No load current of single phase transformer

The no-load current $I_0$ has two components: the core-loss (active) component $I_c$, in phase with $V$, and the magnetising component $I_m$, lagging $V$ by 90°. $I_0 = \sqrt{I_c^2 + I_m^2}$.

Core-loss component

$$\begin{aligned} \text{Volume of iron} &= l_i A_i = 2.7\times130\times10^{-4} = 0.0351\ \text{m}^3\\ \text{Weight} &= 0.0351\times7500 = 263.25\ \text{kg}\\ P_i &= 2.1\times263.25 = 552.8\ \text{W}\\ I_c &= \frac{P_i}{V_1} = \frac{552.8}{6600} = 0.0838\ \text{A} \end{aligned}$$

Primary turns

$$T_1 = \frac{V_1}{4.44fB_mA_i} = \frac{6600}{4.44\times50\times1.2\times0.013} = 1905.8 \approx 1906$$

Magnetising component

Maximum mmf needed:

$$\begin{aligned} AT_{iron} &= at\times l_i = 650\times2.7 = 1755\\ AT_{joints} &= \frac{B_m}{\mu_0}l_g = \frac{1.2}{4\pi\times10^{-7}}\times1\times10^{-3} = 954.9\\ AT_{max} &= 1755 + 954.9 = 2709.9 \end{aligned}$$

Assuming sinusoidal magnetising current, the rms value is

$$I_m = \frac{AT_{max}}{\sqrt2\,T_1} = \frac{2709.9}{\sqrt2\times1906} = 1.0054\ \text{A}$$

No-load current

$$\begin{aligned} I_0 &= \sqrt{I_c^2 + I_m^2} = \sqrt{0.0838^2 + 1.0054^2} = 1.0088\ \text{A}\\ \cos\phi_0 &= \frac{I_c}{I_0} = 0.0830 \end{aligned}$$

        Ic = 0.084 A
   O-------->  V (reference)
   |\
   | \  I0 = 1.009 A
   |  \
   v   v
  Im = 1.005 A (lags V by 90 deg)

Answer: no-load current ≈ 1.009 A (core-loss component 0.0838 A, magnetising component 1.005 A, no-load pf ≈ 0.083).