Ky Thuat Do Luong Ch5

7/31/2019 Ky Thuat Do Luong Ch5

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Gio trnh: kthuto lng 1 Dnh cho sinh vin nhm ngnh k thut

Bin son: PGS.TS. Nguyn Hu Cng Khoa in t -i hc K thut CNVersion1 - Thng 8 nm 2009

94

Chng5

O CC I LNG KHNG IN

5.1. Khi nim chung

i lng khng in l nhng i lng vt l nh nhit , p sut, lu lng, nng vt cht, v.v... Cc i lng ny chim a phn trong cc h thng thng tin o

lng v iu khin tng. Thit bo cc i lng khng in c s khi nh

hnh 5-1.

Cc i lng khng in c cc chuyn

i o lng (C) bin i thnh cc i

lng in, sau a vo mch o (M)

tnh ton, gia cng tin tc v a kt qu

ra ch th th hin kt quo (CT).

5.2 o nhit

5.2.1. Khi nim chung

- Phng php chung o nhit l da trn hin tng truyn nhit (nhit

dn, i lu v bc x nhit).

- Cc phng php o nhit c th rt khc nhau tu theo yu cu v k

thut, v di o nhit v c th thc hin bng 2 phng php: o trc tip v o

gin tip.o trc tip l phng php o trong c cc chuyn i nhit in c t

trc tip mi trng cn o. i vi nhit cao thng o bng phng php gin

tip, dng ct ngoi mi trng o v cn gi l phng php khng tip xc.

Bng 5-1 cho bit cc dng c v phng php o nhit vi cc di o khc nhau.

C M CTYX

Hnh 5-1. S khi ca thit bo

C - Chuyn i o lngM-Mch o lng CT- Chth


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95

Bng 5-1

Nhit oCNhit k

-270 0 1000 2000 3000 100000

Nhit in tr

- Platin

- Niken

- ng

- Bn dn

Cp nhit in

- Vt liu qu

- Vt liu thng

Ha quang k

- Bc x

- Mu sc

- Cng sng

5.2.2. o nhit bng phng php trc tip

1) Nhit kcp nhit ngu

a, Nhc li nguyn l

Phng php o nhit bng cp nhit ngu l phng php c s dng

rng ri v thun li. Cu to ca nhit k nhit ngu nh hnh 5-2 gm hai thanh kim

loi khc nhau a, b c hn vi nhau ti mt u l u lm vic t1. Hai u cn

li t0 l u t do.

Khi nhit u t1 khc vi nhit u t0 chng tonn mt sc in ng:

Eab(t1, t0) = Eab(t1) - Eab(t0) (5-1)

Nu gi cho nhit t0 khng thay i v t1 ph

thuc vo mi trng o nhit th ta c:

Eab(t1, t0) = Eab(t1) - C = f(t1) (5-2)

C - hng s

toto

t1

a b

Hnh 5-2. S nhit knhit ngu


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96

Sc in ng Eab ph thuc vo nhit t1 v nhit t0 cng nh ph thuc

vo vt liu ch to cc thanh kim loi a v b.

Bng 5-2 cho bit mt s cp nhit in thng dng.

Bng 5-2: Thng sca mt scp nhit in thng dng

Cp nhitDi nhit

lm vic

Sc in ng

(mV) chnh xc

ng/Constantan

= 1,63 mm-270 370 -6,25 19

(-40oC 100oC) 0,8%

(100oC 350oC) 0,75%

Cromel/Alumeo

= 3,25 mm

-270 1250 -5,35 50,63(0oC 400oC) 3oC

(400o

C 800o

C) 0,75%Cromel/Constantan

= 3,25 mm-276 870 -9,8 66,4

(0oC 400oC) 3oC

(400oC 870oC) 0,75%

Platin-Rodi (10%)

/Platin

= 0,51 mm

-50 1500 -0,23 15,5(0oC 400oC) 2,5%

(600oC 1500oC) 0,4%

Platin-Rodi/Platin-Rodi (30/6)

= 0,51 mm

0 1700 0 12,42 (870oC 1700oC) 0,5%

b, Gii thiu cp nhit trong thc t

- Trong thc t cc cp nhit ngu c ch to thnh mt thit b c cu to

nh hnh 5-3. Trong cc dy kim loi c lng vo ng s cch in, bn ngoi l

v thp khng r v chu c nhit cao. u dy ra c ni vo hp u ni.

Mch o ca nhit k nhit ngu l mt milivonmet nh hnh 5-4.

