HD BTD Đo luong TTCN

7/30/2019 HD BTD o luong TTCN

1/25

TRNG I HC K THUT CNG NGHIP

KHOA IN TB mn: O LNG IU KHIN

BI TP LN

MN HC

O LNG V THNG TIN

CNG NGHIP

Sinh vin : DNG TH PHNG

L TRUNG THNH

Lp : H6

Gio vin hng dn : PHM VN THIM

Thi Nguyn2011


2/25

NHN XT(ca gio vin hng dn)


3/25


4/25

2

DANH MC HNH NHHnh 1. 1 : Tn s met cng hng ................................................................. 45Hnh 1. 2 : S o dng, p, cos, tn s, nng lng tc dng v phnkhng ............................................................................................................... 56Hnh 1. 3 : th vc t ca cng t tc dng ch nh mc ................ 78Hnh 1. 4 : th vc t ca cng t tc dng ch non ti ..................... 89Hnh 1. 5 : th vc t ca cng t phn khng 3 pha 2 phn t c cun dyni tip ph pha C ...................................................................................... 910Hnh 2. 1 : Mux 1 ........................................................................................ 1819Hnh 2. 2 : Mux 2 ........................................................................................ 1920Hnh 2. 3 : Mux 3 ........................................................................................ 2021Hnh 2. 4 : Mux 4 ........................................................................................ 2021Hnh 2. 5 : S h thng truyn dn thng tin ......................................... 2122

DANH MC BNG BIU

Bng 1 : Bng trng thi mux 1 .................................................................. 1920Bng 2 : Bng trng thi mux 2 .................................................................. 1920Bng 3 : Bng trng thi mux 3 .................................................................. 2021Bng 4 : Bng trng thi mux 4 .................................................................. 2122Bng 5 : Bng trng thi tng hp cc mux ................................................ 2122


5/25


6/25

4

Chng I : O LNG

1.1. S o dng, p, cos, tn s, nng lng tc dng, nng lngphn khng cho trm pha cao p

o dng in : S dng 3 Ampe k AA, AB, AC. o in p : S dng 6 Vol k : 3 Vol k VA, VB, VCo in p pha,

3 Vol k VAB, VBC, VCAo in p dy. o cos : S dng Cosk 3 pha. o nng lng tc dng : S dng cng t 3 pha 2 phn t c cun

dng mc pha A v C. Ta c :o Cun dng pha AA ( AI ) : Cun p mc pha A v B ( ABU ).o Cun dng pha C ( CI ) : Cun p mc pha C v B ( CBU ).

o nng lng phn khng : S dng cng t 3 pha 2 phn t ccun dy ni tip ph pha C. Ta c :

o Cun dng pha A ( AI ) : Cun p mc pha B v C ( BCU ).o Cun dng pha B ( BI ) : Cunp mc pha C v A ( CAU ).

o tn s : S dng Tn s met cng hng.

Hnh 1. 1 : Tn s met cng hngTn s met cng hng gm mt nam chm in, to ra bi cun dy

in qun trn li st hnh ch U, mt ming thp nm trong t trng canam chm in, gn cht vo n l cc thanh rung c tn s dao ng ringkhc nhau. Tn s dao ng ring ca hai thanh rung k nhau hn km nhaul 0,25 hoc 0,5 Hz ty theo chnh xc ca thit b. u t do ca ccthanh rung hng ra pha ngi quan st. Khi a in p c tn s fxcn ovo nam chm in, tt c cc thanh rung s dao ng. Tuy nhin thanh rung

Comment [a1]: Vh]a nu ra c m hiu (sseri) ca thit bang bn trn th trng


7/25

5

no c tn s dao ng ring bng tn s fx s dao ng cc i do cnghng ring, cn cc thanh khckhng cng hng th khng dao ng cci.Nh vy ta s c kt qu ti tr s tng ng vi thanh rung cc i.

