Ky Thuat Do Luong Ch1

7/31/2019 Ky Thuat Do Luong Ch1

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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 CN

Version1 - Thng 8 nm 2009

1

Chng 1

KHI NIM CHUNG VO LNG

1.1. nh ngha v phn loi

1.1.1. nh ngha

o lng: l mt qu trnh nh gi nh lng i tng cn o c kt qu

bng s so vi n v.

Vi nh ngha trn th o lng l qu trnh thc hin ba thao tc chnh:

- Bin i tn hiu v tin tc.

- So snh vi n vo hoc so snh vi mu trong qu trnh o lng.

- Chuyn n v, m ho c kt qu bng s so vi n v.

Cn c vo vic thc hin cc thao tc ny ta c cc phng php v h thng o

khc nhau.

Thit bo v thit bmu

- Thit bo l mt h thng m i lng o gi l lng vo, lng ra l i

lng ch trn thit b (l thit bo tc ng lin tc) hoc l con s km theo n v

o (thit bo hin s). i khi lng ra khng hin th trn thit b m a ti trung

tm tnh ton thc hin cc Algorithm k thut nht nh.

- Thit b mu dng kim tra v hiu chnh thit bo vn vo.

Theo quy nh hin hnh thit b mu phi c chnh xc ln hn t nht hai

cp so vi thit b kim tra.

V d: Mun kim nh cng tcp chnh xc 2 th bn kim nh cng tphi

c cp chnh xc t nht l 0,5.


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2

1.1.2. Phn loi

1.1.2.1. Thit bo lng

C nhiu cch phn loi song c th chia thit bo lng thnh hai loi chnh l

thit bo chuyn i thng v thit bo kiu so snh.

Thit bo chuyn i thng

i lng cn o a vo thit b di bt k dng no cng c bin thnh gc

quay ca kim ch th. Ngi o c kt qu nhthang chia v nhng quy c trn

mt thit b, loi thit b ny gi l thit bo cin. Ngoi ra lng ra cn c th

bin i thnh s, ngi o c kt qu ri nhn vi h s ghi trn mt my hoc my

tng lm vic , ta c thit bo hin s.

Thit bo kiu so snh

Thit b so snh cng c th l ch th cin hoc l ch th s. Tu theo cch so

snh v cch lp i lng b (b m ho s tng t) ta c cc thit b so snh khc

nhau nh: thit b so snh kiu tung (i lng o x v i lng b xk lun bin

i theo nhau); thit b so snh kiu qut (i lng b xk bin thin theo mt quy lut

thi gian nht nh v s cn bng ch xy ra ti mt thi im trong chu k).

Ngoi ra cng cn c vo vic lp i lng b ngi ta chia thnh thit b m

ho s xung, tn s xung, thi gian xung. Cn c vo iu kin cn bng ngi ta chia

thnh thit b b khng lch (zero) v thit b b c lch (vi sai).

Cn c vo quan h gia lng ra v lng vo, ngi ta chia thnh: thit bo

trc tip (i lng ra biu th trc tip i lng vo), thit bo gin tip (i lng

ra lin quan ti nhiu i lng vo thng qua nhng biu thc ton hc xc nh),

thit bo kiu hp b (nhiu i lng ra lin quan ti nhiu i lng vo thng qua

cc phng trnh tuyn tnh).

1.1.2.2. Chuyn i o lng

C hai khi nim:

- Chuyn i scp hay cn gi l cm bin, sensor (S: Sensor): C nhim v

bin mt tn hiu khng in sang tn hiu in, ghi nhn thng tin gi tr cn o. C

rt nhiu loi chuyn i scp khc nhau nh: chuyn i in tr, in cm, in

dung, nhit in, quang in...


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3

- Chuyn i chun ho: C nhim v bin i mt tn hiu in phi tiu chun

thnh tn hiu in tiu chun theo chun Quc t (thng thng U = 0 10V; I = 4

20mA).

Vi loi chuyn i ny ch yu l cc b phn p, phn dng, bin in p, bin

dng in, cc mch khuch i c nghin cu kcc gio trnh khc nn ta

khng xt.

