Extension of Range.

Transcription

1 Extenion of ange Shunt are ued for the extenion of range of Aeter. So a good hunt hould have the following propertie:- 1- The teperature coefficient of hunt hould be low 2- eitance of hunt hould not vary with tie 3- They hould carry current without exceive teperature rie 4- They hould have theral electrootive force with copper * Manganin i ued for DC hunt and Contantan a AC hunt. Aeter:- PMMC i ued a indicating device. The current capacity of PMMC i all. t i ipractical to contruct a PMMC coil, which can carry a current greater than 100 A. Therefore a hunt i required for eaureent of large current. = nternal reitance of oveent (coil) in Ω h = eitance of hunt in Ω = f = Full cale deflection current of oveent in Apere h = Shunt current in Apere = Current to be eaured in Apere Since the hunt reitance i in parallel with the eter oveent, the voltage drop acro hunt and oveent ut be ae. = h h h = = h We can write of hunt 1 = = 1 + h h h h = ( ) = i known a ultiplying power = eitance of hunt h ( 1) Or h = 1 1

2 Multi ange Aeter:- Let 1, 2, 3, 4 be the hunt ultiplying power for current 1, 2, 3, 4. = 1 h 1 = ( ) 1 ( 1) 2 h 2 = ( 1) 3 h 3 = ( 1) 4 h 4 Volteter:- For eaureent of voltage a erie reitor or a ultiplier i required for extenion of range. = Deflection current of oveent = nternal reitance of oveent = Multiplier reitance V = Full range voltage of intruent V = + ( ) V V = = * For ore than 500 V ultiplier i ounted outide the cae. Multi ange Volteter: V = 1 1 V = 2 2 V = 3 3 V = 4 4 * For average value divide the reading by For peak value ultiply the voltage by To get peak-to-peak ratio ultiply the reading by ** Therocouple and hot wire intruent are ued for eaureent of true power and r value of voltage & current. Volteter & Aeter by Moving Coil ntruent:- Sae proce a applied in PMMC. 2

3 Electrodynaic type Volteter & Aeter:- Shunt i connected acro the circuit for aeter and ultiplier reitance i connected in erie for volteter. Extenion of ange Nuerical Exaple 1:- A oving coil aeter ha a full cale deflection of 50 µ Ap and a coil reitance of 1000 Ω. What will be the value of the hunt reitance required for the intruent to be converted to read a full cale reading of 1 Ap. Solution 1:- Full cale deflection current ntruent reitance = 1000 Ω Total current to be eaured = 1 A eitance of aeter hunt required h 6 = 50 *10 A 1000 = = *10 Exaple 2:- The full cale deflection current of an aeter i 1 A and it internal reitance i 100 Ω. f thi eter i to have cale deflection at 5 A, what i the value of hunt reitance to be ued. Solution 2:- Full cale deflection current = 1 A = A ntruent reitance = 100 Ω. Total current to be eaured = 5 A 100 eitance of aeter hunt required h = = h = Ω Exaple 3:- The full cale deflection current of a eter i 1 A and it internal reitance i 100 Ω. f thi eter i to have full-cale deflection when 100 V i eaured. What hould be the value of erie reitance? Solution 3:- ntruent reitance = 100 Ω 3 Full-cale deflection current = 1 A = 1*10 A Voltage to be eaured V = 100 V V 100 equired erie reitance e = = 100 = 99, 900 Ω 3 1*10 3

4 Exaple 4:- A PMMC intruent give full cale reading of 25 A when a potential difference acro it terinal i 75 V. Show how it can be ued (a) a an aeter for the range of A (b) a a volteter for the range of V. Alo find the ultiplying factor of hunt and voltage aplification. Solution 4:- Potential drop acro ter in al 75*10 ntruent reitance = = ntruent current 25*10 (a) Current to be eaured = 100 A Multiplying power of hunt = 100 = *10 = Shunt reitance required for full cale deflection at 100 A h = = = = 7.50 *10 = Ω (b) Voltage to be eaured V = 750 V V 750 e = = 3 = 29, 997 Ω 3 25*10 Voltage aplification = *10 = An. 3 3 = 3Ω Exaple 5:- A oving coil intruent give full cale deflection of 10 A and potential difference acro it terinal i 100 V. Calculate (a) hunt reitance for full-cale deflection correponding to 200 A (b) Serie reitance for full reading correponding to 1000 V. Solution 5:- 3 Potential drop acro ter in al 100*10 ntruent reitance = = = 10Ω 3 ntruent current 10*10 (a) Shunt reitance required for full cale deflection correponding to 200 A 10 4 h = = = *10 Ω *10 (b) Serie reitance required for full cale deflection correponding to 1000 V V 1000 e = = 10 = 99, 990 Ω 3 10*10 Exaple 6:- A oving coil intruent having internal reitance of 50 Ω indicate full cale deflection with a current of 10 A. How can it be ade to work a (i) a volteter to read 100 V on full cale (ii) an aeter of 1 A, on full cale? Solution 6:- eitance of the intruent coil = 50 Ω Current flowing through the intruent for full-cale deflection = 10 A = A 4

