Correct Answer - Option 1 : 598.5 rpm
Concept:
The EMF equation of a DC Machine is
\({E_b} = \frac{{NPϕ Z}}{{60A}}\)
In a dc motor, back emf is
Eb = V – IaRa
Where,
N is the speed in rpm
ϕ is the flux per pole
P is the number of poles
Z is the number of conductors
V is the terminal voltage
Ia is the armature current
Ra is the armature resistance
Calculation:
Given V = 440 V
Ia = 100 A
Ra = 0.3 Ω
P = 6
A = P = 6 (LAP winding)
Z = 822
ϕ = 0.05 wb
Eb = V – IaRa
Eb = 440 - 100 × 0.3 = 410 V
\({E_b} = \frac{{NPϕ Z}}{{60A}}\Rightarrow N =\frac{E_b\times 60\times A}{Pϕ Z}\)
\(N = \frac{410 \times 60 \times6 }{6\times 0.05\times 822}=598.54\;rpm\)
∴ The speed is 598.5 rpm