bÁo cÁo kttp

C C

Thc hnh k thut thc phm GVHD: Nguyn Hu Quyn

Bi 1: C C1. C S L THUYT

1.1 . Cc khi nim trong c c.

Khi nim c c: C c l qu trnh lm tng nng ca dung dch bng cch tch mt phn dung mi nhit si, dung mi tch ra khi dung dch bay ln gi l hi th.

Mc ch ca qu trnh c c:

Lm tng nng ca cht ha tan trong dung dch.

Tch cht rn ha tan dng rn (kt tinh).

Tch dung mi dng nguyn cht (ct nc).

Cc phng php c c:

C c p sut kh quyn: L phng php n gin nhng khng kinh t.

C c p sut chn khng: Dng cho cc dung dch c nhit si cao, d phn hy v nhit,

C c p sut d: Dng cho cc dung dch khng phn hy nhit cao, s dng hi th cho cc qu trnh khc.

1.2 . Cn bng vt cht trong h thng c c 1 ni.

Theo phng trnh cn bng vt cht ta c:

G = Gc + W

GX = GcXc (theo hm lng cht kh c trong dung dch)

Trong :

G khi lng nguyn liu, [kg]; kg/s

Gc khi lng sn phm, [kg]; kg/s

W lng hi th, [kg]; kg/s

X - nng cht kh trong nguyn liu, [phn khi lng]

1.3 . Cn bng nhit lng trong h thng c c 1 ni.

Theo nh lut bo ton nhit .

Trong :

Q1: Nhit do dung dch mang vo

Q2: Nhit do hi t ngng t

Q3: Nhit do dung dch sau c c mang ra

Q4: Nhit do hi th mang ra

Q5: Nhit do nc ngng mang ra

Q6: Nhit do qu trnh c c

Q7: Nhit tn tht ra mi trng

G.c.t + D.i = Gc.cc.tc + W.i + D.cn.tn + Qc + QmtVi:

t : Nhit nguyn liu, []

tc ; Nhit sn phm, [].

tn : Nhit nc ngng, []

c : Nhit dung ring nguyn liu, [J/kg.]

cc : Nhit dung ring sn phm, [J/kg.]

cn : Nhit dung ring nc ngng, [J/kg.].

i : Hm nhit trong hi t, [J/kg]

i: Hm nhit trong hi th, [J/kg]

Lng hi t tiu tn :

Tnh b mt truyn nhit .

Theo phng trnh truyn nhit.

Q = K.F.(.(hi = D.(i - cn.tn).

Trong :

Q: lng nhit truyn, [J].

K: H s truyn nhit,[ W/m2.].

F: Din tch b mt truyn nhit, [m2].

(: Thi gian c c,[s].

(hi : Hiu s nhit hu ch , [].

Rt ra b mt truyn nhit :

[m2]

1.4 . S thit b

H thng c c gm cc thit b chnh sau:

Ni c c hai v c cnh khuy.

My khuy trn

Thit b ngng t ng xon.

Bnh cha nc ngng.

Bm chn khng loi vng nc.

p k o chn khng.

Nhit k in t.

H thng in

X nha cha dung dch u

1.5 . Tin hnh th nghim

1.5.1. Chun b

- Lm quen vi h thng thit b, tm hiu cc van v tc dng ca n.

- Lm quen vi thit b o nhit , cc v tr o v cch iu chnh cng tc o nhit .

- Lm quen vi thit b o hm lng ng

- Xc nh cc i lng cn o.

- Chun b dung dch sa em c c

- Chun b bng s liu th nghim

1.5.2. Cc bc thc hin.

Bc 1: Ra thit b

Kim tra cc van; m van 6, 10, cc van cn li ng.

M cng tc tng.

Chun b 20 lt nc sch trong x nha

Ht chn khng khi kim p k ch 0.6 at th tt bm.

M van 1 ht ht nc sch vo trong ni.

Kim tra mc nc trong v o bng cch m van 5.

M cng tc in tr

M cng tc khuy trn.

Khi nhit nc trong ni t 60oC th x nc bng cch x chn khng van 1 sau m van 4 x nc trong ni ra ngoi.

Tt my khuy trn.

Ra lp ln 2

Bc 2: C c dung dch.

Pha 5lt dung dch sa 15% .

Kim tra cc van , van 6 m, cc van cn li ng.

M cng tc tng.

Ht chn khng bng cch bm chn khng v m van 10 khi kim p k ch 0,6 0,8 at th tt bm v kha van 10.

M van 1 ht ht 5lt dung dch vo trong ni.

M van 9 nc vo ng xon ngng t hi th.

M cng tc khuy trn( 5 pht khuy 1 ln, mi ln 1pht).

K t lc dung dch trong ni si (60oC), th c 10pht ly mu 1 ln o Brix, ly nc ngng t ra o th tch v ghi nhn cc nhit .

Cch ly mu : M van 2 trong thi gian 5giy sau ng van 2 li, m van 3 ly mu.

Cch ly nc ngng t : ng van 6, m van 7, van 8, ly nc ngng xong thao tc van ngc li tr v trng thi ban u( ch trong lc nuc ngng t khng c ht chn khng).

Khi dung dich trong ni t 50 Brix tr ln th dng qu trnh c c.

M van 1 thng p kh tri.

M van 4 x dung dch sau c c ra ngoi cn khi lng.

Tt my khuy trn.

Bc 3: V sinh thit b

Kim tra cc van: van 6, 10 m, cc van cn li ng.

M cng tc tng.

Chun b 20 lt nc sch trong x nha .

Ht chn khng khi chim p k ch 0.6 at th tt bm.

M van 1 ht ht nc sch vo trong ni.

M cng tc khuy trn trong thi gian 3 pht.

M van 4 x nc trong ni ra ngoi.

Ra lp ln 2.

Tt my khuy trn.

Tt cng tc tng.

2. CU HI CHUN B

1. Cc phng php o nng ca dung dch ng?

Phng php t trng

Phng php pha long

Phng php chit quang

2. Tnh p sut tuyt i da vo ch s p k?

=3. Brix l g?

Mt b phn trong cn Brix; t l bi trng lng ca cc cht rn c th ha tan trong xiro nhit .

Nng cht kh c biu din bng % hay cn gi l nng Brix. Brix biu th t l % khi lng cc cht ha tan so vi khi lng nc ma hay dung dch ng.

4. u nhc im ca phng php c c gin on?

u im:

Gi c cht lng, tnh cht sn phm, hay cc cu t d bay hi

Nhp liu v tho sn phm n gin, khng cn n nh lu lng.

Thao tc d dng

C th c c n cc nng khc nhau

Khng cn phi gia nhit ban u cho dung dch

Cu to n gin gi thnh thp

Nhc im:

-Qu trnh khng n nh, tnh cht ha l ca dung dch thay i lin tc theo nng , thi gian.

-Kh gi c chn khng trong thit b.

3. X L S LIU.BNG S LIU

Thi gian

(pht)Nng

dd ng

(Brix)

Lng nc

ngng

V(ml)Nhit

nc vo

(0C)Nhit

nc ra

(0C)Nhit

dung tch

(0C)Nhit

hi th

(0C)

01500000

101663049326665

201927046326263

302034046325961

402338046325961

502524046326061

602727046326062

703239045325859

803612547326565

904225047326463

100437047326463

Khi lng sn phm:

GII TRNH TNH TON

Tnh nng phn khi lng ca dung dch dng nhp liu v i th tch lng nc ngng thu c sang m3Sau khi pha dung dch ng nhp liu ta tin hnh o nng dung dch ng bng Brix k ta c Bx ban u, t suy ra nng phn khi lng nhp liu

Dung dch ng lc u o c l 15 Bx th v

Ta c bng s liu sau:

Thi gian (pht)Nng phn khi lng

ng nhp liu xLng nc ngng thu c V(l)

00,150

100,160,63

200,190,27

300,200,34

400,230,38

500,250,24

600,270,27

700,320,39

800,360,125

900,420,25

1000,430,07

Tnh khi lng dung dich ng nhp liu:

Vi:

+

+ Nng dung dch ng lc u o c l 15; s dng bng tra khi lng ring ca ng theo Brix ta c:

EMBED Equation.3 Suy ra:

Tnh khi lng nc ngng thc t: W*

Vi: + 0+0,63+0,27+0,34++0,38+0,24+0,27

+0,39+0,125+0,25+0,07

= 2,965 (l)

+ ( nhit 300C)

Suy ra: W* = (kg)

Tnh cn bng vt cht v cc i lng cha bit trong phng trnh:

Vi khi lng sn phm :

Suy ra nng cht kh trong sn phm:

Tm sai s gia l thuyt v thc nghim.

