Delta Transformer Calculations

Delta Transformer Calculations

Extracted from Mike Holt’s Illustrated Guide to Electrical Calculations

To correctly specify a transformer with delta winding connections, you need to know how to do the related calculations.

The names for transformers configurations such as delta and wye derive from the way the windings are connected inside the transformer. These connections determine the way the transformer will behave, and so they also determine the methods of calculation required for properly applying a given transformer.

Delta-connected transformers have the windings of three 1Ø transformers connected in series with each other to form a closed circuit. The line conductors connected where two 1Ø transformers meet. This configuration is a “delta system,” because in an electrical drawing it looks like a triangle (Greek symbol Dfor the letter “delta”). Many call it a high-leg system because the voltage from Line 2 to ground is higher than that of the other legs. For example, a 120V delta transformer will have a 208V leg (Figure 12–1).

Graphics are not included in this newsletter.

Definitions

To avoid confusion, we need to clarify some key terms:

Line: The ungrounded (hot) conductor(s). See Figure 12–2 and Figure 12–3.

Line Current: The current on the ungrounded conductors (Figure 12–3, B1 and B2). In a delta system, the line current is greater than the phase current by the Square Root of 3 (which is approximately 1.732). In a wye system, the line current equals the phase current.

Line Voltage: The voltage between any two line (ungrounded) conductors (Figure 12–3, A1 and A2). In a delta system, the line voltage equals the phase voltage (Figure 12–11). But, the delta transformer also has a high-leg.

Phase Current: The current flowing through the transformer winding (Figure 12–3, D1 and D2). In a delta system, the phase current is less than the line current by the Square Root of 3. In a wye system, the phase current equals the line current.

Phase Load: The load on the transformer winding.

Phase Voltage: The internal transformer voltage generated across any one winding of a transformer. For a delta secondary, the phase voltage equals the line voltage. In a wye system, the phase voltage is less than the line voltage by the Square Root of 3 (Figure 12–3, A2 and C2).

Ratio: The number of primary winding turns divided by the number of secondary winding turns.

Unbalanced Load: (Neutral Current). The load on the secondary grounded (neutral) conductors.

With these terms defined, we are ready to tackle transformer calculations. We will look specifically at delta transformers.

Delta transformer current.

In a delta transformer, the line current does not equal the phase current (as it does in a wye transformer). Because each line from a delta configured transformer is connected to two transformer phases, the line current from a 3Ø load will be greater than the phase current by the Square Root of 3. Note these formulas:

• ILine = IPhase x square Root of 3
• ILine = VALine/(ELine x Square Root of 3)
• IPhase = ILine/Square Root of 3
• IPhase = VAPhase/EPhase

You can use the delta triangle (Figure 12–18) to calculate delta 3Ø line and phase currents. Place your finger over the desired item, and the remaining items show the formula to use.

If you plug some numbers in, you can more clearly see the effects of the delta configuration on currents. Let’s try this with a 240V, 36 kVA, 3Ø load (Figure 12–15).

Line: Total line power = 36 kVA

• ILine = VALine/(ELine x Square Root of 3)
• ILine = 36,000 VA/(240V x Square Root of 3)
• ILine = 87A, or
• ILine = IPhase x 3 x Square Root of 3
• ILine = 50A x 1.732 = 87A

Phase: Phase power = 12 kVA (per winding)

• IPhase = VAPhase/EPhase
• IPhase = 12,000 VA/240V = 50A, or
• IPhase = ILine/Square Root of 3
• IPhase= 87A/1.732 = 50A

We can also use the formula: ILine = VALine / (ELine x Square Root 3). What is the secondary line current for a 480 to 240/120V, 150 kVA, 3Ø delta transformer (Figure 12–13)?

