Making plastic trash bags requires a fast, reliable drive to operate a sealer at the end of a blown-film line. Unfortunately, one bag manufacturer’s system lacked both speed and reliability. After a retrofit, the system now has both speed and reliability. The higher speed alone increased productivity by 33%.

Previous system

To seal and perforate plastic trash bags, a 3-hp dc drive, through a 15:1 gearbox and a clutch-brake unit, rotated cams that move a 600-lb platen up and down, Figure 1. The platen measures 3 ft × 1.5 ft × 4 in..

The dc motor ran continuously while the clutch-brake engaged and disengaged to start and stop the cams. When the clutch is engaged, the cams start rotating, driving the platen down to seal and perforate a plastic bag. The cams then allowed the platen, which is forced up by springs and air cylinders, to return to the start position. The clutch disengaged, and the brake engaged, stopping the cams.

A photo-sensor detected a point on the plastic material and triggered the clutch-brake to start the cams moving. The cams were stopped when a proximity sensor detected that a cam completed a full 360-deg rotation.

Different plastic materials require different line speeds, which were set by the operators. In turn, the line speed determined the dc motor speed.

The machine was designed to operate up to 60 cycles per minute. However, when operated at 54 cpm, the clutchbrake failed about every 30 days, incurring a $700 repair cost for each failure.

Replacement system

Criteria for the new system called for an adjustable-speed drive that can change speed as the line speed changes and can cycle on and off to eliminate the clutch-brake.

The dc drive, clutch-brake, and worm gearbox were replaced with an adjustable- frequency inverter equipped with dynamic and dc-injection braking, and an integral gear 5-hp ac brakemotor. The brake is used only for emergency stopping.

Because the high-cyclic duty produces motor heating, the brakemotor also contains a separately powered blower that continuously moves a constant velocity of air over the motor regardless of motor speed.

The dc injection braking applies dc to the ac motor during the last eighth of the deceleration ramp. (For more information on this braking technique, see the article, “Selecting nonfriction stopping methods,” in this issue.)

The new system uses the same sensors. The photo-sensor initiates the sealing cycle, and the proximity sensor determines cam position to initiate the stopping sequence.

System sizing

To size a drive package for this application, requires determining the required torque, which in this application is based on both load inertia and the torque needed to compress the springs and air cylinders. However, calculations involving cam systems are complex, because the load inertia and load torque change during the cam’s rotation. To conserve time for this retrofit application, it is often beneficial to approximate some of the values.

(For more information on establishing inertia, see the Power Transmission Design Handbook.)

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This application can be broken into six operating segments, Figure 2, with the typical times (in seconds) for each:
1. Starting (accelerating) — 0.1 sec.
2. Running the platen down — 0.15.
3. Running the platen up, aided by springs and air cylinders — 0.15.
4. Decelerating during the dynamicbraking stage — 0.1.
5. Stopping the cams at the top with dc-injection braking, which stays on for a fraction of a second after the cams stop — 0.1.
6. Dwell time — 0.4.

Segment 1

The first step is to estimate the worst case load inertia at the motor shaft. To approximate this value, the following equation uses the maximum linear velocity and the motor shaft speed to reflect the inertia to the motor shaft. Measurements of the platen during operation, shows that the platen moves at a maximum speed of 110 fpm. For a trial value, it was assumed that a 1750 rpm motor would be operated at 80 Hz, producing 2300 rpm.

If you want to determine the inertia at the output shaft of the speed reducer, use the speed at that point. In this case it is 2300 ÷ 16.54 = 139 rpm, because a new reducer with a different reduction ration replaced the 15:1 reducer originally installed.

To determine accelerating torque requires determining load and motor inertia and resistance (frictional) torque. The maximum load (platen) inertia was previously calculated. To obtain a value for motor inertia, we assumed a 5-hp motor, which has a rotor inertia of 0.139 lb-ft2.

To ensure that the accelerating torque value is a conservative estimate, the load inertia including the platen (at the motor shaft) is included during the motor accelerating process. This assumption inflates the calculated acceleration torque because the cams do not move the platen while the motor is accelerating. This conservative assumption should assure that the accelerating torque is not understated, which could result in an undersized drive package.

The system uses rolling element bearings, therefore the friction torque at start-up is relatively nonexistent. To calculate the accelerating torque, use equation 2:

Often overlooked, the larger the motor, the more the rotor inertia. Thus, some applications are better served by a smaller motor with less rotor inertia. Therefore, sizing becomes an iterative process. A rule of thumb — and a starting point — calls for matching the load inertia (at the motor shaft) to the motor inertia. As with any guideline, this should not be followed blindly.

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Other segments

To determine the torque requirements while the platen compressed the springs and air cylinders, Segment 2, we used the existing application. We slowed the dc motor and monitored its current during this phase, then used the motor’s torque per ampere to establish the required torque.

Again, because the friction torque is very low, the torque values in Segments 2 and 3 are the nearly the same, except Segment 3 torque is negative. Any differences will become apparent only during a detailed analysis.

The torque in Segment 5 — produced while dc braking — is controlled by inverter programming and is negative, because it stops and holds the motor.

In summary, Table 2 lists the estimated torque requirements and typical time for each application segment. All torque values are at the motor shaft.

With the exception of Segment 5, which is the dc-injection segment, the dominate torque requirements are in Segments 1 and 4 when the system is accelerating and decelerating. This value determines the motor size.

Motor size verification

Motor torque and thermal capacities determine if a motor is adequately sized. The torque capacity involves both continuous (steady state) and peak. In this application, the continuous torque, which defines the required motor heating capacity is the root-mean-squared (rms) value of the torque values listed in Table 1:
Trms = 106 lb-in.
Tpeak = 156 lb-in.

The 5-hp motor selected as the first unit it try in this iterative process has continuous torque capacity of 140 lb-in. operating at 80 Hz and 210 lb-in. peak. Thus, the 5-hp motor has sufficient capacity to power the plastic bag sealer. However, the peak torque requirement is close to the peak motor torque. To ensure extra controller torque capacity, a 7.5-hp inverter was selected to provide more available peak current and, hence, more available peak torque.

Because of the high duty cycle and relatively large rms torque requirement, the motor is equipped with a continuous speed external cooling fan. This fan forces air over the motor even while the motor is operating at low speeds or at stall.

Operating results

The new drive package increased the production rate to 72 cycles per minute (a 33% increase) and has operated for several months without any downtime. Also other blown-film lines in the plant have been similarly retrofitted.

Mark A. Jones, is product training manager for SEW-Eurodrive, Inc., Lyman, S. C.