htnl dc xang chk compatibility mode 9983

H thng nhin liu v t ng iu chnh tc C T

H THNG NHIN LIU V T NG IU CHNH

TC NG C T TRONG

Ngi dy: TS. KHNG V QUNG

B mn ng c t trong Vin C kh ng lc Trng HBK H Ni---------*****---------

Bi ging:


Mc ch mn hc:

Trang b nhng kin thc c bn v cu to, nguyn l hot ng, kim tra iu chnh, m hnh tnh i vi h thng nhin liu v b iu tc ca ng c xng v ng c diesel.

i tng mn hc:

Sinh vin chuyn ngnh ng c t trong nm th 4, sau khi c hc mn nguyn l ng c t trong


Vai tr ca HTNL trong ng c:

c lm vic m th tc tng p sut (p/)min, tng tnh kinh t v tng tnh hiu qu th nhin liu (h2 chy) cung cp cho c phi m bo chy ng lc v chy hon ton.

Ne th hin lng nhin liu cung cp i vi c diesel cn c xng c tnh theo lng hn hp.

Tnh kinh t e tng (ge gim) ni nn kh nng chuyn nhin liu thnh cng c ch (e-xng = 3320 ge = 260380 g/kW.h; e-diesel = 4330 ge = 200285 g/kW.h).

HTNL ng vai tr quan trng v c c chy ng lc, ht, v m hay khng l do HTNL. V vy, cn phi nghin cu HTNL m bo cc yu cu trn.

T ng iu chnh tc c: m bo cho c lm vic n nh 1 tc nht nh khi ti ngoi thay i.


100% 21% 15%

Conventional ICE Vehicle

Theo hng Ford

Ch 15% nng lng c dng quay bnh xe


- C xng ( ch nm trong khong t 0,6 n 1,2), v hn hp kh c hnh thnh t bn ngoi xylanh (tr ng c phun xng trc tip). Do vy, iu chnh ti trng phi dng phng php iu chnh lng hn hp cung cp bng bm tit lu hay cn gi l bm ga trn ng np.

- C diesel ( rt rng t 1,210), do hn hp c hnh thnh t bn trong xylanh, nn hn hp ca diesel khng ng nht, nn iu chnh ti, ngi ta dng phng php iu chnh cht thay i thnh phn kh hn hp bng cch ch thay i lng nhin liu cung cp.


Ni dung mn hc

HTNL C Xng HTNL C Diesel TC TC

Khi qut v to hn hp trong C xng

HTNL C xng dung CHK

B CHKB CHK hin iCu to, thit k b

CHKCc cm khc ca

HTNL dng CHKHTNL C phun

xng (phun xng nhiu im iu khin bng c kh hoc iu khin bng in t, phun xng n im)

Nhim v, s h thng

Bm cao p (bm dy, bm phn phi)

Vi phunMt s loi vi

phun v bm cao p khc

Vi phun bm P-TCc cm khc ca

HTNLHTNL Common

Rail

Tnh n nh trong ch lm vic ca CT

iu kin cn lp T trn CT

Phn loiT c kh trc tipT gin tipc tnh tnh ca phn t

cm bin b TCc thng s nh gi

trng thi tnh ca b TTnh ton tnh hc ca b

b T c khng hc ca b T c

kh trc tipH thng t ng iu

chnh tc ng c


Ti liu:

1. Gio trnh chnh:

H thng nhin liu v t ng iu chnh tc ng c t trong, Nguyn Tt Tin, V Th Lt.

2. Ti liu tham kho:

- Nguyn l ng c, Nguyn Tt Tin, Nh xut bn Gio dc, H Ni 2000

- Kt cu v tnh ton ng c t trong (tp III), Nguyn c Ph, Nguyn Tt Tin

- H thng phun xng in t dng trn xe du lch, Hong Xun Quc, Nh xut bn Khoa hc k thut

- Cc bi bo trong nc v quc t

- Sch ting anh.


nh gi kt qu hc tp:

- thc hc tp (thi gian ln lp) 10%

- Kim tra gia k 20%

- Bi tp hoc bo co 10%

- Kim tra cui k 60%

Nhim v sinh vin:

- D lp y (ng gi, i mun c truyn thng coi nh b vng)

- c ti liu


CHNG 1: H THNG NHIN LIU NG C XNG

1.1. Khi qut v to hn hp trong ng c xng

1.1.1. Yu cu i vi h2 v cc p2 to h2 ca ng c xng

- Cht lng qu trnh chy, c/s, h/s ng c ph thuc nhiu vo qu trnh to hn hp.

- C thnh phn hn hp phi ph hp vi tng ch lm vic ca ng c (: h s d lng khng kh (); = GKK-tt/Gnl*L0.

- Hn hp phi ng nht trong 1 xylanh v ng u gia cc xylanh.

- p ng cc yu cu trn th hin nay i vi ng c xng c cc phng php hnh thnh hn hp nh sau:

C xng dng CHK, hoc loi phun xng trn ng np th hnh thnh hn hp bn ngoi.

C phun xng trc tip th to hn hp bn trong (GDI).

- Cc phng php trn u c thit b v bin php c th m bo cht lng hn hp.


1.1.2. Nhng yu t nh hng n cht lng hn hpa) Thi gian

Nu thi gian di th qu trnh hnh thnh hn hp cng tt, hin nay c xu hng tng Ne bng vic tng nc, nh vy s gim thi gian hnh thnh hn hp nhng do tc lu ng ca dng kh np mnh lit hn (sng p sut) dn n to iu kin ci thin c qu trnh hnh thnh hn hp nc cao.

b) Nhit mi trng

- Nu nhit ln s to iu kin co nhin liu bay hi thun li cho qu trnh hnh thnh h2. Tuy nhin khi nhit ln qu cng khng tt v lm nh hng n lng kh np thc t vo xylanh.

- C th ly v d: xe my dng xng 92 v 95, cc thi tit khc nhau.

- i vi c xng vic sy nng kh np mi nhin liu d bay hi bng cch sau:

Sy nng bng kh thi Sy nng bng nc lm mt, khi nc lm mt qua lm mt cho

ng c vi nhit 90950C


c) Kt cu ng ng np v CHK

- ng ng np phi thanh thot v m bo cc ng ng t CHK vo cc xylanh phi tng ng nhau.

