4 What is the worst case terminal voltage of a 750 HP SCIM w

4) What is the worst case terminal voltage of a 750 HP SCIM with 70 % eff and 80 % pF and a N/P terminal voltage of 2300 Vac running at full load if on the same bus a 500 HP, SCIM with a locked rotor code of G, 70 % eff and 80 % pF and a N/P terminal voltage of 2300 Vac is started from a transformer rated 7000 kVA transformer, 69kV to 2400 Vac and an impedance of 5-1/2%? Due to very short distances, cable impedance is not of any concern here.

Solution

Rated Power drawn by SCIM 1 is P_o = 750 HP or 559.275 kW .

Given efficiency of SCIM 1 is eff = 70% and Power factor of SCIM 1 is P.F = 80% .

Using efficiency input power supplied to SCIM 1 is P_in = P_o / eff ;

P_in = 559.275/0.7 = 798.964 kW;

Current drawn by SCIM 1 is given I₁ = P_in / (1.732 * V_L-L * P.F)

I₁ = 798.964*10³ / (1.732 * 2300 * 0.8) =  250.704(at angle(-36.86⁰)) A;

Now SCIM 2 was connected with

P_o = 500 HP or 372.85 kW ; eff = 0.7 ; P.F = 0.8 ;

Input power drawn by SCIM 2 is P_in = P_o / eff = 372.85 / 0.7 = 532.642 kW;

Current drawn by this motor is

I₂ = P_in / (1.732 * V_L-L * P.F) = 532.642*10³ / (1.732 * 2300 * 0.8) = 167.136(at angle(-36.86⁰)) A;

The two SCIM 1 and SCIM 2 are supplied from a transformed of rating

7000 kVA and 69 kV / 2400 V and impedance of Z = j0.055 .

Voltage drop due impendance of transformer is V_d = I*Z ;

Where I = I₁ + I₂ = 250.704(at angle(-36.86⁰)) + 167.136((at angle(-36.86⁰))) = 417.84(at angle(-36.86⁰)) A;

V_d = 417.84(at angle(-36.86⁰)) * j0.055 = 22.98(at angle(53.14⁰))

Terminal of transformer will be

V_s - V_d = 2400 - 22.98(at angle(53.14⁰)) = 2386.28(at angle(-0.44⁰))

Magnitude of terminal voltage of transformer is V = 2386.28 V which will be the worst case when two SCIM loaded at a time.