Q. Two shunt wound generators running in parallel supply a total load current of 4000 A. Each generator has an armature resistance of 0.02? and a field of 40?. The fiels are excited so that the emf induced in one generator is 480 V and in another 490 V, find the bus - bar voltage and kilowatt output of each machine.

 

Sol.            V = bus bar voltage

 

                  I₁ = current output of generator 1

 

                  I₂ = current output of generator 2

 

             I₁ + I₂ =  4000

               I_sh = V/R_sh

 

       Bus - bar voltage

 

       In each machine

 

       Bus bar voltage + voltage

 

       Drop in armature = EMF generated

 

       for  generated 1, E₁ = V + I_a1 R_a

 

                 480 = V + (I₁ + V/40) × 0.02

 

               (I_a1 = I₁ + I_sh)

 

      for  generated 2 490 =  V + (I₂ + V/40) × 0.02

 

      Subtrating both with each other we get,

 

            0.021₂ - 0.021₁ = 10

 

  or      I₂ - I₁ = 500

 

  or      (4000 - I₁) - I₁ = 500

 

             I₁ = 1750A and I₂ = 4000 - 1750 = 2250A

 

Now          480 = V + (I₁ + V/40) × 0.02    or      490 =  V + (I₂ + V/40) × 0.02

 

                On solving V = 444.8 volts

 

           Kilowatt output

 

               kW output of gen. 1, VI₁/1000 = 444.8 × 1750/1000 = 778.4kW

 

               kW output of gen. 2, VI₁/1000 = 444.8 × 2250/1000 = 1001kW 

Posted Date: 7/23/2012 1:54:07 AM | Location : United States

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