Problem: An elevator control room houses 4 10-horsepower motors with 92% efficiency, 30 feet of linear LED lighting, drawing 5 watts per foot, and 1 compute...
Given: 30 feet of linear LED lighting, drawing 5 watts per foot, and 1 computer workstation, drawing 300 watts; 1.08 CFM Del...
Approach: The room is supplied with 74 degree air using a transfer fan.
Calc: They're just supplying return air using a transfer fan, so it's 74 degrees.
Calc: And since this room would otherwise get hotter than 80 degrees, it's cold enough to do the job.
Result: And that works out to 1,494 CFM, which is closest to answer choice B.
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Student questions asked in live office hours about this problem
OH 73: HVAC: HVAC-7
Q: For HVAC-7, I accidentally used the section 9.1.4 electric motor formula (for motors outside the conditioned space) and got a very high heat gain — when should that equation actually be used?
A: The 9.1.4 formula applies when the motor is outside the conditioned space — in that case, only the heat losses (the inefficiency portion) enter the space, so efficiency goes in the denominator. When the motor and all driven equipment are inside the conditioned space, 100% of the electrical input eventually becomes heat in that space, so you don't divide by efficiency.
OH 113: HVAC: HVAC-7
Q: For HVAC-7, why can't I use equation 9.1.4 with efficiency in the denominator — and since the motor and machine are in the conditioned space similar to problem 17, why isn't 100% of the load used?
A: This is a concept that trips up almost everyone — the key is where both the motor and its driven load are located relative to the conditioned space. If both are inside, all electrical input becomes heat in the room; if only the driven equipment is inside, you account for that shaft work plus any loss from the motor separately.