V ta cn o cc thng s trn pha cao p ca trm bin p nn taphi dng bin dng (BI) v bin p (BU).S o nhHnh 1. 2Hnh 1. 2:

*

*

*

*

* ** *

* *

kWhkVARh

A B C

abc

MY BINP

Cao p

H p

BI

BU

AA

AC

VA VBC VCA HzA

AB

VABVB VC

o

* **

Cos

R R

HzB HzC

Hnh 1. 2 : S o dng, p, cos, tn s, nng lng tc dng v phn khng1.2. Chn thang o cho cc thit b trn s (Hnh 1. 2Hnh 1. 2)1.2.1.Chn h s bin dng in (KI)

Dng in nh mc pha s cp l :

(A)96,225.63

1000

U3

SI

dm

dmdm

Do ta chn bin dng in sao cho :(A)5I

(A)96,225I

tcdm

scdm

Vy ta chn BI c5

100K I

1.2.2.Chn h s bin in p (BU)in p nh mc pha s cp l : (KV)6Udm

Do ta chn bin in p sao cho :(KV)0,1(V)100U

(KV)6U

tcdm

scdm

Formatted: Font: 14 pt, Not Bold

Comment [a2]: Khng c nit khi c dytrung tnh

Formatted: Font: (Default) Times New Roman,Bold, Not Italic

Comment [a3]: Udm ly pha scp


8/25

6

Vy ta chn BU cU

6000K

100

1.2.3.Chn thang o cho Ampe kDng innh mc pha th cp ca bin dng l 5 (A) nn ta chn

ampe k c thang o DA = 5 (A).1.2.4.Chn thang o cho Vol k

in ppha th cp ca bin in p l 100 (V) nn ta chn vol k cthang o DV =100 (V).1.2.5.Chn thang o cho Cos k 3 pha

Cos k c cun dng mc vo th cp bin dng, cun p mc vo

th cpbin in p nn(V)100U

(A)5I

dmcos

dmcos

. Thang o cos k t -1 n 1.

1.2.6.Chn thang o cho tn s met cng hngTn s met cng hng mc vo th cp bin in p nn

(V)100Udmtsm . Thang o tn s met t 0 n 50 Hz.

1.2.7.Chn thang o cho cng t tc dngCng t tc dng c cun dng mc vo th cp bin dng, cun p

mc vo th cpbin in p nn :

(V)100U(A)5I

dmcttd

dmcttd

Thang o ca cng t tc dng : maxdmcttddmcttdcttd .t.cos.I.U3D vi

Udmcttd=100 (V), Idmcttd=5 (A), cos=1, tmax=24.30= 720 (h). Suy ra :

)(538,623)(623538720.1.5.100.3Dcttd kWhWh

1.2.8.Chn thang o cho cng t phn khngCng t phn khngc cun dng mc vo th cp bin dng, cun p

mc vo th cpbin in p nn :

(V)100U

(A)5I

dmcttd

dmcttd

Thang o ca cng t phn khng : maxdmctpkdmctpkctpk .t.sin.I.U3D

vi Udmctpk=100 (V), Idmctpk=5 (A), sin=1, tmax=24.30= 720 (h). Suy ra :

kWh623,538Wh62353820.100.5.1.73Dctpk

Comment [a4]: 0.5 n 1

Comment [a5]: 45 n 55 Hz

Comment [a6]: Chn thang o ch c nghatrong vic tnh sai s ca php o


9/25

7

1.3. Tnh s ch ca mi cng t trong mt thng, xc nh costb1.3.1.Tnh s ch ca cng t tc dng

Cng t tc dng 3 pha 2 phn t c cun dng pha A v C, tac: ( AI

, ABU ) v ( CI

, CBU ).

Xt cng sut tc thi trong mch 3 pha :p3pha = uAiA + uBiB + uCiC.

y l mch ba pha ba dy, v khng c dy trung tnh nn dng intrung tnh bng khng, ngha l : iA + iB + iC = 0 hay iB = - (iA + iC)

Vy : p3pha = uAiA - uB(iA + iC) + uCiC= iA(uA - uB) + iC(uCuB)= iAuAB + iCuCB

Suy ra cng sut tc dng ba pha l :

)I,Ucos(IU)I,Ucos(IUP CCBCCBAABAAB3pha

=>Nng lng tc dng 3 pha l :

.tI,U.cos.IUI,U.cos.IU.KK

1W CCBCCBAABAAB

IU

cttd

= Wcttddm + Wcttdnt

1.3.1.1. S ch cng t tc dng ch nh mcTa c : tdm = 0,8.tmax = 0,8.720 = 576 (h) ; IAdm = ICdm = Idm 96,225 (A)UAB = UCB = Udm = 6 (KV); Cosdm= 0,95 => dm

o18,19 . th vec t :