1.1.2.3. Thp thit bo

Vi mt thit b c th (mt knh):

+ Chuyn i o lng: bin tn hin cn o thnh tn hiu in.

+ Mch o: thu nhn, x l, khuch i thng tin.... bao gm: ngun, cc mch

khuch i, cc b bin i A/D, D/A, cc mch ph

+ Ch th: thng bo kt qu cho ngi quan st, thng gm ch th s v ch th

cin, ch th t ghi, v.v...

1.1.2.4. Vi h thng o lng nhiu knh

Trng hp cn o nhiu i lng, mi i lng o mt knh, nh vy tn

hiu o c ly t cc sensor qua b chuyn i chun ho ti mch iu ch tn hiu

mi knh, sau sa qua b dn knh (multiplexer) c sp xp tun t

truyn i trn cng mt h thng dn truyn. c s phn bit, cc i lng o

trc khi a vo mch dn knh cn phi m ho hoc iu ch (Modulation - MOD)

theo tn s khc nhau (th d nh f10, f20...) cho mi tn hiu ca i lng o.

Ti ni nhn tn hiu li phi gii m hoc gii iu ch (Demodulation

DEMOD) ly li tng tn hiu o. y chnh l hnh thc o lng t xa

(Telemety) cho nhiu i lng o.

Lng vo Mch o Ch th Lng ra

C s

cp

Hnh 1.1. Cu trc h thng o mt knh


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4

H thng o lng nhiu knh cn gi l h thng thng tin o lng v c

phn thnh nhiu nhm: h thng o lng, h thng kim tra tng, h thng chn

on k thut v t hp o lng tnh ton.

Ngy nay h thng thng tin o lng pht trin mnh m thnh cc h thng

chuyn dng nh: h thng o lng, gim st SCADA; h thng iu khin phn tn:

DCS; h thng thng tin tch hp:IIS.

1m

10

Tn hiu o 1 MOD

i0im

Tn hiu o i MOD

fnm n0

Tn hiu o n MOD

202mTn hiu o 2 MOD

V2

V1

Vi

VN

MB chun hotnhiu

X2

X1

XI

XN Phn knh theo tn

DEMOD

DEMOD

DEMOD

DEMOD

nm

im

2m

f1m

X2

X1

XN

XInm

im

2m

f1m

phtthu

B thu nhn chbin tn hiu

gii iu ch

S

S

S

S

Hnh 1.2. V d v mt h thng o lng nhiu knh


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1.2. S cu trc thit bo lng

1.2.1. H thng o bin i thng

Trong h thng o bin i thng, i lng vo x qua nhiu khu bin i trunggian c bin thnh i lng ra z. Quan h gia z v x c th vit:

z = f(x)

trong f(.) l mt ton t th hin cu trc ca thit bo.

Trong trng hp quan h lng vo v lng ra l tuyn tnh ta c th vit:

z = S.x (1-1)

y S gi l nhy tnh ca thit b.

Nu mt thit b tuyn tnh gm nhiu khu ni tip th quan h gia lng vo

v lng ra c th vit:

z =n

ii 1

S .x=

(1-2)

trong Si l nhy ca khu th i trong thit b.

1.2.2. H thng o kiu so snh

Trong thit bo kiu so snh i lng vo x thng c bin i thnh ilng trung gian yx qua mt php bin i T:

yx= T.x.

Sau yx c so snh vi i lng b yk.

Ta c: x ky y y = .

C th cn c vo thao tc so snh phn loi cc phng php o khc nhau.

T

y

yk

x yX

Hnh 1.3. H thng o kiu so snh


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1.2.2.1. Phn loi phng php o cn cvo iu kin cn bng

a) Phng php so snh kiu cn bng (Hnh 1.4)

Trong phng php ny, i lng vo so snh: yx

= const; i lng b yk

=

const.

Ti im cn bng: y = yx yk 0 .

b) Phng php so snh khng cn bng(Hnh 1.5)

Cng ging nh trng hp trn song 0y .

