kg 과 dry air 에포함된 water vapor 의 total volume H vh[ m / kg dry air] = T( + ) = ( HT ) ( tip) PV = nrt = ( m / M) RT 에서 3 V RT PV T 22.4 m / kgmol 1 = = ( ) STP = TK ( ) m PM T PM 273 K M kg / kgmol 9. Drying 11

12 8. Total enthalpy of an air-water vapor mixture (H y ) T 0 가 air & water vapor 의 datum temp. 라면 total enthalpy 는두성분의현열과 T 0 에서 water vapor 의증발잠열 λ O [kj/kg water vapor] 의합 H y [kj/kg dry air]= c S (T-T 0 )+Hλ O =( H) (T-T 0 )+Hλ O 9. Drying 12

13 T c s λo To 9. Drying 13

14 습도표 1 atm abs P 에서공기 - 수증기혼합물의물성을편리하게표현한도표 9. Drying 14

15 9. Drying 15

16 9. Drying 16

17 Solution H=0.0225kgH 2 O/kg dry air H P =14% c S = H= X0.0225=1.047kJ/kg dry air v H =(22.41/273)T(1/28.97+H/18.02) =(22.41/273)*(273+60)*(1/ /18.02) =0.977m 3 /kg dry air 9. Drying 17

18 Adiabatic Saturation Temperature( 단열포화온도 ) 궁극적으로습구온도와근사한온도로서습도측정에사용 에너지수지 c S (T-T 0 )+Hλ 0 +mc P (T S -T 0 )= c S (T S -T 0 )+H S λ 0 에서 T 0 를 T S 로취하면 c S (T-T S )+Hλ S = H S λ S c S (T-T S )=-(H-H S )λ S H HS cs H = = (9.3 11) T T λ λ S S S 습도표에서 adiabatic saruration curve 의식 => H 가포함되어있어정확한직선은아님. 9. Drying 18

19 Wet Bulb Temperature( 습구온도 ) Adiabatic sat. temp. ~ 많은양의물을유입, 기체와접촉시켜얻는정상상태의온도 Wet bulb temp. ~ 적은양의물을단열조건하에서흐르는기체와접촉시켜얻는정상상태비평형온도 정상상태에서물은기체흐름으로증발증발에너지 기체흐름 T 에서적셔진천 (T W ) 으로대류에의한열전달 9. Drying 19

20 증발에너지는[ kj /] s q = N M Aλ (9.3 12) 여기서 A A W N kgmol H O evaporated m s 2 A [ 2 / ] M [ kg H O / kgmol H O] A 2 2 λ 는에서물의증발잠열 T [ kj / kg H O] W W 2 N = k ( y y) A y W 여기서는표면에서수증기의몰분율 y W y는에서수증기의몰분율 bulk air 9. Drying 20

21 y nho H/ M n + n H/ M + 1/ M 2 A = = W H O air A B 2 H가적은값이면 H/ MA y = = 1/ M y = H W M B M A B HM M A B (9.3 14) HwMB HMB MB NA = ky( yw y) = ky( ) = ky ( Hw H) M M M A A A 9. Drying 21

22 MB q = N M Aλ = k ( H H) M Aλ M A A W y W A W A = k M ( H H) Aλ = ha( T T ) H H T T y B W W W W W hkm y B = (9.3 18) λ W h/k v M B 는습도비 (Psychrometric ratio) 실험값이대략 0.96~1.005 이값은 c S (= H) 에서 H 가적을때와거의같은값따라서 (9.3-11)=(9.3-18) 즉단열포화선을습구선으로사용가능습구온도 ===> 공기 - 수증기의습도결정에사용 9. Drying 22

23 Ex9.3-4 건구온도 60, 습구온도 29.5 일때습도를구하시오 9. Drying 23

24 9.4 Equilibrium Moisture Content of Materials 물질의건조과정은평형관계와속도관계의관점에서접근 일정습도 H 와온도를갖는공기와젖은고체가충분히접촉하여평형에도달하면고체는한정된수분함량을갖는다. 수분함량 = kg H 2 O/100 kg dry solid ( 평형수분함량 ) 9. Drying 24