- in trng dy c qui nh l 5, khi in trng dy c gi tr nh

hn 5 ngi ta dng in tr iu chnh R/c b vo in trng dy. Trong

qu trnh o nhit bng cp nhit ngu thng gp nhng nguyn nhn gy nn sai

s l:


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97

Hnh 5-3.Cu to ca cp nhit ngu

1- Dy kim loi 2- ng scch in 3-u hn 4- V thp 5-u ni ra

Rd - in trng dy R/c - in triu chnh

- in trmch o thay i khi nhit mi trng thay i. T hnh 5-4 ta

thy in p o c trn milivonmet c tnh

UMN = I.RV (5-3) \

Vi RV - in trca milivonmet

Hnh 5-4. Mch o ca cp nhit ngu vi chth


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98

Hnh 5-5. S

mch

o c

a c

p nhi

t

ngu c thit bb nhit

M I =VABc/dd

t

RRR2R

E

+++(5-4)

RAB - in trcp nhit

T (5-4) v (5-3) ta c:

UMN = VVABd/cd

t RRRR2R

E.

+++(5-5)

UMN = VVmd

t RRR

E.

+(5-6)

Ta cng c

UMN =

V

md

t

RR1

E

+(5-7)

Vi Rmd = 2Rd + R/c + RAB - in trmch o

Mt khc ta c: UMN = IRV = Et - IRm (5-8)

in p UMN lun nh hn sc in ng Et do c in p ri trn in tr

mch o. T cng thc (5-7) ta thy rng mun cho sai s php o gim th RV cng

ln cng tt. Thng thng in tr ca milivonmet ln hn (40 50) ln in tr

mch o v sai s ch yu do in trca milivonmet thay i.- B nhitu tdo: nhit u t do (to) c duy tr nhit chun

0oC nhng trong thc t nhit mi trng (to) thng khc vi nhit u t do

(to) do sc in ng Eto c thng nh hn

sc in ng Et chun. Khc phc sai

s trn ngi ta thc hin n nh nhit

u t do bng phng php gi

nhit u t do vng nc ang

tan hoc dng thit b b nhit u

t do nh hnh 5-5.

Thit b b l mt mch cu gm

c cc in tr, R1, R2, R3 lm bng

Maganin, RT l in trng. Cu c

hiu chnh cn bng nhit

chun to = 0oC. Khi nhit mi

trng khc vi nhit chun in tr


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99

RT thay i lm cho mch cu mt cn bng. Ti hai im cd sut hin in p Ucd.

in p ny b vo sc in ng nhit b thay i khi to to v ta c:

EAB(t1, to) = EAB(t1, to) + Ucd (5-9)

2) Nhit knhit in tr

a, Nhc li nguyn l

Nhit k nhit in tr l nhit k s dng chuyn i nhit in tr. Chuyn

i nhit in tr l mt thit b bin i nhit thnh s thay i thng sin tr

R: RT = f(t) (t - nhit ).

Nhit in trc phn thnh: nhit in trkim loi v nhit in trbn

dn.

b, Gii thiu nhit knhitin trtrong thc t

Nhit in trkim loi

Nhit k kim loi c ch to di dng dy kim loi hc mng mng kim loi

c in trsut thay i theo nhit . Cc nhit in trkim loi thng lm bng

platin, Niken v ng. Chng c ch to theo hai dng: qun dy trn li cch in

hoc t vo mt ng thp khng g, chu c nhit cao v c cch in bng

cc ng s (hnh 5-6).


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100

Hnh 5-6

a) Nhitin trdy qun1- Dy nhit in tr

2- ng s cch in

3- Bt xit nhm

4- V bc

b) Nhitin trcng nghip

1- Dy nhit in tr

2- ng thp bo v

3-

4- Hp u ra


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101

Quan h gia nhit v in trc biu din bng biu thc:

RT = R0(1 + t) (5-10)

R0 - in trnhit chun

- h s nhit ; t- nhit mi trng- i vi ng: cu = 4,3.10

-3 (1/oC)

- i vi niken: Ni = 5.10-3 (1/oC)

- i vi Platin: pt = 4,3.10-3 (1/oC)

Di nhit s dng:

- T -195oC 260oC i vi Ni

- T -260oC 1400oC i vi Pt

- T -50o

C 200o

C i vi CuHnh 5-7 l mt nhit k b mt dng o nhit trn

b mt vt rn. Vt liu s dng l Ni, Fe - Ni hoc Pt. Chiu dy

cvi m v c din tch c1cm2.