- CHK i vi 1 bung hn hp cp cho nhiu xylanh s khng m bo s ng nht v thh phn hn hp gia cc xylanh, do phi dng nhiu bung hn hp.

d) Thnh phn, tnh cht nhin liu

Nhin liu c nhiu thnh phn chng ct nh, d bay hi, to hn hp ng u, hm lng hi cao, dn n thnh phn hn hp s tt.


a) Yu cu

1.1.3. Yu cu v phn loi

Cung cp vi thch hp vi tng ch lm vic ca c. Phn ln nhin liu trong hn hp dng hi xng, phn cn li

c x ti dng ht c kch thc rt nh.

H s gia cc xylanh phi ng u nhau.

b) Phn loi

HTNL c xng dng ch ho kh (cacbuarat)

HTNL phun xng in t.


1.2. H thng nhin liu ng c xng dng ch ha kh (CHK)1.2.1. Nhim v

-D tr, lc sch v cung cp nhin liu

-Chun b v cung cp hn hp xng v khng kh m bo s lng v thnh phn hn hp ph hp vi tng ch lm vic ca c

1. ng xng, 2. phu xng, 3. ng thng kh, 4. thng xng, 5. thc o mc xng, 6. kha xng, 7. ng dn xng, 8. bnh lc xng, 9. bm chuyn xng, 10. cc lng v li lc tinh, 11. bnh lc khng kh, 12. bnh gim m, 13. b ch ha kh, 14. b hn ch tc cc i ca c, 15. phao ch mc xng, 16. nt tho xng, 17. u ng ht.

1.2.2. S


Zicl, chi tit c ch to chnh xc c th tit lu nh lng lu lng xng ht ra ng nh thit k.

Lng hn hp i vo c c iu chnh nh bm ga 7.Vnl=56 m/s; Vkk=2530 m/s.


1.3.1. Yu cu v phn loi i vi BCHK

- Cung cp hn hp m bo lng v thnh phn ph hp vi tng ch lm vic ca c.

V cht ch ti nh v trung bnh th yu cu lm vic tit kim nht ge min.

Khi ton ti phi t c cng sut Ne max (hn hp m) ch khng ti phi m bo ng c lm vic n nh

- Trong mi iu kin mi trng v p sut, nhit mi trng khc nhau phi d khi ng.

- D iu chnh theo trng thi k thut v iu kin s dng c.

- Cu to n gim chc chn, s dng, bo dng v sa cha d dng.

1.3. B CHK

a) Yu cu


- CHK khng c bung phao: loi ht, loi phun dng cho c lm vic cc v tr khc nhau. V d nh ng c my bay, my ca tay

- CHK c bung phao; hin nay c s dng ph bin, ch yu l loi ht xng (ht xung v ht ln, ht ngang)

b) Phn loi


1.3.2. BCHK n gim

a) S v cu toBao gm: bung phao (bu xng) cha nhin liu. Phao xng (bng nha hoc ng) cng van kim m bo mc xng trong bu xng khng thay i v mc y xng lun m bo cch vi phun 1 on h = 58 mm, xng khng t tro ra khi BCHK khng lm vic.

4

6

7 5

Zchl l chi tit chnh xc, n m bo quan h n nh gia lng nhin liu i qua zchl v chnh p trc v sau ca zchl.

Vi phun (ng ng dn nhin liu) ming vi phun t hng.

Hng l tit din nh nht c nhim v to chn khng (ph = po - ph) ht xng (ph > x.nl.g) vi x = h.

Bm gi ch lm nhim v khi khi ng

Bm ga; v tr ca bm ga chnh l lng hn hp v chnh l ch ti ca ng c.


b) Nguyn l lm vic

ht c nhin liu th ph > ct p ng vi chiu cao x.nl.g, tc l ph > x.nl.g y x = h vi ph = p0 ph

Trong :

g: gia tc trng trng

nl: trng lng ring ca nhin liu

x = h: l khong cch t mc nh/l bung phao v ming v/ph

ph: l p sut tng i

ph: l p sut tuyt i

Nu ph cng tng th lng nh.l ra cng nhiu

X ti nhin liu to thun li cho vic hnh thnh hn hp. Qua kt qu kho nghim thy, mun x ti nh/l th Vkk > Vnl khong 56 ln. C th Vnl = 56 m/s v Vkk qua hng = 2530 m/s th nh/l c x ti hon ton.

Ha trn hn hp: nhin liu x ti s bay hi ha trn vi khng kh to thnh hn hp n cui hnh trnh nn th kt thc.


c) c tnh ca BCHK n gin

- Thnh phn ca hn hp th hin qua , s thay i theo ch lm vic ca c. nh gi s hot ng ca BCHK khi thay i ch lm vic ca c cn phi ti c tnh ca BCHK. K c i vi BCHK n gin cgn cn phi xt ti c tnh ca n.

- nh ngha:c tnh ca BCHK l quan h ca h s d lng khng kh vi mt trong nhng thng s c trng cho lng hn hp np vo ng c (lu lng khng kh Gk, d chn khng ti hng ph, cng sut c ch ca c)

Tc l:

= f(ph) chn khng

= f(Gk) lng kk np vo

= f( m bm ga)

= f(Ne) cng sut

= f(ph ti c)


- Xy dng c tnh = f(ph)

+) Theo nh ngha

= Lng kk thc t np vo c/lng kk l thuyt t chy nh/l

= Gk/Gnl.L0

Gkk: Lng kk thc t np vo c

Gnl: Lng nh/l

L0: Lng kk l thuyt cn thit t chy ht 1 kg nh/l

L0: (1/0,23).[(8/3).C + 8H O] (kg/kg nh/l)

+) c = f(ph) th cn phi xc nh Gk, Gnl = f(ph)

Khi gi thit dng mi cht khng chu nn, chy n nh v lin tc


+) Xc nh Gk:

