Problem: A 20,000 square foot factory uses LED lighting with a power density of 0.7 watts per square foot for the LED strips plus 15% for drivers.
Given: 7.5 horsepower motors operating at an average of 70% of their rated capacity; 3.412 BTU per hour per watt; 3.4 BTU pe...
Approach: So we have quite a bit of information here, but it's all laid out pretty nicely for us to apply, so let's just break it down an...
Calc: There are 120 factory employees performing light machine work.
Calc: It says it's a 20,000 square foot factory and we've been given the power density in watts per square foot.
Result: We'd probably be looking at about a thousand times 7,7, and we might end up, call it 288, maybe 289, even if we were 290, we're much closer to answ...
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Student questions asked in live office hours about this problem
OH 21: HVAC 17
Q: For HVAC-17 (motor heat gain), why don't you divide horsepower by efficiency — section 9.1.4 shows that formula for motors in the conditioned space?
A: When the motor and all its driven machinery are both inside the conditioned space, every watt of electrical input eventually becomes heat in that room — so you don't divide by efficiency, because none of the energy escapes the space. The 9.1.4 formula with efficiency in the denominator is for the case where the motor is outside but the driven load is inside.
OH 77: HVAC: HVAC-17
Q: For HVAC-17, why don't we divide by motor efficiency — shouldn't we use the formula for motors operating in a conditioned space?
A: Great catch — this trips up almost everyone. When the motor and all its driven machinery are inside the conditioned space, 100% of the electrical input becomes heat in that space, so you do not divide by efficiency.
OH 86: HVAC: HVAC-17
Q: I understand that motors in the room contribute their full load — but if the motors were in a different location, why would we only count 7% (the inefficiency), rather than just ignoring the motor entirely?
A: If the motor is outside but the driven load (like a fan) is inside, only the heat losses from the motor — roughly 7% for a 93%-efficient motor — leak into the conditioned space. The shaft work going into the fan inside the room then also becomes heat in that space and must be accounted for separately.
OH 90: HVAC: HVAC-17
Q: How can we use total motor power as the heat load when only 7% is wasted as heat — what's the difference between a motor and an electric unit heater in this context?
A: When the driven equipment is also in the conditioned space, 100% of the motor's electrical input eventually becomes heat in that space — the mechanical work just converts to heat through friction, fan air resistance, and other losses inside the same room. An electric heater converts electricity directly to heat; a motor converts electricity to motion, but that motion ends up as heat in the same space.
OH 100: HVAC: HVAC-17
Q: For HVAC-17, shouldn't the motor heat gain include efficiency in the denominator as shown in the reference handbook formula?
A: The distinction is what you're starting with — if you're given brake horsepower (shaft output), and the fan it drives is in the conditioned space, that BHP all becomes heat in the room. The handbook formula with efficiency in the denominator converts from rated input power to heat gain; if you already have shaft power, don't divide again.
OH 101: HVAC: HVAC-17
Q: Why didn't we multiply by 0.93 motor efficiency in HVAC-17, but we did include the 0.7 operating capacity factor?
A: The motor and all the machinery it drives are inside the conditioned space, so all heat — including the 93% that becomes mechanical work — ends up in the room. The 0.7 operating capacity just means the motors are running at 70% of rated power, so you reduce the input accordingly — those are two completely separate adjustments.