Calculate Transformer Regulation & Losses (As per Transformer Name Plate)
June 22, 2020 8 Comments
Calculate Transformer Regulation and Losses for following Transformer Name Plate Details
- KVA rating of Transformer(P)=16000VA
- Primary voltage(Vp)=11000V
- Secondary voltage(Vs)=433V
- No load losses(W0)=72Watt
- No load current(I0)=0.59Amp
- Full load losses(W)=394Watt
- Impedance voltage(Vi)=480Volt
- LV resistance(Rs) =219.16 miliΩ
- HV resistance(Rp) =215.33 Ω
- Amb temperature(c)=30 Deg C
- Total Connected Load on Transformer(Pl)=10000VA
Calculation:
- % Loading of Transformer=Pl/P
- % Loading of Transformer=10000/16000 = 63%
I2R Calculation:
- HV Full load current (Ip) =P/Vpx1.732
- HV Full load current (Ip) =16000/11000×1.732=0.84 Amp
- LV Full load current (Is)=P/Vsx1.732
- LV Full load current (Is)==16000/433×1.732=21.33 Amp
- HV Side I2R losses= IpxIpxRp
- HV Side I2R losses= 0.84×0.84×215.33=227.8 Watt—–(A)
- LV Side I2R losses= IsxIsxRs
- LV Side I2R losses==21.33×21.33×219.16=149.63 Watt—(B)
- Total I² R losses @ Amb temp(Ir)=A+B
- Total I² R losses @ Amb temp(Ir)=227.8+149.63=377.43 Watt
- Total Stray losses @ Amb temp (Ws) =Full Load Losses-I2R Losses
- Total Stray losses @ Amb temp (Ws) =394-377.43=16.57 Watt
- I² R losses @75° c temp =Irx310/235xc =149.63×310/235×30 =441.52Watt
- Stray loses @ 75° c temp =(Wsx(235+c))/310
- Stray loses @ 75° c temp =(16.57x(235+30))/310=14.16 Watt
- Total Full load losses at @75° c=441.52+14.16=455.69 Watt
- Total Impedance at ambient temp (Ax)=Vix1.732/Ip
- Total Impedance at ambient temp(Ax)=480×1.732/0.84=989.94Ω
- Total Resistance at amb temp (Ar)=Ir/IpxIp
- Total Resistance at amb temp (Ar)=377.43/0.84×0.84=535.15Ω
- Total Reactance (X)=√AxxAx+ArxAr
- Total Reactance (X)=√989.98×989.94+535.15×535.15=832.82Ω
- Resistance at@ 75° c (R)= (310xAr)/(235+c)=310×535.15/235+30 = 626.03Ω
- Impedance at 75° c (X1)=√2X+2R=√2×626.03+2×832.82 = 1041.88Ω
- Percentage Impedance = (X1x0.5774xIpx100)/Vp
- Percentage Impedance = (1041.88×0.5774×0.84×100)/11000=4.59%
- Percentage Resistance (R%) =(Rx0.5774xIpx100)/Vp
- Percentage Resistance(R%) =(626.03×0.5774×0.84×100)/11000=2.76%
- Percentage Reactance(X%) = (Xx0.5774xIpx100)/Vp
- Percentage Reactance(X%) = (832.82 x0.5774×0.84×100)/11000=3.67%
Regulation
- Regulation at Unity P.F =2.76
- Regulation at Unity at 0.8 P.F =((R% x cosØ)+(X% x SinØ))+(0.005x((R% x SinØ)+(X% x CosØ)))
- Regulation at Unity at 0.8 P.F =((2.76 x 0.8)+(3.67 x 0.6))+(0.005x((2.76 x0.6)+(3.67 x 0.8)))=4.43
Results
- Total I² R losses @ Amb. temp(Ir)= 377.43Watt
- Total Stray losses @ Amb. temp (Ws) =16.57 Watt
- Regulation at Unity P.F =2.76
- Regulation at Unity at 0.8 P.F =4.43
JIGNESH PARMAR
I have a design for an autotransformer which shall be added with an output coil to be wound at both end terminal wires of the autotransformer. My idea is such as to produce an additional output aside from the tapped output of the autotransformer. My question is, is my design possible and will it produce the additional power output aside from the tapped power output of the autotransformer?
In total losses at 75 temperature hv side losses are not considered.
Thank you!! Jignesh Parmar.
I² R losses @75° c temp =Irx310/235xc =149.63×310/235×30 =441.52Watt
Stray loses @ 75° c temp =(Wsx(235+c))/310
Stray loses @ 75° c temp =(16.57x(235+30))/310=14.16 Watt
1) Please explain how do you arrive 310 & 235?
Total Reactance (X)=√AxxAx+ArxAr
Total Reactance (X)=√989.98×989.94+535.15×535.15=832.82Ω
2) Reactance = sqrt of (square of Z -Square of R) but in your answer you have mentioned plus sign for reactance formula?
Impedance at 75° c (X1)=√2X+2R=√2×626.03+2×832.82 = 1041.88Ω
3) Why do you use 2 instead of Square?
Percentage Impedance = (X1x0.5774xIpx100)/Vp
4)how do you arrive 0.5774 in above formula?
Regulation at Unity at 0.8 P.F =((R% x cosØ)+(X% x SinØ))+(0.005x((R% x SinØ)+(X% x CosØ)))
5) What about 0.005 in above formula?
Please correct me if my Perspective of thinking is wrong in the above observations
Can any one tell tentative loss of 10 MVA transformer at 25% loading. 66kv/11kv., Delta/star
2500kva transformer and 6600 V primary and 420 V secondary.
Can you please help me to solve
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