Magnetic circuits and transformers
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| flux/flʌks/ | từ thông |
| magnetomotive force | lực từ động |
| reluctance | điện kháng từ |
| air gap | air gap |
| transformer/trænsˈfɔːmə/ | Máy biến áp |
| open-circuit test | open-circuit test |
| short-circuit test | short-circuit test |
One law, mirrored
- A coil of $N$ turns at current $I$ is a flux pump: magnetomotive force $F = NI$ (ampere-turns) drives flux $\Phi$ through reluctance $R_{m} = l/(\mu A)$ — the magnetic mirror of Ohm's law, $\Phi = F/R_{m}$.
- Density $B = \Phi/A$ is what iron cares about: power cores live between 1.0 and 1.5 T.
A core: $l = 0.5$ m, $A = 10$ cm², $\mu_{r} = 4000$, 200 turns at 0.5 A. $F = 100$ At, $R_{m} = 99\,472$ At/Wb, $\Phi = 1.005$ mWb, $B = 1.005$ T ✓ both routes ($H$-route and $R_{m}$-route) agree.
{exercise:uni-ee0911-rm}
A core has l = 0.5 m, A = 10 cm², µr = 4000, wound with 200 turns at 0.5 A. The flux is about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
F = 100 At over Rm = 99,472 At/Wb gives Φ = 1.005 mWb and B = 1.005 T — the normal power-core range.
The air gap punchline
- Air has $\mu_{r} = 1$: four thousand times weaker than this core. A 0.5 mm gap adds $R_{gap} = 397\,887$ At/Wb — four times the whole core.
- The same 100 At then drives only 0.201 mWb: one fifth of the flux, from a gap a hair thick.
{exercise:uni-ee0912-gap}
Cutting a 0.5 mm gap into that core changes the flux by about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The gap alone has R = 397,887 At/Wb, four times the core, so total reluctance quintuples and Φ falls to 0.20 mWb.
Two tests, two branches
- The ideal transformer trades volts for amps: $V_{1}/V_{2} = N_{1}/N_{2}$, $I_{1}/I_{2} = N_{2}/N_{1}$; power across is conserved.
- The open-circuit test (rated volts, LV side) reads the shunt branch: 400 V, 0.7 A, 240 W gives $R_{c} = 667\ \Omega$, $X_{m} = 1108\ \Omega$. The short-circuit test (rated current, HV side) reads the series branch: 80 V, 5 A, 300 W gives $R_{eq} = 12$, $Z_{eq} = 16$, $X_{eq} = 10.6\ \Omega$.
{exercise:uni-ee0913-tests}
An open-circuit test reads 400 V, 0.7 A, 240 W. The core-loss current is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Ic = P/V = 240/400 = 0.6 A; the remaining 0.36 A is the quadrature magnetising current.
Pay the losses, find the peak
- Full load, unity power factor: $\eta = 10\,000/(10\,000 + 240 + 300) = 94.9\%$. At 0.8 pf it falls to 93.7% — fixed core loss weighs more as output drops.
- Efficiency peaks where copper loss equals core loss: $x = \sqrt{240/300} = 0.894$ of full load.
{exercise:uni-ee0914-eta}
That 10 kVA transformer at full load, unity pf, has copper 300 W and core 240 W. Its efficiency is about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
η = 10,000/(10,000+240+300) = 94.9%. Peak efficiency sits at x = √(240/300) = 0.894 of full load.