Nhit in trbn dn

l in trch to t hn hp cc oxit bn dn a tinh th

nh: MgO; MgAl2O4; Mn2O3; Co2O3; NiO; ZnTiO4. Cc bt oxit

ny c trn vi nhau theo mt t l thch hp, sau nn nh

dng v thiu kt nhit 1000oC. Cc nhit in trc ch to vi hnh dng

khc nhau (a, hnh tr, vng, v.v...) nh hnh 5-8.

c tnh quan trng ca loi in trny l c nhy nhit rt cao, gp hng

chc ln nhit in trkim loi v chia thnh 2 loi.

- Nhit in trc h s nhit dng.

- Nhit in trc h s nhit m.

Di nhit lm vic c th thay i t vi tuyt i (oK) n 300oC.

Quan h gia in trv nhit c tnh theo biu thc:

RT = R0exp

oT

1

T

1B

R0 l in trnhit tuyt i T0B l h s nm trong khong 3000 5000K

Hnh 5-7


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102

Hnh 5-9.Mch o nhit in tr

Hnh 5-8. Nhit in trbn dn

nhy nhit c dng: R = -B/T2 (5-12)

c, mch o

- Mch o s dng vi cc nhit in tr thng l cc mch cu kt hp vi

cc mch khuch i tng nhy ca thit b. Hnh 5-9 l s ca mch o

thng dng trong cng nghip.

- trnh sai s do nhit mi trng thay i lm cho in trng dythay i khi o ngi ta thng s dng mch cu 3 dy. Hnh 5-10 l nguyn

l mch cu khng cn bng, ch th

lgmmt tin, trong R1, R2, R3,

RT l cc in tr ca mch cu (RT-

nhit in tr).

R0, R0 - in trca lgmmt

R4 - in trb nhit ;

Rd - in trng dy.

Mch cu c iu chnh cn

bng nhit chun (t0). Khi o RT

thay i theo nhit mi trng v cu

mt cn bng. Ch th lgmmt c khc theo nhit . S trn c kh nng

loi trc sai s khi in p ngun cung cp thay i.


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103

5.3. Cc phng php o lc, ng sut v p sut


- Qu trnh o lc, ng sut v p sut chim mt lng tng i ln trong cc

x nghip c h thng o v iu khin tng cng nh trong qu trnh nghin cu

cl tnh ca cc vt chu lc trong ch to my.

- Phm vi o lc rt rng t 10-12N 108N. Lc c th l nhng i lng

khng i (lc tnh) n nhng lc c tc bin thin nhanh (lc ng). Thng lc

c chia thnh nhng gii o khc nhau, mi gii o c th s dng cc phng php

v thit b khc nhau.

- i vi p sut, gii o t 0 1010N/m2. Ngng nhy ca thit bo t 0,1

0,01 N/m2

. Cc dng co p sut c chia thnh nhiu di o khc nhau. o ngsut ca cc loi vt liu thng nm trong khong o 0 150.107 N/m2 .Mc ch l

xc nh sc bn ca vt liu.

Phng php o lc, ng sut v p sut c th thc hin theo hai cch: o trc

tip v o gin tip.

+ Phng php o trc tip: l phng php s dng cc chuyn i c i

lng vo tng ng vi lc, ng sut cn o. i lng ra c bin i thnh cc

Hnh 5-10. S mch o vi chthlgmmt

a)o trc tip b) Dng mch cu


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104

tn hiu in. Mch o v ch th cho kt quo khng thng qua h dn truyn trung

gian.

+ Phng php o gin tip: trong s dng cc phn tn hi, cc h dn

truyn, bin lc, t suy ra i lng cn o.Cc phng php trn c s dng rng ri. S dng phng php no l tu

thuc yu cu v nhim v thc hin chng.

o lc, ng sut v p sut phn ln c tha v php o l hocl

lv:

El

l =

; Trong :

l - chiu di ca i tng o

l - bin thin theo chiu di - ng sut cca vt liu

E - mun n hi.

Mch o thng l mch cu, kt hp vi cc tng khuch i v chnh lu. Ch

th l cc dng co cin, t ghi, in t v cc dng c s.

5.3.2. Cc phng php v dng co lc

1.o lc bng phng php o trc tip

a,Lc kkiu p in

Chuyn i p in c dng rng ri l thch anh. Tinh th thch anh c

ct thnh tng phin c rng Y v chiu dy X (hnh 5-11).

Khi tc ng 1 lc vo trc X, in tch xut hin c tnh theo biu thc:

q = d1Fx d1 - h s p in (5-13)

Nu tc ng lc theo trc Y ta c:

q = - d1(Y/X).Fy (5-14)

Y, X - kch thc ca chuyn i.

Thch anh c: - Hng sin mi = 39,8.10-12 F/m.- H s p in d1 = 2,1.10

-12 C/N.