Do c hot ng c tnh chu k, nn lu ng ca kk qua hng v xng qua vi phun ca BCHK c tnh dao ng, v thc cht l dng chy n nh. Khi chuyn t ng c 4 k sang ng c 2 k hoc tng s xylanh ni vi 1 BCHK s gim bt tnh dao ng ca dng chy. Nu 4 xylanh ca c 4 k hoc 2 xylanh ca c 2 k ni vi BCHK s khng thy r tnh dao ng ca dng chy. V vy c th coi dng chy ca xng v kk trong BCHK nh 1 dng chy n nh. Mt khc chn khng ti hng BCHK ph thng < 2000 mm ct nc ( 20 kPa 0,02 MPa 0,2 at) khi c hot ng tc cc i v m ht bm ga. Nh vy vi Ph bin ng t 0 n 20 kPa c th b qua tnh chu nn ca kk v coi lu ng ca kk nh l cht lng khng chu nn. Vy ta c th gi thit chuyn ng ca dng kh l n nh v lin tc, v khng chu nn, nn ta c:


+) Xc nh Gk: (tip)

Theo phng trnh Becnuly vit cho dng mi cht kh qua mt thong bu xng v mt hng khuych tn.

khhhk pfG .2

Trong :

h: l h s lu lng, khi kk qua hng c tit din l f hng.

Vi h = v.b; v: h s tc , b: h s bp dng,

vi

h

hb f

f min

fh: tit din hng ht, k: khi lng ring ca kk


+) Xc nh Gnl:

Gi thit nh/l qua hng l n nh v lin tc. Cng s dng ph/trnh becnuly vit cho dng chy qua mt thong bu xng v mt zichl.

Kt qu:

nlnlhddnl gxpfG ).(2

fd: tit din ca zchl, d: h s lu lng ca xng qua zchl

Vy t Gk v Gnl theo ph thay vo biu thc tnh ta c:

nlnlhkh

d

h

d

h

gxp

p

f

f

L

2

1

0

ghpp

f

f

L nlh

h

nl

k

d

h

d

h

2

1

0

Ta c:

Vi h = x


shf

f

LK

nl

k

d

h

0

1

Kghp

p

nlh

h

d

h

2

t

Do :

Do ph.nl.g qu nh v th c th coi 1

ghp

p

nlh

h

Kd

h

Vy Qua y ta thy c theo ph th cn phi c quan h nh sau:

hd

h pf

Qua thc nghim ta xc nh c quan h ca

hd

hh

pf

pf

(th hin lng kk np vo c)

(th hin lng nhin liu phun ra)


hd

ph

h

d

h/d

+) Gii thch h = f(ph)

Tng ph tng cng sut tc l tng lng hh np vo c.

Ban u h tng nhanh sau tng chm. L do lc u cc lp kk chuyn ng c tng

dn (lp kk dnh vo thnh). Khi ph tng ti mt mc no th ht lp hh chnh. Do

khi ph tng na th h vn bng hng s.

Khi ph tng kh cao th h gim 1 t l do tc ca dng kh khng ph hp vi tit din hng (to xoy) tng sc cn h cng ln ngha l Gk cng ln


+) Gii thch d = f(ph)

Ban u ph tng th d tng nhanh sau d tng chm do khi ph tng dn ti lp dnh chuyn ng tng ln sau lng ht. d khng gim v khng c xoy. Ch : d th hin lng nh/l Gnl ph thuc vo kch thc,hnh dng hnh hc, nhn bng ca zchl, v th ring i vi zchl nh/l sau khi ch to phi:

Kim tra kch thc

o lu lng trn thit b o hn hp

Nhn xt: Nu nh t s l/d = 12 th s nh hng ca hnh dng hnh hc, nhn bng ca zchl, ti d tc l Gnl l n nh. Nu l/d khc 12 th cc nhn t trn nh hng ti lu lng l kh ln buc chng ta phi kim tra kch thc v hnh dng.

Tm li ta c quan h:

h/d = f(ph) v = C.(h/d) tc l = f(ph)


Nhn xt: Khi tng ph tc l tng ga (tng ph ti, tng cng sut)

dn n gim ngha l hh m dn. Nh vy c tnh BCHK n

gin khng th p ng yu cu s

dng c v ch ph ti ln kh hh

mi m cn ch ti nh v

ph ti trung bnh th hh rt long.

Nu iu chnh BCHK to c hh c thnh phn cn thit cc ch

ph ti ln th khi cho c chy cc ch khng ti hoc t ti kh hh s

rt long, dn ti c s cht my v ngc li nu iu chnh BCHK to

c hh c thnh phn cn thit cc ch t ti, th khi c chy ch

ton ti kh hh s m, vt ra ngoi gii hn chy ca nh/l, c s cht my,

v vy cn c ng c tnh khc?

Gk

ph

Gk

Gnl.L0

Gnl.L0

=1


1.3.3. c tnh l tng ca BCHK

a) nh ngha

ng c tnh l tng ca BCHK l hm s biu th quan h gia h s d lng khng kh ti u (tt nht) ca kh hn hp v lu lng (Gk) ca khng kh np vo ng c t/ = f(Gk).

b) Xy dng c tnh

-Mun xy dng c ng c tnh l tng ca b CHK, trc tin phi xc nh cc h/s th hin mi quan h gia cng sut v sut tiu hao nhin liu ca c (ch tiu k thut v kinh t) theo thnh phn kh hh khi c chy s vng quay nht nh v khng thay i v tr tay ga: Ne = f(); ge = f().

fg

fN

e

e c xc nh ly trong phng th nghim vi iu kin:

nc = const

Bm ga c nh

Mun thay i bng cch thay i tit din zchl.

-


- Xc nh c +) Lc tc dng P (gy ti cho c)

+) S vng quay ca c

Xc nh c cng sut ng c

- o tiu hao nhin liu Gnl, v lng kk np Gkk

ge = Gnl/Ne

= Gkk/Gnl.L0

- Khi thay i v tr bm ga ta c v s cc ng

fg

fN

e

e

Nhn xt:

+) Khi bm ga m hon ton Ne-max < 1; ge-min > 1

+) Khi ng dn bm ga (gim ti) Ne-max gim; ge-min gim

+) Khi ni cc im Ne-max (1,2,3) th ta c ng a, ng a l ng iu chnh CHK c c Ne-max mi v tr bm ga.


