SUMPNER’S TEST - Computer Programming

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Monday, April 4, 2011

SUMPNER’S TEST


SUMPNER’S TEST

AIM :
To predetermine the efficiency and regulation of a given single phase Transformer by conducting back-to-back test and also to find the parameters of the equivalent circuit.

APPARATUS REQUIRED:

S. No.
Name of the Apparatus
Range
Type
Quantity
1
Auto Transformer
(0-270) V
-
2
2
Wattmeter
300 V, 10A
75 V, 5 A
LPF
UPF
1
1
3
Ammeter
(0-2) A
(0-20) A
MI
MI
1
1
4
Voltmeter
(0-75) V
(0-150) V
MI
MI
1
1
5
Connecting Wires
2.5sq.mm
Copper
Few

PRECAUTIONS:
1.    Auto Transformer whose variac should be in zero position, before switching on the ac supply.
  1. Transformer should be operated under rated values.

PROCEDURE:
1.    Connections are made as shown in the circuit diagram.
2.    Rated voltage of 110V is adjusted to get in voltmeter by adjusting the variac of the Auto Transformer which would be in zero before switching on the supply at the primary side.
3.    The readings of voltmeter, ammeter and wattmeter are noted on the primary side.
4.    A voltmeter is connected across the secondary and with the secondary supply off i.e switch S is kept open.  The voltmeter reading is noted.
5.    If the reading of voltmeter reads higher voltage, the terminals of any one of secondary coil is interchanged in order that voltmeter reads zero.
6.    The secondary is now switched on and SPST switch is closed with variac of auto transformer is zero.
7.    After switching on the secondary the variac of transformer (Auto) is adjusted so that full load rated secondary current flows.
8.    Then the readings of wattmeter, Ammeter and voltmeter are noted.
9.    The Percentage Efficiency and percentage regulation are calculated and equivalent circuit is drawn.















FORMULAE:
                                                                   W1
Core loss of each transformer Wo = ----- Watts
                                                                 2
                                                                                   W2
Full load copper loss of each transformer Wc = ------ Watts.
                                                                                    2

                                                                  Wo                             Io
Wo = V1I1 Cos jo               jo = Cos-1 ---------                  I1 = ---- A
                                                                 V1 I1                           2

Iw  =  I1 CosFo                     =  I1 CosF                         V2 = Vs/2 x A

Ro = V1 / Iw                        Xo  =  V1 / Iμ                          Ro2   =  Wc / I22      Zo2  =  V2 / I2

Xo2 = Ö Zo22 – Ro22

Copper loss at various loads = I22 Ro2



PERCENTAGE REGULATION:

1.    Upf : I2 / V (Ro2 CosFo) X 100

2.    Lagging pf : I2 / V (Ro2 CosFo + Xo2SinFo) X 100

3.    Leading pf :  I2 / V (Ro2 CosFo - Xo2SinFo) X 100  

Output Power    (1) Upf : 3Kw
                           (2) Pf   : 3Kw CosFo

 Input Power = Output Power + Core loss + Cu loss

         Output power
Efficiency h%   =  -------------------------- X 100%     
                                              Input Power






EQUIVALENT CIRCUIT:
 








MODEL GRAPHS:














RESULT:  
Thus the efficiency and regulation of a given single phase Transformer is carried out by conducting back-to-back test and the equivalent circuit parameters are found out.

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