- ng sut cho php = 70 10 N/mm2

- in trsut = 1012m

- Mdun n hi E = 80.109 m.m-2


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105

- Hnh 5-12 trnh by cu to ca lc k p in thch anh. Trong phin

thch anh hnh ch nht c mt cnh c ct song song vi trc X, b mt c ct

vi gc 35o so vi trc Z (hnh 5-12a). S khi ca chuyn i nh hnh 5-12b,

trong khuch i c phn hi dng. Tinh th thch anh dao ng vi tn s cbn

fo v lch tn sf khi chu lc F.

Hnh 5-12. p in thch anh

f = FD

nkf2o (5-15)

F- lc tc ng; k - hng sD- kch thc chuyn i; n- h s

tng nhy, chuyn i c thc ghp bng nhiu phin song song vi

nhau.

Hnh 5-11. Chuyn i pin


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106

b,Lc kin trlc cng (in trtenx)

in trlc cng l loi chuyn i m nguyn l lm vic da trn s thay i

in trkhi b bin dng. Cc vt liu c tnh cht trn l mt s kim loi hoc bndn. Vi cc vt liu kim loi thng l constantan, nichrome, Platin - Vonfram.

Quan h gia bin dng v s thay i in trc biu din bng biu thc:

l

lK

R

R =

(5-16)

K - h s ph thuc vo vt liu v hnh dng ca chuyn i

K = 2,1 i vi constantan

K = 2,5 vi NichromeK = 4,1 vi Platin - Vonfram

Hnh 5-13 l cu to ca chuyn i in tr lc cng dy mnh. Dy in tr

c ng knh = 0,02 0,03 mm c un thnh hnh rng lc c chiu di tc

dng lo = 8 15 mm, rng ao = 3 10mm. in tr thay i trong khong 10 -

150.

- Vi bn dn thng l silic n tinh th

phan tp c ct thnh tng ming nh hnh 5-14,

chiu di l = 0,1mm n vi milimet, chiu dy c10-2mm

Hnh 5-15 l mch o s dng in trlc cng loi dy mnh trong thanh n

hi o lc c dn hai in tri xng. Khi c lc tc dng, thanh n hi b bin

dng lm cho cc in trb bin dng theo v in trca chng b thay i. Cc in

Hnh 5-13.Chuyn i in trlc cngdy mnh

1- Tm giy mng; 2- Dy in tr;

3- Dy dn; 4-u ct

Hnh 5-14.Chuyn i in trlc cng

Dng ct


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107

trny c mc trn mch cu hai nhnh hot ng . in p ra ca cu t l vi lc

tc ng. Gii hn o ca thit b trn vi lc tc ng t 103 104N. Sai s c tht

ti 0,2 0,5%.

Trong :R0 - in trcnh; R/c - in trcn bng v bin

R1 + R, R2 - R : in trlc cng.

Rc, C - in trv in dung cn bng pha.

1 - Khuch i; 2- Chnh lu (Mch o)

2.o lc bng phng php gin tip- Cc chuyn i s dng o lc theo phng php gin tip l cc loi

chuyn i in cm, h cm, in dung, in tr, v.v...

- Di tc ng ca lc o cc phn ng ca cc chuyn i b di chuyn v

bin thnh tn hiu in, sau ta o tn hiu in xc nh lc.

Hnh 5-16 v s cu trc ca lc k ch th s s dng chuyn i in cm.

Hnh 5-15. Mch o lc dngchuyn i lc cng

Hnh 5-16. S cu trc ca lc kch

ths


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108

Hnh 5-17. Chuyn i in cmmc vi sai

1- Mch t;

2- Li thp phn ng;3- Cun dy in cm

My pht tn s (MFTS1 v MFTS2) l loi my pht LC trong hai chuyn

i in cm L1 v L2c mc theo kiu vi sai

hnh 5-17. Khi cha c

lc tc ng ln li thp phn ng 2 ca chuyni in cm ca chng bng nhau L1 = L2 v 2

my pht tn su pht cng tn s 40 kHz. Khi

c lc tc ng ln phn ng 2 khe hkhng kh

1 v 2 ca chuyn i 1 v 2 thay i do in

cm ca cun dy L1 khc vi in cm L2.

My php tn s ph thuc vo in cm

L m L1 v L2 khc nhau nn tn s pht ca

MFTS1 khc vi MFTS2. (V d: MFTS pht tn

s 35kHz, MFTS2 l 45 kHz). Chng c a vo biu ch tn s (C). Tn s ra

sau biu ch l tn s tng 80 kHz v tn s hiu 10 kHz. Tn s 10 kHz l tn s

phn nh lc o c a qua b lc v b nhn. B nhn tn tng tn s vi mc

ch lm gim sai s trong qu trnh o, tn s nhn c a vo b to xung v b

m ch th kt quo.