6

5

4

3

2

1b

Ne

10

a

I/Ne

II/

III/

9

8

I 100%

II 70%

III 40%

1


+) Khi ni cc im (4,5,6) ta c ng b, l ng iu chnh CHK c lm vic kinh t nht.

+) Do vy ta nn s dng vng gia 2 ng a v b v nu ngoi ng th Ne gim v getng. Vic la chn ng a hoc b ty thuc vo lnh vc khai thc c.

+) Tm li: s dng c tt nht (tnh hiu qu v tnh kinh t)

Th ti nh v trung bnh cn c lm vic kinh t nht ge-min, tcc l ng b.

Khi m hon ton bm ga (ton ti) yu cu c pht ra cng sut ln nht Ne-max ng a.

ng c l ng quan h t = f(Ne) khi nc = const v ng vi 1 ng Ne = f() th ch c 1 Gk nn hon ton c th i c thnh = f(Gk)


c

b

Ne

a

Gk

Gk

n31

2n2

n1

n3 < n2


Khi thay i nc ta s c v s ng = f(Gk)

nhnh ti nh ng = f(Gk) st nhau

Ti ln khi tng nc dn ti tng hh nht dn

T ta v ng bao s 2 l ng iu chnh CHK c lm vic ch tit kim nht (ge-min) mi v tr bm ga v mi v tr s vng quay. Cn ng s 1 l ng iu chnh CHK c lm vic hiu qu nht Ne-max mi v tr tay ga v mi s vng quay.

n gim th chng ta c 1 ng trung bnh, tc l c tnh l ca BCHK hot ng cc tc khc nhau.

Ta thy: ti nh v trung bnh (bm ga m nh v trung bnh): c Gktng (ti tng) dn ti tng (hn hp nht dn)

ti ln (bm ga m ln) Gk tng (ti tng) dn ti gim (hh nht dn)


c) So snh c tnh l tng c tnh ca BCHK n gin

Gk

L tng

1

n gin

- BCHK n gin khng th chun b hh cho c vi thh phn tt nht mi ch lm vic ca c

- c c ng c tnh st vi ng c tnh l tng th trn BCHK n gin cn phi b sung thm mt s c cu v h thng m bo cc yu cu sau:

+) Ch khng ti: hh m = 0,40,8, xng phun ti, phn b u.

+) Khi bm ga m tng i rng, cn hh long, = 1,071,15

+) Khi bm ga m hon ton (Ne-max), cn hn hp m = 0,750,9

+) Khi khi ng, cn hn hp m = 0,30,4

+) Khi tng tc, phi cung cp thm nh/l trnh hin tng hh b nht


1.3.4. Cc h thng ca BCHK

A) Cc h thng phun chnh

HTPC nhm cung cp hh cho c lm vic kinh t nht trong vng lm vic ph bin ca c, trn c tnh l tng ca bchk htc phi cung cp hh nht dn khi ti trng tng. Trong thc t c nhiu loi htc, ph bin l 1 s loi sau:

- htpc gim chn khng sau gicl chnh.

- htpc c gicl b sung

- htpc c iu chnh tit din zchl chnh kt hp vi HT khng ti

- htpc iu chnh chn khng hng


1) H thng phun chnh c gim chn khng sau zchl chnh

H thng chnh gim chn khng sau zchl chnh

1

2

kk

p0y H

h

ph3

Ngoi zchl nhin liu 1 cn c thm zchl khng kh 2


- Nhn thy:

+) chn khng ti hng ph = p0 ph

+) chn khng sau zchl chnh pd = p0 pd (pd p sut tuyt i sau zchl chnh).

+) Lng nhin liu qua hng vi tit din f

pfGnl 2

: h s lu lng

p: chnh p sut trc v sau tit din

: mt ca mi cht (nhin liu)

- Nguyn l lm vic: chia lm 3 giai on

+)Giai on 1: Khi ph x.g.nl ; nhin liu khng ht ra c, tc l Gnl = 0, =


+)Giai on 2: Khi ph c gi tr: x.g.nl < ph (H+x).g.nl; khi ny HTPC lm vic nh b CHK n gin, tc l ph tng dn ti gim, lc Gkkv Gnl tng

C th: khi ph tng th nhin liu giai on 2 qua zchl 1 (t ng 3) qua vi phun v vo hng, lm cho y tng, khi khng kh qua 2 vo 3 v coi p3 p0 nh vy pd = y.g.nl.

211

)( 2 nlddII

nl gyfG

Nhin liu qua zchl 1 c tnh theo cng thc sau:

d1: h s lu lng ca nhin liu khi qua zchl 1 c tit din l fd1

Cui giai on 2: th y = H lc p3 p0 (pd = H.g.nl), v lng nhin liu qua zchl 1 s l:

211

)( 2 nlddIIC

nl gHfG


+)Giai on 3:

ph (H+x).g.nl;

u giai on 3: chn khng ph tng t ngt, ht mnh lm cho khng kh khng kp in vo qua zchl 2 lm cho p sut sau zchl chnh gim t ngt lm cho pd tng t ngt, t pdg gii hn. Lm cho Gnl tng t ngt, lc

nlnldgddnl gHpfG )(2

3

Trong giai on 3: khi ph tng th Gnl1 tng v khng kh vo qua ng 2, cun theo cng nhin liu v vo hng, c hin tng bt xng. Nh vy Gk2 lm cn tr s tng pd.

(pd tng chm i so vi CHK n gin, tc l khi ph tng dn n Gnl tng chm, cho nn tng (theo c tnh l tng ca BCHK)


pd =k.ph trong

222

2

2

)()(

)(

ddhh

hh

ff

fk

fd2: zchl khng kh

Nhn xt:

+) Nu fd2 = 0, ngha l bt kn zchl 2, k = 1, dn n pd = ph, HTPC lm vic nh CHK n gin.

+) Nu fd2 = , ngha l b zchl 2, k = 0, dn n pd = 0, pd = p0, khi lng nhin liu cung cp ph thuc vo cao H. v lng nhin liu c xc nh nh sau:

constgHfG nlddIIInl

2111 2

+) Ta c th hon ton la chn fd2 c pd hp l, Gnl.L0. v c theo mun.