Nng cui ca qu trnh:

Lng nc ngng thu c trong qu trnh c c:

Vi:

(kg)

Suy ra:

th.

+ th biu din quan h gia ch s Bx v thi gian c c :

Brix

015

1016

2019

3020

4023

5025

6027

7032

8036

9042

10043

+ th biu din mi quan h lng nc ngng thu c v thi gian c c :

Lng nc ngng thu c V(l)

00

100,63

200,27

300,34

400,38

500,24

600,27

700,39

800,125

900,25

1000,07

BI 2: CHNG CT

1. C S L THUYT Mc ch th nghim

Kho st nh hng ca lu lng hon lu v v tr mm nhp liu ln sn phm, trng thi nhit ng ca nhp liu trn trn s mm thc, hiu sut ca mt ct chng ct v tinh khit ca sn phm.

L thuyt chng ct

1.1. M hnh mm l thuyt l m hnh ton n gin nht da trn cc c s sau:

a. Cn bng gia hai pha lng hi cho hn hp hai cu t

b. iu kin ng lc hc lu cht l tng trn mm l tng cho hai pha lng hi l:

Pha lng phi ha trn hon ton trn mm

Pha hi khng li cun cc git lng t mm di ln mm trn v ng thi c nng ng nht ti mi v tr trn tit din

Trn mi mm lun t s cn bng gia hai pha

1.2. Hiu Sut

chuyn t s mm l thuyt sang s mm thc ta cn phi bit hiu sut mm. C ba loi hiu sut mm c dng l: Hiu sut tng qut, lin quan n ton thp; Hiu sut mm Murphree, lin quan n mt mm; Hiu sut cc b, lin quan n mt v tr c th trn mt mm

Hiu sut tng qut Eo: l hiu sut n gin khi s dng nhng kn chnh xc nht, c nh ngha l t s gia s mm l tng vs mm thc cho ton thp

Eo = = Hiu sut mm Murphree: l t s gia s bin i nng pha hi qua mt mm vi s bin i nng cc i c th t c khi pha hi ri mm cn bng vi pha lng ri mm th n

EM = trong :

yn: nng thc ca pha hi ri mm th n

yn+1: nng thc ca pha hi vo mm th n

y*n: nng pha hi cn bng vi pha lng ri ng chy chuyn mm th n

Ni chung, pha lng ri mm c nng khng bng vi nng trung bnh ca ph lng trn mm nn dn n khi nim hiu cc b

i. Hiu sut cc b c nh ngha nh sau:

EM = trong :

yn: nng pha hi ri khi v tr c th trn mm n

yn+1: nng pha hi mm n ti cng v tr

yen: nng pha hi cnbng vi pha lng ti cng v tr

1.3 . Mi quan h gia hiu sut mm Murphree v hiu sut mm tng qut

Hiu sut tng qut ca thp khng bng vi hiu sut trung bnh ca tng mm. Mi quan h gia hai hiu sut ny ty thuc trn dc tng i ca ng cn bng v ng lm vic. Khi mG/L>1 hiu sut tng qut c gi tr ln hn v mG/L 50 suy ra:

Gi tr C ph thuc vo (dng phng php ni suy)

Re.10-3688,97710

C19,52729,9333

Suy ra:

+ Dng lnh:

Ch chy mng: Re2 < 2320

Nn:

V L/D > 50 suy ra:

Chun s Gr:

Suy ra:

Tnh h s cp nhit

Tra bng 39 Bng tra cu qu trnh c hc truyn nhit-truyn khi ta c:

Dng nng:

(W/m2*)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Vic chn nhit tng ph hp, t yu cu (sai s < 0,05)

1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m*)

bi th nghim ny, ta xem lp cu l khng ng k, do vy xem nh loi b nh hng ca lp cu bm trn b mt ng, tc l rb / db 0

Vi: (W/m.): h s dn nhit ca vt liu inox

Nn: (W/m.)

TRNG HP 21. Tm tt:

1.1. Tnh nhit lng Q theo cng thc v tn tht nhit.

Dng nng:

Nhit trung bnh ca dng nng:

Tra bng 39 Bng tra cu qu trnh c hc truyn nhit-truyn khi:

(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:

Nhit trung bnh ca dng lnh:


(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)

1.2. Tnh hiu nhit trung bnh :

Hiu nhit :

chnh lch nhit trung bnh logarit:

EMBED Equation.3 1.3. Tnh h s truyn nhit di thc nghim.

p dng cng thc:

Suy ra:

(W/m.)

1.4. Tnh h s cp nhit , .

Xc nh ch chy ca lu cht bng chun s :

Dng nng:

nhit , tra bng 39 Bng tra cu qu trnh c hc truyn nhit-truyn khi ta c:

Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn

(Vi nhit ca dng nng vo)

Ta c:

(vi: nhit vo ca dng lnh)

- Xc nh chun s Prandlt ca lu cht v ca tng.

Chun s Pradlt c th tra bng hoc tnh theo cng thc:

Ta tin hnh tra bng 39 Bng tra cu qu trnh c hc truyn nhit-truyn khi ta c:

- Xc nh chun s Nu: ty thuc vo Re m ta tnh c Nu.

+ Dng nng:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3688,90910

C19,52729,72733

Suy ra:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-32,533,4004

C4,467,7210,3

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Vic chn nhit tng khng ph hp, khng t yu cu (sai s > 0,05), v vy qu trnh tnh c lp li vi gi tr mi.

Chn nhit tT1 = 81,9990C


Ta c:





+ Dng nng:

Ch chy chuyn tip:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

Tnh h s cp nhit

Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 820C), nn ta ly nhit tng c sai s nh hn: tT1 = 81,9990C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)

TRNG HP 31. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :



p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn


Ta c:






+ Dng nng:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3688,86510

C19,52729,59533

Suy ra:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3455,0546

C10,315,515,71619,5

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:




Ta c:






+ Dng nng:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3688,86510

C19,52729,59533

Suy ra:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3455,0546

C10,315,515,71619,5

Suy ra:

Tnh h s cp nhit

Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 820C), cn gim nhit xung th sai s li tng ln nhiu hn, nn ta ly nhit tng c sai s nh hn: tT1 = 81,9990C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)

BNG TM TTLu lng dng nng lnhReKLKL*

Dng nngDng lnh

3 38977.221683.0385.1317.575

3 68909.55340088.64239.638

3 988655054.9593.9153.567

KLNngLnh

85.138977.221683.03

88.6428909.553400

93.9188655054.95

KL*NngLnh

17.5758977.221683.03

39.6388909.553400

53.56788655054.95

TRNG HP 41. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :



p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:


Nn:

V L/D > 50 suy ra:

Chun s Gr:

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Vic chn nhit tng ph hp, t yu cu (sai s < 0,05)

1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m*)



Nn: (W/m.)

TRNG HP 51. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :



p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-32,533,4334

C4,467,86210,3

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Vic chn nhit tng khng ph hp, khng t yu cu (sai s > 0,05), v vy qu trnh tnh c lp li vi gi tr mi.(Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-32,533,4334

C4,467,86210,3

Suy ra:

Tnh h s cp nhit

Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 820C), cn nu h nhit tng xung th sai s cng ln hn, nn ta ly nhit tng c sai s nh hn: tT1 = 81,9990C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)

TRNG HP 61. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :


1.3. Tnh h s truyn nhit di thc nghim.

p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3455,1506

C10,315,516,119,5

Suy ra:

Tnh h s cp nhit


Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Vic chn nhit tng khng ph hp, khng t yu cu (sai s > 0,05), v vy qu trnh tnh c lp li vi gi tr mi.( Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3455,1506

C10,315,516,119,5

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 820C), cn nu h nhit tng xung th sai s s cng ln, nn ta ly nhit tng c sai s nh hn: tT1 = 81,9990C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)


Dng nngDng lnh

6 318231.9591702.128130.88517.07

6 618138.4623433.38141.55048.84

6 918138.4625150.07153.62871.68

KLNngLnh

130.88518231.9591702.128

141.55018138.4623433.38

153.62818138.4625150.07

K*LNngLnh

17.0718231.9591702.128

48.8418138.4623433.38

71.6818138.4625150.07

TRNG HP 71. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :



p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:


Nn:

V L/D > 50 suy ra:

Chun s Gr:

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2*)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:


( Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:


Nn:

V L/D > 50 suy ra:

Chun s Gr:

Suy ra:

Tnh h s cp nhit

Dng nng:

(W/m2*)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 820C) v nhit tng 2 phi ln hn 37,5; cn nu h nhit tng 1 xung th sai s s ln hn, v vy th ta ly nhit tng c sai s nh hn: tT1 = 81,420C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)TRNG HP 81. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :



p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.

Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-32,533,4724

C4,468,0310,3

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Vic chn nhit tng khng ph hp, khng t yu cu (sai s > 0,05), v vy qu trnh tnh c lp li vi gi tr mi.( Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-32,533,4724

C4,468,0310,3

Suy ra:

Tnh h s cp nhit

Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 810C), cn nu h nhit tng xung th sai s s cng ln, nn ta ly nhit tng c sai s nh hn: tT1 = 80,9990C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)

TRNG HP 91. Tm tt:


Dng nng:



(J/kg0C) Tnh :

(kg/s)

Nhit lng :

(W)

Dng lnh:



(J/kg0C Tnh :

(kg/s)

Nhit lng :

(W)

Tn tht nhit :

(W)


Hiu nhit :



p dng cng thc:

Suy ra:

(W/m.)



Dng nng:


Tnh:

(m/s)

Suy ra:

Dng lnh:


Tnh:+ (m/s)

+

Vi: (m2)

Nn:

Suy ra:

Xc nh chun s Nu:

- Tnh nhit tng.Chn


Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3455,2086

C10,315,516,33219,5

Suy ra:

Tnh h s cp nhit


Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:




Ta c:






+ Dng nng:

Ch chy ri:

Nn:

+ Dng lnh:

Ch chy chuyn tip:

Nn:

V L/D > 50 suy ra:


Re.10-3455,2086

C10,315,516,33219,5

Suy ra:

Tnh h s cp nhit

Dng nng:

(W/m2.)

(W/m2)

Dng lnh:

(W/m2)

(W/m2)

So snh:

Sai s c gim nhng khng ng k, nu tip tc tng nhit tT ln na th khng c (v tT1 phi nh hn nhit tv1 = 800C), cn nu gim nhit tng xung th sai s s cng ln, nn ta ly nhit tng c sai s nh hn: tT1 = 71,990C1.5. H s truyn nhit di l thuyt c tnh theo cng thc:

(W/m.)



Nn: (W/m.)


Dng nngDng lnh

9 327584.421714.286152.02228.82

9 627230.773472.34166.91953.673

9 927011.4655208.511171.39781.31

KLNngLnh

152.02227584.421714.286

166.91927230.773472.34

171.39727011.4655208.511

K*LNngLnh

28.8227584.421714.286

53.67327230.773472.34

81.3127011.4655208.511

BI 4: SY I LU

1. C S L THUYT Mc ch th nghim

Sinh vin bit cch vn hnh thit b sy.

Xy dng ng cong sy v ng cong tc sy

Xc nh cc thng s sy: tc sy ng tc, m ti hn, m cn bng, thi gian sy ng tc v gim tc

nh gi sai s ca qu trnh sy.

nh ngha sy i lu

Sy i lu l qu trnh tch m ra khi vt liu bng cch cp nhit cho m bay hi. Trong , c hai qu trnh truyn nhit v truyn m u c thc hin bng phng php i lu

1.1 . c trng ca qu trnh sy

Qu trnh sy din ra rt phc tp, c trng cho tnh khng thun nghch v khng n nh. N din ra ng thi 4 qu trnh: truyn nhit cho vt liu, dn m trong lng vt liu, chuyn pha v tch m vo mi trng xung quanh

1.2 . Xc nh tc sy theo cn bng nhit ca qu trnh sy

Lng nhit do dng tc nhn sy cung cp cho khong thi gian d(

(1)

Nhit ny c tiu hao

un nng vt liu:

(2)

Bay hi m v qu nhit hi: [r+Ch(t-th)]dGa

(3)

Trong :

(: h s cp nhit t tc nhn sy vo vt liu sy, W/m2

F: b mt vt liu, m2t, (, th: nhit tc nhn sy, vt liu v hi bo ha,

Go, Co: khi lng v nhit dung ca vt liu sy, kg; j/kg

Ga, Ca: khi lng v nhit dung ca m, kg; j/kg

r: An nhit ha hi ca m, j/kh

Ch: Nhit dung ring ca hi m, j/kg

Lng m bc hi trong thi gian d(:

dGa = d(GoU) = GodU

(4)

U: hm m (hay m) ca vt liu, tnh theo vt liu kh, kg m/kg vt liu kh

T (1), (2), (3) v(4), thit lp cn bng nhit:

(F(t-()d( = (GoCo + GaCa)d( + Go[r + Ch(t - th)]dU

(5)

T (5) rt ra:

y l biu thc tnh tc sy theo cn bng nhit

1.3 . Phng trnh c bn ca ng hc qu trnh sy

Theo phng trnh truyn m t vt liu vo tc nhn sy:

dGa = kpF(pm-p)d(

(7)

Vi:kp: h s truyn m trong pha kh kg/m2.h.(p

Pm, p: p sut ca hi m trn b mt vt liu v trong pha kh, mmHg (at)

Thay Ga=G0U vo (7) v bin i ta c:

(8)

Khi hi m khng b qu nhit (tc t=th) th biu thc (5) c bin i thnh:

(9)

q: cng dng nhit hay mt dng nhit

t:

Vi (o: khi lng ring ca vt liu kh, kg/m3

Vo: th tch vt kh, m3

C: nhit dung ring ca vt liu m, j/kg

Ro: bn knh qui i ca vt liu, m

Khi , nu b qua nhit lm qu nhit hi m, ta c:

(10)

vi : chun s Rebinde c trng cho ng hc qu trnh sy

Biu thc (10) l phng trnh c bn ca ng hc v sy, n cho bit s bin i m ca vt liu theo thi gian. Ta c th nhn c biu thc (10) khi gii h phng trnh vi phn m t truyn nhit truyn m trong vt liu. Nhng ni chung h phng trnh ny khng gii c bng phng php gii tch.

1.4 . Lng nhit cp cho vt liu trong giai on sy gim tc (q2)

Trong giai on sy gim tc, ng cong tc sy c dng ng thng, nn tc sy trong giai on ny c biu din:

(11)

K: h s t l, gi l h s sy. N ph thuc vo tc sy v tnh cht ca vt liu m, 1/s

K chnh l h s gc ca ng cong tc sy giai on sy gim tc, nn:

(12)

: h s sy tng i, ph thuc vo tnh cht vt liu m

Uth: m ti hn

U*: m cn bng

N: tc sy ng tc, kg m/(kg vt liu kh.s)

Tch phn phng trnh (11) ta nhn c:

(13)

Hay logarit ha (8), ta c

(14)

Nh vy, nu bit c h s sy K, c th xc nh c thi gian cn thit thc hin giai on sy gim tc

H s sy tng i c xc nh bng thc nghim v c th tnh gn ng nh sau:

(15)

Vi Uo: m ban u ca vt liu

T ta c:

(16)

Thay (12) v (15) vo phng trnh (11), ta c:

(17)

Thay (17) vo (10) ta c

(18)1.5 . Lng nhit cung cp cho vt liu trong giai on sy ng tc (q1)

Trong giai on sy ng tc, ton b lng nhit cung cp t dng tc nhn bng lng nhit bc hi m v nhit vt liu khng i nn:

1.6 . Cng trao i nhit (q(x))

Nh vy, theo biu thc (20), khi bit chun s Rb s tnh c cng trao i nhit theo m ca vt liu. 1.7 . ng cong sy v ng cong tc sy. th ng cong tc sy

th ng cong sy

1.7.1 ng cong sy

L ng cong biu din s thay i ca m vt liu (U) theo thi gian sy

U = f

(21)Dng ca ng cong sy:

Ph thuc vo nhiu yu t nh lin kt gi m v vt liu, hnh dng kch thc; Cu trc vt liu, phng php v ch sy.

ng cong sy l hm ca qu trnh sy. V vy, tuy ch v phng php sy khc nhau nhng dng ng cng sy l tng t nhau.

1.7.2 ng cong tc sy:

L ng cong biu din mi quan h gia tc sy v m ( hm m) ca vt liu sy:

(22)

T biu thc (21) v (22), r rng ng cong tc sy l o hm ca ng cong sy.

1.8 Cc giai on ca qu trnh sy

1.8.1 Giai on un nng vt liu (AB)

Giai on ny xy ra nhanh vi khong thi gian ngn khng ng k. Ton b nhit do dng tc nhn cp dng un nng vt liu t nhit u ln nhit bu t (t).

