Motor Starters Part 4: Selecting and Sizing StarDelta Parts
This article makes the selection and sizing of the stardelta motor starter parts very easy. Here you will learn how to select and size the fuse, circuit breaker, thermal overload relay, and contactor.
Part 3 of the series discussed the basic theories of the stardelta motor stater, noting the composition of contactors, fuses, circuit breakers, and overload relays. These components come in different sizes and create challenges in selection and sizing. This article will address how to select and size the fuse, circuit breaker, thermal overload relay, and contactor.
A Sample Motor Challenge to Solve
In this article we will walk through the design of a stardelta motor starter to meet the following requirements:
 415 V 3phase motor
 10 HP
 Code A
 Noninductive load
 Efficiency of 80%
 PF of 0.8
 Rotating at 600 RPM.
Motor Torque and Current
To work out the solutions, we must start with the basic calculation of the motor’s torque and current.
The 5252 Conversion Factor
When calculating the motorrated torque or the Full Load Torque (FLT) of the motor, use the constant 5252. The number 5252 represents the point during the revolution range of the motor where the torque and the horsepower cross paths. It is derived from the relationship between the motor speed, power, and torque as follows:
$$1 \text{ HP} = 33000 \text{ ftlb/min}$$
Given that:
$$1 \text{ revolution} = 360° = 2\pi$$
The crossing path factor will be determined by:
$$\frac{33000}{2\pi} = 5252.113 ≈ 5252$$
Formula for the Motor Full Load Torque (FLT)
Let's now caluclate the value of the motor's full load torque:
$$\text{FLT} = \frac{5252\cdot\text{HP}}{\text{RPM}}$$
Let's now input the values from our sample motor design challenge:
$$\text{FLT} = \frac{5252×10}{600} = 88 \text{ lbft}$$
We can now converting FLT from lbft to Nm using the folloowing:
$$1 \text{ lbft} = 1.356 \text{ Nm}$$
Therefore,
$$\text{FLT} = 88 \times 1.356 = 119 \text{ Nm} $$
Calculating the Starting Torque of the Motor
For motors of capacity below 30 kW, then the induction motor starting torque is 3 times the motor full load torque. We can plug in our previously calculated value for the FLT to derive the starting torque:
$$\text{Starting Torque of the Motor} = 3 \times \text{Full Load Torque}$$
$$\text{Starting Torque of the Motor} = 3 \times 119 = 357 \text{ Nm}$$
Calculating the Lock Rotor Current of the Motor
The Lock Rotor Current is in the form of a range that can be attained from the standard table of Locked Rotor Current given the code of the motor as shown below in Table 1. Note that the full table runs from code A to code V. For our sample motor problem, the moder is code A.
Table 1. Locked Rotor Currents
CODE 
MIN 
MAX 
A 
1 
3.14 
B 
3.15 
3.54 
C 
3.55 
3.99 
D 
4 
4.49 
The formula for calculating the locked rotor current (LRC) of the motor is given by:
$$\text{LRC} =\frac{1000 \times \text{ HP} \times \text{ Locked Rotor Current Code Value}}{1.732 \times V}$$
Now, we can calculate the minimum and maximum motor LRC using the horsepower and voltage of the motor:
$\text{LRC}_{min} =\frac{1000 \times 10 \times 1}{1.732 \times 415} = 13.91 \text{ A}$
\$\text{LRC}_{max} =\frac{1000 \times 10 \times 3.14}{1.732 \times 415} = 43.69 \text{ A}$
Calculating the Full Load Current of the Motor
The formulas for the line and phase full load currents are given by equals:
$$FLC_{LINE}=\frac{1000 \times P}{1.732×PF}$$
$$FLC_{phase}=\frac{FLC_{Line}}{1.732}$$
Using those equations and plugging in the values for our sample motor:
$$FLC_{LINE}=\frac{1000×10×0.746}{1.732×415×0.8}=12.97 \text{ A}$$
$$FLC_{phase}=\frac{12.97}{1.732}=7.49 \text{ A}$$
Calculating the Stardelta Starter Motor Starting Current
Finally, the formula for the stardelta starter motor starting current equals:
$$\text{Motor Starting Current} = 3 \times FLC = 3 \times 12.97 = 38.91 \text{ A}$$
After calculating all the basic required values, it becomes easy for engineers and technicians to size the components of the stardelta starter.