Di o ca thit b trn l 20 kN. nhy 1000 Hz/N. Sai s phi tuyn 0,3%, sai s

tr 0,03%.5.3.3. o ng sut v bin dng

o ng sut v o bin dng thc t l mt, l o l hay l/l. o ng sut v

bin dng c th dng cc chuyn i khc nhau, thng thng dng chuyn i in

trlc cng.

S bin dng ca chi tit v ng sut c tnh theo biu thc:

l = kR

(5-17)

Vi: l = l

l; R = R

R

= E1 = E kR

(5-18)

Trong : k - h s; E - mun n hi v bin dng.

Hnh 5-18 l s mch o ng sut v bin dng ca mt dm chu lc v

phng php dn cc chuyn i. Khi i tng o, khng c nhng bin dng nh


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109

nhau v ngc du nhau, hiu chnh sai s nhit c th dn mt chuyn i ln

i tng o, cn chuyn i kia dn ln mt mu cng loi, mu ny t trong cng

iu kin nhit hnh 5-18a.

Trng hp khi o bin dng nh hnh 5-18b ta dn mt chuyn i pha trn

dm cn chuyn i kia dn pha di. Khi c lc tc dng chuyn i pha trn chu

lc ko, chuyn i di chu lc nn. Khi cu mt cn bng in p ra c tr s ln

hn khi mch cu ch c mt chuyn i bin dng v c kh nng loi trc sai s

do nhit mi trng gy nn.

5.3.4. Cc phng php o p sut1. Khi nim v p sut

p sut l t s gia lc tc dng vung gc ln mt mt vi din tch ca n.

p sut l mt i lng cbn xc nh trng thi nhit ng hc ca cc cht. S

vn hnh ca cc qu trnh cng ngh, tnh trng cc thit b cng ngh v ch vn

hnh ca chng c xc nh bng p sut.

n v p sut l Pascal (Pa) (Pa = N/m2)

Di o p sut t 10-12 1011 Pa.

Hnh 5-18. S mch o bin dng v ng sut dng chuyn i in trlc

cng


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110

Hnh 5-19.p kp in

Cc dng co p sut c phn loi theo p sut cn o v nguyn l tc

dng:

- p ko p sut d.

- Kh p ko p sut kh quyn.- p k tuyt i o p sut tnh t 0 tuyt i.

- Chn khng k.

Theo nguyn l lm vic c th phn thnh: o trc tip v o gin tip.

Sau y l mt s dng co p sut c s dng rng ri trong cc qu trnh

cng ngh:

2. o p sut bng phng php trc tip: p kin

Hnh 5-19 l s nguyn l ca p k p in. Di tc dng ca p sut P

mng n hi 4 to nn mt lc nn ln p in thch anh 2 c ng knh 5mm, b

dy 1mm. in tch q xut hin in cc 1 c a vo b khuch i in t 5 c

tng trvo c1013. Quan h gia in tch q v p sut P l:

q = kFP (5-19)

vi F - in tch hu ch ca

mng; k- h s.

gim qun tnh ca thit b ngi ta

gim th tch ca bung 3. V tn s dao ngring ca h thng mng chuyn i khong

hng chc kHz nn c c tnh ng hc cao

v c s dng rng ri o v kim tra p

sut trong h thng c dng chy nhanh. Gii

hn o ca thit b trong khong 2,5 100

MPa. Cp chnh xc 1,5 2.


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111

Hnh 5-20.p khcm vi sai

3. o p sut bng phng php gin tip

l cc thit b di tc dng ca p sut, nhcc khu dn ng p sut bin

thnh di chuyn, cc chuyn i o di chuyn s xc nh c p sut cn o. Cc

phn t dn ng c th l l xo ng, ng xiphng v mng n hi hoc mng do.Hnh 5-20 l s nguyn l ca p k h cm vi sai hay cn gi l p k bin p vi

sai. Thit b gm c khu dn ng 1 l mt l xo ng kim loi c un cong, mt

u gi c nh cn u kia t do.

Chuyn i 2 l mt bin p vi sai vi cun

dy s cp 7 v hai cun th cp 4 v 5.

