White Paper:

DC Drives


Get rid of your old motor-generator set



A solid state DC drive replacement for a Ward-Leonard set can pay for itself in a year or less, reduce downtime, and expand the scope of your system network.

For years the DC drive was the unchallenged leader in industrial applications. Even after AC drives were introduced in the 1970s, the DC drive's smooth operation and huge installed base enabled it to hang onto its dominance in the market.

Then along came vector control and other advancements in AC drive technology, and suddenly DC technology had some real competition. Of course, there are several reasons why a plant may need to stay with DC technology - not the least of which may be a big investment in DC motors that simply aren't ready for the scrap heap. However, if you're still using motor-generator (M-G) sets to power your DC motors (Fig. 1), give serious thought to replacing them with a solid state drive.

Chances are, your venerable M-G set (known as a "Ward-Leonard" from the method of control) has given you years of faithful service. So why change it now? Simply put, it will save you money. The M-G set's inefficiency is costing you a lot in power consumption, and maintaining three rotating machines (motor and generator combination, plus the DC motor) takes more than a few dollars. Each malfunction is likely to cause costly production losses, and reconditioning and/or repairing M-G sets can cost you from $1500 to $5000. Then too, your old-timer lacks precise speed regulation, and is a lonely, isolated device in a world filled with process control governed from a master location. The dollars lost from the inefficiencies the latter is likely to cause can be enormous.

You could get rid of all these drawbacks by replacing the M-G set with a solid state DC drive (Fig. 2). What's more, you probably can recover the cost for this relatively inexpensive retrofit in less than twelve months.



FIG. 1: Motor-generator sets suffer from low efficiency and high maintenance.


FIG. 2: A modern SCR drive can quickly pay for itself.

 

WHEN IS DC DRIVE TECHNOLOGY A MUST?

  1. When you are faced with high power requirements--more than 60 hp.
  2. When snubbing (that is, bleeding energy back across resistors) isn't an acceptable alternative to true line regeneration.
  3. When economics (high purchase price/installation costs) prevent you from replacing an existing DC motor.
  4. When DC drive "culture" is ingrained in your plant's infrastructure (that is, installed base, operators, software, service personnel) and is at odds with a movement to AC technology

To show you how to justify this capital expenditure, let's make a few reasonable assumptions and derive a few rules of thumb. You can then plug in the values for your plant and see how you would fare if you made the switch to solid state.

Let's say you operate eight hour shifts, five days a week, 50 weeks a year. Assume typical efficiencies for 100 hp are 88% for DC motors, 73% for M-G sets, and 96% for solid state DC controllers, and your plant's utility rate is 100/kWh. Remember:

  1. Motors are rated based on their output power.
  2. Efficiency typically is defined as output divided by input.
  3. Energy costs are computed based on the input power to the converter.
  4. Horsepower (hp)= kilowatts/0.746

To get a given horsepower out of a DC motor, the horsepower input to the motor is:

hpin = hpmtr/Effmtr

The horsepower into the DC motor is the same as the horsepower out of the converter. Thus, the input to the two different types of converters can be calculated as follows:
For M-G sets:
hpin-mg= (1/Effmg)(hpmtr/Effmtr)
In terms of kW,

kWin-mg = 0.746 (1/Effmg)(hpmtr/Effmtr)
= 0.746 (1/0.73) (hpmtr/0.88)
= 1.16(hpmtr) kWh/hp

For a solid state DC drive:
hp in-scr = (1/Effscr) (hpmtr/Effmtr)

kWin-scr = 0.746 (1 /Effscr) (hpmtr/ Effmtr)
= 0.746 (1/0.96) (hpmtr/0.88)
= 0.88 (hpmtr) kWh/hp

Subtracting the input power of the solid state drive from that of the M-G set gives the kWs saved by making the change:

kWsaved = kWs = kWmg - kWscr
= 1.16 (hpmtr) - 0.88 (hpscr)
= 0.28 (hpmtr) kWh/hp

We assumed 2000 shift-hr/year. At the utility rate of 10¢/kWh, the annual shift cost/kW is $200. If we multiply the per-shift operating costs by the kWs saved, we can find the annual savings/shift/horsepower:

Savings = S
= (kWs) (operating cost)
= 0.28 (hpmtr) ($200/kW/shift/yr)