1.2.2.2. Phn loi phng php o cn cvo cch to in p b

a) Phng php m ho thi gian

Trong phng php ny i lng vo yx = const cn i lng b yk cho tng t

l vi thi gian t:

yk= y

0.t (y

0= const)

Ti thi im cn bng yx = yk = y0.tx

x

x0

yt

y= . (1-3)

i lng cn o yxc bin thnh khong thi gian tx.

y php so snh phi thc hin mt b ngng

x k

y sign(y y ) = =

0

1 x k

x k

y y

y y

30)

Sai s ngu nhin c tnh:

x = k. (1-20)

trong k l h s, c tra trong s tay k thut (bng hoc ng cong).

Trng hp ny ta khng quan tm nhiu trong k thut.

b) Khi sln o c hn (n 30)

Qu trnh gia cng c tin hnh nh sau:

+ K vng ton hc c ly l trung bnh cng ca n ln o

[ ]

n

ii 1

d

xM x x x

n== = =

. (1-21)

+ Phng sai ca sai s ngu nhin c tnh theo cng thc Bessel

( )n n2 2

i ii 1 i 1

x x

n 1 n 1= =

= =

=

2n ni

ii 1 i 1

xx

n

n 1= =

. (1-22)

Thc t, ta khng th bit chnh xc gi tr ca x m ch ly kt qu l gi trc

lng trung bnh ca n ln o. V vy phng sai s gim i n ln

x

o t

D

M(x)

Hnh 1.11. K vng v tn x ca lut phn bchun


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( )

( )

n 2

ii 1

x

x x

n n 1n=

= =

. (1-23)

+ Sai s ngu nhin c tnh:

x = kst. x

trong kst l h s Student, n ph thuc vo s ln thu thp n v xc xut yu cu p.

H s kst c tra trong cc s tay k thut: kst = f(n,p).

+ Kt quo c tnh:

x =( )

2n nn

i iii 1 i 1i 1

st

1

x xx nx x k

n n n 1= ==

=

. (1-24)

Ch : Trong thc t c nhng ln thu thp s liu cho kt qu khng ng tin

cy(v ta thng gi l nhiu ca tp s liu), ta phi loi b ln o ny nhthut ton

sau:

Sau khi tnh ta so snh cc i vi 3 vi i = 1 n n, nu ln o no c

i3 th phi loi b ln o v tnh li tu vi (n 1) php o cn li. C

th chng minh rng vic loi b m bo tin cy 99,7%.

V d 1.3: Tnh kt quo v sai s ngu nhin vi mt xc sut ng tin p =

0,98 ca mt php o in trbng cu kp vi kt qu nh sau (n v tnh = m).

140,25; 140,5; 141,75; 139,25; 139,5; 140,25; 140; 126,75; 141,15; 142,25;

140,75; 144,15; 140,15; 142,75. Bit sai s ngu nhin c phn b chun.

Bi lm:

Theo gi thit, sai s ngu nhin c phn b chun, v vy cc bc tnh tonc tin hnh nh sau:

K vng ton hc l: [ ]14

i

i 1

RR M R 139,96 (m )

14== = = .

Tip theo tnh( )

214

i

i 1

R R4,03 (m )

14 1=

= =

.


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So snh cc i iR R = vi 3 . Ta thy ln o th 8 phm phi sai lm ln

( 8 8R R 3 = ) nn ta b qua ln o ny v tnh li tu vi 13 ln o cn li. Ta

lp bng sau:Bng 1.1. V d v tnh ton sai s ngu nhin

STT Ri

(m)

i

(m)

2i

(m)2

1 140,25 0,73 0,5329

2 140,5 0,48 0,2304

3 141,75 0,77 0,5929

4 139,25 1,73 2,9929

5 139,5 1,48 2,1904

6 140,25 0,73 0,5329

7 140 0,98 0,9624

8 141,15 0,17 0,0289

9 142,25 1,27 1,612910 140,75 0,23 0,0529

11 144,15 3,71 13,7641

12 140,15 0,83 0,6889

13 142,75 1,77 3,1329

Tng:= 0 Tng: 23,64

Tnh li: [ ]

13

ii 1

RM R R 140,98

13== = =

(m)

64,2313

1

2=

=i

i (m)2


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( )i

22

iR R1, 4 (m )

13 1 13 1

= =

Kim tra cc kt quo, ta thy khng c trng hp no b loi; ta tnh bc tip

theo:

R0, 38(m )

13

= .