Gia cc cun dy t li thp 6 gn vi

u t do ca l xo ng. Cun dy 4 v 5

u ngc chiu nhau. u ra ca cn th

cp c ni vi in trR1 v R2 thay

i gii hn o trong khong 25%. Khi c

dng in I1 chy qua cun scp s to ra

mt t thng mc vng qua 2 cun dy th

cp 4 v 5 lm sut hin cc sc in ng

cm ng e1, e2. ln ca chng ph thuc

vo h cm M1 v M2 gia cun scp v cc cun th cp v ta c:e1 = 2fI1M1 ; e2 = 2fI1M2 (5-20)

Do hai cun th cp u ngc chiu nhau nn:

E = e1 - e2 = 2fl1(M1 - M2) = 2fMI1 (5-21)

Vi R1 v R2 l cc in trcnh, in p ra c tnh:

Ura = 2fI1Mra (5-22)

Mra l gi tr h cm ph thuc vo di chuyn X ca li thp 6:

Mra = Mmax maxXX

(5-23)

Mmax l gi tr h cm ln nht ca chuyn i khi li thp di chuyn ln nht

Xmax. Thay (5-23) vo (5-22) ta c:

Ura = XX

MpfI2

max

max1 (5-24)

Chiu di ton phn ca li t 1,6 4 mm. in p ra ca thit b thay i t

-1V 1V tu theo s thay i pha ca tn hiu, tng ng vi s thay i h

cm M ca chuyn i t -10 10 mH.


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112

Ngoi ra, o p sut ta c th s dng p k kiu in trlc cng (cn gi

l in tr tenzo). Hnh 5-21 l s thit bo p sut kiu in tr lc cng. Cc

mun tenzo l mt mng mng kim loi 4 trn c gn bn in trtenzo silic, to

thnh mch cu khng cn bng. Mun tenzo gn trn 2 chia mi trng o thnhhai phn nhhai mng kim loi 1 v 3. Hiu s p sut o P = P1 - P2 tc dng ln cc

tenzo qua mng n hi v cht lng gia hai phn mi trng lm cho tenzo bin

dng. Tn hiu ra ca mch cu c ly ra tin cc 5 v ni vi khuch i 6.

Cung cp cho mch cu l b ngun 7. Tn hiu ra c chun ha vi dng in t 4

20 mA.

Vi thit b trn c tho c p sut d t 0 103 kPa n 0 60 MPa, p

sut chn khng 10 0 kPa, p sut tuyt i t 0 2,5 kPa hoc 0 2,5 MPa.

Cp chnh xc t c n 0,6; 2,0; 1,5. Thi gian tn hiu ra xc lp khi c s

thay i p sut t 0,5 2,5 giy.

Hnh 5-21.p kdng chuyn i tenzo


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113

5.4. o lu lng v mc cht lng


Lu lng c phn thnh lu lng trung bnh v lu lng tc thi.

Lu lng trung bnh c tnh:Qtb = V(t2 - t1) (5-25)

Trong V - Th tch vt cht

t2 - t1 l thi gian o.

Lu lng tc thi c tnh theo cng thc:

Q = dV/dt (5-26)

Thit bo lu lng c gi l lu lng k.

T cc cng thc trn, ta c th suy ra:

V = 2

1

t

t

Qdt v m = 2

1

t

t

Gdt (5-27)

m - khi lng vt cht

G - lu lng khi tc thi: G = dm/dt

5.4.2. Cc thit bo lu lng v lu tc

1. Lu lng kcnh qut

Nguyn l lm vic ca lu lng k cnh qut da trn s vng quay ca

tuabin trong mt n v thi gian t l vi tc dng chy.

n = k.v (5-28)

k - h s t l

k = 2rFcotg (r - bn knh trung bnh ca cnh tuabin; - gc lch gia

cnh v trc tuabin)

v - tc dng chy qua tit din F ca ng dn.

Lu lng th tch qua cnh qut c tnh bng:Q = v.F (5-29)

T phng trnh (5-28) v (5-29) ta c:

n = F

Q.k(5-30)

Hnh 5-22 l s ca mt lu lng k cnh qut. Trong cnh qut 1 ging

nh cnh tuabin, quay trn gi 2 c gn vo tm 3 trong ng dn. 4 c

tc dng hn ch di chuyn ca cnh qut. Trc cnh qut lm bng vt liu khng


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114

Hnh 5-23. S cu to lulng ktin

dn t trong gn li thp 5 bng vt liu dn t. Bn ngoi ng t nam chm vnh

cu 6 trn n qun dy cm ng 7.

Khi cnh qut quay, t thng ca nam chm s tng ln khi li 5 nm dc trc

ca nam chm v gim i khi li thp 5 nm vung gc vi n.

Khi t thng mc vng trong cun dy cm ng thay i s xut hin mt sc

in ng cm ng E = -Wdt

d. Mi vng quay t thng tng gim hai ln v tn s

cm ng f trong cun dy cng tng gp 2 ln s vng quay ca tuabin.

o tc vng quay bng mt tn s k c th xc nh c lu lng cn o.

