Sizing Gen-Sets For Large Motor Starting

Feb 1, 2008 12:00 PM, By Larry A. Bey, Cummins Onan Corp.

You've lost normal power. Your engine-generator set (gen-set) starts up and reaches speed. Now, you want to start some large motors key to your operation. Suddenly, starter holding coils drop out, starter contacts chatter, and a few motors stall due to insufficient torque for acceleration. Can this happen to you? It sure can, if you haven't sized your gen-set properly.

We all know that motors draw a high inrush current during starting:typically six times full load current. But, inrush currents for the high-efficient motors specified today are almost double that amount. Motors with high inertia loads can also require up to three times rated power during starting.

Yes, it's common for motor starting kVA requirements to determine the size of the set. However, the following factors also play a key role in sizing gen-sets:

• Harmonics caused by variable frequency drives.
• Use of high-efficiency motors.
• Sequential starting of motors.

What's involved. When starting motors, large voltage and frequency dips may occur if the generator set isn't sized properly. Other loads connected to the generator output may be more sensitive to voltage and frequency dips than the motor or motor starter, and this may cause problems. For example, a rate of change greater than 1 Hz/sec in generator frequency may cause some static UPS units to malfunction.

If the load on the generator set is a single large motor, particularly one requiring high starting torque, a number of problems can occur. They include: sustained low-voltage operation that can cause overheating; extended load acceleration times; opening of circuit breakers or motor protective devices; engine-generator protection shutdowns; and more.

Your gen-set's ability to start large motors without excessive voltage and frequency dip is a function of the complete system. This includes:

• The engine power available;
• The generator's capacity;
• The response of the generator excitation system;
• The energy stored in the rotating inertia of the gen-set; and
• The acceleration of the motor and its load.

You must consider all these factors for proper gen-set sizing. Here's a simple rule for estimating the size of an engine-generator set for motor starting: 1kW of generator set rating per each 3/4 to 1 hp of motor nameplate.

Let's look more closely at a motor start. Induction motors have typical starting characteristics. The curve of motor current versus speed shows that during starting, the motor draws approximately six times its full load current; this current remains high until the motor reaches about 80% of speed. This high inrush current causes a dip in generator voltage. The electric power initially required by the motor (with the motor at standstill) is about 150% of rated power. The power required by the motor peaks at about 300% of rated power and 80% of speed with full voltage applied. But, the generator set supplies less than 300% power because starting voltage is lower than full voltage during acceleration, and because the generator set's rotating inertia transfers energy to the motor.

The motor must develop greater torque than required by the load. The motor's torque curve at full voltage is above the load's torque curve. The difference between the torque developed by motor and the torque required by the load determines the rate of acceleration. Since torque is proportional to voltage, any reduction in voltage means a proportional reduction in torque.

A properly sized generator set will support the high starting kVA requirements of the motor, and maintain sufficient output voltage for the motor, so it can develop adequate torque to accelerate the load to rated speed.

All standby gen-sets use synchronous generators with exciters. Many are available with permanent magnet generator (PMG) excitation systems. The PMG provides excitation power independent of the generator terminal voltage. As such, it can maintain full excitation:even during transient loading, such as motor starting. Full excitation power results in a less extensive voltage dip and improved recovery times.

Using reduced-voltage starting. Though a voltage dip often causes various problems, a controlled reduction in voltage at motor terminals can be beneficial, but only when reduction in motor torque is acceptable. Reducing motor starting kVA can reduce the required size of the gen-set, reduce the voltage dip, and provide a softer start for the motor loads. When sizing gen-sets, you must first determine the acceptable level of motor torque required during starting, or the loads will accelerate slowly, or even fail to reach full speed:ultimately causing motor damage.

Using solid-state starters. Solid-state starters can adjust the starting torque, acceleration ramp time, and current limit for controlled acceleration of a motor when it starts. For the purpose of sizing a gen-set, the current limit adjustment reduces the inrush current and may be used to reduce the starting kW and kVA requirement on the generator. The range of available current limit settings is typically from 150% to 600% of full-load current. A 600% current limit setting on the solid-state starter results in a gen-set sizing that's the same as an across-the-line starting. A 300% current limit setting reduces starting kVA by 50%.

Use of the current limit setting also reduces motor torque available to the load. From a gen-set sizing perspective, an extended acceleration ramp time and low current limit setting (if appropriate for the motor and the mechanical load) would result in the least voltage and frequency excursions.

One downside to using solid-state motor starters is their integral SCRs (silicon-controlled rectifiers) will cause voltage distortion. To compensate, you'll have to oversize the generator. The recommendation: two times the running kW load, except where you're using an automatic bypass. If the solid-state starter does have an automatic bypass, the SCRs are only in the circuit during starting. Once the motor is running, the bypass contactor closes and shunts the SCRs. In this case, you can ignore the voltage distortion during starting, and you don't have to add generator capacity.