- c im:

+) Vi loi HTPC ny l HTPC duy nht m ta c th thc hin c phun bt xng, phun ti, bay hi tt mi ph ti v mi s vng quay.

+) BCHK ny n gin, chc chn, d iu chnh, s dng hu ht trn cc BCHK.


H thng chnh c zchl b sung

H

y

ph

kk

12

2) H thng phun chnh c zchl b sung


Nhn thy:

- Zchl chnh 1 v vi phun 4 lm vic nh CHK n gin

Gnl1.L0 = f(ph)

- Zchl b sung 2 v vi phun 5 v ng kh 3 chnh l HTPC c iu chnh chn khng trong trng hp b zchl khng kh

Gnl = Gnl1 + Gnl2 Gnl.L0 = Gnl1.L0 + Gnl2.L0

- Nguyn l lm vic:

+) Giai on 1:

0nlnlh Ggxp


+) Giai on 2:

nlhnl gxpgxH Khi nay, ht lm vic nh BCHK n gin, tc l khi ph tng dn ti Gnl2tng v ch ph thuc vo y

V do 222203 2 nlddnl gyfGpp

Cui giai on 2:

nlddnl gHfGppHy 222203


+) Giai on 3: nlh gxHp

0003 dd pppppHy

constgHfG nlddnl 2

222 2

Cui II

nlnlhddhnl gxpfLpfLG 211001C hnlnlnl pfLGLGLG 02010

Nh vy, khi tng ph s dn ti tng v Gkk tng nhanh

Nhn xt:

+) Do c zchl b sung 2 nn c sai s ph ca CHK

+) iu chnh tit din ca 2 zchl fd1 v fd2 c c theo mun l rt kh khn.

+) Kt cu phc tp, nhiu chi tit

Loi ny c c s dng nhng khng rng ri nh loi trn


1

2

3

4

5 6

3) HTPC thay i tit din gicl kt hp vi h thng khng ti:

a) Dn ng bng c kha) Dn ng bng c kh

HTPC: Zchl chnh 1, kim diu chnh zchl 6, l xo v cn iu chnh

HTKT: Zchl khng ti 3, zchl khng kh 4, ng dn chn khng sau bm ga


- Nguyn l lm vic

+) i vi HTPC

L xo lun c xu hng y kim s 1 i xung

Khi ph tng (tng m bm ga), qua tay n (cn iu khin) nng kim s 1 ln, lm cho fd1 tng, tc l Gnl1 tng, dn ti hh m dn ( gim xung)

+) i vi HTKT

Nhim v: Ch m bo cho c lm vic ch khng ti. Trong trng hp ny HTKT lm vic sut cng vi HTPC.

ch khng ti bm ga ng kn, chn khng sau bm ga ln ht xng qua zchl 3 v kh qua zchl 4 to thnh bt xng v phun vo sau bm ga, dn ti hh m ( nh)

Khi lm vic c ti (cng vi HTPC) th khi ph tng (tng m bm ga) lm cho pg gim, dn ti Gnh-5 gim, hh nht dn ( tng dn).

Kt hp li ta c, khi tng ph (tng m bm ga) th tng.


Nhn xt:

V tr kim 2 (tit din ca zchl 1) ch ph thuc vo v tr bm ga m khng ph thuc vo tc c. xt ti nh hng ca tc c ta xt loi dn ng bng chn khng.

b) Dn ng bng chn khngb) Dn ng bng chn khng

HTPC: Zch l chnh 1, Kim iu chnh 2, Pittng 8, l xo 9, ng dn chn khng sau bm ga 10.

HTKT: Ging nh phn dn ng c kh)


- Nguyn l lm vic:

+) Nu cng 1 v tr bm ga th khi lm vic s vng quay nh, chn khng sau bm ga s nh (pg nh), l xo s 9 y pittng 8 i ln, kim 2 i ln, dn ti fd1 tng, lm cho Gnl1 tng, do nh (hn hp m).

+) Nu s vng quay ln, pg tng, thng c sc cng l xo 9, pittng 8 i xung, kim 3 i xung, tit din fd2 gim, Gnl gim, dn ti tng (hn hp long).

- Nhn xt:

Khi tng nc tng lm cho (hn long dn), c th xy dng c c tnh l tng nhng nu nc tng qu ln th lm cho pittng 8 v kim 2 i xung

qu lm cho fd2 nh qu, qu ln (hh qu nht). khc phc hin tng ny phi c cn 4 dn ng t bm ga hn ch nhc im ny. Nh vy

c th kt hp dn ng bng c kh v chn khng


c) Dn ng bng c kh v chn khngc) Dn ng bng c kh v chn khng

- Vi tc lun ln ta dng c kh.

- Vi nc thay i th cn kt hp c 2.

- Sau mt thi gian lm vic, kim s 2 mn, cn phi iu chnh li.


4) H thng chnh iu chnh chn khng hng:

a) b) c)

Khi bm ga m ti mc no nhm tng ti, chn khng hng khuch tn ln s m cc l l xo hoc van b sung khng kh lm cho hn hp nht dn.

-a thm khng kh vo hng

-Thay i tit din ca hng


a) Loi a thm khng kh vo sau hnga) Loi a thm khng kh vo sau hng

- Khi m bm ga cn nh tc khng kh

pk cn nh, ct p tc nh khng thng c

sc cng ca l xo, vn ng. Lc HTPC lm

vic nh BCHK n gin, ngha l tng m

bm ga (ph ti tng), lm cho gim.

- Khi m bm ga ln ti 1 mc th tc

pk ln, ct p tc ln thng c lc cng

l xo, y l xo 1 m, khng kh s qua hng v

xung quanh hng, lm cho ph tng nhng tng chm, tc l lng nh/l c ht ra khi vi

phun cng tng chm, dn ti tng (hn hp nht).

phphg


a) Loi thay i tit din ca hnga) Loi thay i tit din ca hng

- Nu nh m bm ga nh, vn tc ca pk nh, ct p tc khng thng c l xo, lc tit din hng l min lm vic nh CHK n gin.

- n khi tng m bm ga, tng, n lc m ln, vn tc kh ln, ct p tng ln, l xo l b nn li, tit din hng tng lm cho ph tng chm li, dn ti Gnl tng chm li, tng.