Hnh 1. 3 : th vc t ca cng t tc dng ch nh mc

18,19oBI

AU

CU

CBU

BU

AI

CI

ABU

18,19o

18,19o

30o


10/25


11/25

9

Hnh 1. 5 : th vc t ca cng t phn khng 3 pha 2 phn t c cun dy ni tipph pha C

Mch bapha c ngun i xng, xt tng phn t, ta tnh c momenquay nh sau :

q2q1q

CACCCABBCAq2

BCCCBCAABCq1

MMM

U,IcosIU,IcosIKUM

U,IcosIU,IcosIKUM

3phaq

CCCBBBAAAq

CCCC

BBCCCCAAdq

C0

C

B0

BC0

CA0

Adq

CACC

CABBBCCCBCAAdq

Q3KM

sinUIsinUIsinUI3KM

)sin2

1Icos

2

3I

sinIsin2

1Icos

2

3Isin(IKUM

]30cosI

90cosI150cosI90cos[IKUM

]U,IcosI

U,IcosIU,IcosIU,Icos[IKUM

M men quay t l vi cng sut phn khng trong mch ba pha, vy sch ca cng t t l vi nng lng phn khng trong mch ba pha.

Cng sut phn khng ba pha l :

AU

CU

BCU

BU

AI

CI

CAU

A

BI

BC


12/25


13/25

11

0,927206,279509,833

509,833

WW

W

QP

Pcos

222ctpk

2cttd

cttd

22tb

1.4. Tnh sai s tuyt i v tng i ca cc php o trong 2 trnghp ph ti

Gi s c mt php o gin tip i lng y thng qua cc php o trctip x1, x2,, xn : y = f(x1, x2,, xn).

Ta c :

n

n

2

2

1

1

dx

x

y...dx

x

ydx

x

ydy

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

*xx

yx

x

y...x

x

yx

x

yy

n

1i

2

i

i

2

n

n

2

2

2

2

1

1

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

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

y

yy . Nu tnh theo %, ta c : %100.% yy .

Vi hm y = x1.x2 th ta c : **xxx

x

yx

x

yy

2

221

2

122

2

2

2

2

1

1

xx

Vi hm y = x1.x2.x3 th ta c :

2322

21

2

223

21

2

123

22

2

3

3

2

2

2

2

1

1

xxx

xx

yx

x

yx

x

yy

xxxxxx

***

1.4.1.Sai s ca cc php o ch nh mc1.4.1.1. Sai s ca php o dng in

Dng in qua cc pha A, B, C (Ba pha nh nhau) c xc nh theocng thc : Idm = KI.IAdmvi IAdml dng m Ampe k o c ch nh

mc : (A)4,813

5

100

96,255

K

II

I

dmAdm .

Sai s tuyt i ca Ampe k l :

Comment [a7]: Chia lm hai trng hp non tiv nh mc


14/25

12

IAdm = A.DA = 2%.5 = 0,1 (A)Sai s tuyt i ca bin dng in l :

KI = BI. KI = 1%.5

100= 0,2 (A)

p dng cng thc (**) ta c : sai s tuyt i ca php o dng in ch nh mc l :

(A)2,220

.0,1)5

100(.4,8130,2

I.K.IKI

2222

2

Adm2I

2Adm

2

Idm

Sai s tng i ca php o dng in ch nh mc l :

2,307%.100%96,225

2,220.100%

I

I

dm

dm%Idm

1.4.1.2. Sai s ca php o in pin p dy UAB = UBC = UCA = Ud = Udm. Ta li c Udm = KU.UVVi UV l in p m Vol k VAB, VBC, VCA o c ch nh

mc, UV = 100 (V) => Ud = KU.UV.

in p pha UA = UB = UC = Upha =3

Udm => Upha =3

UK VU .

in p dy o c :

Ud = KU.UV =1

60.100 = 6000 (V)

in p pha o c :

Upha =3

UK VU =3.1

60.100 3463,102 (V)

Sai s tuyt i ca Vol k l :UV = V.DV = 2,5%.100 = 2,5 (V)Sai s tuyt i ca bin in p l :

KU = BU. KU = 1%.1

60= 0,6 (V)

p dng cng thc (**) ta c : sai s tuyt i ca php o in p dy ch nh mc l :