5 (i) Serie reitance required to eaure 100 V V 100 e = = 50 = 9950 Ω 0.01 (ii) Shunt required to eaure 1 A current 50 h = = = Ω Exaple 7:- A oving coil intruent ha a reitance of 2 Ω and it read upto 250 V when a reitance of 5000 Ω i connected in erie with it. Find the current range of the intruent when it i ued a aeter with the coil connected acro a hunt reitance of 2 illi Ω. Solution 7:- eitance of the intruent coil = 2 Ω Current flowing through the intruent for full-cale deflection Full cale reading = + Serie rei tan ce 250 = = A = A Shunt reitance = 2*10 Ω h *10 * 2 Current through hunt reitance = = = *10 Current range of intruent = Full cale deflection current = + = = 50 A h 98 Exaple 8:- A oving coil aeter give full cale deflection with 15 A and ha a reitance of 5 Ω. Calculate the reitance to be calculated in (a) Parallel to enable the intruent to read upto 1 A (b) Serie to enable it to read up to 10 V. [UPTU 2002] A Solution 8:- ntruent reitance = 5 Ω Full cale deflection current = 15 A = 15*10 (a) Current to be eaured = 1 A Shunt reitance to be connected in parallel 5 h = = = Ω *10 (b) Voltage to be eaured V = 10 V Serie reitance required 3 A 5

6 e V 10 = = 5 = *10 Ω ntruent Tranforer Baic Why intruent tranforer? n power yte, current and voltage handled are very large. Direct eaureent are not poible with the exiting equipent. Hence it i required to tep down current and voltage with the help of intruent tranforer o that they can be eaured with intruent of oderate ize ntruent Tranforer Tranforer ued in conjunction with eauring intruent for eaureent purpoe are called ntruent Tranforer. The intruent ued for the eaureent of current i called a Current Tranforer or iply CT. The tranforer ued for the eaureent of voltage are called Voltage tranforer or Potential tranforer or iply PT. ntruent Tranforer: Fig 1. Current Tranforer Fig 2. Potential Tranforer Fig 1. indicate the current eaureent by a C.T. The current being eaured pae through the priary winding and the econdary winding i connected to an aeter. The C.T. tep down the current to the level of aeter. Fig 2. how the connection of P.T. for voltage eaureent. The priary winding i connected to the voltage being eaured and the econdary winding to a volteter. The P.T. tep down the voltage to the level of volteter. Merit of ntruent Tranforer: 1. ntruent of oderate ize are ued for etering i.e. 5A for current and 100 to 120 volt for voltage eaureent. 6

7 2. ntruent and eter can be tandardized o that there i aving in cot. eplaceent of daaged intruent i eay. 3. Single range intruent can be ued to cover large current or voltage range, when ued with uitable ulti range intruent tranforer. 4. The etering circuit i iolated fro the high voltage power circuit. Hence iolation i not a proble and the afety i aured for the operator 5. There i low power conuption in etering circuit. 6. Several intruent can be operated fro a ingle intruent tranforer. atio of ntruent Tranforer: Soe definition are: 1. Tranforation ratio: t i the ratio of the agnitude if the priary phaor to econdary phaor. Tranforation ratio: for a C.T. for a P.T. Noinal atio: t i the ratio of rated priary winding current (voltage) to the rated econdary winding current (voltage). for a C.T. Turn ratio: Thi i defined a below. for a P.T. 7

8 for a C.T. for a P.T. Burden of an ntruent Tranforer: The rated burden i the volt apere loading which i periible without error exceeding the particular cla of accuracy. Current Tranforer equivalent circuit: X1 = Priary leakage reactance 1 = Priary winding reitance X2 = Secondary leakage reactance 8

9 Z0 = Magnetizing ipedance 2 = Secondary winding reitance Zb = Secondary load Note: Norally the leakage fluxe X1 and X2 can be neglected Current tranforer, iplified equivalent circuit: Current tranforer: Phae diplaceent (δ) and current ratio error (ε): Current Tranforer Baic: Current Tranforer (CT ) can be ued for onitoring current or for tranforing priary current into reduced econdary current ued for eter, relay, control equipent and other intruent. CT that tranfor current iolate the high voltage priary, perit grounding of the econdary, and tep-down the agnitude of the eaured current to a tandard value that can be afely handled by the intruent. 9

10 atio :The CT ratio i the ratio of priary current input to econdary current output at full load. For exaple, a CT with a ratio of 300:5 i rated for 300 priary ap at full load and will produce 5 ap of econdary current when 300 ap flow through the priary. f the priary current change the econdary current output will change accordingly. For exaple, if 150 ap flow through the 300 ap rated priary the econdary current output will be 2.5 ap (150:300 = 2.5:5). Current Tranforer: Caution: npect the phyical and echanical condition of the CT before intallation. Check the connection of the tranforer requireent for the intruent or the yte requireent before connecting the CT. npect the pace between the CT phae, ground and econdary conductor for adequate clearance between the priary and econdary circuitry wiring. Verify that the horting device on the CT i properly connected until the CT i ready to be intalled. The econdary of the CT ut alway have a burden (load) connected when not in ue. NOTE: A dangerouly high econdary voltage can develop with an open-circuited econdary. Contruction of Current Tranforer: Current tranforer are contructed in variou way. n one ethod there are two eparate winding on a agnetic teel core. The priary winding conit of a few turn of heavy wire capable of carrying the full load current while the econdary winding conit of any turn of aller wire with a current carrying capacity of between 5/20 apere, dependent on the deign. Thi i called the wound type due to it wound priary coil. Wound Type Contruction of Current Tranforer: 10

