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.
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)
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)
- 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:
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 Powermoved/Powerinput. With a 250 ton chiller, this machine is capable of moving 250 tons of cooling, therefore the Powerinput would be:
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)
- 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 /ft3*hour* °F)(6,750 ft3/min)(21 °F)
q= 153,090 BTU/hour
Eric A. Woodroof, Ph.D., is the Chairman of the Board for the Certified Carbon Reduction Manager (CRM) program and he has been a board member of the Certified Energy Manager (CEM) Program since 1999. His clients include government agencies, airports, utilities, cities, universities and foreign governments. Private clients include IBM, Pepsi, GM, Verizon, Hertz, Visteon, JP Morgan-Chase, and Lockheed Martin.