Transformer Tests (Open, Short-circuit test)

TRANSFORMER TESTS

Performance of a transformer can be calculated on the basis of its equivalent circuit which contains four main parameter, the equivalent resistance 'R_o1' as referred to primary and 'R_o2' referred as secondary, the equivalent leakage reactance X_o1, X_o2 as referred to primary and secondary respectively, the core loss conductors G₀ and the magnetizing susceptance B_o.

These constants (or) parameters can be easily determined by two tests.

(i) Open-circuit test

(ii) Short-circuit test

These tests are very economical and convenient, because they required information without actually loading the transformer.

3.3.8.1 Open Circuit Test (or) No-Load Test (OC Test)

The purpose of this test is to determine no-load loss (or) core loss and no-load I₀ which is useful in finding X_o and R_o.

The transformer primary is connected to AC supply through ammeter, wattmeter. The secondary of transformer kept open. Usually low voltage side is called as primary and high voltage side is called as secondary.

The primary is excited by rated voltage, the wattmeter measured input power. The armature measures input curved and voltmeter gives the applied primary voltage.

As transformer secondary is open, it is on load so current drawn by the primary is no load current I_o. The two components of the no load current are

I_m = I_o sin ϕ_o

I_c = I_o cos ϕ_o

where cos ϕ_o = no load power factor and hence power input can be written as

W_o = V_oI_o cos ϕ_o

As secondary is open I₂ = 0. Thus its reflected current on primary is also zero. So primary current I₁ = I₀. The transformer no load current is always very small, handly 2 to 4% of its fall load value.

Total copper losses in OC test are negligibly small. As output power is zero and copper losses are very low, the total input power is used to supply iron losses. This power is measured by the wattmeter (ω₀). Hence the wattmeter in OC test gives iron losses which remain constant for all the loads.

ω_o = P_i = iron losses

ω_o = W_o/V_oI_o cos Ɵ

cos Ɵ_o = W_o/V_oI_o (No load power factor)

The value of cos Ɵ₀ is known then we can obtain,

I_c = I_m cos Ɵ_o

I_m = I_o sin Ɵ_o

The we determine the circuit parameters

R_o = V_o/I_c and X_o = V_o/I_m

3.3.8.2 Short Circuit Test (SC Test)

The fig shows the short circuit testing of transformer. The secondary is short circuited with the help of thick copper wire (or) solid link. As secondary is shorted, its resistance is very small and on rated voltage it may draw very large current. Such large current can cause over heating and burning of the transformer. To limit this short circuit current, primary is enough to cause rated current to flow through primary which can be observed on ammeter. Voltage applied is low which is a small fraction of the rated voltage. The iron losses are function of applied voltage so the iron losses in reduced voltage test are very small. Hence wattmeter reading is the power loss which is equal to full load copper losses as iron losses are very low.

W_sc = (P_cu) F.L

W_sc = V_sc I_sc cos Ɵ_sc

cos Ɵ_sc = W_sc/ V_scI_co

W_sc = I_sc² R_le

R_le = W_sc/I_sc²

Z_le = V_se/I_se = √R_le² + X_le²

X_le = √Z_le² - R_le²

Thus we get the equivated circuit parameters R_le, X_le and Z_le.

Example: 20

A 220/440 V single phase transformer has the following test results

OC test: 230 V, 1 A, 60 watts, LV side

SC test: 18 V, 18 A, 80 watts, HV side

Obtain the equivalent circuit parameters of transformer refered to LV side.

Solution:

OC test:

V_o = 220 V, I_o = 1A, W_o = 60 W

W_o = V_oI_o cos Ɵ_o

No-load input power factor cos Ɵ_o = W_o/V_oI_o

= 60 / 220×1

cos Ɵ_o = 0.27

sin Ɵ_o = 0.961

I_c = I_o cos Ɵ_o

= 1 × 0.27

I_c = 0.27

R_o = V_o/I_w = 220/0.27

R_o = 814 Ω

I_m = I_o sin Ɵ_o

= 1 × 0.961

I_m = 0.961

X_o = V_o/I_m = 220/0.961

X_o = 228.92

SC test

I_c = 0.27 R_o = 814Ω

I_m = 0.96 X_o = 228.92 Ω

SC voltage V_sc = 18 V

SC current I_sc = 18 A

SC power W_sc = 80 W

Z_o2 = V_sc/I_sc = 18/18 = 1Ω

W_sc = I_sc² R_o2

R_o2 = W_sc/I_sc² = 80/(18)² = 0.246Ω

Transformation ratio K = V₂/V₁ = 440/220 = 2

Z_ol = Z_o2/K² = 1/2²

= 0.25 Ω

R_o1 = R_o2/K² = 0.246/2² = 0.0617 Ω

X_o1 = √Z_A² - R_o1²

= √(0.25)²- (0.0617)²

X_o1 = 0.242 Ω