11 Another very coon type of contruction i the o-called window, through or donut type current tranforer in which the core ha an opening through which the conductor carrying the priary load current i paed. Thi priary conductor contitute the priary winding of the CT (one pa through the window repreent a one turn priary), and ut be large enough in cro ection to carry the axiu current of the load. Window-type Contruction of Current Tranforer: Another ditinguihing feature i the difference between indoor and outdoor contruction. 15kV Outdoor CT 15kV ndoor CT 11

12 Contruction of Current Tranforer ndoor Type Outdoor Type indoor unit are protected due to their being ounted in an encloure of oe kind Not equired Not equiored The indoor type ut be copatible for connection to bu type electrical contruction The outdoor unit ut be protected for poible containated environent outdoor unit will have larger pacing between line and ground, which i achieved by the addition of kirt on the deign. For outdoor type the hardware ut be of the non-corroive type and the inulation ut be of the non-arctracking type. outdoor type are norally on the poletop intallation. Circuit connection for current and power eaureent uing C.T. 12

13 Equivalent Circuit of C.T. Fig. 1 Equivalent circuit of C.T. Phaor diagra 13

14 Fig 2. Phaor diagra A ection of Phaor diagra elationhip in current tranforer: Fig 3. A ection of Phaor diagra Fig 1 repreent the equivalent circuit and Fig 2 the phaor diagra of a current tranforer. n= turn ratio = (No. of econdary winding turn)/(no. of priary winding turn) r = reitance of the econdary winding; x = reactance of the econdary winding; re = reitance of external burden i.e. reitance of eter, current coil etc. including lead; xe = reactance of external burden i.e. reactance of eter, current coil etc. including lead; Ep = priary winding induced voltage ; E = econdary winding induced voltage; Np = No. of priary winding turn; N = No. of econdary winding turn; V = Voltage at the econdary winding terinal; = econdary winding current; p = priary winding current; ϴ = phae angle of tranforer; Φ = working flux of the tranforer; δ = angle between econdary winding induced voltage and econdary winding current = phae angle of total burden including ipedance of econdary winding 1 x + xe tan ( ) r + r e 14

15 = phae angle of econdary winding load circuit i.e. of external burden 1 xe tan ( ) r o = exiting current; = agnetizing coponent of exciting current, e = lo coponent of exciting current, α = angle between exciting current o and working flux ϕ Conider a all ection of the phaor a hown in Fig. 3. We have bac = 90 -δ-α, ac = o, oa = n and oc = p. bc= o in (90 -δ-α) = o co (δ+α), ab = o co (90 -δ-α) = o in (δ+α). Now (Oc) 2 = (oa + ab) 2 + (bc) 2 or p 2 = [n + o in (δ+α)] 2 + [o co (δ+α)] 2 = n in 2 (δ+α) + 2 noin (δ+α)+ o 2 co 2 (δ+α). = n no in (δ+α) + o 2 p = [n no in (δ+α) + o 2 ] ½ (1) Tranforation ratio = p / = [n2 o in (δ+α) + o2]½ / (2) Now in a well deigned current tranforer o << n. Uually o i le than 1 percent of p, and p i, therefore, very nearly equal to n. Eqn. (2) can be written a e δ + α + 0 δ + α [ n 2n in( ) in ( ) = n + 0 in( δ + α) 0 = n + in( δ + α)(3) The above theory i applicable to cae when the econdary burden ha a lagging power factor i.e., when the burden i inductive. Eqn. (3) can be further expanded a: n + 0 (in δ coα + coδ in α) in δ + e co δ n + 15

16 A co = 0 αand in e = 0 α Phae angle The angle by which the econdary current phaor, when revered, differ in phae fro priary current, i known a the phae angle of the tranforer. +ve if econdary revered current lead the priary current -ve if econdary revered current lag behind the priary current. The angle between and p i θ. Therefore, the phae angle i θ. Fro the phaor diagra, A θ i very all, we can write Now o i very all a copared to n, and, therefore we can neglect the ter o in(δ+α) 0 (4) bc bc 0 co( δ + α ) tan θ = = = ob oa + ab n + in( δ + α ) θ = 0 co ( δ + α ) n + in ( δ + α ) 0 ra d. (5) co( ) 0 δ + α (6) θ = rad n 0 coδ coα 0 inδ inα coδ e inδ rad n n c o in ( δ e δ ) d eg ree Π n (7) (8) Error in current tranforer: Turn ratio and tranforation ratio are not equal. The value of tranforation ratio i not contant. t depend upon: 1. Magnetizing and lo coponent of exciting current, 2. The econdary winding load current and it power Thi introduce coniderable error into current eaureent 16