Vi phng php trn sai st c t 1 0,3%, phm vi o 0,3 m3/h 36000 m3/h

vi cht kh v 0,01 m3/h 14000 m3/h vi cht lng.

2. Lu lng ktin

Nguyn l hot ng ca lu lng k tin

da trn nh lut cm ng in t. Khi c dy dn

chuyn ng trong t trng, trong dy dn cm ng

mt sc in ng t l vi tc di chuyn ca dydn. Nu ta dng mt cht lng dn in chy qua

gia hai cc ca mt nam chm, sc in ng sinh

ra trong cht lng s t l vi tc dng chy hay t

l vi lu lng cn o.

Hnh 5-22. S cu to lu lng kcnh qut


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115

Hnh 5-23 l s ca lu lng k tin. ng dn dung dch 1 c ch to bng

vt liu khng dn t c cht lng dn in chy qua. T trng bin thin do nam

chm in 2 to nn xuyn qua dng cht lng cm ng mt sc in ng v c

ly ra trn hai in cc 3 v 4 a vo thit bo (TB). ln ca sc in ng c tnh:

E = kBdv (5-31)

Trong : k - h s; - tn s gc ca t thng do nam chm to ra.

B - cm ng t; d - ng knh trong ca ng.

v - tc trung bnh ca cht lng.

Sc in ng biu din qua lu lng:

E = dk4 BQ (5-33)

vi Q =4

dv 2l lu lng dng chy

Lu lng k trn c u im l khng c qun tnh do c th o lu lng

bin thin theo thi gian. Ch th ca dng c khng ph thuc vo thng s vt l ca

cht lng (p sut, nhit , mt , nht) v khng ph thuc vo sc cn phi

vi dng chy. Thit b cho php o lu lng trong phm vi rng t 1 2500 m3/gi

cho cc ng dn c ng knh t 10 1000 mm vi tc chuyn ng t 0,5 - 10

m/s. Cp chnh xc t c t 1 2,5.

5.4.3. o mc

1)o mc bng phng php o p sut thu tnh

Thc cht ca phng php l o p sut thu tnh P ca ct cht lng c

cao h v t trng khng i.

P = gh (5-34)

o p sut thu tnh c th thc hin theo hai cch:+ p kt mc thp nht ca cht lng.

+ p k vi sai vi mt pha t mc thp nht ca cht lng v pha kia ni

vi phn cha khng kh pha trn mc cht lng.

Hnh 5-24a l so mc bng p k. p sut o c t p k lin quan n

chiu cao h ca mc: P = gh.

Hnh 5-24b l so mc bng p k vi sai trong cc b h. p k vi sai 1

c ni qua ng y 2 ni vi b cha v bnh so snh 3. Bnh cn bng 3 dng b


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116

Hnh 5-24.o mc bng phng php op sut thu tnh

a)o mc bng p kb)o mc bng p kvi sai trong bhc)o mc bng p kvi sai vi bc p sut

p sut tnh to ra do ct cht lng hl trong ng y. Trong qu trnh o bnh cn bng

gi khng i van 4 dng duy tr mc khng i trong bnh 3. Khi trng lng

ring khng i vi h1 = h2 ta c gim p sut o c bng p k vi sai l:

P = gh (5-35)

Khi o mc trong cc thit b chu p sut ta s dng s hnh 5-24c. Bnh

cn bng 3 t tng ng vi mc cao nht v ni vo thit b. p sut tnh P c

a vo c hai ng dn v gim p sut c tnh nh sau:

P = ghmax - Lgh

Khi h = 0, P = Pmax, khi h = hmax th P = 0

2.o mc bng in dung

Hnh 5-25 l s thit bo mc bng chuyn i in dung, trong chuyn

i in dung 1 l thanh kim loi thng, ph lp chng n mn ha hc t gia thng

kim loi ng cht lng 2. Khi thng rng, in dung ca thng l 8pF, lc y cht

lng in dung tng ln n 30pF. Chuyn i in dung c mc vo mt nhnh ca

mch cu khng cn bng (MC), nhnh th hai gm t C1 c in dung 8pF v t C2

mc song song vi C1 c in dung 22pF qua kha K. Hai nhnh khc ca mch cu l

cc cun dy th cp ca my bin p. Cu c cung cp bng mt my pht cao tn

(MF) c tn s 1 10 MHz. in p ra ca mch cu c chnh lu qua b chnh

lu.