(V)161,555

.2,5)1

60(.1006,0

U.K.UKU

2222

2

V2U

2V

2

Ud

Comment [a8]: Sai s chp nhnc


15/25

13

Sai s tng i ca php o in p dy ch nh mc l :

2,693%.100%6000

161,555.100%

U

U

d

d%Ud

Sai s tuyt i ca php o in p pha ch nh mc l :

(V)948,531

.2,5)1

60()

3

100.(6,0

U.K)3

U.(KU

2222

2

V2U

2V2

Upha

Sai s tng i ca php o in p pha ch nh mc l : 4,445%.100%

3463,102

153,948.100%

U

U

pha

pha

%Upha

1.4.1.3. Sai s ca php o nng lng tc dngNng lng tc dng ca ph ti ch nh mc l :

(kWh)547215.456,0135

100.

1

60.W.KKW cttddmIUtddm

Sai s tuyt i ca cng t tc dng ch nh mc l :Wtddm= cttd.DcttddmVi :

(kWh)498,830(Wh)498830

76.100.5.1.53.t.cos.IU3D dmdmcttddmcttdcttddm

=> Wtddm= cttd.Dcttddm= 1,5%.498,830 7,482 (kWh)p dng cng thc (***) ta c:sai s tuyt i ca php o nng lng

tc dng ch nh mc l :

(kWh)297,18531

482,7.5

100.)

1

60(013,456..0,2)

1

60(013,456.

5

100.6,0

W.K.KW.K.K.W.KKW

22

2

222222

2

2

2

cttddm2I

2U

2cttddm

2

I2U

2cttddm

2I

2

Utddm

Sai s tng i ca php o nng lngtc dng ch nh mcl:

2,166%.100%547215

11853,297.100%

W

W

tddm

tddm%Wtddm

1.4.1.4. Sai s ca php o nng lng phn khngNng lng phn khng ca ph ti ch nh mc l :


16/25


17/25


18/25

16

(kVARh)67766.56,4725

100.

1

60.W.KKW ctpkntIUpknt

Sai s tuyt i cacng t phn khng ch non ti l :Wpknt= ctpk.DctpkntVi :

(kVARh)124,707

(VARh)124707

44.100.5.1.13

.t.sin.IU3D ntdmctpkdmctpkctpknt

=> Wpknt= ctpk.Dctpknt= 1,5%.124,707 1,871 (kVARh)p dng cng thc (***) ta c : sai s tuyt i ca php o nng

lng phn khng ch non ti l :

(kVARh)184,2441

871,1.5

100.)

1

60(472,56..0,2)

1

60(472,56.

5

100.6,0

W.K.KW.K.K.W.KKW

22

2

222222

2

2

2

ctpknt2I

2U

2ctpknt

2

I2U

2ctpknt

2I

2

Upknt

Sai s tng i ca php o nng lng phn khng ch non ti l:

%602,3.100%67766

2441,184.100%

W

W

pknt

pknt

%Wpknt

1.4.3.Sai s ca php oCostbTa c :

2ctpk

2cttd

cttdtb

WW

Wcos

Sai s tuyt i ca cng t tc dng l :Wcttd= cttd.Dcttd = 1,5%.623,538 9,353 (kWh)Sai s tuyt i ca cng t phn khng l :Wctpk= ctpk.Dctpk= 1,5%.623,538 9,353 (kWh)p dng cng thc (*) ta c : sai s tuyt i ca php o cos tb l :

Comment [a9]: Tnh cho hai trng hp


19/25

17

006,0

353,9.

279,206833,509

279,206.833,509353,9.

279,206833,509

279,206

W

WW

.WW-W

WW

W

WW

cosW

W

coscos

2

322

2

322

2

2

ctpk32

ctpk2cttd

ctpkcttd

2

cttd32

ctpk2cttd

2ctpk

2

ctpk

ctpk

tb

2

cttd

cttd

tbtb

Sai s tng i ca php o costb l :

%647,0.100%0,927

006,0.100%

cos

cos

tb

tb%cos tb


20/25

18

Chng II : H THNG THNG TIN CNG NGHIP

2.1. Thit k cc thnh phn ca h thng truyn dnH thng truyn dn c khong cch 2km nn l h thng truyn

thng xa. H thng s dng m hnh c cc knh thu thp tn hiu song songni tip.