17 n power eaureent it i neceary that the phae of econdary winding current hall be diplaced by exactly 180 fro that of the Priary current. Here, phae difference i different fro 180 by an angle θ. Hence due to C.T. two type of error are introduced in power eaureent. Due to actual tranforation ratio being different fro the turn ratio. Due to econdary winding current not being 180 out of phae with the priary winding current. atio error and phae angle error atio Error i defined a: Percentage ratio error = ((noinal ratio actual ratio)/(actual ratio))x100 Kn = 100 (9) Phae angle c o in ( δ e δ ) d e g Π n r e e Approxiate forula for error: The uual intruent burden i largely reitive with oe inductance. Therefore, δ i poitive and all. Hence, in δ = 0 and co δ = 1. Therefore equation (4) and (8) can be written a: But, And (10) and (11) e 180 = n + θ = deg ree and, Π n therefore, eqn. (10) and (11) can be rewritten a n p n e n + = n 1 + p θ = d e g Π p e p r e e Proble No.1 Two current tranforer of the ae noinal ratio 500/5 A, are teted by Silbee ethod. With the current in the econdary of the tranforer adjuted at it rated value, the content in the iddle conductor = 0.05e-j126.9 A expreed with repect to current in the econdary of tandard tranforer a the reference. t i known that tandard tranforer ha a ratio correction factor (CF) of and phae error +8. Find CF and phae angle error of tranforer under tet. 17

18 Solution: Noinal ratio = 500/5 =100, = 5 A Since i the reference, = 5+j0 = 0.05e-j126.9 = 0.05(co j in ) = j 0.04 Now, current in the econdary of tet tranforer x=- = 5 +j 0 (0.03 j 0.04) = j 0.04 or x 5.03 A Angle between x and = 0.04/5.03 rad = Phae angle between x revered and p= = atio correction factor CF of tandard tranforer = Therefore, actual ratio of tandard tranforer = CF noinal ratio = = Priary current p= = = A Actual ratio of tranforer under tet x= p/ = /5.03 = atio correction factor of tet tranforer CF = Actual ratio/noinal ratio = 99.55/100 = Proble on CT 1. A current tranforer ha a ingle turn priary and 200 turn econdary winding. The econdary winding upplie a current of 5 A to a non inductive burden of 1Ω reitance. The requiite flux i et up in the core by an f of 80 A. The frequency i 50 Hz and the net cro ection of the core i 1000 q.. Calculate the ratio and phae angle of the tranforer. Alo find the flux denity in the core. Neglect the effect of agnetic leakage, iron loe and copper loe. Solution: f we neglect the agnetic leakage, the econdary leakage reactance becoe zero. Therefore, the econdary burden i purely reitive and the ipedance of the burden i equal to the reitance of the econdary winding. pedance of econdary circuit = 1 Ω Voltage induced in the econdary winding E = Current in the econdary ipedance of econdary winding = 5 1 =5 V. A the econdary burden i purely reitive, the econdary current i in phae with econdary induced voltage and the econdary power factor i unity or δ = 0. The lo coponent of no load current i to be neglected and, therefore, e = 0. = e 18

19 Exciting Current = = agnetizing coponent and α = 0. Magnetizing coponent of no load current =(agnetizing f)/(priary winding turn) =80/1=80A Secondary winding current = 5 A eflected econdary winding current = n =200 5 =1000A Fro the phaor diagra hown in Fig. 1 n S P θ S Φ Fig. 1 = A. Actual tranforation ratio = p/r = /5 =2000. Phae angle θ = tan 1 (/n) = tan 1 (80/1000) = 4 34 We Know that, E = 4.44fΦ N Φ = E/(4.44 f N) = 5/( ) = wb. Area of core = 1000 Sq.. = 1000 µ q. Maxiu flux denity B=(0.1126wb)/1000 µ q. = Wb/Sq.. Proble No. 2 = p

20 A current tranforer with a bar priary ha 300 turn in it econdary winding. The reitance and reactance of econdary circuit are 1.5Ω and 1.0 Ωrepectively including the tranforer winding. With 5 A flowing in the econdary winding, the agnetizing f i 100 A and the iron loe i 1.2W. Deterine the ratio and phae angle error. Solution: Priary winding turn Np= 1; Secondary winding turn N = 300; Turn ratio = N/Np = 300/1 = 300. Secondary circuit burden ipedance = 2 2 (1.5) + (1.0) = 1.8 Ω For econdary winding circuit: co δ = 1.5/1.8 = and in δ = 1.0/1.8 = Secondary induced voltage E = = 9.0 V. Priary induced voltage Ep = E/n = 9.0/300 = 0.03 V. Lo coponent of current referred to priary winding e = iron lo/(ep) = 1.2/0.03 = 40 A. Magnetizing current = (agnetizing f)/(priary winding turn) = 100/1 = 100 A Actual ratio = = 300+( )/5 = n the abence of any inforation to the contrary we can take noinal ratio to be equal to the turn ratio, or Kn = n = 300 Percentage ratio error Kn = 100 = ( )/317.6 = -5.54%. Phae angle θ = = 180/π (( )/(300 5)) = Proble No. 3 A 100/5 A, 50 Hz CT ha a bar priary and a rated econdary burden of 12.5 VA. The econdary winding ha 196 turn and a leakage inductance of 0.96 H. With a purely 20

21 reitive burden at rated full load, the agnetization f i 16 A and the lo excitation require 12 A. Find the ratio and phae angle error. Solution: Secondary burden = 12.5 VA. Secondary winding current = 5 A Secondary circuit ipedance = 12.5/5 2 = 0.5 Ω. Secondary circuit reactance = 2π = 0.3 Ω Phae angle of econdary circuit δ = in /0.5 = in Therefore, in δ = 0.6 and co δ = (1) 2 + (0.6) 2 = 0.8. Priary winding turn Np = 1. Secondary winding turn N = 196. Turn ratio n = N/Np = 196. Noinal ratio = Kn = 1000/5 = 200 Magnetizing current = (agnetizing f)/(priary winding turn) = 16/1 =16 A Lo coponent e = (excitation for lo/priary winding turn). = 12/1 = 12 A. Actual ratio = = (( )/5) = atio error = ((noinal ratio actual ratio)/(actual ratio)) 100 = (( ) / ) 100 = +0.08% Phae angle θ = 180/π (( )/(196 5)) =0.327 = Potential Tranforer Baic Potential tranforer are norally connected acro two line of the circuit in which the voltage i to be eaured. Norally they will be connected L-L (line-to-line) or L-G (line-to-ground). A typical connection i a follow: 21