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117

iu chnh cho ch th c gi tr cc i thc hin bng cch ng kha K v

iu chnh in trR/c. Ch th c th l mt miliampemt, in th k tng hoc

dng c s. Thit b c tho c mc nhit 100oC v p sut ca bnh t 0

106 N/m2. Sai s ca thit bt c tuyn tnh trong khong o ln, gii o t

mm 5m.

5.5. o vn tc v gia tc5.5.1. o tc quay

1. My pht tc

My pht tc l dng co tc quay ca ng c. l cc my pht in

mt chiu v xoay chiu, trc quay ca my pht c ni vi trc quay ca ng c.

Khi ng cquay my pht pht in mt chiu to ra sc in ng t l vi tc

quay.

Trong my pht tc xoay chiu, quan h gia sc in ng v tc quay

ging my pht in mt chiu nhng in p ra l in p xoay chiu c tn s t lvi tc quay:

f =an

60(5-36)

Trong : a - si cc; n - tc quay; f - tn s.

o in p U hoc tn s f c th xc nh c tc . Do chnh xc thp

v cu to phc tp nn phng php ny c s dng khng nhiu.

Hnh 5-25.S thit bo mc dng chuyn i in dung


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118

2.o tc bng phng php m xung

Nguyn l lm vic ca phng php l bin tc quay ca i tng o

thnh cc xung. Tn s xung ph thuc vo tc quay ca i tng, o tn s ta c

th xc nh c tc quay.Hnh 5-28 l s thit bo tc quay bng chuyn i cm ng. B bin

i tc - tn s gm c bnh rng 1, cun dy cm ng 2. l phn t nhy vi s

lng rng l p = 1; 6; 60; 180; 200; 250 v 600 c tn s thp nht l 10 Hz.

Khi bnh rng quay, phn t nhy to thnh cc xung.

Tn s ln nht c th nhn c khi o l

fmax =

60

n.p max(5-37)

Trong s rng khng nh hn

p = 600/nmin.

Vi: nmax - tc quay ln nht

nmin - tc quay nh nht

Theo s nh hnh 5-28 vi P = 60, thi gian o l 1 giy thit b c th ch

th trc tip tn s quay thnh tc . Ch th l dng c s trong cc xung tn s tphn t nhy c a qua b khuch i vo thit b tnh v a ra ch th s. Sai s

ca bm l 1 xung.

5.5.2. o gia tc v bin rung (chn ng)

Cc dng co chn ng c gi l chn ng k, o gia tc gi l gia tc

k.

1.Chn ng kkiu cm ng

Hnh 5-29 l s cu to ca chn ng k cm ng. Thit b gm c gi

bng kim loi khng dn t 1 gn ln i tng o. Trn gi c thanh dao ng 2,

nam chm 3 v l xo 4 to thnh khi qun tnh. Bnh thng, di tc dng ca trng

lng, khi qun tnh v lc y ca l xo trng thi cn bng, cun dy cm ng 5

t gia khe hkhng kh ca nam chm ng yn.

Hnh 5-28. S thit bo tc dng

chuyn i cm ng


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119

Hnh5-30.S gia tc kcon lc

Khi c chn ng, nam chm v cun dy di chuyn tng i vi nhau lmcho t thng mc vng qua cun dy thay i, to ra sc in ng cm ng. Sc in

ng ny t l vi bin rung v c a vo khuch i v ch th. Sc in ng

cm ng c th b nhiu do t trng bn ngoi, kh nhiu ngi ta qun thm

cun dy b 6 c s vng bng cun dy 5 nhng c u vi chiu ngc li. Sc

in ng do nhiu sinh ra trong cun dy 5 v 6 s kh ln nhau v u ra ch cn li

sc in ng do cun dy cm ng 5 to ra khi di chuyn trong khe hkhng kh.

Nhb vi phn v tch phn, thit b c tho c rung vi bin X, tc X

v gia tc X .

2. Gia tc k

o gia tc khng thay i ngi ta s dng gia

tc k con lc. Hnh 5-30 l s nguyn l ca gia tc k

con lc vi chuyn i in dung. Di tc dng ca gia

tc, con lc c chiu di l v khi qun tnh m b lch i

mt gc , bn cc gia ca chuyn i in dung mc

kiu vi sai di chuyn mt khong do in dung cahai tin c gi tr khc nhau. Mch cu (gm hai na

cun dy th cp ca bin p v hai tin C1, C2) mt

cn bng. in p ra ca mch cu t lc vi gia tc o.

Gia tc k con lc dng o gia tc nh hn 0,1g (g - gia

tc trng trng). Vi gia tc ln hn, gc lch qu ln

gy sai s do phi tuyn.

Hnh 5-29.Chn ng kcm ng


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