Industrial Utility Efficiency

# Pneumatic vs. Electric Tool Calculations and Considerations

Pneumatic tools are powered by compressed air, while electric tools are powered by electricity as the motive power. Pneumatic tools are commonly used throughout chemical process industries, construction, woodworking, metalworking and many other applications. Compressed air systems are a necessary part of most plant operations. However, according to the U.S. Department of Energy Compressed Air Challenge compressed air systems are(1):

• Often the least efficient source of energy in a plant
• Often the biggest end use of a plant’s electricity
• Frequently used inappropriately

To operate a 1-horsepower compressed air motor, approximately 7 horsepower of electrical power requirement is put into the air compressor. Even more power is needed when the pressure is higher than a typical pressure of 90 psi.

For this example, 0.5-horsepower pneumatic and electric tool energy uses are compared. A tool run time of 400 hours/year is used—and all the calculations are presented so the savings for other cases can be easily substituted when considering other variations.

Figuring out the energy savings for the switch from pneumatic to electric tools requires an estimate of energy use for each case. The effect of replacing a few tools in a large compressed air system may be too small to detect using power monitoring on the air compressors. However, it is still a good practice, and when part of a larger program to reduce air consumption, the combined efforts will amount to something measurable. Another positive aspect may be that reduced compressed air use frees up needed air compressor capacity.

### Calculating Compressed Air Tool Energy Use

Some assumptions are made in this process. One is that the tools run at capacity—full-rated flow in the pneumatic case, and full-power draw in the electric case. Obviously that is not true all the time, and if a closer load profile is known, then the energy use for each load at its annual hours of operation can be added together to obtain a better estimate.

In addition, whenever considering load reductions, there is the issue of part-load air compressor efficiency, and where the facility operates on the air compressor efficiency curve. While this example uses a value close to typical full-volume efficiency, the actual results could be much different. If the site merely moves down the efficiency curve of a modulated inlet vane air compressor, the savings would be modest. However, if enough compressed air-powered tools are taken out of the system to allow a lightly loaded air compressor to shut off completely, then the savings are much larger.

In a recent plant energy audit, the recommendation was made to replace pneumatic tools with electric. The site had two 125-hp, load/no-load air compressors, and the second air compressor ran unloaded most of the time. The site only had dryer capacity to serve one air compressor. In addition to the basic savings recommendation (assuming full-load efficiency), it was emphasized that changing out enough tools to allow the second air compressor to remain off would bring larger-than-calculated savings and bring the benefit of allowing the site to use dry air everywhere without an additional capital expenditure.  Savings after that point would be a little less than calculated.

Analysis

The following equation can be used to calculate the annual electrical demand of the existing pneumatic tool:

AEDPneumatic = EDPneumatic x t

Where:

AEDPneumatic = Annual electricity use of the pneumatic tool in kWh/year

EDPneumatic = Electrical demand of the pneumatic tool in kW

t = Annual hours of operation of the pneumatic tool (400 hours/year, in this particular plant)

*t can be considered 20 percent utilization for one shift (2,000 hours/year) operation, 10 percent utilization for a two-shift (4,000 hours/year) operation, 6.7 percent utilization for three shifts (6,000 hours/year), and so on.

The following equation can then be used to calculate the electrical demand of the existing pneumatic tool:

EDPneumatic = CFMPneumatichair comp

Where:

EDPneumatic = Electrical demand of the pneumatic tool in kW

hair comp= Compressed air generation efficiency (0.16 kW/cfm) (2)

CFMPneumatic = cfm usage of the pneumatic tool

The average cfm of a typical 0.5-hp pneumatic tool is calculated in the following table:

 Rated CFM of Typical Pneumatic Tools During Operation Pneumatic Tool Make/Model CFM Speedaire/2YPR1 (3) 18 Chicago Pneumatic/CP9288 (4) 24 Desoutter/DR350 P20000 (5) 20 Average 20.7

The following calculations reveal the electrical demand of a 0.5-horsepower pneumatic tool consuming an average of 20.7 cfm.

EDPneumatic = 20.7 cfm x (0.16 kW/cfm) = 3.31 kW

The following calculation reveals the annual electricity use to supply compressed air to this pneumatic tool:

AEDPneumatic = 3.31 kW x (400 hours/year) = 1320 kWh/year

### Corded Electric Tool Energy Use and Savings

Some applications are not suitable for conversion from pneumatic to electric tools. One reason is higher powered tools use larger electric motors that can get too heavy for frequent use without a tool balancer. Another reason is the durability between electric and pneumatic may also be an issue. Frequently used hand tools may get replaced every few months, so in those cases it may be wise to track the tool cost and replacement frequency to create a total operating cost comparison, including the compressed air and electricity use. While corded electric tools will always win on the energy front, the total cost of operation may present a different story—if the electric tools have a different lifespan and cost than the pneumatic tools.