Trong giai on ny lng m tch ra khng ng k, m vt liu gim khng nhiu v tc sy nhanh ln vi tc cc i (N). thng giai on ny b qua trong tnh ton.

1.8.2 Giai on sy ng tc (BC)

Trong giai on ny, tc khuch tn m t trong lng vt liu ra b mt ln hn tc bc hi m t b mt vt liu, nn b mt vt liu lun bo ha m.

Tan b lng nhit cung cp bc hi m b mt (m t do) v b mt bc hi l b mt ngoi ca vt liu khng i nn cc thng s sy sau y s khng i: Nhit b mt vt liu v tc sy; v m vt liu gim nhanh.

Thi gian sy trong giai on ny l (thi gian sy ng tc c xc nh t:

- (23)

Nn tch phn (23) ta c

(24)

Vi l m cui giai on sy ng tc.

1.8.3 Giai on sy gim tc (CD)

Do bc hi ht m b mt ch cn m lin kt, nn b mt bc hi b co hp li dn i su vo trong lng vt liu.

Tc khuch tn m trong vt liu chm lm gim tc chung.

Nhit ca vt liu tng dn t nhit bu t (t) n nhit dng tc nhn (t) nhit bu kh.

Lc ny; Trong vt liu xut hin 3 vng : m, bc hi v kh.

Trong giai on ny, nu ng cong tc sy c dng ng thng (hoc qui i sang ng thng N2 =ax +b) th ta c th phn tch tnh thi gian sy giai on sy gim tc ny () :

Vi U: m cn bng, m kt thc giai on sy gim tc.

1.9 . Thi gian sy vt liu

Thi gian sy vt liu c tnh bng tng thi gian ca 3 giai on sy: t nng vt liu . Sy ng tc v sy gim tc . C th b qua giai on t nng vt liu, v giai on ny xy ra rt nhanh. Biu thc tnh thi gian sy nh sau:

(26)

Vi m vt liu cui qu trnh sy. Tng ng vi .U2> U v thng c ly : U2 = U + 2(%)

1.10 S thit b

1. Ca kh vo

5. Bung sy

2.Qut ly tm

6. Khy sy

3.Caloriphe

7. Ca ra kh thi

4. Cn

1.11 Tin hnh

Chun b th nghim:Xc nh khi lng vt liu kh ban u (G0) ca vt liu:

. M ca bung sy ra, t cn thn

. c gi tr cn (G0)

Lm m vt liu:

Sau khi cn xong, ly vt liu ra v nhng nh nhng (trnh rch vt liu) vo chu nc. Ch khong 30 giy cho nc thm u, ly vt liu ln v ro nc sau xp vo gi.

Chun b ng h eo tay o thi gian.

Kim tra h thng:

. Lp li ca bung sy.

. M ht cc van ca hai ca kh vo ra

. Chm y nc vo bu t (pha sau h thng)

Khi ng h thng:

Khi ng qut: bt cng tc ca qut ht dng tc nhn vo v thi qua caloriphe gia nhit dng tc nhn.

Khi ng caloriphe, bt cng tc Caloriphe.

Ci t nhit cho Caloriphe nhit th nghim.

Tin hnh th nghim

Ch h thng hot ng n nh khi: nhit ca Caloriphe t gi tr mong mun (). Tin hnh sy vt liu nhit kho st.

o s liu trong ch th nghim.Cc s liu cn o: Khi lng, nhit bu kh, bu t v thi gian.

Cch c:

. Khi lng (gam) khi t gi vt liu sy, c s hin th trn cn.

. Nhit (0C) : Nhn nt tng ng cc v tr cn o v c s trn ng h hin s.

1.12 Chuyn ch th nghim:

M ca bung sy, ly vt liu ra lm m tip (lp li nh ban u).

Ci nhit Caloriphe gi tr tip theo cho ch sy mi.

Ch h thng hot ng n nh.

Lp li trnh t nh ch u.

1.13 Kt thc th nghim

Tt cng tt ca in tr Caloriphe.

Sau khi tt Caloriphe c 5 pht, tt qut cho Caloriphe ngui.

2. CU HI CHUN B

1. K tn mt vi thit b sy?

Theo phng php np nhit, cc my sy c chia ra loi i lu v tip xc.

Theo dng cht ti nhit : khng kh, kh v hi.

Theo tr s p sut trong phng sy: lm vic p sut kh quyn v cht ti nhit trong cc my sy i lu: cng chiu, ngc chiu v vi cc dng ct nhau.

Theo kt cu: phng ng hm, bng ti, sy tng si, sy phun, thng quay, tip xc, thng hoa, bc x nhit.

2. ng cong sy?

L ng cong biu din s thay i ca m vt liu (U) theo thi gian sy (()

U=f(()

Dng ca ng cong sy:

Ph thuc vo nhiu yu t nh lin kt gi m v vt liu, hnh dng kch thc; cu trc vt liu, phng php v ch sy.

ng cong sy l hm ca qu trnh sy. V vy, tuy ch v phng php sy khc nhau nhng dng ng cng sy l tng t nhau.

ng cong tc sy ?

ng cong tc sy l dng cong biu th mi quan h gia tc sy( N) v m (W) ca vt liu sy.

ng cong tc sy l o hm ca ng cong sy.

Sy l g ? S khc nhau gia sy v c c ? Sy l qu trnh tch m ra khi vt liu bng phng php nhit.Nhit c cung cp cho vt liu m bngdan nhit, i lu, bc x hay nng lng in trng c tn s cao.

S khc nhau gia sy v c c.

Sy :

L lm gim khi lng vt liu,tng bn v bo qun sn phm

Theo phng php dn nhit,bc x,nng lng in trng

C c: L lm tng nng dung dch,tch rn ha tan d kt tinh hay tch dung mi dang nguyn cht

Theo phng php p sut kh quyn,chn khng v p sut d

3. X L S LIUBng s liu th

STTCh sy 500CCh sy 600C

(ph)G(g)T(0C)Tk(0C)(ph)G(g)T(0C)Tk(0C)

10154452535301548575958

25144452535351347555656

3101244525252101147585656

4151144525353151146555658

5201144525252201146555656

6251144525353

7301144525353

8351144525353

Khi lng ca vt liu sy ban u: G0 = 10gGii trnh tnh ton

STTCh sy 500C

(ph)G(g)T(0C)Tk(0C)

101544525353

251444525353

3101244525252

4151144525353

5201144525252

6251144525353

7301144525353

8351144525353

Tnh theo thc nghim

m vt liu (tnh theo vt liu kh):

(%)

Thi im bt u sy lng m bay ra l 0 (g)

C th nh sau:

(%)

(%)

Tng t, ta tnh c: (%) ; (%)

Tc sy:

(%/h)

C th nh sau:

(%/h)

(%/h)

(%/h)

(%/h)

X l s liu c trnh by thnh bng nh sau:

i(ph)Gi(g)Wi (%)NTkvTvPb(mmHg)Ph(mmHg)

101550053446558

25144012553446558

310122025052446559

415111012553446558

5201110052446559

6251110053446558

7301110053446558

8351110053446558

Cch tra nh sau: dng gin Ramdzim

T nhit T ta ko di (theo ng xin 300 so vi trc honh) gp ng th dng li, ti im (im *) ta ko thng xung trc Y gp ng p sut hi ring phn (l ng xin 450 so vi trc honh) dng li, ti ko thng qua phi song song vi trc honh ta c gi tr Pb tng ng.

Tip theo, ti im (*) ko thng song song vi ng H (xin 1200 so vi trc honh), v t nhit Tk ta ko thng gp ng H trn th dng li, tip tc ko thng xung gp ng p sut, ko qua phi (tng t nh trn) ta c Ph tng ng.

D

Tnh theo l thuyt.

+ Cc s liu:

Tc tc nhn sy: Vk = 1,6 m/s

Khi lng bnh em sy: G0 = 10 (g)

Din tch b mt giy lc: (m2)

+ Tnh ton:

Cng m:

(kg/m2h)

Vi: (kg/m2.h.mmHg)

B: p sut phng sy , B = 760mmHg

Pb(tb) = mmHg

Ph(tb) = mmHg

Suy ra: (kg/m2h)

Tc sy ng tc:

(%/h)

m ti hn:

(%kg m/kgkkk)

Vi: Wc : m cn bng = 3%

Thi gian sy:

Thi gian sy ng tc:

(h)

Thi gian sy gim tc:

EMBED Equation.3 (h)

Thi gian tng cng qu trnh sy gn ng:

(h)

Lp bng so snh: so snh cc thng s: m ti hn, tc sy ng tc, thi gian sy v sai s.

