Last week I was in Paris at an International Energy Agency meeting, where several experts shared their thoughts on what is needed to advance energy management. Many ideas were discussed, but the following are deemed most important:

- The need for effective communication skills (to get approval for energy management projects)
- The inclusion of ancillary benefits when evaluating energy programs
- The need for more training.

Items 1 & 2 are related because technical people need marketing skills to craft their message and sell the benefits of an energy project to multiple levels in an organization. In my experience, a project must yield wins for the people in both the boiler and board rooms.

Item 3 surprised me because I know there are many great learning opportunities already available today. In addition, with the amount of information on the web, an individual can learn a lot on their own. However, perhaps the need for training is related to an influx of new people into the career of energy management, which for decades has not been the most popular career field. In fact, a labor shortage of qualified energy managers has been predicted for some time. This makes sense because many of my colleagues are thinking about retiring. If new people are entering the field, how do we ensure they have the necessary training to become proficient or qualified?

**What is a Qualified Energy Manager?**

The definition depends on the geographic location and specific field. For example, data center managers are usually not experts at managing energy in a food processing facility. Energy managers in Hawaii (>28 cents per kWh) may take a totally different approach than those in Oklahoma (<10 cents/kWh).

However, there is one basic certification for energy managers that is acknowledged worldwide. The four-hour Certified Energy Manager (CEM) exam is known to be difficult, but it is a mark of distinction for technical energy managers. Once an individual has attained the CEM, they meet the prerequisites for many jobs, and can pursue additional training to become a specialist in lighting, motors, boilers, etc.

**Sample CEM Test Questions**

Below are some sample CEM test questions to give you a sense of what is involved. Note that there are actually 140 multiple choice questions on the real CEM exam. If you are already a CEM, you may review these questions to quiz your competency level. Either way, I hope you find these questions thought-provoking and useful. Good luck. Answers appear on the following page.

- First, let’s try a question about energy purchasing decisions, as the correct answer can save your company a lot of money.

An energy manager is evaluating the operating costs of two water heaters. One is natural gas fired with an efficiency of 68%, another is electric with an efficiency of 97%. Natural gas costs $0.92/therm, electricity costs $0.095/kWh. Determine the cost per MMBTU for heating water for each system.

- Now, consider power factor penalties, which can be a line item on your electric bill. If it is there, you will want to eliminate that penalty using the knowledge below.

A facility with a demand of 700 kW has a power factor of 0.72 and a facility load factor of 80%. The facility is charged a penalty due to the power factor being less than 0.82. To ensure the facility will never again pay a power factor penalty, what size of capacitor is needed to correct the power factor to 85%?

- Do you have a demand ratchet on your rate structure? If you do, the problem below shows how much it could be costing you.

If you pay $10 per kW per month and have a 80% demand ratchet for 11 months, how much extra will you pay (for next 11 months) if you had a 400kW spike (above normal demand) during last month?

- Savings can be significant from variable speed drives, as the question below demonstrates.

If a motor has a VSD installed that slows down a 100 HP motor by 20%, what are HP savings?

- The coefficient of performance or energy efficiency ratio can be important to estimating building loads.

A 250-ton chiller has COP of 4.2. If the chiller is running at full capacity, what is the load in kW of the chiller?

- The qualitative problem below is about how lighting retrofits can influence energy cost savings.

Use of occupancy sensors in a typical conference room application will typically:

- Reduce monthly lighting operating hours
- Increase lamp calendar life in hours (not operating hours)
- Require group relamping of fixtures
- 1 and 2
- 1, 2, and 3

- HVAC is a common area for energy savings. The problem below is about basic HVAC design.

Air at 69 degrees F. dry bulb and 50% relative humidity flows at 6,750 cubic feet per minute and is heated to 90 degrees F. dry bulb. How many BTU/hr is required in this process? Assume no duct losses. PageBreak

** Answers **

- To compare fairly, we must get the fuel prices in the same $/unit. We choose $/MMBTU, and we know that 1kWh=3412BTU and then incorporate the efficiency information.

$/MMBTU_{electric}.

If we use 1 kWh, that would be 3412 BTUs. Then the $/MMBTU in fuel cost would be:

= ($0.095/3,412 BTU)*(1,000,000BTU/1 MMBTU)*(1/.97)

= $28.7/MMBTU

$/MMBTU_{gas}:

We know that 1 Therm equals 100,000 BTUs. Then the $/MMBTU in fuel cost would be:

= ($0.92/Therm)*(1 Therm/100,000BTU)*(1,000,000BTU/1 MMBTU)(1/.68)

= $13.53/MMBTU

- There are a few ways to solve power factor correction problems. One way is to look up the PF correction tables (internet) and then multiply the appropriate table factor by the Real Power (kW). For this example, to go from 72% to 85% Power Factor, then the table factor is 0.343.

If the load is 700kW, then the capacitor size would be:

= (700 kW)(0.343 kVAR/kW)

= 240 kVAR

*Another way to solve is to use the relationships that PF=kW/kVA and the Pythagorean Theorem to find the initial and final kVAR… The difference in kVAR would be the size of the capacitor to correct the PF to the new level.*

- Ratchet penalties are effective measures to encourage energy consumers to reduce kW spikes, because if you have a ratchet penalty, you will pay the ratchet percent for the next 11 months (even if you don’t use any demand during that period). In this case, if you set a new peak 400kW above your normal operating load (old peak), you will pay:

= (400 kW)(0.8)(11 months)($10/month*kW)

= $35,200 in extra charges (over the next 11 months)

- Use the “Fan Laws” HP2=HP1 (RPM2/RPM1)3. Therefore, if the new speed (RPM2) is 80% of the old speed (RPM1), the new power load (HP2) will be:

HP_{2}=HP_{1} (.8)^{3}

HP_{2}=100HP (0.512)

HP_{2}=51.2 HP

*This may seem “too good to be true”, but the fan laws are real. As the Fan Laws are similar to a “theoretical maximum”, in practice you may not completely achieve this level of savings, but you can come very close, which is still amazing. *

- The Coefficient of Performance is the Power
_{moved}/Power_{input}. With a 250 ton chiller, this machine is capable of moving 250 tons of cooling, therefore the Power_{input}would be:

4.2=Pm/P_{i}

4.2=250 tons/P_{i}

Pi = 59.52 tons

*But “tons” must be converted into kW to answer the question. A “ton” is a rate of energy flow equal to 12,000 BTUs per hour and we also know that 1 kW = 3412 BTUs/hour. Thus the amount of kW can be found by: *

*=(59.52 tons)(12,000 BTU/hr*ton)(kW/3412BTU/hr)*

*=209 kW*

- Answer = 4. Because occupancy sensors will reduce kWh, but cannot be guaranteed to reduce demand (the lights may come on during a peak period). If the lights are “on” less hours per day, they should last longer (assuming the lights are not turned “off” and “on” rapidly). Implementing occupancy sensors does not mandate a group relamping strategy.

- Use the equation for “sensible heat transfer” (actually a “short cut” equation endorsed by ASHRAE):

q = 1.08(CFM)∆T [q units are in BTU/hour]

q= 1.08 (BTU*min /ft^{3}*hour* °F)(6,750 ft^{3}/min)(21 °F)

q= 153,090 BTU/hour

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