What factors need to be considered in determining the size of the electric motor required to drive a fan with 40% static efficiency?
The size of the electric motor needed to drive a fan with 40% static efficiency depends on factors such as the fan's capacity, static pressure, and static efficiency. In this case, the fan has a capacity of 60,000 ft3/hr at 60°F and a barometer of 30 in Hg, with a static pressure of 2 in WG on full delivery. To calculate the power required for the electric motor, we need to consider these parameters and apply the relevant formula.
Calculating the Power Required
To determine the size of the electric motor required to drive the fan, we need to consider the fan's capacity, static efficiency, and static pressure. The correct answer is: O 0.8 HP.
The capacity of the fan is given as 60,000 ft3/hr, which indicates the volume of air the fan can deliver in one hour. The static pressure is stated as 2 in WG (inches of water gauge), which represents the pressure the fan can generate against resistance.
The static efficiency of the fan is mentioned as 40%, which indicates the ratio of the actual work performed by the fan to the input power.
To calculate the power required to drive the fan, we can use the following formula:
Power (HP) = (Flow Rate * Static Pressure) / (Fan Efficiency * 6356)
Converting the given capacity of 60,000 ft3/hr to cubic feet per minute (CFM), we get:
Flow Rate (CFM) = 60,000 ft3/hr / 60 min/hr = 1000 ft3/min
Substituting the given values into the formula, we have:
Power (HP) = (1000 * 2) / (0.4 * 6356) ≈ 0.793 HP
Rounding off to the nearest option, the size of the electric motor required to drive this fan is 0.8 HP.
Therefore, the correct answer is: O 0.8 HP.