22 Fig. 2 elationhip in a Potential Tranforer: The theory of a potential tranforer i the ae a that of a power tranforer. The ain difference i that the power loading of a P.T. i very all and conequently the exciting current i of the ae order a the econdary winding current while in a power tranforer the exciting current i a very all fraction of econdary winding load current. Fig 3. And Fig 4. how the equivalent circuit and phaor diagra of a potential tranforer repectively. = econdary winding current, r = reitance of econdary winding Φ = working flux in wb. = agnetizing coponent of no load (exciting) current in A, e = iron lo coponent of no load (exciting) current in A, 0 = no load (exciting) current in A, E = econdary winding induced voltage V = econdary winding terinal voltage, N p = priary winding turn N = econdary winding turn x = reactance of econdary winding r e =reitance of econdary load circuit x e = reactance of econdary load circuit = phae angle of econdary load circuit = tan -1 x e /r e E p = priary winding induced voltage, p = priary winding current r p = reitance of priary winding x p = reactance of priary winding Turn ratio n = N p / N = E p / E Secondary voltage when referred to priary ide are to be ultiplied by n. When econdary current are referred to priary ide, they ut be divided by n 22

23 Fig. 3 Equivalent circuit of a P.T. Fig. 4 Phaor diagra of P.T. 23

24 Fig. 5 Enlarged and concie phaor diagra of a P.T Actual Tranforation ratio An enlarged concie phaor diagra i hown in Fig. 5. θ = phae angle of the tranforer = angle between V P and V S revered = phae angle of econdary load circuit β = phae angle between P and V S revered. Now oa = V P co θ Fro Phaor diagra oa = n V S +n S r S co + n S x S in + P r P co β+ P x P in β Or V P co θ = n V S +n S r S co + n S x S in + P r P co β+ P x P in β =n V S +n S (r S co + x S in ) + P r P co β+ P x P in β..(i) Phae angle θ i very all and, therefore, both V P and V S revered can be taken perpendicular to Φ and, hence ocd = (approxiately) and ecd = (approxiately). Thu P co β = e + ( S /n ) co P in β = + ( S /n ) in Now θ i very all uually le than 1 and therefore, co θ =1 and hence we can write: V P co θ = V P Subtituting the above value in (i), we have: V P = n V S +n S (r S co + x S in ) + ( e + ( S /n ) co ) r P + ( + ( S /n ) in ) x P 24

25 = n V S + S co (n r S + r P /n) + S in (n x S + x P /n) + e r P + x P..(ii) = n V S + ( S /n) co (n² r S + r P ) + ( S /n) in (n² x S + x P ) + e r P + x P = n V S + ( S /n) co P + ( S /n) in X P + e r P + x P = n V S + ( S /n) ( P co + X P in )+ e r P + x P..(iii) Here P = equivalent reitance of the tranforer referred to the priary ide = n² r S + r P and X P = equivalent reactance of the tranforer referred to the priary ide = n² x S + x P Actual tranforation (voltage) ratio = V P / V S (( S /n) ( P co + X P in )+ e r P + x P ) = n+ (iv) V S Eqn (ii) ay be written a: V P = n V S +n S co (r S + r P /n²) + n S in (x S + x P /n²) + e r P + x P V P = n V S +n S co S + n S in X S + e r P + x P = n V S +n S ( S co + X S in ) + e r P + x P (v) Where S = equivalent reitance of tranforer referred to econdary ide = r S + r P /n² X S = equivalent reactance of tranforer referred to econdary ide = x S + x P /n². Actual tranforation (voltage) ratio =V P / V S n S ( S co + X S in )+ e r P + x P = n+ (vi) V S Uing eqn. (iii) and (v), the difference between actual tranforation ratio and turn ratio i: (( S /n) ( P co + X P in )+ e r P + x P ) -n =.(vii) V S = n S ( S co + X S in )+ e r P + x P θtan V S.(viii) ab PxP co PrP in β + nx co n r in = = oa nv + n r co + n x in + r coβ + x in β P P P P P 25

26 The ter in the denoinator involving P and S are all and, therefore, they can be neglected a copared with nv S. PxP co PrP in + n x co n r in = nv xp e + co rp + in + nx co nr in n n = nv = = xp rp co + nx in + nr + exp r n n nv co n 2 2 ( ) ( ) x + n x in n r + n r + x r nv P P e P P S Since θ i all, θ = tan θ θ = n ( co in ) X + x r P P e P P nv x r = ( X co in ) + rad V e P P nv rad. Error in potential tranforer atio error (Voltage Error): The actual ratio of tranforation varie with operating condition and the error in econdary voltage ay be defined a, % atio Error = K n 100 Phae angle error: n an ideal voltage tranforer, there hould not be any phae difference between priary winding voltage and econdary winding voltage revered. However, in an actual tranforer there exit a phae difference between V P and V S revered. ( X P co P in ) + exp rp θ = n rad. nv 26 x r = ( X co in ) + rad V e P P nv

