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Monday, April 4, 2011

SWINBURNE’S TEST


SWINBURNE’S TEST
AIM:

            To conduct Swinburne’s test on DC machine to determine efficiency when working as generator and motor without actually loading the machine.

APPARATUS REQUIRED:

S.No.
Apparatus
Range
Type
Quantity
1
Ammeter
(0-20) A
MC
1
2
Voltmeter
(0-300) V
MC
1
3
Rheostats
1250W, 0.8A
Wire Wound
1
4
Tachometer
(0-3000) rpm
Digital
1
5
Resistive Load
5KW,230V
-
1
6
Connecting Wires
2.5sq.mm.
Copper
Few



PRECAUTIONS:
The field rheostat should be in the minimum position at the time of starting and stopping the motor

PROCEDURE:
1.    Connections are made as per the circuit diagram.
2.    After checking the minimum position of field rheostat, DPST switch is closed and starting resistance is gradually removed.
3.    By adjusting the field rheostat, the machine is brought to its rated speed.
4.    The armature current, field current and voltage readings are noted.
5.    The field rheostat is then brought to minimum position DPST switch is opened.







TABULAR COLUMNS:

AS MOTOR:
S. No.
V
(Volts)
IL
(Amps)
Ia
(Amps)
Ia2Ra
(Watts)
Total Losses
W (Watts)
Output Power
(Watts)
Input Power
(Watts)
Efficiency
h%
























AS GENERATOR:

S. No.
V
(Volts)
I1
(Amps)
Ia
(Amps)

Ia2Ra
(Watts)

Total Losses
 (Watts)
Output Power
(Watts)
Input Power
(Watts)
Efficiency
h%

























TABULAR COLUMN:
S.No.
If (Amps)
Io (Amps)
V (Volts)












 DETERMINATION OF ARMATURE RESISTANCE:
















PROCEDURE:
1. Connections are made as per the circuit diagram.     
2. Supply is given by closing the DPST switch.
3. Readings of Ammeter and Voltmeter are noted.
4. Armature resistance in Ohms is calculated as Ra = (Vx1.5) /I



TABULAR COLUMN:

S.No.
Voltage
V (Volts)
Current
I (Amps)
Armature Resistance
Ra (Ohms)











FORMULAE:
Hot Resistance  Ra = 1.2 X R Ω
            Constant losses       = VIo – Iao2 Ra watts
            Where Iao       = (Io – If) Amps

AS MOTOR:
Load Current IL         = _____ Amps (Assume 15%, 25%, 50%, 75% of
     rated current)
            Armature current Ia = IL – If Amps
            Copper loss               = Ia2 Ra watts
            Total losses               = Copper loss + Constant losses
           Input Power               = VIL  watts
           Output Power            = Input Power – Total losses
          Output power      
Efficiency h%           =      ---------------------- X 100%
                                                 Input Power
AS GENERATOR:
            Load Current IL         = _____ Amps (Assume 15%, 25%, 50%, 75% of
    rated current)
            Armature current Ia = IL + If Amps
            Copper loss               = Ia2 Ra watts
            Total losses               = Copper loss + Constant losses
           Output Power            = VIL  watts
            Input Power              = Input Power +Total losses
      Output power     
Efficiency h%           = ----------------------- X 100%
                                            Input Power

MODEL GRAPH:





 











RESULT:
Thus the efficiency of the D.C machine is predetermined by Swinburne’s test.

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