Rule of thumb #1

S = $56/ (hpmtr)/shift/year/hp

Your utility might rebate 6¢ for each kWh saved in the first year of operation, or 50% of the installed cost, whichever is less. The rebate based on energy savings/horsepower/shift is:

kWhs = (kWs) (hrs)
= 0.28 (hpmtr) (2000 hrs/shift/year)
= 560 (hpmtr)/hp/shift/year

Rebate = $0.06 (560) (hpmtr)/shift/year
= $33.6 (hpmtr)/hp/shift/year

Rule of thumb #2

Utility rebate = $33.6 (hpmtr)/hp/shift/year

To see how these rules pan out, let's apply them to a 100 hp M-G set application. A plant is operating two shifts. The cost of a 100 hp 240 VDC drive is about $12,000.

Using rule of thumb #1:

Savings = [$56 (hpmtr)/hp/shift/yr]
x 2 shifts x 100 hp
= $11,200/year

Using rule of thumb #2:

Utility Rebate
= $33.6 (hpmtr)/hp/shift/yr
x 2 shifts x 100 hp
= $6,720

Since we assumed the utility will rebate 6¢/kWh or 50% of the installed cost, whichever is less, let's assume installation costs are 25% of equipment costs.

$6,720 > Rebate < 0.5 x $12,000 x 1.25

$6,720 > Rebate < $7,500.

During the first year the plant spent $15,000 for new equipment and installation. The company reduced its operating cost by $11,200 and the utility company sent it a check for $6,720. It's $2,920 better off than if it hadn't replaced its M-G set. Starting with the second year of operation, the company will be $11,200 better off each year.

Some caveats, of course, go with this example. If the SCR drive operates frequently at low speeds and power factor penalties are applied, actual dollar costs may be less than calculated. The table gives a breakdown of expected savings under various conditions. Note that it assumes a power cost of 5¢/kWh, rather than the 10¢/kWh used in the previous example.

In addition, the foregoing has assumed that you will use the original DC motor of the old Ward-Leonard system. The M-G set provided a smooth DC supply, unlike the output of the solid state controller, which has ripple.

 

A SAVINGS EXAMPLE

A DC motor is rated at 100 hp and is 88% efficient. The machine operates 10 hours per day, six days per week, 50 weeks per year. The cost of electricity is $0.05/kWh.
Operating cost =

$ / kWh x hrs x motor hp x 0.746 kW
Motor efficiency x converter efficiency
    = 0.05 x     3000  
x       (100/0.88) x  
0.746
x       1/converter efficiency  
    = $12,716 x   1/converter efficiency

The average efficiency of a motor-generator set is 73%, vs 93% for Reliance's solid-state V*S drive (based on run/stop and variable speed duty cycle). Therefore:

Operating cost of the motor-generator =
$12,716 x 1/0.73=$17,419

Operating cost of the V*S drive=
$12,716 x 1/0.96 = $13,246.

Operating cost savings for the retrofit = $4173
Approximate cost to recondition a 100 hp M-G set = $4000.

If the M-G set motor has a square laminated frame design, it was designed for rectified power. If the motor is a Super T with an "A" after the frame designation, it's most likely OK on rectified power. If it's neither of these, you may need a DC choke in the armature circuit (Fig. 3). To find out for sure, contact the motor manufacturer.

Specify the inductance of the choke at two times the motor inductance (mH), and the current rating (amps) at equal to the full load armature amps of the motor. While you're at it, you might also want to change from solid to split carbon brushes, to improve commutation.

The DC drive used in the retrofit should be easily interfaced with (and, thus, able to become an integral part of) a networked process control system. To do this, you fit the drive with a combination option board, which provides network communications. In some cases, the network offered will be proprietary; in others, it will be open.

Some networks allow you to upload and download enormous amounts of information. Each physical drop or network station can be monitored, and will report its status to the master network station. In addition, each physical drop can have its performance parameters (e.g., acceleration and deceleration rate) changed. All of this is a far cry from the old M-G set mumbling to itself in its remote comer of the plant.

FIG. 3: If the DC motor will not run well on the pulsating DC from an SCR drive, an armature choke may be needed.

 


Reprinted from INSTRUMENTATION & CONTROL SYSTEMS July, 1996




Document D-5858


Note:

This material is not intended to provide operational instructions. Appropriate Reliance Industrial Company instruction manuals and precautions should be studied prior to installation, operation, or maintenance of equipment.