Vi xc sut yu cu p = 0,98; n = 13; theo ph lc 3, ta chn kst l trung bnh

cng ca hai trng hp ng vi n=12 v n=14:

st

2, 72 2, 65k 2,6852

+= =

st RR k . 1,04 (m ) = .

Kt quo l:

R R R 140,98 1,04 (m )= = .

1.4.2. Tnh ton sai s gin tip

Trong thc t c nhiu phng php o m kt quc tnh t php o trc tip

khc, ngi ta gi php o l php o gin tip.

Gi s c mt php o gin tip i lng y thng qua cc php o trc tip x1,

x2, xn : y =f(x1,x2,xn).

Ta c:

1 2 n1 n2

y y ydy dx dx ... dx

x xx

= + +

. (1-25)

Sai s tuyt i ca php o gin tip c nh gi :2 2 2 2n

1 2 n i1 2 n ii 1

y y y yy x x ... x x

x x x x=

= + + + =

(1-26)

Hoc c thnh gi:

n

1 2 n i1 2 n ii 1

y y y yy x x .... x x

x x x x=

= + + + =

(1-27)


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Trong : x1, x2, xn l sai s tuyt i ca php o cc i lng trc tip

x1, x2,xn.

Ch : Vic tnh sai stuyti theo(1-26) v (1-27) c thcho kt qu khc nhau do

vic sdng cc chun khc nhau; tuy nhin trn thc t, ngi ta thng nh gi

sai stuyti theo(1-26)

Sai s tng i ca php o gin tip c tnh l:

y

y

y

= (1-28)

Nu tnh theo %, ta c :

y(%)=y.100% (1-29)Trong trng hp nh gi sai s tuyt i theo cng thc (1-26) th sai s tng

i c tnh :

2 2 2 21 n

y1 n

x y x y...

y x y x

= + +

(1-30)

Bng 1.2. Bng tnh sn sai s tuyt i v tng i

ca mt s hm y thng gp

Hm y Sai s tuyt i y Sai s tng i y y / y =

x1 + x2 ( ) ( )2 2

1 2x x + ( ) ( )

2 21 2

21 2

x x

(x x )

+

+

x1.x

2 ( ) ( )

2 22 2

1 2 2 1x x x x +

2 21 2

1 2

x x

x x

+

1

2

x

x ( ) ( )

2 22 21 2 2 1

42

x x x x

x

+

2 21 2

1 2

x x

x x

+

xn n 1nx x x

nx


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V d 1.4: Ngi ta s dng Ampemet v Voltmet o in trbng phng

php gin tip. Ampemet c thang o 10A, cp chnh xc 1,5; Voltmet c thang o

250V, cp chnh xc 2,0. Tnh sai s tuyt i v tng i cho hai trng hp c kt

quo sau:

a) Kt quo vi s ch Ampemet l 8,5A, Voltmet l 220V

b) Kt quo vi s ch Ampemet l 10A, Voltmet l 250V

Bi lm

Ta p dng theo cng thc tnh sai s theo gi tr trung bnh, bnh phng(1-26).

Tu bi, ta c :DI = 10A DU = 250V

I% = 1,5 U% = 2,0

+ Sai s tuyt i ca Ampemet l :

1,5. % 10. 0,15( )

100I II D A = = =

+ Sai s tuyt i ca Voltmet l :

2,0. % 250. 5( )

100U UU D V = = =

a) Kt quo vi s ch Ampemet l 8,5A, Voltmet l 220V

+ Gi trin trtheo php o l :

22025, 88( )

8,5

UR

I= = =

+ Sai s tuyt i ca php o in tr:

R

Hnh 1.12. V d v tnh ton sai sgin tip

U100k V

A


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2 2 2 2 2 2 2 2

2 4. . .

R R U U I U U IR U I I

U I I I I

+ = + = + =

2 2 2 2

48,5 .5 220 .0,15 0,7447( )8,5+= =

+ Sai s tng i ca php o in tr:

0,7447.100% .100% 2,8775

25,88RR

R

= = = (%)

b) Kt quo vi s ch Ampemet l 10A, Voltmet l 250V

+ Gi trin trtheo php o l :250

25( )10

UR

I= = =


2 2 2 2 2 2 2 2

2 4. . .

