TWO WATTMETER METHOD AIM: To measure power input to a three phase balanced load by two wattmeter method. APPARATUS: 1.
Three phase lamp bank
2.
A.C. voltmeter-(0-600V)
3.
A.C. Ammeter (0-10A)-3-Nos
4.
Wattmeter (10A,600 V) -2 Nos
THEORY: The power input to a three phase star or delta connected balanced or unbalanced load can be measured with two wattmeter method. In this method the current coils of the two wattmeters are connected in any two lines while voltage coil of each wattmeter is connected between its own current coil terminal and a line without a current coil. Here the current coils are inserted in the lines R and Y ,then the pressure coils are connected between R-B for one wattmeter and Y-B for other wattmeter as shown in circuit diagram. The connections are same for star or delta connected load. When two wattmeters are connected in this way, the algebraic sum of two wattmeter readings gives the total power dissipated in the three phase circuit. If W1 and W2 are the two wattmeter readings then total power is
Proof- Consider R.M.S. values of currents and voltages. For balanced load , IR = IY = IB = Iph; VR= VY=VB= Vph; VRB = VYB=VL. Now W1= IR × VRB ×cos (IR^ VRB) W2= IY × VYB ×cos (IY^ VYB) To find angle between (IR and VRB) and(IY and VYB) see the phasor diagram IR^ VRB = 30 - ф IY^ VYB = 30 + ф W1= IR × VRB ×cos(30 - ф) and W2= IY × VYB ×cos (30 + ф) W1= IL × VL ×cos(30 - ф) and W2= IL× Vl ×cos (30 + ф) W1+ W2= IL × VL[ cos(30 - ф)+ cos (30 + ф)] = IL × VL[cos(30)cos( ф)+sin (30)sin( ф)+ cos(30)cos( ф)-sin (30)sin(ф)] =2 IL × VL[cos(30)cos( ф)] = 2 IL × VL[√3/2 cos( ф)] W1+ W2=√3 IL × VL cos ( ф)=Total power PROCEDURE:1. Make the connections as per circuit diagram 2.
Switch on the supply and adjust the line voltage to 420V.
3.
Switch on lamp load and keep it balanced. Take readings
of voltmeter , ammeter and wattmeter.
CONCLUSION:
QUESTIONS:1. State advantages and disadvantages of two wattmeter method . 2. Explain how active power can be measured with the help of single wattmeter. 3. Two wattmeter method is used to measure power in a 3-phase balanced load. Find the power factor ifi) the two readings are equal and have the same sign ii) the two readings are equal and have the opposite sign iii) the reading of one wattmeter is zero. iv) the reading of one wattmeter is half of the other.
OPEN CIRCUIT AND SHORT CIRCUIT TEST ON 1 - PHASE TRANSFORMER O
AIM: To determine efficiency and regulation of a transformer from equivalent circuit parameters determined by conducting open circuit and short circuit test on transformer. APPARATUS: 1. Single phase transformer -1 KVA ,115/230 V 2. A.C ammeter –(0-2.5A), (0,5A) 3. A.C.voltmeter (0-150V) 4. Wattmeter (5A,150V)
THEORY: In this method , the actual load is not used on transformer. The equivalent circuit parameters of a transformer are determined by conducting 1. Open circuit test (O.C.test)- The transformer primary is connected to
a.c supply through ammeter, wattmeter and dimmerstat.The secondary of transformer is kept open. Usually L.V.side is used as primary and H.V. side is used as secondary to conduct O.C.test.With the help of dimmerstat, rated voltage is given to primary. The wattmeter measures input power. The ammeter measures input current. The voltmeter gives the value of rated primary voltage applied at rated frequency. A voltmeter is connected across secondary to measure secondary voltage which is V2=E2. when primary is supplied with rated voltage. As voltmeter resistance is very high, though voltmeter is connected, secondary is considered as open circuited. Therefore I2=0. So secondary copper losses are zero. And primary current I1= I0 (No load current) .This current is only 2 to 4 % of its full load value,
hence copper losses on primary are also very low. Thus the total copper losses in O.C. test are negligibly small. Against this the input voltage is rated at rated frequency hence flux density is maximum. Hence there are iorn losses. As output power is zero and copper losses are very small, the total input power is used to supply iron losses. This power is measured by wattmeter W0. Hence the wattmeter on O.C test gives iron losses which remain constant for all the loads. 2. Short circuit test (S.C.Test)- In this test, primary is connected to a.c.
supply through dimmerstat,ammeter,voltmeter as shown in circuit diagram. The secondary is short circuited with the help of thick copper wire. Generally H.V.side is connected to supply and L.V.side is shorted. As secondary is shorted ,its resistance is very small and on rated voltage it may be draw very large current . Such large current can cause overheating and burning of the transformer. To limit this S.C. current ,primary is supplied with low voltage, which is just enough to cause rated current to flow through primary which can be observed on an ammeter. The low voltage can be adjusted with the help of dimmerstat. The currents flowing through the windings are rated currents hence the total copper loss is full load copper loss. Now the voltage applied is low which is fraction of rated voltage. The iron losses are function of applied voltage. So they are very small. Hence the wattmeter reading gives full load copper losses.
OBSERVATION TABLE: Open circuit test
V0 (Volts)
I0 (Amps)
W0(Watts)
Vsc (Volts)
Isc (Amps)
Wsc(Watts)
Short circuit test
CALCULATIONS:Open circuit test We know that, W0 =V0I0cosф Cosф= W0 / V0I0 = no load power factor=_________ Once cosф is known we can obtain Ic=Io cos ф =________
(core loss component)
Im=Io sin ф =________
(magnetizing component)
Now we can determine exciting circuit parameters R0=V0/Ic=______ X0=V0/Im=______
Short circuit test
We know that, Wsc=VscIsccosфsc cosфsc= Wsc / VscIsc =short circuit power factor=_________ Wsc= Isc2 R1e
R1e= Wsc/ Isc2=______
Z1e= Vsc/ Isc =√ R1e2+ X1e2 =___________
X1e= √ Z1e2- R1e2 =________
Efficiency and Regulation From O.C test, we get From SC test, we get
W0= Pi = Iron losses Wsc= Pcu (F.L) = copper losses
%η on F.L = (V2 I2(F.L) cosф)/[( V2 I2(F.L) cosф)+ W0+ Wsc]×100 %η at any load = (nV2 I2 cosф)/[( nV2 I2 cosф)+ W0+ n2Wsc]×100 Where n=fraction of full load , I2= n I2(F.L) %R =(I1 R1ecosф ± I1 X1esinф)/V1×100 %R at F.L.=___________ Where I1, I2 are rated currents at F.L. For any other load I1 , I2 will change by fraction n
RESULT TABLE Sr.No. 1 2 3 4 5 6
% of full Load 20% 40% 50% 60% 80% 100%
%η
%R
CONCLUSION:
QUESTIONS:1. Explain working principle of single phase transformer. 2. Derive emf equation of single phase transformer. 3. Explain significance of O.C. ,S.C. test performed on single phase transformer. 4. Draw and explain significance of all parameters of an exact equivalent circuit of a single phase transformer .Also derive the approximate equivalent circuit from this by stating the rules of transfer of various parameters from one side to the other. 5.Enlist various losses in transformer. How these losses can be minimized