Nhn xt:

- Dng loi ny th dn n tn tht ct p thng c lc cng ca l xo n n lm gim h s np v.

- la chn lc cng ca cc l xo c ph hp l kh khn, sau thi gian lm vic, l xo khng chun na.

c)


B) Cc h thng ph

1) H thng lm m (HT tit kim)

- S cn thit:

+) Khi m hon ton bm ga (ton ti), cn pht ra cng sut ln nht (Ne-max), tc l cn phi cung cp thm nhin liu hh m ( = 0,80,9).

+) Thng cung cp t 80% m bm ga. T tng ph s lm gim.

- p ng c yu cu trn, BCHK cn c h thng lm m

+) Dn ng bng c kh

+) Dn ng bng chn khng


1 2

34

5

H thng lm m dn ng c kh

m bmhn hp

%

a) H thng lm m dn ng kiu c kh

c ch ti nh v trung bnh, bg 5 m cha ln, nn ch c h thng chnh lm vic (hh nht dn khi ti tng).

Khi bg m ln, qua n dn ng 4, kim iu chnh 3 c nng ln, lm tng tit din lu thng qua gicl 1 b sung thm nhl cho htchnh lam vic.

HTLD dn ng kiu c kh c u im l n gin, nhng c nhc im l thi im bt u lm m ch ph thuc vo m bm ga m khng ph thuc vo tc vng quay ca c. Do vy nh hng n c tnh ti ca c ch ny.

Cng sut ca c tng do lm m ti 80% m bm ga tr i. Khi tc ln cng sut tng nhanh theo m bm ga nn lm m 80% l hp l (ng 1), cn ch nh, cng sut ca c tng chm (ng 2) nn khi lm m 80%, cng sut tng rt t, Do vy nn lm m sm hn (khong 50%).


b) H thng lm m dn ng kiu chn khng

1 2 3

4

5

6

7

H thng lm m dn ng chn khng

- Khi bg m nh, chn khng sau bg ln, tc dng thng sc cng l xo 7, ko piston 6 i ln, kim 4 di xung v gicl lm m 2 ng nh. Khi ch c htc cung cp hn hp nht dn.

- Khi bm ga m ln, chn khng sau bg gim, l xo 7 y piston 6 i xung, van 4 i ln, gicl lm m 2 b sung thm nhl vo htc.

- chn khng sau bg khng nhng ph thuc vo m bg m cn ph thuc vo n. Khi n tng, chn khng sau bg tng. Do , thi im bt u lm m khng ch ph thuc vo m bg m cn ph thuc vo n.

- Trong trng hp n thp, khi bm ga m ln, chn khng sau bg nh, piston 6 i xung, gicl lm m 2 b sung thm nhl cho htc. y l u im ca ht, tuy nhin n nh ca ht ny km. V vy mt s ht kt hp c 2 ht trn.


2) H thng khng ti

1

3

6

7

2

4

5

1. l cung cp kh hh, 2. l chuyn tip, 3. vt iu chnh hh, 4. ng hh, 5. gicl khng kh,

6. gicl nhin liu, 7. vt knh ga.

- HTKT m bo sao cho ng c lm vic n nh ch khng ti.

- Khi c lv khng ti, bm ga ng kn, lu lng kk qua hng khuch tn nh, chn khng ti y nh, kh nng ht xng v ho trn xng vi kk km. Do ht khng c kh nng cung cp hn hp cho c chy khng ti. Trong khi chn khng sau bm ga ln nn c tn dng ht xng ra hng khuch tn v to hn hp cho c chy khng ti.

- Xng ht qua gicl 6 cng khng kh qua gicl 5 to thnh hn hp trong ng 4 (dng nh tng) v a vo hng qua l s 1. Khi c chuyn t ch khng ti sang ch c ti, bm ga m to dn. chn khng sau bm ga gim, dn ti lng hn hp qua ht khng ti gim.

- Khi ht chnh cha lm vic (v chn khng ti hng cn nh - lm c b cht my). khc phc, trong ht khng ti c l chuyn tip 2 (khi l 2 sau bm ga). Khi ch khng ti l 2 c tc dng b sung thm kk)


3) H thng khi ng

H thng khi ng

1

2

3

45

6

Khi khi ng, n thp (50100 v/ph), dn n tc dng kh qua hng nh, nhl phun vo t v cht lng phun km. Hn na khi c ang trng thi lnh nn xng kh bay hi v to thnh lp mng trn ng ng np, hn hp thc t to thnh rt long v c kh khi ng. V vy, khi ng c d dng phi cung cp thm nhl lm m hn hp.

- Khi khi ng bm gi 5 ng, chn khng ti hng v sau bm ga ln, nn c htc v htkti u lm vic, v vy hn hp cung cp ch khng ti l m.

- Khi c hot ng bm gi 5 m, nhng trc th van 4 m b sung khng kh, trnh cho hn hp qu m.


4) H thng tng tc

H thng tng tc

12

3

4

567

8Khi cn thit phi tng nhanh n hay ti trng c phi m t ngt bm ga. Khi y, lng khng kh vo c tng nhanh nhng lng nhl khng tng kp do qun tnh ca xng ln hn nhiu so vi qun tnh ca khng kh nn hn hp nht i t ngt c th lm cht my.

-Khi bg m t ngt, qua n dn ng 2 v l xo 6 y piston 4 i xung. p sut di piston 4 tng ln t ngt, van 3 ng li, nhl khng tr li bung phao m nng van 8 ln ri phun vo hng khuych tn qua vi phun 7, b sung cng bc mt lng nhl cho qu trnh tng tc c.- Khi tng ti t t, bg m chm, nhl lt qua van bi 3 tr li bng phao, qu trnh bm tng tc khng xy ra. Khi ng bg, piston 4 i ln, nhl qua van bi 3 np vo khng gian bn di piston 4.- Trong qu trnh m t ngt bm ga, l xo 6 b nn li. Khi qu trnh ny kt thc, l xo s gin ra t t c tc dng ko di qu trnh phun nhl mt thi gian na. Do c th trnh c hin tng c r my ln t ngt ri cht my do hn hp li nht i t ngt v htc cha kp cung cp nhl theo yu cu ca c.


