How to determine copper loss in a transformer

• The amount of heat produced in the windings and connections. • The amount of heat produced in the iron core. • How effectively the heat can be removed from the transformer when the thermal rating of the transformer is reached. At this point, the heat being produced must equal the heat being removed or dissipated – thermal equilibrium. Transformer Heat, Copper and Iron Losses (on photo courtesy of Siemens: Geafol-Cast-resin transformer) Copper load losses” or “ I 2R losses“. As the transformer is loaded, heat is produced in the primary and secondary windings and connections due to I 2R. At low loads, the quantity of heat produced will be small but as load increases, the amount of heat produced becomes significant. At full load, the windings will be operating at or near their design temperature. Figure 1 shows the relationship between load-current and the heat produced in transformer windings and connections. Figure 1 – Relationship between Load and Heat Produced in Transformer Windings Iron (or Core) Losses The iron loss is due to stray The majority of the flux is as indicated in the following Figure 2, flowing around the core. Figure 2 – Circulating Core Flux Some of the flux however, will try to flow at angles to the core and will cause eddy currents to be set up in the core itself. The term eddy is used because it is aside from the main flow. To combat this effect, the core is laminated as illustrated in Figure 3. The laminations provide small gaps between the plates. ...

I wonder if anyone can help me understand copper losses in transformers a bit better. I'm having problems understanding the highlighted part on the attached picture half way down the page. I sure its a simple maths thing which escapes me but why the \$\left( \frac \right)^2\$ ? Specifically, what's inside the brackets? \$\begingroup\$ Load power could be seen as a function of current, because the voltage doesn't change. But the copper loss has the variable voltage too, with the current, with the same changing proportion. You know, power is voltage * current. Then the power proportion would be its square because of same proportions. \$\endgroup\$ The iron losses are 3 kW regardless of load. The copper losses are equal to the iron losses when the load (current) is 80% of full load. Since copper losses are proportional to the square of current, to scale up from 80% load to 100% load multiply the losses at 80% load by the square of the 100/80 increase in load. Transformer maximum efficiency occurs when the Iron losses (Fe Loss) are equal to the Copper Losses (Cu Loss). i.e. Cu Loss = I2R = 3kW = 3000 W At max efficiency, transformer delivers 80% full load. P=VI*cosØ Transformer Parameters V1/V2 = Vp/Vs = 440/110 Full load kVA = 40 • S = VI I = S/V At full load, Ifl =40000/110 = 363.6364 A At 80% full load, I0.8fl = 0.8*363.6364 = 290.9091 A At max efficiency, R = 3000/I2 (Ω) = 3000/290.90912 = 0.03545 Ω Cu Loss(@full load) = (Ifl)2 *R = 363.63642 * 0.03545 = 4.6875 kW ≈ 4.7 kW...

Three-phase power transformers are used in power grids worldwide for efficient electric power transmission. Though they offer significant advantages over single-phase transformers in terms of capacity, load balance, and efficiency, the computation of losses isn’t as straightforward. Using the COMSOL Multiphysics® software, we can reliably compute the losses in the core, coils, and carpentry, as well as important lumped parameters like primary and secondary inductance. The safety and reliability of transformers depend largely on how well the design is able to dissipate its losses. Negligence in this regard would warrant penalties and could result in large mishaps. Power Transformers: An Introduction The efficiency of electric power transmission from the source (such as a power generation plant) to the destination (such as the customer) is calculated by comparing the power generated and power received. For maximizing the transmission efficiency, energy losses during transmission need to be minimized. While transmitting power over long distances, this is achieved by reducing the currents flowing through the transmission network by increasing the voltage prior to transmission and decreasing it at the receiving end, generally in a substation. A power station with three-phase transformers in Bruchsal, Germany. Image by Ikar.us — Karlsruhe:Datei:Kändelweg NE.jpg, own work. Licensed under For AC power, this “stepping up” and “stepping down” can be done based on a surprisingly simp...

Copper loss calculator: Enter the primary current and resistance, secondary current and resistance. Then press calculates button to get copper loss in watts. If you need to find single-phase transformer copper loss, then take the first results or if you need three-phase copper loss then consider the second result. Enter Primary Current: Amps Enter Primary Resistance Ohms Enter secondary Current: Amps Enter secondary Resistance: Ohms Result – Single Phase Copper Loss: Watts Result – Three Phase Copper Loss: Watts Copper loss calculation formula: Copper loss Pc in Watts is equal to the resistance R (Ω) in ohms times of the square of the current I (A) in Amps. Copper loss Pc = I 2 x R Watts But the transformer has two winding such as primary winding and secondary winding. Hence we need to calculate the copper loss for both primary and secondary windings. Therefore, transformer copper loss Pc in watts is equal to the sum of the primary copper loss and secondary copper loss. The formula can be written as Pc = Primary copper loss + Secondary Copper loss Pc = I (P-A) 2 x R1 (Ω) + I (S-A) 2 x R2 (Ω) Here, I (P-A) = Primary Current in Amps. I (S-A) = Secondary Current in Amps. (Ω) = Primary Resistance in ohms. R2 (Ω) = Secondary resistance in ohms. Three-phase Transformer copper loss calculation: Three-phase copper loss Pc in watts is equal to 3 times of the single-phase copper loss. Hence we can write it as, Pc = 3 * (I (P-A) 2 x R1 (Ω) + I (S-A) 2 x R2 (Ω)) Watts