Vi: (%/h)

Vi: (%)

V th

Ch sy 600C

(ph)G(g)T(0C)Tk(0C)

01548575958

51347555656

101147585656

151146555658

201146555656

Tnh theo thc nghim

m vt liu:

(%)

Thi im bt u sy lng m bay ra l 0 (g)

C th nh sau:

(%)

(%)

(%)

(%)

Tc sy:

(%/h)

C th nh sau: i 5 pht = 5/60 = 0,083 h

(%/h)

(%/h)

(%/h)

(%/h)

X l s liu c trnh by thnh bng nh sau:

i(ph)Gi(g)Wi (%)NTkvTvPb(mmHg)Ph(mmHg)

101550059488885

25133025056478475

310111025056478475

4151110056466960

5201110056466960

Cch tra nh trn.

Tnh theo l thuyt.

+ Cc s liu:

Tc tc nhn sy: Vk = 1,6 m/s

Khi lng bnh em sy: G0 = 10 (g)

Din tch b mt giy lc: (m2)

+ Tnh ton:

Cng m:

(kg/m2h)

Vi: (kg/m2.h.mmHg)

B: p sut phng sy , B = 760mmHg

Pb(tb) = = 78,8 mmHg

Ph(tb) = mmHg

Suy ra: (kg/m2h)

Tc sy ng tc:

(%/h)

m ti hn:

(%kg m/kgkkk)

Vi: Wc : m cn bng = 3

Thi gian sy:

Thi gian sy ng tc:

(h)

Thi gian sy gim tc:

(h)

Thi gian tng cng qu trnh sy gn ng:

(h)

Lp bng so snh: so snh cc thng s: m ti hn, tc sy ng tc, thi gian sy v sai s.

Vi: (%/h)

Vi:

V th

BAI 5: CT CHM

1. C S L THUYT

Mc ch th nghim

Kho st c tnh ng lc hc lu cht v kh nng hot ng ca ct chm bng cch xc nh:

- nh hng ca vn tc dng kh v lng ln tn tht p sut ( gim p) khi i qua ct.

- S bin i ca h s ma st ct kh fck theo chun s Reynolds (Re) ca dng kh v suy ra cc h thc thc nghim.

- S bin i ca tha s ( lin h gia gim p ca dng kh qua ct kh v qua ct t theo vn tc dng lng.

- Gin gii hn kh nng hot ng ca ct (gin ngp lt v gia trng).

L thuyt ct chm

1.1. gim p ca dng kh

gim p (Pck ca dng kh qua ct ph thuc vo vn tc khi lng G ca dng kh qua ct kh (khng c dng chy ngc chiu). Khi dng kh chuyn ng trong cc khong trng gia cc vt chm tng dn vn tc th gim p cng tng theo. S gia tng ny theo ly tha t 1,8 n 2,0 ca vn tc dng kh.

(1)

Vin = 1,8 2,0

Khi c dng lng chy ngc chiu, cc khong trng gia nhng vt chm b thu hp li. Dng kh do di chuyn kh khn hn v mt phn th tch t do gia cc vt chm b lng cht lng chim c . Khi tng vn tc dng kh ln, nh hng cn tr ca dng lng tng u n cho n mt tr s ti hn ca vn tc kh, lc gim p ca dng kh tng vt ln. im ng vi tr s ti hn ca vn tc kh ny c gi l im gia trng. Nu tip tc tng vn tc kh qu tr s ti hn ny, nh hng cn tr h tng gia dng lng v dng kh rt ln, (Pc tng mau chng khng theo phng trnh (1) na. Dng lng lc ny chy xung cng kh khn, ct im lt.

ng biu din log((Pc/Z) ( gim p sut ca dng kh qua mt dn v chiu cao ca phn chm trong ct) d kin nh trnh by trn hnh 1.

Hnh 1: nh hng ca G v L i vi gim p ca ct (Pc

1.2. H s ma st fck theo Rec khi ct kh

Tr lc thp kh:

Trong :

h - chiu cao lp m, m

wo- vn tc pha kh

a - b mt ring, m2/m3( - xp, m3/m3(k khi lng ring ca khng kh, kg/m3fck - h s ma sat ca dng chy qua lp ht, ph thuc vo Rek

Khi Rek40:

1.3 gim p (Pc khi ct t

S lin h gia gim p ct kh (Pck v ct t (Pc c th biu din nh sau:

(Pc = ((Pck

(6)

Do c th d kin fc = (.fck

(7)Vi(: h s ph thuc vo mc xi ti ca dng lng L, kg/m2s.

Leva ngh nh hng ca L ln ( nh sau:

( = 10(L

(8)

haylog ( = (L

(9)

Gi tr ( ty thuc vo loi, kch thc, cch thc sp xp vt chm (xp ngu nhin hay theo th t) v ln ca lu lng lng L. Th d vi vt chm l vng s Raschig 12,7 mm, chm ngu nhin, xp ( = 0,586; gi tr ca L t 0,39 n 11,7 kg/m2s v ct hot ng trong vng di im gia trng.

( = 0,084

Mt s ti liu cn biu din s ph thuc gia t s vi h s xi ti nh sau:

Khi A < 0,3 cho vt chm bng s c d < 30 mm, ta c:

(10)

(11)

1.4 im lt ca ct chm

Khi ct chm b ngp lt, cht lng chim ton b khong trng trong phn chm, cc dng chy b xo trn mnh lit, hin tng ny rt bt li cho s hot ng ca ct chm. Gi gi tr ca GL tng ng vi trng thi ny l GL*.

Hnh 2: Gin lt ca ct chm

Zhavoronkov kt lun rng trng thi ngp lt xy ra khi hai nhm s sau c s lin h nht nh vi nhau cho mi ct.

(12)

v

(13)

Vifck: h s ma st ct kh.

v: vn tc di ca dng kh ngay trc khi vo ct, m/s.

: nht tng i ca cht lng so vi nc. , nu cht lng l nc th .

Do s lin h gia (1, (2 trn gin log(1 log(2 s xc nh mt gin lt ca ct chm, phn gii hn hot ng ca ct chm di ng ny.

1.5 S nguyn l thit b

I-My thi kh

1,2-Van iu chnh lu lng dng kh

II-Lu lng k dng kh

3-Van x nc ng trong ng kh

III-Ct chm

4,6-Van iu chnh lu lng dng lng

IV-Bn cha

5-Van to ct lng ngn kh

V-Bm

7-Van iu chnh mc nc trong ct chm

VI- Lu lng k dng lng

8-Van x nhanh khi lt ct chm

D-lp m vng s Raschig

9-Van x y bn cha

Trnh t th nghim

1) Kha li tt c cc van lng (t 4 n 8).

2) M van 2 v kha van 1, 3.

3) Cho qut chy trong 5 pht thi ht m trong ct. Tt qut.

4) M van 4 v 7. Sau cho bm chy.

5) M van 5 v t t kha van 4 chnh mc lng y ct ngang bng vi ng nh mc g. Tt bm v kha van 5.

6) o gim p ca ct kh:

Kha tt c cc van lng li. M van 1 cn 2 vn ng. Cho qut chy ri t t m van 2 chnh lu lng kh vo ct.

ng vi mi gi tr lu lng kh chn ta c (Pck trn p k U theo mmH2O. o xong tt qut, ngh 5 pht.

7) o gim p khi ct t:

M qut v iu chnh lu lng kh qua ct khong 15 20%.

M van 4 v cho bm chy. Dng van 6 ti lu lng k chnh lu lng lng. Nu 6 m ti a m phao vn khng ln th dng van 4 tng lng lng.

ng vi lu lng lng chn c nh, ta chnh lu lng kh v c gim p (Pc ging nh (Pck trc . Ch l tng lng kh n im lt th thi.

2. CU HI CHUN B

1. Cc yu t nh hng n gim p ca ct kh?

gim p ca dng kh qua ct ph thuc vo vn tc khi lng G ca dng kh qua ct kh (khng c dng chy ngc chiu) .Khi dng kh chuyn ng trong cc khong trng gia cc vt chm tang dn vn tc th gim p cng tang theo.Khi c dng lng chy ngc chiu ,cc khong trng gia cc vt chm b thu hp li dng kh do di chuyn kh khan hn v 1 phn th tch t do gia cc vt chm b lng cht lng chim c.Khi tang vn tc dng kh ln nh hng cn tr ca dng lng tng u n cho n 1 tr s ti hn ca vn tc kh, lc gim p ca dng kh tang vt ln.

2. u v nhc im ca vt chm bng s?