S lng knh l 12 knh :o Knh 1 (S1) : Tn hiu dng IAt ampe k AA.o Knh 2 (S2) : Tn hiu dng IBt ampe k AB.o Knh 3 (S3) : Tn hiu dng ICt ampe k AC.o Knh 4 (S4) : Tn hiu in p UAt vol k VA.o Knh 5 (S5) : Tn hiu in p UBt vol k VB.o Knh 6 (S6) : Tn hiu in p UCt vol k VC.o Knh 7 (S7) : Tn hiu in p UABt vol k VAB.o Knh 8 (S8) : Tn hiu in p UBCt vol k VBC.o Knh 9 (S9) : Tn hiu in p UCAt vol k VCA.o Knh 10 (S10) : Tn hiu gc pha cos t cos met ba

pha.

o Knh 11 (S11) : Tn hiu tn s pha A t tn s met HzA.o Knh 12 (S12) : Tn hiu tn s pha B t tn s met HzB.

Thit k cc Mux : Mux s dng y l CD4051B.o Mux 1 : 2 bit, dng cho 3 tn hiu dng in

S1 D0 S2 D1 S3 D2

MUX 1

CLK

0Q

0Q1D

1A0A

0D

2D

Hnh 2. 1 : Mux 1


21/25

19

Bng 1 : Bng trng thi mux 1Chn a ch Chn clock u raA0 A1 CLK Q0X X 0 0

0 0 1 D00 1 1 D11 0 1 D2X X 1 0

o Mux 2 : 3 bit, dng cho 6 tn hiu in p S4 D3 S5 D4 S6 D5 S7 D6 S8 D7 S9 D8

MUX 2

CLK

1Q

1Q4D

4A3A

3D

5D

6D

7D

8D

2A

Hnh 2. 2 : Mux 2

Bng 2 : Bng trng thi mux 2Chn a ch Chn clock u ra

A2 A3 A4 CLK Q1

X X X 0 0

0 0 0 1 D30 0 1 1 D4

0 1 0 1 D50 1 1 1 D6

1 0 0 1 D7

1 0 1 1 D8

X X X 1 0


22/25


23/25

21

Bng 4 : Bng trng thi mux 4Chn a ch Chn clock u raA7 A8 CLK Q4X X 0 0

0 0 1 D120 1 1 D131 0 1 D14X X 1 0

Bng 5 : Bng trng thi tng hp cc mux

Chn a chChn

Clock u ra

A0 A1 A2 A3 A4 A5 A6 A7 A8 CLK Q1 Q2 Q3 Q4

X X X X X X X X X 0 0 0 0 0

0 0 X X X X X 0 0 1 D0 0 0 Q1

0 1 X X X X X 0 0 1 D1 0 0 Q1

1 0 X X X X X 0 0 1 D2 0 0 Q1

X X 0 0 0 X X 0 1 1 0 D3 0 Q2

X X 0 0 1 X X 0 1 1 0 D4 0 Q2

X X 0 1 0 X X 0 1 1 0 D5 0 Q2

X X 0 1 1 X X 0 1 1 0 D6 0 Q2

X X 1 0 0 X X 0 1 1 0 D7 0 Q2

X X 1 0 1 X X 0 1 1 0 D8 0 Q2

X X X X X 0 0 1 0 1 0 0 D9 Q3

X X X X X 0 1 1 0 1 0 0 D10 Q3

X X X X X 1 0 1 0 1 0 0 D11 Q3

X X X X X X X X X 1 0 0 0 0

2.2. S h thng truyn dnHnh 2. 5 : S h thng truyn dn thng tin


24/25

22

KT LUNQua qu trnh thc hin bi tp di, ta c th nm vng c

cc cch mc cc thit b o, cc phng php o dng in, inp, cos, tn s, nng lngtc dng v phn khngtrong li inba pha cao th; cch tnh cc sai s tuyt i, tng i ca ccphp o.Bn cnh , ta bit c cch xy dng h thng truyndn thng tin i xa, phc v cc mc ch o lng, bo v chotrm bin p.


25/25

23

TI LIU THAM KHO

1. Nguyn Hu Cng, Nguyn Vn Ch; Gio trnh k thut olng, Nh xut bn i hc Quc gia H Ni, 2008.

HD BTD Đo luong TTCN

Documents

Transcript of HD BTD Đo luong TTCN