Three Phase System. c 2 ω Fig.1a -----(1) Eb1b2 ) -----(2) Ea1a Ea1a2, Eb1b2 & Ec1c2 are RMS values. Ec1c2. Fig.

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1 Three hase System Generation of three hase system: Consider following hase simle loo generator shown in figure a. ts generated voltage will be as shown in figure b. c a c ω b b 4 Fig.a e Sin( ) -----() aa m t bb ec c msin( t ) -----() eb b msin( t 4 ) -----() aa, bb & cc are S values. aa cc Fig.c: hasor Diagram Double Subscrit notation: -A=+A A A A=-A A A=A- Fig. reared by: Nafees Ahmed

2 hase Sequence: t is the sequence in which current or voltages in different hases attain their maximum values. hase sequence may be -- or -- as shown bellow Fig.a: hase Sequence eaa ebb ecc Fig.b: hase Sequence nterconnection of -hase system: The six terminals of three hase winding can be connected to form any of the below.. Star or W () connected -ϕ System. esh or Delta( ) connected - ϕ System Star or W () connected -ϕ System: hase oltage ine Current ine oltage ( ) N Fig.a reared by: Nafees Ahmed

3 , & are called hase voltages., & are called hase currents., & are called line voltages. From figure it is clear that ++= ine current () = hase Current () Fig.b: ine voltage from above hasor diagram ( ) ts magnitude cos6 For balanced -hase system = = = (hase oltage) So cos6 Similarly ( ) cos6 ( ) cos6 reared by: Nafees Ahmed

4 Hence for balanced system = = = = f ϕ is the angle between hase voltage and hase current then Active ower of hase = Cos ϕ = Cos ϕ = Cos ϕ W eactive ower of hase = Sin ϕ = Sin ϕ = Sin ϕ A Aarent ower of hase = = = A esh or Delta ( ) connected - ϕ System hase oltage ine oltage Fig.4a, & are called hase voltages., & are called hase currents. From figure it is clear that ++= ine oltage ine oltage () = hase oltage () Fig.4b reared by: Nafees Ahmed

5 Aly KC at node ( ) ts magnitude from above hasor diagram cos6 For balanced -hase system = = = (hase Current) So cos6 Similarly ( ) cos6 ( ) cos6 Hence for balanced system = = = = f ϕ is the angle between hase voltage and hase current then Active ower of hase = Cos ϕ = Cos ϕ = Cos ϕ W eactive ower of hase = Sin ϕ = Aarent ower of hase = = Sin ϕ = Sin ϕ A = A xamle : f the hase voltage of a three hase star connected alternator is. What will be the line voltage? i. When the hases are correctly connected? ii. When the connection to one hase is reversed? Solution: i. ine voltage () = hase oltage () 5 reared by: Nafees Ahmed

6 ii. et connation of hase is reversed - = - = Fig.5a: hasor Diagram when hases are correctly connected Fig.5b: hasor Diagram when connection of hase is reversed ine voltage from above hasor diagram ( ) ts magnitude cos cos Similarly ( ) cos6 ( ) cos xamle : Three identical of ohm are connected in star to a 45, -hase, 5 Hz suly. Calculate i. Total ower taken by load ii. ower consumed in the resistance if they are connected in delta to same suly. iii. f one of the resistance is oen circuited in each case calculated the ower consumed. Solution: i. ower in star 6 reared by: Nafees Ahmed