Make use of this free transformer calculator to instantly estimate voltage, load currents, various losses, and other related parameters. Let’s move on towards ten calculations of the idea; and read; transformer. What Is A Transformer? In the field of electrical technology: “A passive component that transfers electrical energy among various electrical circuits is called a transformer” Transformer Symbol: Carefully watch the symbolic diagram of the transformer. Later on, we will elaborate it in detail for you as well. Another add is that our best transformer calculator will also let you estimate each and every element related to transformer functionality. Types of Transformer: Various types of transformer are there that are used for particular purposes in different fields. These include: Step Up Transformer: This kind of transformer has a secondary voltage greater than that of the primary voltage. This transformer type is used in a locality or area where the voltage rating is very low and the public needs to use appliances that operate on higher voltages. Step Down Transformer: In this transformer, the primary voltage is higher than the secondary voltage. Step down transformers are typically used in commercial or residential localities where consumers use different devices that operate on lower voltages. This online step down transformer calculator also signifies the working of this kind. Single phase Transformer: This transformer operates on only single phase power systems....

The transformer is an electric device that has a completely static structure, meaning it has no moving parts. It performs the crucial function of transferring electricity from one circuit to the other at a required frequency. But do you know a considerable amount of power is lost in this transferring process? Yes, you hear us right. What’s more notable is that these losses lead to the heating up of the transformer, which affects its efficiency. Therefore, it becomes important to calculate transformer losses to keep its efficiency under check. Now you must be thinking, what are transformer losses, and why do they occur? Don’t worry, we have got you covered! In this blog, you will learn all the important aspects that relate to how you can calculate transformer losses. So, let’s dive in! What is the Function of a Transformer? Before we start with transformer losses, let’s understand how a transformer works. You may have seen compact structures around your house in the electric power grid. These structures constantly convert high voltage power into low voltage to meet the requirements of different residential or commercial spaces. It is important because most power grids supply 220 kV or 440 kV across different power stations. Such a high-power electric current can cause potential harm to various electronic devices. Therefore, transformers are used to step down the electric voltage for safe power consumption. What are Transformer Losses? To calculate transformer losses, it is ...

When the alternating flux cuts the steel core, an EMF is induced in each lamination, causing a current (called an eddy current) to flow in the closed electrical circuit of the lamination. This eddy current flows through the resistance in each lamination, causing heat to be generated in the laminations and therefore in the core as a whole. Although eddy-current losses are effectively reduced by using laminations for the core, they are never entirely eliminated. Hysteresis The alternating flux also causes changes in the alignment of the magnetic domains in the magnetic core, with the magnetic polarity reversing 100 times a second. This change is energy consuming and heat is produced within the core. The energy loss is referred to as hysteresis loss, the degree of loss being dependent on the nature of the material used for the laminations. Silicon steel has low hysteresis losses, making it suitable for electrical laminations. Figure 1 shows a comparison of two hysteresis curves for different materials. It can be seen that the silicon steel curve has a smaller area, representing a lower energy loss and reduced heat production. Figure 1 Hysteresis curves The total iron losses represent the power absorbed by the iron core and so are proportional to I e, the energy component of I 0 in Figure 2. The mutual flux Φ remains fairly constant from no load to full load, therefore it follows that the excitation current I 0 producing that flux, and so I e, will also be constant. The iron l...

The total losses that take place in the winding resistance of a transformer are known as the ‘Copper losses’. These losses in a transformer should be kept as low as possible to increases the efficiency of the transformer. To reduce the copper losses, it is essential to reduce the resistance of primary and secondary winding coils of the transformer i.e. size of the winding conductor is selected very carefully. These are also known as the variable losses as these are dependent on the square of load current. To determine the copper losses, The total copper losses in transformer are: = I 1 2R 1 + I 2 2R 2 = I 1 2R 01 = I 2 2R 02 Where , I 1 , I 2 = primary and secondary currents respectively, R 1, R 2 = primary and secondary resistances respectively, R 01, R 02 = Iron Losses in Transformer The power losses that take place in its iron core are known as the ‘Iron losses’. These losses occur due to alternating flux set up in the core. In a transformer, flux set up in the core remains constant from no load to full load. Hence these power losses are independent of load and also known as constant losses of a These losses have two components named hysteresis losses and eddy current losses. To determine the iron losses, Hysteresis Power Losses in Transformer When a magnetic material is subjected to reversal of flux, power is required for the continuous reversal of molecular magnets. This power is dissipated in the form of heat and is know as ‘Hysteresis Loss’. The hysteresis loss of a...

This article needs additional citations for Please help Find sources: · · · · ( January 2021) ( Copper loss is the term often given to winding loss is often preferred. The term load loss is used in Calculations [ ] Copper losses result from Copper Loss ∝ I 2 ⋅ R where t is the time in Effect of frequency [ ] For low-frequency applications, the power loss can be minimized by employing conductors with a large cross-sectional area, made from low- With high-frequency currents, the Reducing copper loss [ ] Among other measures, the electrical energy efficiency of a typical industrial References [ ]