27 Proble on PT Proble No. 1 A potential tranforer, ratio 1000/100 volt, ha the following contant: Priary reitance = 94.5Ω, econdary reitance=0.86 Ω Priary reactance = 66.2 Ω, total equivalent reactance = 110 Ω No load current = 0.02 A at 0 4 p.f. Calculate: (i) Phae angle error at no load (ii) burden in VA at UPF at which phae angle will be Zero. Solution: No load power factor co α = 0.4 Sin α = (1) 2 (0.4) 2 = Therefore, e = 0 co α = = 0.008A = 0 in α = = A. Turn ratio n = 1000/100 = 10, Phae angle At no load S = 0 Therefore ϴ = ( e x P r P )/(nv S ) = ( )/( ) rad = (ii) At UPF co = 1 and in = 0. Therefore ϴ = ( S /n) X P - e x P r P nv S For ϴ = 0, ( S /n) X P - e x P r P = 0. Or S = (n/x P ) ( r P - e x P ) = 10/100 ( ) = A Burden = V S S = = 10.9 VA Proble No.2 A potential tranforer rated 6900/115 Volt, ha turn in the priary winding and 375 turn in the econdary winding. With 6900 volt applied to the priary and the econdary circuit open circuited, the priary winding current i 0.005A lagging the voltage by With a particular burden connected to the econdary, the priary winding current i A lagging the voltage by Priary winding reitance = 1200Ω, Priary winding reactance = 2000Ω, econdary winding reitance = 0.4 Ω, econdary winding reactance = 0.7 Ω. (i) Find the econdary current and terinal voltage uing the applied priary voltage V P = j0 a reference. Find the load burden alo. (ii) Find the actual tranforation ratio and alo the phae angle. 27

28 (iii) f the actual ratio = the noinal ratio under above condition, what change hould be ade in the priary turn? Solution: Noinal ratio K n = 6900/115 = 60 Priary winding turn N P = Secondary winding turn N S = 375 Turn ratio = n = 22500/375 = 60 No load current 0 = 0.005A No load p.f. = co 73.7 = 0.28 and in 73.7 = 0.96 Priary current P = A, Priary power factor= co53.1 =0.6 Now priary voltage V P i taken a reference and, therefore, we can write: V P = j0, P = (0.6-j0.8) = j = 0.005(0.28-j0.96)= j { phaor S /n i the phaor difference of P and 0 } Therefore S /n = ( j0.01)-( j0.0048)= j S (revered) = n ( S /n) = 60( j0.0025) = j r S = -(0.366-j0.312) = j0.312 Therefore, S = 0.48A Priary induced voltage E P = V P P Z P = 6900 j0 ( j0.01)( j2000) = 6900 j0 (9 +j3) = 6891 j3 V. Secondary induced voltage E S (revered) = E S /n = (6891 j3)/60 = j 0.05 V E S = -( j 0.05) = j 0.05V Secondary terinal voltage = V S = E S - S Z S = j 0.05V ( j0.312)(0.4+j0.7)= j0.18 or V S = V Secondary burden = V S S = = 55 VA (ii) Actual ratio = V P / V S = 6900/ = V S (revered)= - ( j0.18) = j0.18V Angle by which V S (revered) lag V P = tan -1 (0.18/114.49) (0.18/114.49) rad = (0.18/114.49)(180/Π) = 0.90 = (iii) The olution to the proble lie in reducing the turn ratio (decreaing priary nuber of turn) o that the actual ratio equal the noinal ratio. New turn ratio = (60/60.27) 60 Therefore new value of priary turn = (60/60.27) 60 N S = (60/60.27) = Therefore, reduction in priary turn = = 101 Here, change in voltage drop caued by change in turn ratio i neglected and hence the olution i only approxiate. 28

29 Teting of ntruent Tranforer Method for finding ratio and phae angle error experientally are broadly claified into two group: 1. Abolute ethod: n thee ethod the tranforer error are deterined in ter of contant i.e., reitance, inductance and capacitance of the teting circuit. 2. Coparion ethod: n thee ethod, the error of the tranforer under tet are copared with thoe of a tandard current tranforer whoe error are known. Each of the two ethod can be claified, according to eaureent technique eployed a 1. Deflection Method: Thee ethod ue the deflection of uitable intruent for eauring quantitie related to the phaor under conideration or to their deflection. The required ratio and phae angle are then found out fro the agnitude of deflection. Thee ethod ay be ade direct reading in oe cae. 2. Null Method: Thee ethod ake ue of a network in which the appropriate phaor quantitie are balanced againt one another. The ratio and phae angle error are then found out fro the ipedance eleent of the network. The ethod ay be ade direct reading in ter of calibrated cale on the adjutable eleent in the network. Teting of Current Tranforer There are three ethod: 1. Mutual nductance ethod: Thi i an abolute ethod uing null deflection. 2. Silbee Method: Thi i a coparion ethod. There are two type; deflectional and null. 3. Arnold Method: Thi i a coparion ethod involving null technique. Silbee Method: The arrangeent for Silbee deflectional ethod i hown in Fig.1. Here the ratio and phae angle of the tet tranforer X are deterined in ter of that of a tandard tranforer S having the ae noinal ratio. 29