VFDs require larger generators. All versions of variable frequency drives (VFDs) are current limiting and reduce starting kW and kVA. The current drawn by these drives is nonlinear (having harmonics), which causes a distorted voltage drop across the reactance of the generator. Since VFDs are nonlinear, you must include an additional generator capacity sizing factor to keep voltage distortion to a reasonable level of approximately 15% total harmonic distortion (THD) or less. The larger the generator, the greater the reduction in impedance of the power source (generator), which in turn, reduces the effects caused by harmonic current distortion.

For six-pulse VFDs, a typical generator sizing factor would be twice the running kW of the drive. This offsets any reduction in starting kW and kVA. If it is the pulse width modulated (PWM) type (or includes an input filter to limit current distortion to less than 10%), then you can reduce the sizing factor down to 1.4 times the running kW of the drive.

Using a step starting sequence. The starting sequence of loads can have a significant effect on the size of a gen-set. One commonly used approach is to assume all connected loads will start in a single step. This results in the largest gen-set selection. Unless you do something to add load incrementally (such as multiple transfer switches with staggered time delays, or a step load controller), then you should use a single-step load for sizing purposes.

In multiple step applications, you start the largest motor first, to minimize the gen-set size. Once placing all loads on line with the gen-set, you can stop and start load equipment with automatic controls. Here, you'll have to size the gen-set by assuming the largest motor starts last, with all other connected loads already on line.

Examples of sizing gen-sets. You can size a gen-set with manual calculations (using a worksheet) or with PC software available from most major gen-set manufacturers. The basic process is the same. It's always best to use actual data (if known).

If this information isn't available, using PC software is the best option, since much of the required information on typical load characteristics is available as default information. If you use the manual sizing procedure, it should result in a recovery voltage of at least 90% of rated voltage and a starting instantaneous voltage dip of approximately 20% to 40%.

The instantaneous voltage dip and frequency dip will likely vary from manufacturer to manufacturer, based on equal ratings of gen-sets. For a closer estimation of transient (starting instantaneous voltage) performance, use the manufacturer's sizing software.

Using the manual sizing procedure.

Step 1: Gather information. You'll need to know the following for each motor load:

• Nameplate hp,
• Running kilowatts (RkW),
• Running kilovolt-amperes (RkVA),
• Running motor power factor (PF),
• Starting motor PF, and
• Locked rotor kVA/hp.

You can use the following equation to calculate RkW and RkVA for motors: RkW = [(Nameplate hp) x (0.746kW/hp)] / Efficiency (eq. 1)

RkVA = RkW / Running motor PF (eq. 2)

To calculate starting kilovolt-amperes (SkVA) and starting kilowatts (SkW) for motors, use these equations:

SkVA = (Nameplate hp) x (Locked rotor kVA/hp) (eq. 3)

SkW = (SkVA) x (Starting motor PF) (eq. 4)

Step 2: Total the RkW, RkVA, SkW, and SkVA numbers for all the loads.

Step 3: Select the gen-set by comparing the RkW, RkVA, SkW, and SkVA to the ratings on the manufacturer's specification sheets (after appropriate derating for ambient temperature and altitude).

Example One calculation. Determine gen-set size for three loads started across-the-line in a single step. Here's pertinent information:

• Two 200 hp motors, Code G, 92% running efficiency, 0.25 starting PF, 0.91 running PF.
• Total 100kVA of fluorescent lighting, starting PF of 0.95, and running PF of 0.95 (Note: We're using the terms starting and running PF for the lighting load here for clarification when adding the motor loads. Actually, the ballast for the lighting load has a constant PF of 0.95.)

Step 1: Information gathering and calculations. 200 HP motor:

RkW = (200 hp x 0.746 kW/hp) / 0.92 = 162.2kW

RkVA = 162.2kW / 0.91 PF = 178.2kVA

SkVA = 200 hp x 5.9 kVA/hp41180kVA

SkW = 1180kVA x 0.25 PF = 295kW

Florescent Lighting:

RkW = 100kVA x 0.95 PF = 95kW

RkVA = 100kVA

SkVA = 100kVA

SkW = 100kVA x 0.95 PF = 95kW

Step 2: Totals.

Load.......... | RkW | RkVA | SkW | SkVA

200hp Motor | 162.2 | 178.2 | 295 | 1180

200hp Motor | 162.2 | 178.2 | 295 | 1180

Lighting....... | 95.... | 100... | 95. | 100

Totals (kVA). | 420... | 457.. | 685 | 2460

Step 3: Selection. At a minimum, you'll have to size the gen-set to supply the maximum starting (surge) demands and the steady-state running loads of the connected load equipment. In this example (using one manufacturer's published data), you would select a 750kW generator set with 2944 SkVA available at 90% recovery voltage to supply the total load SkVA of 2460. The load totals for RkW, RkVA, and SkW are well within the rating of the 750kW (938kVA) gen-set you selected. The running kilowatt load of 420kW is 56% of the 750kW gen-set standby rating.