R R U U I U U IR U I I

U I I I I

+ = + = + =

2 2 2 2

4

10 .5 250 .0,15

0, 625( )10

+

= =


0,625.100% .100% 2,5(%)

25RR

R

= = =

Nhn xt : sai s tng i ph thuc vo quan h gia gi tri lng o v

thang o ca cc thit bo trc tip.

V d 1.5: Tnh sai s tuyt i v tng i cho hai trng hp c kt quo

nh v d 1.4 nhng s dng cng thc tnh sai s (1-27).

Bi lm

Ta p dng theo cng thc tnh sai s (1-27).

Tu bi, ta cng c :

DI = 10A DU = 250V

I% = 1,5 U% = 2,0


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22


1,5. % 10. 0,15( )

100I II D A = = =


2,0. % 250. 5( )

100U UU D V = = =

a, Kt quo vi s ch Ampemet l 8,5A, Voltmet l 220V


220

25, 88( )8,5

U

R I= = =


2. . .

R R U UR U I I

U I I I

= + = +

2

8,5.5 220.0,151,04( )

8,5

+= =


1,04.100% .100% 4,03%

25,88RR

R

= = =

b, Kt quo vi s ch Ampemet l 10A, Voltmet l 250V


25025( )

10

UR

I= = =


2. . .R R U U

R U I IU I I I

= + = +

2

10.5 250.0,150.875( )

10

+= =


0,875.100% .100% 3,5(%)25R

R

R

= = =


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V d 1.6

Tnh ton sai s gin tip khi th nghim o in trRx v tng tr zx bng

phng php ngun xoay chiu nh Hnh 1.13.

Bit: Wattmet c thang o 100W; s ch 85W; cp chnh xc1.

Volmet c thang o 250V; s ch 240V; cp chnh xc 0,5.

Ampemet c thang o 5A; s ch 4,8A; cp chnh xc 0,2.

Bi lm

Theo u bi, ta c:

DU = 250 V; U%= 0,5; UV = 240(V)

DI = 5 A; I%= 0,2; IA = 4,8(A)

DP= 100 W; P%= 1; PW = 85(W)

Ta tnh ton sai s theo cng thc trung bnh bnh phng(1-26)


0,2. % 5. 0,01( )100 = = =I II D A


0,5. % 250. 1,25( )

100 = = =

U UU D V

+ Sai s tuyt i ca Wattmet l :

1,0. % 100. 1( )

100 = = =

P PP D W

Hnh 1.13. Hnh vcho v d 1.6

Rx

A

V

*W*

u zx


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2 2

853,6892( )

4,8= = = W

x

A

PR

I

+ Gi tr ca tng trtheo php o l :

24050( )

4,8= = = V

x

A

Uz

I

+ Sai s tuyt i ca php o tng tr: ta p dng cng thc theo bng 1.2

2 2 2 2 2 2

4

2 2 2 2

4

. .

4, 8 .1, 25 240 .0, 01 0,2805( )4,8

+ = + = =

+ =

x xx

z z I U U Iz U I

U I I

+ Sai s tng i ca php o tng tr:

0,2805.100% .100% 0,5610%

50

= = =

x

xZ

x

z

z

+ Sai s tuyt i ca php o in tr: trong trng hp ny, quan h hm cha c

tnh sn trong bng 1.2, v vy ta phi tnh trc tip sai s tuyt i t cng thc

(1-26)2 2 2 2

2 3

2.. . .

= + = +

x x w wx w

w

R R P PR P I I

P I I I

2 2

2 3

1 2.85.0,010,0460( )

4,8 4,8

= + =


0,0460.100% .100% 1,2482%3,6892

= = =x

xR

x

RR

.


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1.5. n vo, chun v mu(Tham kho)

1.5.1. Khi nim chung

n vo l gi trn v tiu chun v mt i lng o no c

Quc t quy nh m cc quc gia u phi tun th. Trn th gii, ngi ta ch to

ra nhng n v tiu chun c gi l cc chun.