30 Fig. 1 Silbee deflectional ethod Procedure: The two tranforer are connected with their priarie in erie. An adjutable burden i put in the econdary circuit of the tranforer under tet. An aeter i included in the econdary circuit of the tandard tranforer o that the current ay be et to deired value. W 1 i a watteter whoe current coil i connected to carry the econdary current of the tandard tranforer. The current coil of watteter W 2 carrie a current which i the difference between the econdary current of the tandard and tet tranforer. The voltage circuit of watteter are upplied in parallel fro a phae hifting tranforer at a contant voltage V. The phaor diagra i hown in Fig. 2 30

31 Fig.2. Phaor diagra of Silbee ethod 1. The phae of the voltage i o adjuted that watteter W 1 read zero. Under thee condition voltage V i in quadrature with current. The poition of voltage phaor for thi cae i hown a V q. eading of watteter, W 1 W 1q = V q co 90 = 0. eading of watteter, W 2 W 2q = V q coponent of current in phae with V q = V q q = V q x in (θ x θ ) Where θ x = phae angle of C.T. under tet, θ = phae angle of tandard C.T. = W 1p V x co (θ x θ ) W 1p V x A (θ x θ ) i very all and, therefore, co (θ x θ ) = 1 For above, V x = W 1p - W 2p. Actual ratio of tranforer under tet x = p / x. Actual ratio of tandard tranforer = p /.. The phae of voltage V i hifted through 90 o that it occupie a poition V P and i in phae with. eading of watteter W 1, W 1p = V p co θ = V p. 31

32 eading of watteter W 2, W 2p = V p coponent of current in phae with V p = V p p = V p [ x co (θ x θ )] f the voltage i kept ae for both et of reading, then V= V p = V q. We have W 2q = V x in (θ x θ ), W 1p = Vi W 2p = V[ x co (θ x θ )]= V V x co (θ x θ )] W 2 p x = 1+ W 1p W2 q V W W W in( θx θ ) = co( θx θ ) = = Vx Vx Vx W W 2q 2q tan( θx θ ) = or ( θx θ ) = rad W W W1 p W2 p 1p 2 p 2 p 1p 2 p Or phae angle of tet tranforer, W2 q W2q θ = + θrad rad W W W + θ 1p 2 p 1p a W 2p i very all. Hence if the ratio and phae angle error of tandard tranforer are known, we can copute the error of the tet tranforer. W 2 ut be a enitive intruent. t current coil ay be deigned for all value. t i norally deigned to carry about 0.25A for teting CT having a econdary current of 5A. Proble No.1 Two current tranforer of the ae noinal ratio 500/5 A, are teted by Silbee ethod. With the current in the econdary of the tranforer adjuted at it rated value, the content in the iddle conductor = 0.05e -j126.9 A expreed with repect to current in the econdary of tandard tranforer a the reference. t i known that tandard tranforer ha a ratio correction factor (CF) of and phae error +8. Find CF and phae angle error of tranforer under tet. Solution: Noinal ratio = 500/5 =100, = 5 A Since i the reference, = 5+j0 = 0.05e -j126.9 = 0.05(co j in ) = j 0.04 Now, current in the econdary of tet tranforer x = - = 5 +j 0 (0.03 j 0.04) = j 0.04 or x 5.03 A Angle between x and = 0.04/5.03 rad =

33 Phae angle between x revered and p = = atio correction factor CF of tandard tranforer = Therefore, actual ratio of tandard tranforer = CF noinal ratio = = Priary current p = = = A Actual ratio of tranforer under tet x = p / = /5.03 = atio correction factor of tet tranforer CF = Actual ratio/noinal ratio = 99.55/100 = Power Factor Meter Power factor of a ingle phae circuit i given by co Φ = P/V. By eauring power, current and voltage power factor can be calculated uing the above equation. Thi ethod i not accurate. t i deirable to have intantaneou indication of power factor. Power factor eter indicate directly the power factor of the circuit. Power factor eter have : 1. Current Coil 2. Preure Coil. 1. The current circuit carrie the current whoe PF i to be eaured. 33

34 2. The preure circuit i connected acro the circuit whoe PF i to be eaured and i uually plit into two path. 3. The deflection of the intruent depend upon the phae difference between the ain current and current in two path of preure coil i.e., the power factor of the circuit. 4. The deflection i indicated by a pointer. Type of power factor eter There are two type: 1. Electrodynaoeter Type 2. Moving ron Type Single phae Electrodynaoeter Power Factor Meter Fig.1. Single Phae electrodynaoeter type power factor eter Contruction of electrodynaoeter type power factor eter. Contruction i hown in Fig.1 t conit of two coil 1.Fixed coil which act a current Coil. 2. Moving coil or preure coil. Current coil: 1. Split into two part and carrie the current of the circuit under tet. 2. The agnetic field produced by thi coil i proportional to the ain current. Preure coil: 34

