As is probably intuitively obvious, things never work perfectly and without friction in the real world.

What this means is that we can calculate theoretically how much energy is required to, say, accelerate still air to flowing air with a fan (technically, we are adding kinetic energy to the air) but, in practice, you really need to use more energy because the fan and the motor that drives the fan are not 100% efficient and waste some energy.

In a earlier post, I explained that the power required by a fan can be calculated with the simple formula:

BHP = (CFM * Pressure) / 6356 * Eff_fan)

Where BHP = brake horsepower; CFM = cubic feet of air per minute at standard conditions; Pressure = pressure increase in inches water; 6356 is a complicated constant that converts terms to deliver power input as brake horsepower; and Eff_fan = fan efficiency.

So if you set Eff = 1, you can calculate the minimal amount of energy that would theoretically be needed to deliver the CFM and pressure specified.

I should add that the BHP can be directly converted to kW with the conversion factor of 0.746 kW/BHP. For a motor that is not 100% efficient, the calculation is simply:

kW = (0.746 kW/BHP * BHP) / Eff_motor

So what, right? But actually, you can figure out (very roughly) some interesting things with this. For instance.

My main campus is roughly 2,000,000 square feet.

Generally speaking, air handling systems deliver about 1.5 CFM of ventilation air per square foot of space.

Air handlers might discharge at a static pressure of 4″ water. So:

2,000,000 SF * 1.5 CFM/SF = 3,000,000 CFM (approximate ventilation flow for 2M square feet)

3,000,000 CFM * 4″ / (6356) = 1,888 BHP (power requirements in BHP, theoretical minimum)

3,000,000 CFM * 4″ / (6356 * 0.65) = 2905 BHP (approximate power requirements in BHP, real world)

We can now really easily come up with a rough idea how much power is required to ventilate a 2M square foot facility. Let’s assume the motor running the fan is 95% efficient:

( 2,905 BHO * 0.746 kW/BHP ) / 0.95 = 2,280 kW (roughly)

We can do a reality check by looking at an real building. I have a property at 250,000 square feet that has about 375 HP of supply fan motors:

(250,000 SF * 1.5 CFM/SF * 4″) / (6356 * 0.65) = 363 BHP

So yeah, we’re at least in the realm of reason in our calculation results.

We can also do an amazing high level, but reasonably accurate, estimate of how much energy would be saved by resetting the fan discharge pressure downward by 2/10ths of an inch across the campus:

3,000,000 CFM * 0.2″ / (6356 * 0.65) = 145 BHP (approximate power SAVED in BHP, real world)

Which isn’t chicken feed. If we assume our fan systems run 18 hours per day, we can even calculate the value of static discharge reset. Assume cost per kWh is $0.12.

(( 145 BHP * 0.746 kW/BHP ) / 0.95) * 18 Hr/Day * 365 days = 748,000 kWh

748,000 kWh * $0.12/kWh = $90,000 per year savings.

Not bad. Interesting what you can estimate with only a smidgin of information and a few simple rules of thumb…