1. 416A
2. 360A
3. 180A
4. 144A

Answer: (b) 360A

• ILine = VALine/(ELine x Square Root of 3)
• ILine = 150,000 VA/(240V x 1.732) = 360A

You can calculate the phase current of a delta transformer winding by dividing the phase VA by the phase volts:

IPhase = VAPhase/EPhase

1. The phase load in VA of a 3Ø, 240V load = line load/3 (1/3 of load on each winding
2. The phase load in VA of a 1Ø, 240V load = line load (all on one winding)
3. The phase load in VA of a 1Ø, 120V load = line load (all on one winding)

Now, let’s work an example problem. What is the secondary phase current for a 480 to 240/120V, 150 kVA, 3Ø delta transformer (Figure 12–14)?

1. 416A
2. 360A
3. 208A
4. 104A

Answer: (c) 208A

• Phase power = 150,000 VA/3 per phase = 50,000 VA per phase
• IPhase = 50,000 VA/240V
• IPhase = 208A

Try running these numbers with a 10A load and then with a 75A load. You’ll gain a greater appreciation for what happens in a delta system.

Delta transformer balancing

To properly size a delta/delta transformer, the transformer phases (windings) must be balanced. You can do that with a two-step process:

Step 1. Determine the VA rating of all loads.

Step 2: Balance the loads on the transformer windings as follows

• 3Ø Loads: one-third of the load on each of the phases.
• 240V, 1Ø Loads: 100% of the load on Phase A or B. You can place some 240V, 1Ø load on Phase C when necessary for balance.
• 120V loads: 100% of the load on C1 or C2.

To size the panelboard and its conductors, you must balance the loads in amperes. Why balance the panel in amperes? Why not take the VA per phase and divide by phase voltage? Answer: Line current of a 3Ø load is calculated by the formula:

• IL = VA/(ELine x Square Root of 3)
• IL = 150,000 VA/(240V x 1.732) = 208A per line.

If you took the per-line power of 50,000 VA and divided by one line voltage of 120V, you would come up with an incorrect line current of 50,000 VA/120V = 417A.

Delta transformer sizing

Once you balance the transformer, size it to the load of each phase. Size the “C” transformer using two times the highest of “C1” or “C2.” The “C” transformer is actually a single unit. If one side has a larger load, that side determines the transformer size.

Let’s try a practice problem. What size 480 to 240/120V transformer is required for the following loads: one 240V, 36 kVA, 3Ø heat strip; two 240V, 10 kVA, 3Ø loads; three 120V, 3 kVA loads (Figure 12-19)?

1. three 1Ø, 25 kVA transformers
2. one 3Ø, 75 kVA transformer
3. a or b
4. none of these

Answer: (c) a or b

 Phase A (L1 and L2) Phase B (L2 and L3) C1 (L1) C2 (L3) Line Total One 240V, 36 kVA, 3Ø 12 kVA 12 kVA 6 kVA 6 kVA 36 kVA Two 240V, 10 kVA, 1Ø 10 kVA 10 kVA 20 kVA Three 120V, 3 kVA, 1Ø 6 kVA* 3 kVA* 9 kVA 22 kVA 22 kVA 12 kVA 9 kVA 65 kVA

* Indicates neutral (120V) loads

• Phase winding A = 22 kVA
• Phase winding B = 22 kVA
• Phase winding C = (12 kVA of C1 x 2) = 24 kVA

Now that you understand some transformer calculation basics and specifics on delta transformer calculations, you will be able to size delta transformers correctly. You find delta-delta transformers most often in special applications. The most common configuration is the delta-wye. In the case of a delta-wye transformer, you now know how to size the primary side. In our next Calculations article, we’ll address the wye transformer calculations and you will then be able to size any combination of delta and wye.

Mike Holt’s Comment: This is a very technical article and I know the graphics would make it easier to understand, but it’s a business decision I’m made not to give “everything away”. If you have any questions with this newsletter or if you feel I have made an error, please let me know.

Copyright © 2003 Mike Holt Enterprises,Inc.
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