35 1. Two identical coil A & B pivoted on a pindle. 2. Coil a ha a non inductive reitance connected in erie with it. 3. Coil B ha a highly inductive choke coil L connected in erie with it. 4. The two coil are connected acro the voltage of the circuit. 5. The value of & L adjuted to carry the ae current at noral frequency. Working Principle: 1. Current in coil i in phae with the circuit voltage. 2. Current through coil B lag the voltage by an angle 90 ( ). 3. The angle between the plane of the coil i ade equal to. 4. There i no controlling torque. 5. Miniu control effect uing ilver or gold ligaent for connecting oving coil. Auption ade: Current through coil B lag voltage by exactly 90. Angle between the plane of the coil i exactly 90. Now, there will be two deflecting torque: 1. Torque acting on coil A. 2. Torque acting on coil B. The coil winding are arranged in uch a way that the torque due to two coil are oppoite in direction. Therefore the pointer will take up a poition where thee two torque are equal. Conider the cae of a lagging power factor of co φ. Deflecting torque acting on coil a i: TA = KV Max co φinθ Where θ = angular deflection fro the plane of reference. & Max = axiu value of utual inductance between the two coil. Thi torque act in clockwie direction. Deflection torque acting on coil B i: TB = KV Max co (90 -φ) in(90 + θ) = KV Max in φ co θ. Thi torque act in the anticlockwie direction. The value of Max i the ae in both the expreion, due to iilar contruction of coil. The coil will take up a poition where the two torque are equal. or KV Max co φinθ =KV Max in φ co θ or θ = φ Therefore the deflection of the intruent i a eaure of phae angle of the circuit. The cale of the intruent can be calibrated directly in ter of power factor. 35

36 Fig. 2. Phaor diagra of Fig.1 Weton Frequency Meter: 1. Thi eter conit of two coil ounted perpendicular to each other. 2. Each coil i divided into two ection. 3. Coil A ha a reitance A connected in erie with it and reactance LA in parallel with it. 4. Coil B ha a reactance coil LB in erie and reitance B in parallel. 5. The oving eleent i a oft iron needle. 6. Thi needle i pivoted on the pindle which alo carrie a pointer and a daping vane. Fig. 3 Weton Frequency Meter 36

37 7. There i no controlling force. 8. The erie reactance coil L uppree higher haronic in the current of the intruent. 9. Miniize the wavefor error in it indication. Working Principle The eter i connected acro the upply. The two coil carry the current. The current et up two agnetic field which are at right angle to each other. The agnitude of the field depend upon the agnitude of current flowing in the coil. Both thee field act upon the oft iron needle. The poition of the needle depend upon the relative agnitude of the two field and hence of the current. The eter i deigned in uch a way that the value of variou reitance and inductance are uch that for noral frequency of upply, the value of voltage drop acro LA and B end equal current through coil A & B. Therefore, the needle take up a poition which i at 45 to both the coil. Hence, the pointer i at the centre of the cale. f the frequency increae above the noral value reactance of LA and LB increae. eitance A and B reain contant. Thi ean the voltage acro coil A increae with the increae in frequency copared to that of coil B. Hence, the current in coil A increae while it decreae in coil B. (ncreae in the value of LB).Thu the field of coil A becoe tronger than that of coil B. The tendency of the needle i to deflect in the direction of tronger field and therefore it tend to et itelf in line with the axi of Coil A. Thu the pointer deflect to the left. Oppoite happen for freq. Phae Sequence ndicator Ued to find the phae equence of three phae upplie. There are two type (i) otating type (ii) (ii) tatic type. (i) otating Type: Principle of operation iilar to three phae nduction otor. They conit of three coil ounted 120 apart in pace. The three end of the coil brought out and connected to three terinal arked YB. The coil are tar connected. They are excited by the upply whoe phae equence i to be deterined. An aluiniu dic i ounted on the top of the coil. The coil produce a rotating agnetic field and eddy ef are induced in the dic. 37

38 The eddy ef caue the eddy current to flow in the dic. A torque i produced due to the interaction of eddy current with the rotating field. The dic rotate becaue of the torque and the direction depend on the phae equence of the upply. An arrow indicate the direction of rotation of the dic. f the direction of rotation i the ae a indicated by arrow head, the phae equence of the upply i the ae a arked on the terinal of the intruent. Otherwie, the phae equence i oppoite. Y B DSC Fig. 4. Phae equence indicator Static Type: One arrangeent conit of two lap and an inductor. When the phae equence i YB, lap 1 will be di and lap 2 will glow brightly. f the phae equence i BY, lap 1 will glow brightly and lap 2 will be di. Principle of operation: Auing the phae equence YB, Phaor relation of voltage a VY, VYB and VB. Fig. 5 VY = V(1+j0), VBY = V(-0.5-j0.866) and VB = V(-0.5+j0.866) 38

39 Lap 1 Lap 2 Y r r Y Y nductor L nductor L B B B V B V Y V YB Fig. 5. Circuit of a Static phae equence indicator Aue the current direction a hown in Fig.5 i.e., +Y+B = 0. Now fro Fig. 5b and 5c VY+Yr-r = 0 VYB + jxl Yr = 0 Solving for and Y we have, V r Y = + Y and Y V = YB V Y jx r ( r + 2 jx ) L L Then, ( 1+ j2 X L / r) ( ) = 1+ V / V jx / r Y YB Y L But, VYB/VY = -0.5 j0.866 f the indicator i deigned o that XL = at the line frequency, /Y = j0.232 or /Y = Thu the voltage drop acro lap 1 ( r ) i only 27% of that acro lap 2 (i.e., Y = 0.27 rr ). Thu if the phae equence i YB lap 1 glow dily while lap 2 glow brightly. 39