Example Two calculation. Assume you have the same three loads as in Example One, but now you're using an autotransformer type reduced voltage starter that is set at the 65% starting voltage to start the two motors. This tap setting will reduce the starting kVA by the square of the voltage (0.65)squared, or 0.42 times the starting kVA.

Step 1: Calculations

200 HP motor:

RkW = (200 hp x 0.746 kW/hp) / 0.92 = 162.2kW

RkVA = 162.2kW / 0.91 PF = 178.2kVA

SkVA = 200 hp x 5.9 kVA/hp = 1180 x (0.65)squared = 495kVA

SkW = 495kVA x 0.25 PF = 124kW

Florescent Lighting:

RkW = 100kVA x 0.95 PF = 95kW

RkVA = 100kVA

SkVA = 100kVA

SkW = 100kVA x 0.95 PF = 95kW

Step 2: Totals

Load.......... | RkW.. | RkVA | SkW | SkVA

200hp Motor | 162.2 | 178.2. | 124. | 495

200hp Motor | 162.2 | 178.2. | 124. | 495

Lighting...... | 95..... | 100... | 95... | 100

Totals (kVA) | 420... | 457... | 343. | 1090

Step 3: Selection. Using one manufacturer's published data, you would select a 450kW gen-set to supply the required starting kVA. The running kilowatt load of 420kW is 93% of the gen-set's standby rating. So, if you want a margin for future load additions, you would select a 500kW gen-set running at 84% of rated standby power.

Sidebar: Here's What Causes Dip in Starting Voltage

When you start a motor across-the-line with a gen-set, the motor represents a low impedance load while at locked rotor or stalled condition. This causes a high inrush current. The high motor inrush current (I ms) flows through the generator armature windings and is affected by the reactance. This causes a drop in generator voltage. Impedance controls the flow of current in AC circuits. But, the generator armature reactance is such a large part of its total impedance that resistance is ignored.

The generator terminal voltage drops instantaneously when the motor starter contacts close at time t40, as a function of the subtransient reactance (X"d). Generally, the larger the generator, the lower its reactance. So, one way to minimize the instantaneous voltage dip is to increase the generator size.

The generator terminal voltage may drop further, depending on response of the generator's automatic voltage regulator and the power capability of the excitation system. (Most gen-set automatic voltage regulators include underfrequency protection.)

During momentary overloads, the engine speed may also dip. If it does, the automatic voltage regulator reduces excitation power to the main field, which lowers the generator terminal voltage. This, in turn, reduces the load on the engine, allowing it to recover to rated speed. Typically, a maximum generator terminal voltage dip of 30% will not cause coils to drop out. (This allows for approximately 5% additional voltage drop in the conductors between the generator and the motor).

Although the voltage dip, due to under frequency protection, may extend the voltage recovery time, it also allows the engine to be sized closer to the steady-state running load rather than starting load. This is particularly important with diesel engines, which should not run for an extended duration at less than 30% of rated load. (Extended light-load operation of a diesel engine can result in the accumulation of unburned fuel in the exhaust system, due to incomplete combustion from low combustion temperatures, called wet stacking. Light load operation can also result in engine damage from fuel and water contaminating lubricating oil.)

After the initial voltage dip, it's important the generator restore voltage to a minimum of 90%-rated value while supplying the motor starting kVA. At least 90% recovery voltage is necessary for the motor to develop adequate torque to accelerate its load to rated speed.

A motor starting a high starting torque load, such as a loaded compressor, requires higher recovery voltage than one starting an unloaded compressor. As the motor comes up to speed, the voltage will rise, as the starting kVA input decreases. Once the motor is up to speed, the voltage should return to rated value, if the gen-set is sized properly.

Sidebar: How Inertia Affects Gen-Set Sizing

The moment of inertia of a rotating mass offers resistance to acceleration. The load connected to the motor shaft has its moment of inertia, and in practical situations for specific equipment, this may or may not be available information.

Fortunately, for the purpose of sizing a gen-set, or more specifically to determine the engine power needed to start and accelerate a rotating motor load, the motor load's moment of inertia need only be broadly categorized as low or high inertia.

High inertia loads are characterized by high breakaway torque requiring prolonged acceleration. Low inertia loads are characterized by low starting torque at standstill, with increasing torque as motor speed increases resulting in rapid acceleration to rated speed.

Starting low inertia loads will reduce the normal starting kW needed. Look for more information on this is in the sample calculations within this article.

Sidebar: Examples of High and Low Inertia

High inertia loads include:

• Single- and multi-cylinder pumps
• Single -and multi-cylinder compressors without unloading valves
• Crushers
• Hydraulic elevators without unloading valves

Low inertia loads include:

• Fans, centrifugal and blower
• Compressors starting unloaded
• Centrifugal pumps
• Motor-generator elevators

Note: Pumps starting into high head pressure and large diameter fans or fans starting into high restriction areas should be classified as high inertia loads.