Selecting and Sizing the Fuse
Table 2 contains data on standard fuse current ratings as per NEC 43052, which will make fuse selection and sizing easy.
Table 2. Sizes of fuses per NEC 43052
MOTOR TYPE 
TIME DELAY FUSE 
NONTIME DELAY FUSE 
Single phase 
300 
175 
3phase 
300 
175 
Synchronous 
300 
175 
Wound rotor 
150 
150 
Direct current 
150 
150 
The motor being powered is a 3phase motor. The fuse time delay will be 300% of FLC and the nontime delay will be 175% of the FLC.
Time delay fuse maximum size = 300%×FLC = 300%×13 = 39A
Non  time delay fuse maximum size = 175%×FLC = 175%×13 = 22.75A≅23A
Selecting and Sizing the Circuit Breaker
The table below represents NEC 43052 circuit breaker sizes.
Table 3. Sizes of Circuit Breakers

CIRCUIT BREAKERS AS PER NEC 43052 

MOTOR TYPE 
INSTANTANEOUS TRIP IN % 
INVERSE TIME IN % 
Single Phase 
800 
250 
ThreePhase 
800 
250 
Synchronous 
800 
250 
Wound Rotor 
800 
150 
Direct Current 
200 
150 
Calculate the instantaneous trip circuit breaker using the formula below:
Instantaneous Circuit Breaker Maximum Size = 800%×FLC = 800%×13 = 104A
Now calculate the size of the inverse trip circuit breaker as shown below:
Inverse Trip Circuit Breaker Size = 250%×FLC = 250%×13 = 32.5A≅32A
Selecting and Sizing of Thermal Overload Relay
As the sizing of the DOL starter demonstrated, the thermal overload relay has an upper and lower value.
The Phase Thermal Overload Relay
Lower Limit of the Range = 70%×FLC_{phase} = 70%×7 = 4.9A≅5A
Upper Limit of the Range = 120%×FLC_{phase} = 120%×7 = 8.4A≅9A
The Line Thermal Overload Relay
For the startdelta starter, the thermal overload can be placed in the windings or the line.
When placed in line
Supply > O/L > Main Contactor
Thermal Overload Relay = 100%×FLC_{LINE} = 100%×13 = 13A
When placing O/L Relay in windings
Main Supply > Main  Contactor  Delta Contactor > Overload Relay
Thermal Overload Relay = 58×FLC_{LINE} = 58%×13 = 7.54A≅8A
Selecting and Sizing the Contactor Type
Main and Delta Contactor
Here the contactors are smaller as compared to the one used in DOL Starter. This is because the contractor here is only used to control the currents in the winding only.
The winding current = 1.723×58%×Line current
They are rated at 58% of the motor current rating.
Star Contactor
This is the third contactor and it only carries the star current. The current here is 58 % of the current in the delta contactor, which is 33% of the motor rating currents.
As in the DOL calculation, the same types of contractors exist. From the chart of standard types, the type of contactor is AC1.
Breaking Contactor Capacity = Value in the Chart×FLC_{LINE} = 1.5×13 = 19.5A≅19A
Main Contactor Size = 58%FLC_{LINE}= 58%×13 = 7.54A≅8A
Star Contactor Size =3 3%× FLC_{LINE} = 33%×13 = 4.29A≅4A
Delta Contactor Size = 58 % FLC_{LINE} = 58%×13 = 7.54A≅8A
Key Takeaways of Selecting and Sizing StarDelta Motor Starter Parts
In brief:
 Torque conversion factor is 5252
 The size of the fuse is 39 A for maximum and 23 A for the minimum size
 Instantaneous circuit breaker size is 104 A and the inverse trip circuit breaker size is 32 A
 The lower limit current of the thermal overload relay is 5 A and the upper limit is 9 A
 The line thermal overload relay is 13 A while the windings thermal overload relay is 8 A
 Main contactor is rated 8 A, star contactor 4 A while the delta contactor 8 A
Featured image used courtesy of Adobe Stock