V d: Chun Ampe l dng in c th gii phng 0,001118 gam bc khi

dung dch nitrat trong thi gian 1 giy.

Cp chnh xc ca cc chun Quc t c0,001%.

1.5.2. H thng n v

1.5.2.1.n vcbn: c th hin bng cc n v chun vi chnh xc cao nht

m khoa hc k thut hin i c th thc hin c.

1.5.2.2.n vko theo: l n v c lin quan n cc n vo cbn th hin qua

cc biu thc.

Ngy nay cc nc thng s dng h thng n v thng nht, l h thng

n v quc t SI; l h thng c thng qua hi ngh Quc t nm 1960. Trong

c 7 n v cbn l: mt(m) (chiu di), kilgam(kg) (khi lng), giy(s) (thi

gian), ampe(A) (cng dng in), K(nhit ), mol(n v s lng vt cht),

Cd(cng nh sng).

Ngoi by n v cbn trn cn c cc n v ko theo trong cc lnh vc c, in,

t, quang hc.... v c gii thiu trong bng 1.3.

1.5.3. Cc chun cp 1 Quc gia ca cc n v cbn h thng SI

Chun cp 1 l chun m bo to ra nhng i lng c n v chnh xc nht

ca mt quc gia.

1.5.3.1. Chun n v di

n v di (m). Mt l qung ng nh sng i c trong chn khng

trongkhong thi gian 1/299792458 giy ( CGPM * ln th 17, 1983. *CGPM tn vit

tt ting Php ca i hi cn o quc t).


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1.5.3.2. Chun n vkhi lng

Kilogam(kg) l n v khi lng bng khi lng ca mu kilogam quc t

t ti trung tm mu v cn quc tPari.

1.5.3.3. Chun n vthi gian

n v thi gian giy (s) l khong thi gian ca 9192631770 chu k pht x,

tng ng vi thi gian chuyn gia hai mc gn nht trng thi cbn ca nguyn

t xe-si 133.

1.5.3.4. Chun n vdng in

Ampe (A) l dng in khng i khi chy trong hai dy dn thng, song song,

di v hn, tit din dy trn nh khng ng k, t cch nhau 1mt trong chnkhng, s gy ra trn mi mt di ca dy mt lc 2.10-7 niutn ( CGPM ln th 9,

1948 ).

1.5.3.5. Chun n vnhit

n v nhit l Kelvin (K) - l nhit c gi tr bng 1/273, 16 phn

nhit ng ca im th 3 ca nc ( l im cn bng ca 3 trng thi rn, lng v

hi).

1.5.3.6. Chun n vcng nh sng

n v cng nh sng l Candela (Cd) l cng nh sng theo mt

phng xc nh ca mt ngun pht ra bc xn sc c tn s

540 x 1012 hec v c cng bc x theo phng l 1/638 ot trn steradian

( CGPM ln th 16, 1979 ).

1.5.3.7. Chun n vslng vt cht

n v s lng vt cht (mol) l s lng vt cht c s phn t ( nguyn t, ccht) bng s nguyn t cha trong 12C vi khi lng l 0,012 kg.


27/27


Bng 1.3. Mt sn vo cbn v ko theo

Cc i lng Tn n v K hiu

1. Cc i lng cbn di

Khi lngThi gianDng inNhit S lng vt chtCng nh sng

Mt

KilogamGiy

AmpeKelvinMn

Candela

m

kgsAK

molCd

2. Cc i lng chcTc Gia tcNng lng v cngLcCng sut

Nng lng

Mt trn giyMt trn giy bnh phng

JunNiutnWatt

Watt giy

m/sm/s2

JNW

Ws

3. Cc i lng inLng inin p, thin ngCng in trng

in dungin tr

in trringH sin mi tuyt i

CulngVn

Vn trn mt

Faram

m mtFara trn mt

CV

V/m

F

.m

F/m

4. Cc i lng tT thng

Cm ng tCng t trngin cmH s t thm

Vbe

TeslaAmpe trn mt

HenriHenri trn mt

Wb

TA/m

HH/m

5. Cc i lng quang

Lung nh sngCng sng ring

chiu sng

LumenCandela trn mt vung

Lux

LmCd/m2

lx

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