190.MET July 2022 Q.10(b)

 A 500 V, Single-phase synchronous motor gives a net output mechanical power of 7.46 kW and operates and 0.9 pf lagging. Its effective resistance is 0.8 ohm. If the iron and friction losses amount to 500 W, excitation losses are 800 W, estimate the armature current and commercial efficiency.

Given

V = 500V

Pout = 7.46 kW = 7460 W

cos φ = 0.9

Ra = 0.8

Iron & friction loss = 500 W

Excitation loss = 800 W

To find

1. Armature current, Ia

2. Commercial efficiency, ηc

Solution

The motor power flow is given by

Motor input ------Stator Cu loss -------Armature power---------Iron friction loss -------output power

V Ia cos φ -------------- Ia Ra²  --------------Pm ----------------------I & F loss ------------------Pout

Motor input = V Ia cos φ ; Armature Cu loss = Ia Ra²

Power developed in armature is Pm = V Ia cos φ − Ia² Ra

∴ Ia² Ra − V Ia cos φ + Pm = 0

And Pout = Pm - Other total losses

       7460 = Pm - (800 + 500)

          Pm = 8760 W

 Ia² Ra − V Ia cos φ + Pm = 0

It is in the form of  ax²  + bx + c = 0

So x = Ia , a = Ra , b = - V cos φ and c = Pm

So Armature current Ia = 20.2 A

Efficiency = {output / input } x 100

Input = V Ia cos φ

          = 500 x 20.2 x 0.9

          = 9099 W

Output = 7460 W

Efficiency = [ 7460 / 9099 ] x 100

                  = 82.06 %