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Separation of losses in a dc shunt motor
- 1. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 1 Separation Of Losses In A Dc Shunt Motor
- 2. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 2 Contents SL. NO TITLE PAGE NO. 01 INTRODUCTION 03 02 CIRCUIT DIAGRAM 04 03 APPARATUS REQUIRED 04 04 PRECAUTIONS 05 05 PROCEDURE 05 06 TABULAR COLUMN 05 07 CALCULATIONS 06 08 GRAPH 09 09 CONCLUSION 09
- 3. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 3 INTRODUCTION- At a given excitation, friction losses and hysteresis are proportional to speed. Windage losses and eddy current losses on the other hand are both proportional to square of speed. Hence, for a given excitation (field current) we have, Friction losses = AN Watts Windage losses = BN2 Watts Hysteresis losses = CN Watts Eddy current losses = DN2 Watts Where, N = speed. For a motor on no load, power input to the armature is the sum of the armature copper losses and the above losses. In the circuit diagram, Power input to the armature = V*Ia watts. Armature copper losses = Ia2*Ra watts V*Ia – Ia2*Ra = (A + C)N + (B + D)N W/N = (A+C) + (B+D)N. The graph between W/N & N is a straight line, from which (A+C) and (B+D) can be found.In order to find A, B, C and D seperately, let the field current be changed to a reduced value IIf, and kept constant at that value. If, voltage is applied to the armature as before, we now have , W/N = (A+C1) N + (B+D1) N2 (at the reduced excitation, function and windage losses are still are AN and BN2, but hysteresis losses become C1N and eddy current losses become D1N2. We can now obtain (A+C) and (B+D) as before. Now, C/C1 = (flux at normal excitation/flux at reduced excitation) D/D1 = (flux at normal excitation/flux at reduced excitation) So, if we determine the ratio (flux at normal excitation/flux at reduced excitation) we can find A, B, C, D, C1, & D1
- 4. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 4 CIRCUIT DIAGRAM- APPARATUS- SL.NO NAME TYPE RANGE QUANTITY 01 DC Voltmeter M.C 0-300V 1 02 DC Ammeter M.C 0-2.5/5A 0-2.5/5A 1 04 Variable Rheostat -- -- 138Ω/8.5A 700Ω/1.5A 1 1 03 Tachometer -- -- 1
- 5. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 5 PRECAUTIONS- 1. Keep armature rheostat (38Ω, 8.5A) at its maximum resistance position (cut in position) 2. Keep field rheostat (700 Ω, 1.5A) at its minimum resistance position (cutout position) PROCEDURE- 1. Connections are made as per circuit diagram. 2. Keep armature rheostat (38Ω, 8.5A) at its maximum resistance position and keep field rheostat (700 Ω, 1.5A) at its minimum resistance position 3. Switch on the main supply. 4. Start the motor with the help of 3 point starter 5. Keep the field current to certain constant value with the help of shunt field rheostat. (say 1.0 Amps) 6. Gradually cutout the (38Ω, 8.5A) rheostat connected in series with the armature of the motor in suitable steps and at each step of increase, note down the voltmeter reading and speed of the motor. 