25 결합수 비결합수 9. Drying 25

26 결합수 (bound water) 주어진물질의평형수분함량을연장하여 100% 습도선과교차하였을때의수분함량동일온도에서물이나타내는증기압보다낮은증기압 비결합수 (Unbound water) 100% 습도선과교차될때나타나는물보다많은량함유동일온도에서물이나타내는증기압만큼높은증기압이과잉의수분은고체의공극에갇혀있다. 유리수분 (Free moisture) 평형수분함량이상의수분주어진비교상대습도 (H R ; percentage relative humidity) 하에서제거될수있는수분 9. Drying 26

27 9.5 Rate of Drying Curve 수분함량 1 건조기수분함량 2 설계목표 1. 건조기의크기 2. 사용된공기의습도와온도에대한다양한운전조건 3. 건조시간 선행결정사항 1. 평형수분함량 2. 건조속도 건조속도의실험적결정 건조조건과유사 ( 유사한단구조, 건조비건조면적비와층의두께, 습도, 온도, 속도, 공기흐름방향 ) 9. Drying 27

28 건조곡선 (Drying Curve) 건조속도곡선 (Rate of Drying Curve) 9. Drying 28

29 W : 고체건조고체와수분의합 ( ) W: 건조고체의무게 X t S W WS kg total water = (9.5 1) W kg dry solid t S kg equilibrium moisture X *: 평형수분함량 kg dry solid kg free moisture X : free moisture content kg dry solid X = X X * (9.5 2) LS R=- A R LS A dx dt kg H O m hr 2 : 건조속도[ 2 / ] : 건조고체 [ kg] 2 : 노출된건조표면적 [ m ] (9.5 3) 9. Drying 29

30 9.6 Drying in the Constant-Rate Period 건조고체무게 : L S = 399kg 건조면적 : A=18.58m 2 ===> L S /A=21.5 건조곡선, 건조속도곧선 : 그림 (a), (b) Ex Time of Drying from Drying Curve 유리수분함량 X 1 =0.38 X 2 =0.25 요구되는시간 9. Drying 30

31 정속건조기간중건조속도곡선이용방법 LS dx R = A dt t2= t L X2 S dx t = dt = t1= 0 A X1 R LS t = ( X1 X2) (9.6 2) AR C Ex Drying Time from Rate of Drying Curve Ex 9.2-1을식 (9.6-2) 와그림 9.5-1b를이용하여풀어라 LS 21.5 t = ( X1 X2) = ( ) = 1.85[ hr] AR 1.51 C 9. Drying 31

32 9.6B Method Using Predicted Transfer Coefficients for Constant-Rate Period q = ha( T T ) = N M Aλ W A A W MB NA = ky( yw y) = ky ( HW H) M A q ht ( TW ) RC = NAMA = = = km y B( HW H) Aλ λ W W 9. Drying 32

33 h R kgh O h m = T T 2 C[ 2 / ] ( W) λw h value 1. parallel flow to the dryin surface G = ρv kg m h 2 [ / ] ; 2450 ~ 29,300 vm [ / s];0.61 ~ 7.6 o T[ C] ; 45 ~ 150 h = G W m K [ / ] 2. perpendicular flow to the dryin surface G = ρv kg m h 2 [ / ] ; 3900 ~ 19,500 vm [ / s];0.9 ~ [ / ] h = G W m K L L λ ( X X ) L ( X X ) t = ( X X ) = = R A Ah( T T ) Ak M ( H H) S S W 1 2 S C W y B W 9. Drying 33

34 Ex Prediction of Constant-Rate Drying A= m Hot air ; parallel, v=6.1m/s, T=65.6 H=0.01 kgh 2 O/kg dry air Estimate R C 9. Drying 34

35 h R kgh O h m = T T 2 C[ 2 / ] ( W) λw h = G W m K [ / ] ( T, H) ( T, H ) = (28.9, H ) λ = 2433[ kj / kg] G= ρv에서 W W W W 1+ H 1+ H = = = HT ρ 3 3 vh ( ) 1.037kg 6.1m 3600s G = = kg m h 3 m s hr [ / ] h= = W m K [ / ] R [ kgh O / h m ] = C = [ 2 / ] kg m [ / ] W kg kj J / s 3600s ( ) K m K 2433 kj 1000J W h kgh O h m 9. Drying 35