Analysis

The following equation can be used to calculate the annual energy savings associated with replacing pneumatic tools with electric tools:

AESElectric = AEDPneumatic – AEDElectric

Where:

AESElectric = Annual electrical energy savings in kWh/year

AEDPneumatic = Annual electrical energy use of pneumatic tool in kWh/year

AEDElectric = Annual electrical energy use of the electric tool in kWh

The electricity required for a corded electric tool with the same 0.5-horsepower output as the pneumatic tool considers the output and the motor efficiency. The typical motor efficiency for a 0.5-horsepower electric motor is 65 percent (6). The full-load electric demand then becomes the following:

EDElectric= (0.5 hp)(0.746 kW/hp) / 65% = 0.574 kW

The following equation can be used to determine the annual electrical demand of the electric tool:

AEDElectric = EDElectric x tElectric

Where:

AEDElectric = Annual electrical energy use of the electric tool in kWh

EDElectric = Electrical demand of the electric tool (0.574 kW)

tElectric = Annual hours of operation of the electrical tool (400 hours/year)

The following calculation reveals the annual electricity use of an electric tool:

AEDElectric= 0.574 kW x (400 hours/year) = 230 kWh/year

The following calculation reveals the annual electrical energy savings associated with replacing a pneumatic tool with an electrical tool:

AESElectric= (1320 kWh/year) – (230 kWh/year) = 1090 kWh/year

For this example, if the average power cost was \$0.10/kWh, the annual savings would be \$109 per tool.

### Calculating Energy Consumption of Battery-Powered Tools

Battery-powered tools are as efficient as corded tools, but the battery charging system introduces some losses. Battery-powered tools can (but may not) offer greater efficiencies than compressed air and may be preferred in some cases. The Environmental Protection Agency’s Energy Star program publishes the amount of energy lost by the charging system, and this measure qualifies chargers using Energy Star’s published values (7).

Energy Star’s “average energy ratio” is the amount of energy lost by the charging system divided by the useful energy in the battery. The ratio varies from nearly zero to about 15. As a savings example, consider a tool with a charger that has an Average Energy Ratio (ER) of 2.5. The calculation for corded tools above shows the energy required by the tool when motor efficiency and run time are considered, and that is 230 kWh/year. An ER of 2.5 means the amount of non-active energy used in charging is the following:

Non-active Energy = (Energy Ratio)(Useful energy in the battery)

For this example, the non-active energy can be calculated using the following:

(2.5)(230 kWh/year) = 575 kWh

The total energy includes the useful energy in the battery, resulting in the following calculation for total energy use:

(230 kWh of useful energy) + (575 kWh of non-active energy) = 805 kWh

To obtain the annual energy savings as opposed to using pneumatic tools, the compressed air energy of 1320 kWh/year is plugged into the following equation:

(1320 kWh/year) – (805 kWh/year for battery tools) = 515 kWh/year in Energy Savings

If the average power cost is \$0.10/kWh, then the annual savings would be \$51.50 per tool, for this example. Be sure to check the charger efficiency before assuming a battery-powered tool is more efficient than a pneumatic tool.

### Making the Switch to Electrically Powered Tools

A first look at the facility’s opportunity can be evaluated using a spreadsheet. This survey and initial analysis can get the effort started by showing where the efforts of trying new tools will be most productive. An example of this survey is shown in the following chart:

##### While it might not be possible to replace all the pneumatic tools with electric tools, an evaluation as shown in this chart can get the discussion started on where to begin.

Using a corded electric tool in place of a pneumatic tool offers energy savings, and that should be pursued as a best practice wherever possible. A planned approach would identify where pneumatic tools are used, their run time, cost, and the load on the tool if possible. Then replacement electric tools can be selected and tested. The energy use of the replacement tool can be projected and the cost considered to calculate a payback for each tool. The tools that get replaced will bring savings and reduce the amount of compressed air required by the plant.

For more information visit https://new.consumersenergy.com or contact Jerry Zolkowski, tel: (517) 481-2972, email: Gerard.Zolkowski@dnvgl.com.