7. Bring back the armature rheostat to its original position. 8. Switch off the supply. TABULAR COLUMN- Armature resistance=3.33 Ω(Ra) Full excitation: Field current= 1.2Amps Armature voltage, volts (v) 160 170 180 190 Armature current, Amps(Ia) 0.24 0.2 0.2 0.2 Speed, N(rpm) 1280 1350 1440 1528 W = V*Ia – I2a*Ra, watts 33.20 33.86 35.86 37.6 W/N (watts/rpm) 29.85*10-3 25.08*10-3 24.4*10-3 24.78*10-3
- 6. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 6 Reduced excitation: Field current= 0.8Amps Armature voltage, volts (v) 167 170 180 190 Armature current, Amps(Ia) 0.12 0.118 0.12 0.12 Speed, N(rpm) 1454 1477 1566 1618 W = V*Ia – I2a*Ra, watts 19.99 20.01 21.55 22.75 W/N (watts/rpm) 13.74*10-3 13.55*10-3 13.76*10-3 14.06*10-3 CALCULATION- For motor on no load, power input to the armature is the sum of the armature copper losses and the above losses. Power input to the armature=V.Ia watts. Armature copper losses=Ia2.Ra watts. Stray losses, W = V.I−Ia2.Ra = (A+C)N+(B+D)N2 W/N = (A+C)+(B+D)N Friction losses + Hysteresis losses = AN+CN watts = (A+C)N watts From graph, OR = A+C = 9.8*10-3 watts/rpm OP = A+C1 = 10.4*10-3 watts/rpm Windage losses + Eddy current losses = BN2 + DN2 watts = (B+N)N2 watts From graph, line TR Tanϴ = ∆W/N ∆N = B+D = 3∗10 −3 1100 = 2.7272*10-6 watts/rpm2`
- 7. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 7 line SP Tanϴ1 = ∆W/N ∆N = B+D1 = 5∗10−3 500 = 10*10-6 watts/rpm2 Eb = V-IaRa and Eb = ØZN P = K ØN K ØN = Eb, K Ø = 𝐸𝑏 𝑁 = 𝑉−𝐼𝑎𝑅𝑎 𝑁 At 4th reading At full excitation, value of KØ = 𝐸𝑏 𝑁 = 𝑉−𝐼𝑎𝑅𝑎 𝑁 = 190−(0.2∗3.33) 1528 = 0.129145 At reduced excitation, value of KØ1 = 𝐸𝑏 𝑁 = 𝑉−𝐼𝑎𝑅𝑎 𝑁 = 190−(0.12∗3.33) 1618 = 0.1171819 𝐾Ø 𝐾Ø1 = Ø Ø1 =1.1020899 𝐶 𝐶1 = [ Ø Ø1]1.6 C1 = C[ Ø1 Ø ]1.6 = 0.8555958C OR = A+C = watts/rpm− OP = A+C1 = A+C[ Ø1 Ø ]1.6 OP−OR = A+C [ Ø1 Ø ]1.6− A+C = C([ Ø1 Ø ]1.6- 1)
- 8. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 8 C = 𝑂𝑃−𝑂𝑃 ([ Ø 1 Ø ]1.6− 1) = 0.6∗10−3 144 .4042∗10−3 = 0.0041555 A = OR – C =9.8 ∗ 10−3 − 4.155 ∗ 10−3 = 0.005645 𝐷 𝐷1 = [ Ø1 Ø ]2 = 0.823314715 D1 = D[ Ø1 Ø ]2 Tanϴ = B+D watts/rpm Tanϴ1 = B+D1 = [ Ø1 Ø ]2 watts/rpm Tanϴ1 − Tanϴ = B+D [ Ø1 Ø ]2 – (B+D) = D ([ Ø1 Ø ]2- 1) D = Tanϴ1 − Tanϴ ([ Ø 1 Ø ]2− 1) = 7.2728 ∗10−6 176 .6852∗10−3 = 0.000041162 B = Tanϴ − D = 2.7272 ∗ 10−6 − 0.000041162 = 0.000038434 At rated speed (1500rpm) Friction losses = A* 1500 = 8.4675 watts Windage (air friction) losses = B* 15002 = 86.4765 watts Hysteresis losses = C* 1500 = 6.23325 watts Eddy current losses =D* 15002 = 92.6145 watts Total stray losses = total mechanical losses + iron losses = friction losses + Windage losses + Hysterisis losses + Eddy current losses =193.79175 watts
- 9. Separation Of Losses in a DC Shunt Motor Dept Of E&E Engg Page 9 GRAPH- CONCLUSION- From this experiment, hence we dertermined the different losses.