7 =45, = =45 =45/ olts == / =45/ Am Total ower consumed (esistive load F=) =Cos ɸ = *(45/ )*(45/ )*=86.5 Watts ii. iii. ower in Delta =45, ==45 == / =45/ Am Total ower consumed =Cos ɸ = *(45)*(45/)*=58.75 Watts ower when one resistance is oen circuit Star Connected Delta Connected Ω 45 =Ω 45 Ω 45 =Ω 45 eq 45 4 Fig.6a Watt 7.5 Watt Fig.6a easurement of ower in -hase load: There are three methods. One wattmeter method. Two wattmeter method. Three wattmeter method. One wattmeter method : t is used only for balanced load Total ower consumed by -ϕ load C N =x ower consumed by one hase load =x eading of one wattmeter (W) Note: if load is delta connected, voltage terminal () of C will be connected to ground.. Three wattmeter method: For balanced or unbalanced load total ower 7 =W+W+W Fig.7a: -ϕ Star Connected alanced oad reared by: Nafees Ahmed

8 W C N W C W C Note: f load is delta connected, voltage terminal () of C of all wattmeters will be connected to ground.. Two Wattmeter method: a. When load is star connected: Fig.7b W C i v v i i v W C nstantaneous ower given by wattmeter =i.(v-v) ---() nstantaneous ower given by wattmeter =i.(v-v) ---() Adding () & () += i.(v-v)+ i.(v-v) =vi-vi+vi-vi =vi+vi-v (i+i) Fig.7 c =vi+vi+vi i i i i i = Total instantaneous ower in -hase load b. When load is delta connected: i 8 reared by: Nafees Ahmed

9 W i -i C i v v i i v W i -i C Fig.7 d nstantaneous ower given by wattmeter =-v.(i-i) ---() nstantaneous ower given by wattmeter =v.(i-i) ---() Adding () & () += -v.(i-i)+ v.(i-i) =-vi-vi+vi-vi =-(v+v)i+vi+vi =vi+vi+vi v v v v v v = Total instantaneous ower in -hase load Hence a load may be balanced or unbalanced, star connected or delta connected, total instantaneous ower of -ϕ load will be the sum of the instantaneous ower given by the two wattmeters. Since wattmeter measures active ower so total active ower of -ϕ load will be = W+W W= Active ower measured by wattmeter W= Active ower measured by wattmeter Determination of ower factor: Consider a star connected balanced load et = = = hase voltages (S) = = = hase currents (S) eading of wattmeter W=Cos(-ϕ) = Cos(-ϕ) = Cos(-ϕ) ---() 9 reared by: Nafees Ahmed

10 ϕ ϕ ϕ - Fig.b eading of wattmeter W=Cos(+ϕ) = Cos(+ϕ) = Cos(+ϕ) ---() W+W = Cosϕ=Total active ower of -ϕload ---() W-W = Sinϕ ---(4) (4)/() W W W W tan ower factor Sin Cos W W W W tan W W W W W W Cos tan W W ffect of ower factor on wattmeter readings: We know W=Cos(-ɸ) = Cos (-ɸ) --- () W=Cos(-ɸ) = Cos (+ɸ) --- () When ɸ= i.e F Cosɸ=: W= Cos()= ( /) W= Cos()= ( /) oth wattmeters will show equal readings When ɸ=6 i.e F Cosɸ=.5: W= Cos(-6)= ( /) W= Cos(+6)= reared by: Nafees Ahmed

11 One Wattmeter will show zero reading and total ower consumed = W When ɸ=9 i.e F Cosɸ=: W= Cos(-9)= (/) W= Cos(+9)= -(/) Hence we can say when ower factor angle is greater than 6 one wattmeter will give negative reading. For obtaining the reading of that wattmeter either the connection of current coil() or ressure coil(c) should be changed and reading will be taken as negative. xamle : For a certain load one wattmeter reads KW and other 5 KW after the voltage of this wattmeter has been reversed. Calculate ower and ower factor of the load. Solution: ower =W+W=+ (-5)= 5 KW Cos Cos tan W W W W Cos tan 5 Cos tan 5 5 Cos Q. Draw connection diagram fro measurement of ower in -hase connected load using two wattmeter method. n one such exeriment the load sulied was KW at.7 ower factor lagging. Find the reading of each wattmeter. Hint: W+W== KW Cos tan W W W W.7 reared by: Nafees Ahmed