Archived DC motor in series, finding the speed

[KESTRELx]

Homework Statement

when running at its rated load, a 40kW series motor takes 74A and 550V. The rated speed is 750 rev/min. the armature and field resistances are 0.35 Ω and 0.15 Ω respectivly. when the load torque is doubled the rated value, the current is 110A. dertermin the moror speed and the power output at 100% torque overload.

Answers given, worked solution not
ans = 538 Rev/min and 57.4kW

Homework Equations

Work= Torque x (2pi x speed over 60)
torque = K(t) I(a) ψ (k(t) =constant and I(a)= armature current)
EMF =KNψ (k= constant)
EMF= v-IR

The Attempt at a Solution

Work= Torque x (2pi x speed over 60)
40000= T x ((2pi x 750) / 60)
T=509.29


EMF =KNψ (condition 1)
EMF =KNψ (condition 2)

divide both and cancel
k is constant as its the same motor
flux is constant (as flux is not given a value, i assume that its the same)

EMF(1) over EMF(2) = N(1) over N(2)
put both to the power -1 so as to find N(2) easier
EMF(2) over EMF(1) = N(2) over N(1)

((550 - (74 x 0.35+0.15)) over (550 - (74 x 0.35+0.15))) x 750 = N(2)
N(2) = 723.6 rev/min which is not the value of the answer

seeing as i can't get the speed right i can't go onto finding the power developed, however using the speed provided in the answer i have done it and correctly

Work= 2Torque x (2pi x speed over 60)
Work= 2Torque x (2pi x 538 over 60)
=57386W
=57.4kW

 

[cnh1995]

Under rated load:
Back emf E_b=550-I(R_f+R_a)
∴E_b=513V
Mechanical power developed P₁=E_bI=513×74 W
Under 100% overload condition:
E_b=495V..(same formula with I=110A).
Mechanical power developed P₂=495×110=54.45kW.
Taking ratio P1/P2, we get
P₁/P₂=2πNT/2πN₁T₁
∴513x74/495x110=N/2N₁...(since T₁=2T).
∴750/2N₁=0.6971
∴N₁=537.87 rpm≈538 rpm.

 

[cnh1995]

k is constant as its the same motor
flux is constant (as flux is not given a value, i assume that its the same)

Series motor is not a constant flux motor. Field flux varies with load.