1

2

3

5) C cu hn ch s vng quay ca ng c

Khi c lm vic, c th xy ra trng hp vmt l do no sc cn bn ngoi gim hoc mt t ngt (v d gy trc truyn cng sut hoc chn vt ca tu thu nh ln khi mt nc do sng to...). Ngi vn hnh c trong trng hp nh vy cha phn ng kp ng bt bm ga nn c chy khng ti vi tc vng quay rt ln, lm tng mi mn v c th lm h hng cc chi tit chuyn ng do lc qu tnh qu ln. t ng gim tc vng quay trong trng hp ny, mt s b ch ho kh c trang b c cu hn ch n.

Khi n tng vt qu mt gi tr no trong lc bm ga m to, dng khng kh vo c vi n cao nn c ng nng ln s tc dng ln mt vt trn bm ga 2 thng sc cng l xo 1. Bm ga do c ng bt nn n gim. Trong thc t cn c cc c cu hn ch n lm vic trn c s tn hiu v chn khng hay in t...


1

2

3

4

5

6

6) C cu hiu chnh BCHK theo cao

Khi ng c lm vic cao cng ln so vi mc nc bin, mt khng kh cng gim dn n lng khng kh thc t gim i v hn hp cng m ln. Tuy nhin, do lng khng kh np gim dn n gim cng sut ca ng c ng thi lm tng c hi trong kh thi. khc phc hin tng trn, trong mt vi b ch ho kh c trang b c cu hiu chnh theo cao.

Tit din thng qua ca gc l 1 c iu chnh qua ba r mt 3 thng qua kim 2. Khi ln cao, p sut kh tri gim, ba r mt gin n y kim v bn tri ng bt gc l 2, do gim lu lng nhin liu mt cch tng ng vi gim lng khng kh np.


1.4. Cu to, thit k BCHK

1.4.1. Bung hn hp

L b phn quan trong, l khng gian t bm ga ca BCHK

db

- Kch thc bung hn hp (db) l kch thc m ngi ta da vo chn BCHK lp cho ng c v db quyt nh lng hn hp khng kh v nhin liu vo ng c. Chn db v khi chn CHK lp cho ng c th phi xut pht t vtb ca hn hp khi i qua bung hn hp.

)/(750...

....2

smd

nivv

b

vhtb

: h s qut bung chy

: s k

Theo kinh nghim s dng, nu ng c t cc ch tiu kinh t th:

vtb = 4060 m/s (4 xylanh, bung hn hp)

vtb = 2030 m/s (1 n 2 xylanh)


Tnh ton s b: db:

1000.

nVad hnb

an: h s dao ng ca dng chy v ph thuc vo s xy lanh

i 1 2 3 4 5 6

an 24,2 17,1 14,15 13 12,85 11,9Ch :

Khi la chn hoc tnh c vtb m bo kt qu tt vi iu kin ta la chn t s fh/fb. V nu t s ny m ln qu th s lm tng tn tht ca CHK do n gy nh hng ti kh nng phc hi p sut tnh sau hng. Cn nu t s ny nh qu th lm xng kh bay hi do p sut tnh sau hng l lnv cht lng lm vic ca h thng khng ti v ph ti nh v trung bnh l km

fh/fb = 0,40,5 (t) v c th bng 1 i vi ng c c nh


- Chiu di lb: m bo khng gian t bm ga, c gng gim lb, lb = (0,81,3)db.

- Cu to: bung hn hp c th c lin vi thn BCHK, cn phn ln c c thnh cm ring, khi lp vi thn thng c cnh tn nhit

1.4.2. Hng

- Kch thc ca hng phi m bo sao cho xng lu thng c tt (ph).

- ng thi sc cn khng qu ln.

m bo iu kin y, thng c th dng nhiu hng (hng nh phi c fh-min t vi phun) (23 hng).

1 hng: dh = (0,60,8) db

2 hng: dhn = (0,60,8)db

dht = (0,20,3)db

3 hng: dhn = (11,2)db

dhg = (0,40,5)db

dht = (0,20,3)db


- Kt cu ca hng: i vi hng ca BCHK thng dng dng ng phun Lavan: c hnh dng kh ng hc tt nht, n lm ch kh nng chuyn thnh p sut tnh sau hng tt nht (h = 0,940,99) .

- C th c lin hoc c ri vi thn (xem b sung - trang 36 SGK).

1.4.3. Zch l v vi phun

-Kch thc v hnh dng: phi m bo c quan h n nh gia lng nhin liu qua zchl vi p sut chnh lch trc v sau zchl.

-Kt cu: zch l c th dng nt hoc ch to lin vi vi phun (ta xc nh kch thc, o lu lng). Nu l/d 11,2 th khng cn kim tra li.

1.4.4. Phao v bung phao

-m bo chnh 58 mm

-Vt liu: ng l hoc nha.

-Bung phao, bu xng: thng c c lin vi thn b CHK

Kt cu xem b sung thm trong SGK trang 38 v 39)


1.5.1. BCHK ca ng c gii tc ln

m bo phn phi nhin liu u v lng v cht n cc xylanh th bchk c th c 2 bung hn hp BCHK K88 SGK trang 30

Cn ng c c gii tc ln th ngi ta cng dng 2 nhm bng hn hp. Khi tc nh v ph ti nh th 1 nhm bung hn hp lm vic, kh tc cao v ph ti ln th nhm th 2 bt u lm vic, tc l c 2 nhm lm vic song song.

Hnh 33 trang 45 gii thiu s bchk gii tc cao (gm 2 bung hn hp khc nhau)

Bung 1 c 2 hng v y d cc h thng khc

Bung 2 c 3 hng v ch c h thng phun chnh v h thng chuyn tip (h thng khng ti ca bung hn hp)

Ch : m c cu iu khin bm ga ca bung 2 v s chuyn tip t bung 1 n bung 2

1.5. BCHK hin i


1.5.2. BCHK hin i

- C thm cc van in t iu chnh lng khng kh m bo hn hp ph hp vi tng iu kin lm vic ca ng c, cc van in t ny c iu khin t xa (ECU) khi n nhn wocj tn hiu t h thng cm bin.