3. Thp chm lm vic ch no l tt nht? thc t c th vn hnh ch ny hay khng? Ti sao?


HngV(fit3/p)(cmH2O)

S lnS nh

L = 01245,645,5

22,545,844,2

334643,9

43,546,443,6

5446,943,1

64,547,442,7

L= 4(cmH2O)

1245,944,2

22,546,143,9

3346,543,5

43,546,743,4

5447,442,6

64,547,942,1

L= 51245,944,1

22,546,243,7

3346,543,5

43,547,142,9

5447,742,3

64,548,441,7

L = 6124644,1

22,546,243,7

3346,643,3

43,547,142,9

5447,842,1

64,548,741,2

L = 71246,144

22,546,343,6

3346,743,2

43,547,642,3

5448,542,3

64,549,740,2

L = 81246,143,9

22,546,643,4

3347,642,7

43,548,241,8

5449,440,6

64,55138,9

L = 91246,743,6

22,546,943,2

334842

43,549,140,9

5450,339,6

64,55237,9

(S in m l lc b ngp lt)

Cc s liu lin quan n ct chm:

Ct thy tinh:

ng knh d =0,09 m Chiu cao H = 0,805 m. Chiu cao phn chm Z = 0,6 m. B mt ring a =350 m2/m3 xp Tit din rng ca ct.

a. Ct kh

Trc ht ta phi i ct V ra m3/s

1fit3/pht = 2,83.10-2 m3/pht = = 4,717.10-4 (m3/s)

2fit3/pht = 2 . 2,83.10-2 m3/pht = 2 . = 9,43.10-4 (m3/s)

(s ln s nh).10 = (45,6 45,5).10 = 1 (mmH2O)iV (m3/s)

(cmH20)

(mmH20)

19,43.10-40,11

21,179.10-31,616

31,415.10-32,121

41,651.10-32,828

51,887.10-33,838

62,123.10-34,747

Tnh G:

Gi s ct kh ang vn hnh nhit 400K ( Kelvin)

(kg/m3), (kg/m.s)

(kg/m2.s)

Tnh :

(N/m2)

(N/m2.m)

Tnh Re:

Tnh fCK:

iG

N/m2

Log

()LogGRefCKlog fCKLog Re

10,1959,8116,531,218-0,7197,3181,5210,1821,988

20,244156,96261,62,418-0,613121,7721,4540,1632.086

30,293206,01343,352,536-0,533146,2261,4020,1472,165

40,342274,68457,82,661-0,466170,681,3590,1332,232

50,391372,78621,32,793-0,408195,1341,3230,1222,290

60,44461,07768,452,886-0,357219,5881,2930,1122,342

b. Ct t L = 4 (l/ph)

L= 4Hng V(fit3/p)(cmH2O)

1245,944,2

22,546,143,9

3346,543,5

43,546,743,4

5447,442,6

64,547,942,1

Tnh G:

Gi s ct kh ang vn hnh nhit 300K ( Kelvin)

(kg/m3), (kg/m.s)

(kg/m2.s)

Tnh :

(s ln s nh).10 = (45,9 44,2).10 = 17 (mmH2O)

(N/m2)

(N/m2.m)

Bng s liu c tnh ton nh sau:

iV (m3/s)

(cmH20)

(mmH20)

19,43.10-41,717

21,179.10-32,222

31,415.10-3330

41,651.10-33,333

51,887.10-34,848

62,123.10-35,858

Tnh

Tnh fC = fCK = 17.1,521 = 25,857

L = 4(l/ph)

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,260166,77277,952,444-0,5851606,1781725,8571,4133,206

20,326215,82359,72,556-0,4872013,91,3751,9990,3013,304

30,391294,3490,52,691-0,4082415,4441,4292,0030,3023,383

40,456323,73539,552,732-0,3412816,9881,1791,6020,2053,45

50,521470,88784,82,895-0,2833218,5331,2631,6710,2233,508

60,586568,98948,32,977-0,2323620,0771,2341,5960,2033,559

Tnh tng t nh trn ta c:

L = 5 (l/ph)

iV (m3/s)

(cmH20)

(mmH20)

19,43.10-41,818

21,179.10-32,525

31,415.10-3330

41,651.10-34,242

51,887.10-35,454

62,123.10-36,767

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,260176,58294,32,469-0,5851606,1781827,3781,4373,206

20,326245,25408,752,611-0,4872013,91,5632,2730,3573,304

30,391294,3490,52,691-0,4082415,4442,3813,3380,5233,383

40,456412,02686,72,837-0,3412816,9881,52,0390,3093,45

50,521529,74882,92,946-0,2833218,5331,4211,880,2743,508

60,586657,271095,453,04-0,2323620,0771,4261,8440,2663,559

L = 6 (l/ph)

iV (m3/s)

(cmH20)

(mmH20)

19,43.10-41,919

21,179.10-32,525

31,415.10-33,333

41,651.10-34,242

51,887.10-35,757

62,123.10-37,575

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,260186,39310,652,492-0,5851606,1781928,8991,4613,206

20,326245,25408,752,611-0,4872013,91,5632,2730,3573,304

30,391323,73539,552,732-0,4082415,4441,5712,2030,3433,383

40,456412,02686,72,837-0,3412816,9881,52,0390,3093,45

50,521559,17931,952,969-0,2833218,5331,51,9850,2983,508

60,586735,751226,253,089-0,2323620,0771,5962,0640,3153,559

L = 7 (l/ph)iV (m3/s)

(cmH20)

(mmH20)

19,43.10-42,121

21,179.10-32,727

31,415.10-33,535

41,651.10-35,353

51,887.10-36,262

62,123.10-39,595

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,260206,01343,352,536-0,5851606,1782131,9411,5043,206

20,326264,87441,452,645-0,4872013,91,6882,4540,393,304

30,391343,35572,252,758-0,4082415,4441,6672,3370,3693,383

40,456519,93866,552,938-0,3412816,9881,8932,5730,413,45

50,521608,221013,73,006-0,2833218,5331,6322,1590,3343,508

60,586931,951553,253,191-0,2323620,0772,0212,6130,4173,559

L = 8 (l/ph)iV (m3/s)

(cmH20)

(mmH20)

19,43.10-42,222

21,179.10-33,232

31,415.10-34,949

41,651.10-36,464

51,887.10-38,888

62,123.10-312,1121

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,260215,82359,72,556-0,5851606,1782233,4621,5253,206

20,326313,92523,22,719-0,487201,3922,9080,4643,304

30,391480,69801,152,904-0,408241,5442,3333,2710,5153,383

40,456627,841046,43,02-0,341281,6992,2863,1070,4923,45

50,521863,281438,83,158-0,283321,8532,3163,0640,4863,508

60,5861187,011978,353,296-0,232362,0082,5743,3280,5223,559

L = 9 (l/ph)iV (m3/s)

(cmH20)

(mmH20)

19,43.10-43,131

21,179.10-33,737

31,415.10-3660

41,651.10-38,282

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,260304,11506,852,705-0,5851606,1783147,1511,6733,206

20,326362,97604,952,782-0,487201,392,3133,3630,5273,304

30,391588,69812,992-0,408241,5442,8574,0060,6033,383

40,456804,421340,73,127-0,341281,6992,9293,9810,63,45

c. Ct ngp ltL= 9Hng V(fit3/p)(cmH2O)

5450,339,6

64,55237,9

iV (m3/s)

(cmH20)

(mmH20)

11,887.10-310,7107

22,123.10-314,1141

i

G

N/m2

Log

()LogG

Re

fC

log fC

Log Re

10,5211049,671749,453,243-0,283321,8532,8163,7260,5712,508

20,5861383,212305,353,363-0,232362,00833,8790,5892,559

Tnh chun s th nht

Vi: (kg/m3)

g = 9,81 m/s2

a = 350 m2/m3

+ Vn tc dng kh vo ct:

Vi: (m2)

Nn: (m/s)

Suy ra:

Tnh chun s th 2

L = 9 l/ph = (m3/s)

V = 1,887.10-3 (m3/s)

(kg/m3)

Suy ra:

Ta c bng sau:

iV (m3/s)

log

log

11,887.10-39,812.10-32.727.10-3- 2,008- 2,56

22,123.10-36,75.10-32,424.10-3- 2,17- 2,615

V th Ct kh Gin log theo logG

G(kg/s.m2)logG log

(N/m2)/m

0.195-0.711.218

0.244-0.6132.418

0.293-0.5332.536

0.342-0.4662.661

0.391-0.4082.793

0.44-0.3572.886

Gin logfCK theo Re

ReLogfCK

97.3180.182

121.7720.163

146.2260.147

170.680.133

195.1340.122

219.5880.112

Ct t th log theo logGL = 4L = 5

iLog

()LogG

iLog

()LogG

12.444-0,58512.469-0,585

22.556-0,48722.611-0,487

32.691-0,40832.691-0,408

42.732-0,34142.837-0,341

52.895-0,28352.946-0,283

62.977-0,23263.04-0,232

L = 6L = 7

12.492-0,58512.536-0,585

22.611-0,48722.645-0,487

32.732-0,40832.758-0,408

42.837-0,34142.938-0,341

52.969-0,28353.006-0,283

63.089-0,23263.191-0,232

L = 8L = 9

12.556-0,58512.705-0,585

22.719-0,48722.782-0,487

32.904-0,40832.992-0,408

43.02-0,34143.127-0,341

53.158-0,2835

63.296-0,2326

th LogfC theo ReReLogfC

L = 4L = 5L = 6L = 7L = 8L = 9

1606.1781.4131.4371.4611.50433.4621.673

201.390.3010.3570.3570.392.9080.527

241.5440.3020.5230.3430.3693.2710.603

281.6990.2050.3090.3090.413.1070.6

321.8530.2230.2740.2980.3343.064

362.0080.2030.2660.3150.4173.328

Ct ngp lt

th log- log

log

log

- 2.008- 2.56

- 2.17- 2.615

BAI 6: LC KHUNG BN

4. C S L THUYT

Mc ch th nghim

- Lm quen vi cu to v nguyn l hot ng ca thit b lc khung bn.

- Bit ch vn hnh, kim tra trc khi vn hnh thit b.

- Vn hnh th nghim lc huyn ph CaCO3 trong nc vi p sut khng i.

- Xc nh cc h s lc theo s liu th nghim thu c.

C s l thuyt

1.1. Lc cht lng

Lc l qu trnh phn ring cc hn hp nh 1 vt ngn xp, vt ngn xp c kh nng cho 1 pha i qua cn pha kia c gi li nn cn gi l vch ngn lc.

1.2. Nguyn tc lc

To ra trn huyn ph mt p sut P1, pha lng xuyn qua cc mao dn, pha rn b gi li. Chnh lch gia hai vch ngn gi l ng lc qu trnh lc.

(P = P1 P2

C th to ng lc ca qu trnh lc bng cc cch sau:

Ct p thy tnh

Bng my bm hay my nn

Bng bm chn khng

Cn bng vt tcht trong qu trnh lc:

Vh = V0 + V1 = Va + V

Gh = G0 + G1 = Ga + G

Vh, Gh : Th tch v khi lng hn hp huyn ph em i lc.

V0, G0 : Th tch v khi lng cht rn kh.

V1, G1 : Th tch v khi lng nc lc nguyn cht.

Va, Ga : Th tch v khi lng b m

V, G : Th tch v khi lung nc lc cha nguyn cht.

m ca b :

( % kg m / kg vt liu t).

1.3. Phng trnh tc lc

1.3.1. Tc lc v cc yu t nh hng

Lng nc lc thu c trn mt n v din tch b mt vch ngn lc trn mt n v thi gian gi l tc lc.

Trong :

V Th tch nc lc thu c, m3

F Din tch b mt vch lc, m2

( - thi gian lc, s

Qu trnh lc huyn ph ph thuc vo cc yu t sau: Tnh cht huyn ph: nht, kch thc v hnh dng pha phn tn; ng lc qu trnh lc; tr lc b v vch ngn; din tch b mt vch lc.

1.3.2. Theo DAKSI, tc lc c th biu din di dng phng trnh sau:

(5.2)

Trong :

( - nht ca pha lin tc, Ns/m2Rb = 1/(Pb tr lc ca b lc (tn tht p sut qua lp b), 1/m

Rv =1/(Pv tr lc ca vch lc (tn tht p sut qua vch lc), 1/m

1.3.3. Lc vi p sut khng i, (P = const

phng trnh lc c dngq2 + 2.C.q = K(

q = V/F lng nc lc ring

;

- t s gia th tch b m thu c v lng nc lc

r0 tr lc ring theo th tch ca b lc (1/m2)

1.3.4. Lc vi tc lc khng i, W=const

phng trnh lc c dngq2 + Cq =

1.4. S nguyn l thit b

1. Khung khuy huyn ph

2. Bm huyn ph

3. Thit b lc khung bn

4. p k

Trnh t th nghim

- Pha 500 g bt CaCO3 vo 20lt nc c dd huyn ph lc

- ng van V1 v van V2

- Cho dd pha vo bn cha.

- Bt cng tc my khuy, khuy u hn hp dd CaCO3.

- M van V3, V4,V5,V6.

- M bm, iu chnh p sut bng V4 khi ng h p sut ch mc mong mun

- Hng dd lc u C1 v ghi nhn th tch trong mi thi gian 30 giy.

- Lm th nghim vi cc ch p sut khc nhau

2. CU HI CHUN B

1. Nu cc phng php to chnh lch p sut khi lc?

Tng p sut P1: Dng ct p thu tnh, my bm hay my nn. Gim p sut P2: Dng bm chn khng.

2. Cu to ca lc khung bn?

Gm 2 phn chnh:

Phn th nht l b phn lc v phn th 2 l b phn bm ht v nn dung dch lc qua vt liu lc.

Phn th nht ca thit b lc bao gm cc khung v cc tm lc c p li vi nhau nh mt a quay bng tayPhn th hai l b phn ht v nn dung dch lc gm bm nn p sut cao v hai thng cha bng thp khng r.

Nguyn l hot ng:

y l thit b lc p lc lm vic gin on ngha l nhp liu vo liu tc, nc lc tho ra lin tc nhng b c tho ra chu k.

N c cu to ch yu l khung v bn. Khung gi vai tr cha b lc v l ni nhp huyn ph vo. Bn to ra b mt lc vi cc rnh dn nc lc hoc l cc l lc.

Khung v bn thng c ch to dng hnh vung v phi c s bt kn tt khi ghp khung v bn.

Khung v bn c xp lin tip nhau trn gi . Gia khung v bn l vch ngn lc. p cht khung v bn nh c cu ai vt xon nh tay quay. L dn huyn ph nhp liu ca khung v bn c ni lin to thnh ng dn nh ra ghp vi h thng cp liu. Nc lc chy ra t bn qua h thng ng ng v ly ra ngoi. B c gi li trn b mt vch ngn lc v c cha trong khung. Khi b trong khung y th dng qu trnh lc tin hnhra v tho b.

u im:

B mt lc ln

Dch lc trong v loi b c cc cht cn b.

Giy lc c th thay th d dng.

Khng cn ngi c chuyn mn cao.

Nhc im:

Cn nhiu thi gian v sinh.

Phi thay th giy lc theo chu k.

Gi thnh giy lc cao.

Dch chy nhiu,phn b khng u.

Phi tho khung bn khi cn gim p sut.

3. Cc yu t nh hng ti qu trnh lc?

Tnh cht huyn ph: nht, kch thc v hnh dng pha phn tn, tr lc b v vch ngn, ng lc qu trnh lc, din dch v mt vch lc. Kch thc ht lc v s phn b cc c ht trong lp vt liu lc.

Kch thc, hnh dng, trng lng ring, nng v kh nng dnh kt ca cn bn l lng trong nc x l.

Tc lc, chiu cao lp lc, thnh phn ca lp vt liu lc v chnh p lc.

Nhit v nht ca nc.

Vt liu lc


P1 = 0,3at

(s)10101010101010101010

V(l)6,85,85,45,66,04,85,05,24,65.2

P2 = 0,6at

(s)5555555555

V(l)3,23,44,43,04,23,63,23,83,63,0

P3 = 1at

(s)4444444444

V(l)4,04,04,03,83,83,63,64,43,04,2

Kch thc cnh ca khung v bn: a = 20cm

Gii trnh tnh ton.

P1 = 0,3at

(s)10101010101010101010

V(l)6,85,85,45,66,04,85,05,24,65,2

Tnh nng sut ca qu trnh lc Q.

