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According to Pat Murphy, vice president, certification, at NATE, the 2.5-hour, 100-question exams test technicians' knowledge of the installation, service, maintenance, and repair of commercial refrigeration systems.

"The exams cover light commercial equipment with fractional horsepower to 7.5, which encompasses convenience stores, fast-food chains, and similar types of businesses," he said. "Larger equipment that is 7.5 to 80 horsepower is covered in the commercial refrigeration exams.

"The benefits of certification are well documented," Murphy continued. "Airside tests conducted over the last 10 years show a 25 percent reduction in warranty claims. Evidence also shows that a certified technician can add as much as $10,000 in value to a company."

Although anyone can take the exams, NATE recommends that technicians who take the installation exams possess at least one year of field experience working on refrigeration systems. Technicians planning to take the service exam should have two years of experience in the field.

"Each exam is difficult and provides a good measure of a technician's knowledge," Murphy explained. "We're not testing mediocrity. We're testing excellence."

For more information, visit www.natex.org.

Heat transfer is measured in Btuh. This loss or gain calculation is: Area multiplied by a U-value multiplied by a temperature difference (TD) or Btuh = Area x U-value x TD (Btuh = AUTD). The formula demonstrates that when the surface area of a component with a certain thermal conductivity is exposed to a certain temperature difference, then a defined amount of heat will flow through that component. In this case, 160 x 0.09 x 70 = 1,008, therefore 1,008 Btuh flow through this wall.

The first value is the area of the building component, in this case it is a wall. This part of the equation reflects how much surface area will be exposed to heat transfer. The area of the wall is 20 feet by 8 feet. Therefore, the first value in the equation is 160 square feet.

The next value in the formula is the U-value. This represents the heat transfer qualities, or thermal conductivity, of one square foot of a component. In this question the composite U-value is provided, 0.09.

There are many elements in this factor in the equation. First, the composite value means that this value represents one square foot of the whole wall. Second, the U-value, or thermal conductivity, represents how heat is conducted through the component. A sister value is the resistance to heat transfer of a component, or R-value. R-value is the inverse of the U-value, 1/U = R. R-values are used because they can be added together and U-values cannot. These values are compiled to account for their representative proportions (areas with framing members versus areas with insulation) then the inverse is taken to find the composite U-value (1 / R-11 = 0.09 U-value), in this example the composite U-value is 0.09.

The last value in the formula is the TD. This value was given - 70°F. If this value was not provided, then the designer would calculate the difference between the outdoor design temperature (ODT) and the indoor design temperature (IDT). The ODT depends on the weather for the location of the building (e.g., Phoenix is hot in the summer, Boston is cold in the winter). The IDT depends on the heating or cooling season. Manual J requires 70° in the heating mode and 75° for cooling. Other IDTs may be used, but they must be scientifically defensible. A designer that is given an ODT and an IDT can calculate the TD. In Boston the winter TD would be 58° (IDT 70 - ODT 12 = 58), and Phoenix in the summer would have a TD of 33° (ODT 108 - IDT 75 = 33). In this case the TD is given to us, 70°.

Mechanical efficiency in a building can encompass chillers, cooling towers, and overall system operations, said Kip Bagley, vice president of service for Mesa Energy Systems.

"Chiller efficiency involves cooler and condenser approach temperatures, refrigerant contamination losses, cooler and condenser water flow, and chiller control systems.

"Cooling tower efficiency involves fill media cleanliness, approach temperatures, water flow, mechanical condition, the cooling tower control system, and the water treatment system.

"System efficiency involves optimum start and stop times, the interface with air or waterside economizers, pump control, and cooling coil control."

Bagley noted that a typical energy calculation involves:

Run hours x Tonnage x Efficiency x Cost in kW = Cost per year

"Run hours are expressed as full-load equivalent hours. In our part of the world, this is usually between 2,200 and 2,500 hours per year."

Taking what he said as typical California rates, he gave an example:

2,000 hours x 500 tons x 0.7kW per ton x 13 cents/kW = $91,000 annual energy cost

"So a 500-ton chiller has an annual cost of operation of $91,000. If there is a 10 percent energy loss, that adds a $9,100 loss to the bottom line. Can your customer afford that?"

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Technicians
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