36 9.6C Effects of Process Variables on Constant-Rate Period ht ( TW ) RC = = km y B( HW H) λ 1. W Effect of air flow rate 2. Effect of gas humidity H( at fixed T) T ( fr Humid chart) R W R λ ( 2 2) C 2 ht ( T 2)/ km 2 y B HW H W W = = R ht ( T )/ λ km ( H H ) C1 W1 W1 y B W1 1 ( T TW2) λw 1 ( T TW2) RC2 = RC1 RC1 = R ( T T ) λ ( T T ) W1 W2 W1 C 1 C ( HW 2 H2) ( H H ) W Drying 36

37 3. Effect of gas Temperature T T W ( T T ) ( H H ) R = R = R ( ) ( ) 2 W2 W2 2 C2 C1 C1 T1 TW1 HW1 H1 4. Effect of thickness of solid being dried 5. Experimental effect of process var iables 9. Drying 37

38 9.7 Calculation Methods for Falling-Rate Drying Period 8.7A Method Using Numerical Integration Ex Numerical Integration in Falling Rate Period X 1 =0.38kg H 2 O/kg dry solid X 2 =0.04 걸리는시간 Solution X 1 =0.38 X C =0.195 까지는정속건조기간 LS 21.5 t = ( X1 X2) = ( ) = 2.63[ hr] AR 1.51 C X C =0.195 X 2 =0.04 까지는감속건조에해당 L X1 S dx t = A X 2 R 적분항은도식적분으로구해야한다. 9. Drying 38

39 X R 1/R /R X Area=½ ( ) ( )+½ ( ) ( ) =0.199 t= =4.27[hr] 따라서총건조시간은 =6.9[hr] 9. Drying 39

40 9.7B Calculation methods for Special Cases in Falling Rate Period 1. 속도가 X의선형함수인경우 R = ax + b L X1 S dx 1 t = 에서 dx = dr A X2 R a L R1 S dr LS R1 t = = ln (9.7 2) aa R2 R aa R2 이식에서미지수를제거 a R1 = ax1 + b R = ax + b 2 2 R1 R2 a = X1 X2 LS X1 X2 R1 t = ln (9.7 4) AR R R Drying 40

41 2. 속도가원점을통과하는선형함수인경우건조속도가임계수분함량 (X C ) 으로부터원점을지나는직선인경우 R = ax L X1 S dx 1 t = 에서 dx = dr A X2 R a L R1 S dr LS R1 t = = ln aa R2 R aa R R a = C X C C 이고일 X =X 때이므로 R =R 2 1 C 1 C LS XC RC LS XC XC t = ln = ln (9.7 8) AR R AR X 2 C 2 9. Drying 41

Ex 9.7-2 Approximation of Straight Line for Falling-Rate Period Ex 9.

42 Ex Approximation of Straight Line for Falling-Rate Period Ex 에서 X C 에서 X=0 까지 R vs X 관계가원점을통과하는직선이라가정하고다시계산 solution L X R L X X t = ln = ln AR R AR X S C C S C C C 2 C = 21.5 ln = 4.39[ hr] Drying 42

43 예제 L S =2kg A=1 m 2 R C =1.5kg/m 2 hr, R=aX 평형수분 0.5%, 임계수분 62% 수분 66.7% 35% 소요시간? 9. Drying 43

44 Solution Basis : 젖은고체 (water+dry solid) 100kg X t1 =66.7/( )=2 X t2 =35/(100-35)=0.538[kg H 2 O/kg dry solid] X C =62/(100-62)=1.63 X*=0.5/( )=0.005 X 1 = =1.995 X 2 = =0.533 X C = = Drying 44

45 R C =1.5 t 1 t 2 X 1 =0.533 X C =1.625 X 2 = LS 2kg m hr ( ) t1 = ( X 1 X C ) = = [ hr] 2 AR 1m 15. kg C L X R L X X t2 = ln = ln A R R A R X S C C S C C C 2 C = ln = [ hr] 건조시간= =2.91[hr] 9. Drying 45

46 Assignment # Drying 46

47 9. Drying 47

Microsoft PowerPoint - chapter4-2-web [호환 모드]

Microsoft PowerPoint - chapter4-2-web [호환 모드] Chapter4 Principles of Steady-State Heat Transfer 이광남 정상상태열전달의원리 1 차원해석 Buckingham π theorem ~ u개의기본적단위나차원으로주어진 q개의양또는변수의함수관계는 (q-u) 개의독립적군 (π) 으로표현할수있다. Buckingham method 특별한물리적문제에있어서중요한변수들을나열한다음에 Buckingham