C th nh sau:

(m3/s)

Lng nc lc ring:

Vi: (m2)(n: s bn ca thit b lc)Nn: (m)

P1 = 0,3at

V (m3)

6,85,85,45,66,04,85,05,24,65,2

Q (m3/s)

6,8.5,85,45,66,04,85,05,24,65,2

q (m)

34292728302425262326

(s)0101010101010101010

()0-5212-61133

/

-2-5105-1,6671010-3,3333,333

P2 = 0,6at

V (m3)

3,23,44,43,04,23,63,23,83,63,0

Q (m3/s)

6,46,88,86,08,47,26,47,67,26,0

q (m)

16172215211816191815

(s)0555555555

()015-76-3-23-1-3

/

51-0,71483,333-1,667-2,51,667-5-1,667

P3 = 1at

V (m3)

4,04,04,03,83,83,63,64,43,04,2

Q (m3/s)

1010109,59,599117,510,5

q (m)

20202019191818221521

(s)0444444444

()000-10-104-76

/

-4-41-0,57166,667

Xc nh cc h s C (m3/m2), K (m3/s) trong phng trnh lc.

Phng trnh lc vi p sut khng i c dng:

Tnh tng t nh trn, ta thu c bng sau:

CK

0,3atq1 v q2- 0,032- 9,86.10-5

q2 v q3- 0,028- 7,83.10-5

q3 v q4- 0,028- 7,56.10-5

q4 v q5- 0,029- 8,4.10-5

q5 v q6- 0,027- 7,2.10-5

q6 v q7- 0,025- 6.10-5

q7 v q8- 0,026- 6,5.10-5

q8 v q9- 0.025- 5,98.10-5

q9 v q10- 0.025- 5,98.10-5

Trung bnh- 0,027- 7,257.10-5

0,6atq1 v q2- 0,017- 5,44.10-5

q2 v q3- 0,02- 7,48.10-5

q3 v q4- 0,019- 6,6.10-5

q4 v q5- 0,018- 6,3.10-5

q5 v q6- 0,02- 7,56.10-5

q6 v q7- 0,017- 5,76.10-5

q7 v q8- 0,018- 6,08.10-5

q8 v q9- 0,019- 6,84.10-5

q9 v q10- 0,017- 5,4.10-5

Trung bnh- 0,018- 6,384.10-5

1atq3 v q4- 0,02- 9,5.10-5

q5 v q6- 0,019- 8,55.10-5

q7 v q8- 0,02- 9,9.10-5

q8 v q9- 0,019- 8,25.10-5

q9 v q10- 0,018- 7,875.10-5

Trung bnh- 0,019- 8,815.10-5

th. P1 = 0,3atq (m)0,0290,0270,0280,0300,0240,0250,0260,0230,026

/

-2000-500010005000-166710001000-33333333

P2 = 0,6atq (m)0,0170,0220,0150,0210,0180,0160,0190,0180,015

/

50001000-71483333-1667-25001667-5000-1667

P3 = 1atq (m)0,0190,0180,0220,0150,021

/

-4000-40001000-57166667

Cu hi chun b. 1. Nu mc ch bi th nghim?

Tr li:

- Lm quen vi cu to v nguyn l hot ng ca thit b lc khung bn.- Bit ch vn hnh, kim tra trc khi vn hnh thit b.- Vn hnh th nghim lc huyn ph CaCO3 trong nc vi p sut khng i.

- Xc nh cc h s lc theo s liu th nghim thu c.

2. Lc s dng dng lm g? Cho v d?

Tr li: lc s dng phn ring hay tch cc hn hp khng ng nht (lng rn) hay nh tng thnh hai h rn lng khc nhau.

V d: lc nc rau m sau khi xay, lc du sau khi p,3. Nu cc phng php to chnh lch p sut khi lc?

Tr li: cc phng php to chnh lch p sut khi lc l:

Lm dy hay lm mng vch lc bng lp b lc,

Thay i vn tc chy ca lu cht.

To p lc bn lc hay t my ht bn sn phm4. Lc c my ch , c c trng bng i lng no?

Tr li: lc c 2 ch lc: lc chn khng v lc p c t trng bng b mt lc.

Lc chn khng th b mt lc c i mi lin lc (co b lin tc).

Lc p th phi to lp b dy to thnh o lc5. Phng trnh vi phn lc v nghim ca n?

Tr li: phng trnh vi phn lc l:

t :lng nc lc ring (m3/m2).

Phng trnh (5) c vit gn li nh sau: q2+2.C.q=K.(

Vy nghim ca n l q6. Nu s th nghim lc khung bn ?

Tr li:

1. Khung khuy huyn ph

2. Bm huyn ph

3. Thit b lc khung bn

4. p k

7. Nu cu to nguyn l hat ng, u nhc im v phm vi s dng ca lc khung bn?

Tr li: Cu toMy lc khung bn gm c mt dy cc khung v bn cng kch thc xp lin nhau, gia khung v bn c vi lc. Huyn ph c a vo rnh di tc dng ca p sut ri vo khong trng ca khung. Cht lng qua vi lc sang cc rnh ca bn ri theo van ra ngoi. Cc ht rn c gi li to thnh b cha trong khung. Nguyn l hot ng: cho huyn ph vo 1 bn vch ngn ri to ra trn b mt lp huyn ph p sut P1, L dn huyn ph nhp liu ca khung v bn c ni lin to thnh ng dn nh ra ghp vi h thng cp liu. Nc lc chy ra t bn qua h thng ng ng v ly ra ngoi. B c gi li trn b mt vch ngn lc v c cha trong khung. Khi b trong khung y th dng qu trnh lc tin hnh ra v tho b. u im:_ B mt lc ln._ Dch lc trong v loi b c nm men._ Tm c th thay th d dng._ Lc c cn bn._ Khng cn ngi c chuyn mn cao.

Nhc im:_ Cn nhiu thi gian v sinh._ Phi thay th tm theo chu k._ Gi thnh tm cao._ Dch chy nhiu, phn b khng ng u._ Phi tho khung bn khi cn gim p sut.8. Trnh t tin hnh th nghim?

Tr li: Th nghim lc 1 cpBc 1: Pha 510g bt CaCO3 vo 17 lt nc vo x nha c huyn ph CaCO3 3% khi lng.

Bc 2: ng van V1, V2.

Bc 3: cho dung dch pha vo bn cha dung dch.

Bc 4: bt cng tc my khuy, khuy u hn hp CaCO3.

Bc 5: m van V3,V4, V5, V6.

Bc 6: m bm iu chnh V4 khi ng h p sut ch mc mong mun.

Bc 7: hng dung dch lc u C1 v ghi thi gian cho mi 1000ml. c bit ghi ch thi gian khng n nh.

Bc 8: lp li th nghim cho nhiu ln p sut khc nhau (3 ln)

Lc cp 2 tng t nhng m van V3, V4, V5, V8 ng van V6, V7. V hng dung dch lc u C29. K tn mt vi loi thit b lc ngoi lc khung bn?

Tr li:

Thit b lc p s dng ct lc

Thit b lc chn khng dng thng quay

Thit b lc ly tm.

Thit b lc p,

10. Nu cc phng php tng nng sut lc?

Tr li: cc phng php tng nng sut lc l:

Tng p lc lc.

Tng tc lc.

Gia nhit trong qu trnh lc gim nht

11. Nu cc yu t nh hng ti qu trnh lc?

Tr li: cc yu t nh hng ti qu trnh lc:

Vn tc lu cht lc.

p sut lc.

Lp b lc, tnh cht ca vch ngn.

Lp vi lc.

H thng lc hay thit b lc.

Trng thi ca cht lc, tnh cht ca huyn ph.

Nhit lc

12. Trnh by phng trnh lc khi p sut khng i v ngha ca cc i lng?

Tr li: phng trnh lc Khi p sut khng i

Trong

: nht (kg/ms)

V: th tch nc lc (m3)

S: din tch b mt lc (m2)

: thi gian lc c n nh trc

r0: tr lc ring (1/m2), tr lc lp b to thnh (1kg b kh/1m2 b mt)

X0= Va/V0: t s gia lng b m (m3/ lng nc lc (m3))

Rv: tr lc vch ngn (1/m)13. Nu phng trnh lc khi tc khng i v ngha ca cc i lng?

Tr li: phng trnh lc vi tc khng i: W=const (km hiu qu)

(N/m2)

(: nht (kg/ms)

V: th tch nc lc (m3)

S: din tch b mt lc (m2)

: thi gian lc c n nh trc

r0: tr lc ring (1/m2), tr lc lp b to thnh (1kg b kh/1m2 b mt)

X0= Va/V0: t s gia lng b m (m3/ lng nc lc (m3))

Rv: tr lc vch ngn (1/m)

14. Th no l lc p? Lc chn khng?S khc nhau gia lc p v lc chn khng?

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

Dng lnh

Dng nng

T

Tk

H = cont

EMBED Equation.3

P (mmHg)

Pb

Ph

(1

(2

EMBED Equation.3

1

2

3

1

1

2

3

1

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