1. H thng khi ng

Vn dng bm gi, nhng vic m bm gi c iu khin t ng nh ngunn hit kh thi, nc lm mt hoc bng in.

-iu khin bng nhit kh thi, bng nc lm mt: l xo lng kim lun ko bm gi ng li, khi khi ng nhit tng lm nng l xo, l xo gin n, bm gi t t m, v m to., m hon ton.

-iu khin bng in th c 2 tng in tr, tng 1 lm vic khi my cha nng (l xo lng kim gin n nh - bm ga m t). Tng 2 lm vic khi my nng (khi my nng, l xo lng kim gin n nhiu v bm gi m hon ton.

-iu khin bng kh thi v kt hp bng in: khi mi khi ng l xo gin n t, bm gi m t, khi my nng nhiu cng tc cm ng nhit ng mch dng in lm l xo nng ln gin n nhiu bm ga m to.


2. H thng khng ti

Vn lm vic trn c s h thng khng ti chung, nhng n cn c b phn t ng iu chnh ring cho n ph hp. C 3 ch khng ti: khng ti nhanh, khng ti cng bc v khng ti chun

-Khng ti nhanh (nc = 23003000 v/ph) mc ch l nhanh t ti nhit yu cu ( chy nng my.

-Khng ti chun (nc = 750900 v/ph)

-Khng ti cng bc

Khng ti cng bc - Khi nh ga p phanh, bm ga ng li nhanh, lm pg tng, nhin liu h thng khng ti vn ln, do qun tnh ca h thng chnh lm vic lm cho hn hp qu m chnh v vy ng c lm vic khng n nh, cht my, kh thi qu nhiu cht c hi

H thng iu chnh in t ca h thng khng ti rt quan trng


3. H thng lm m

4. H thng tng tc

1.6. Cc cm khc trong HTNL

-Thng nhin liu

-Bm chuyn

-Lc nhin liu

-Lc khng kh


12

34

5

6 7

8

9

1011

12

1314

15

16

20

17 18 19

BCHK iu khin bng in t

1: bm ga, 2: cm bin tc m bm ga, 3: cn y, 4: c cu iu chnh m bm ga kiu in t - chn khng, 5: c cu iu khin ng m bm khi ng, 6: bm khi ng (bm gi), 7: cn y, 8: kim iu chnh tit din thng qua gc l khng kh khng ti, 9: cm bin nhit ng c, 10: tn hiu nhit ng c, 11: tn hiu tc m bm ga, 12: tn hiu v tr mng n hi ca b iu chnh m bm ga kiu in t -chn khng 4 (hay gc m bm ga), 13: tn hiu tc vng quay ng c, 14: tn hiu t cm bin xc nh (xem hnh 10-5), 15: cc tn hiu ra b iu chnh m bm ga, 16: tn hiu ra iu chnh m bm gi, 17: u pht tn hiu ra, 18: b vi x l, 19: u thu nhn tn hiu vo, 20: b iu khin in t.


Gim ge v kh thi cng nh ci thin cht lng lm vic ca ng c mi ch lm vic, ngi ta trang b cc b phn in t cho BCHK. BCHK khi c gi l BCHK in t (Ecotronic ca hng Bosch- Pierburg). V c bn, BCHK in t gm mt BCHK thng thng, mt b iu khin in t 20 v cc c cu iu khin 4 thay i m bm ga 1 v c cu iu khin 5 thay i m bm gi 6.

B iu khin in t 20 gm c cc b phn chnh nh u nhn tn hiu 19, b vi x l trung tm 18 v u pht tn hiu ra 17. Tn hiu vo s c tip nhn v x l. Sau , b iu khin s pht tn hiu ra iu khin cc c cu chp hnh 4 v 5 nhm to ra thnh phn kh hn hp ti u cho mi ch lm vic ca ng c.

Trong c cu iu chnh m bm ga kiu in t - chn khng 4, v tr ca mng n hi c xc nh bi s cn bng gia lc tc dng ln mng do chn khng sau bm ga v lc phc hi ca l xo. chn khng trong khng gian pha trn mng c iu chnh nh hai nam chm in iu chnh cc van thng. Nh cn y 3, mng xc nh m ca bm ga 1. V tr ca mng n hi c ghi nhn v truyn tn hiu v b iu khin in t 20 qua ng 12.


- Khi ng c khi ng, b iu khin ng bm gi v m bm ga mt gc ph hp. H thng chnh v h thng khng ti cng lm vic cho hn hp m khing ging nh b ch ho kh thng thng (xem 4. ca 5.1.2.3).

- Ch tng tc c thc hin nh sau. T tn hiu m t ngt bm ga 11, b iu khin in t s ch th cho c cu 5 ng m rt nhanh bm gi 6. Do hn hp m ln t ngt p ng cho ng c tng tc.

- Khi ng c chy khng ti, b iu khin in t gi cho tc vng quay khng ti nkt = const. Khi v tr ca bm ga v bm khi ng u do b iu khin in t quyt nh. Ngoi ra, cn iu khin 7 - do c cu iu khin 5 dn ng- s tc ng ln kim hiu chnh 8 ng bt gc l khng kh ca h thng khng ti, hn hp s c lm m. Do thnh phn hn hp c iu chnh t ng ph hp vi ch khng ti nn nkt nh, tit kim nhin liu. Cng chnh v vy m ng c khng b cht my trong nhng trng hp sau: khi ng c t khng ti chuyn v ch c ti hoc gi mt s h thng tiu th vo ng c (v d h thng iu ho nhit trn t); hay khi ng c ang chy c ti chuyn v chy khng ti nn nhit ng c gim dn lm tng nht ca du bi trn dn n ma st tng.

- Ch ko, v d nh khi t xung dc, ngi li b chn ga nn bm ga 1 ch do c cu 4 iu khin. Khi , bm ga s m mt mc no sao cho chn khng sau bm ga nh n mc khng ht xng ra h thng khng ti. Ni khc i khi ng c b ko, ng c khng tiu th xng nn tit kim nhin liu v gim nhim mi trng.

htnl dc xang chk compatibility mode 9983

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