# Calculate Transformer Regulation & Losses (As per Transformer Name Plate)

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
• 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:

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 has completed M.Tech (Power System Control), B.E(Electrical). He is member of Institution of Engineers (MIE) and CEng,India. Membership No:M-1473586.He has more than 16 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in “Electrical Mirror”, “Electrical India”, “Lighting India”,”Smart Energy”, “Industrial Electrix”(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.

### 6 Responses to Calculate Transformer Regulation & Losses (As per Transformer Name Plate)

1. 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?

2. A vijay says:

In total losses at 75 temperature hv side losses are not considered.

3. Rajesh Mistry says:

Thank you!! Jignesh Parmar.

4. dileep says:

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

5. Shaishav Shah says:

Can any one tell tentative loss of 10 MVA transformer at 25% loading. 66kv/11kv., Delta/star

6. Shabeeb says:

2500kva transformer and 6600 V primary and 420 V secondary.