Questions: A multifamily dwelling has eight 1,350-square-foot units, each served with a 120/240-volt THWN copper conductor feeder, with the following in each unit: - 3.5 kW, 240 V water heater - 12 kW, 120/240 V range - 1,750 W, 120 V dishwasher - 4 kW, 120/240 V dryer - 120 V, 1/3 hp garbage disposal - 7 1/2 hp, 240 V air-conditioning unit Calculate the feeder demand load for each unit. Round the answer to the nearest whole number.

A multifamily dwelling has eight 1,350-square-foot units, each served with a 120/240-volt THWN copper conductor feeder, with the following in each unit: - 3.5 kW, 240 V water heater - 12 kW, 120/240 V range - 1,750 W, 120 V dishwasher - 4 kW, 120/240 V dryer - 120 V, 1/3 hp garbage disposal - 7 1/2 hp, 240 V air-conditioning unit Calculate the feeder demand load for each unit. Round the answer to the nearest whole number.
Transcript text: (id, 1089 A multifamily dwelling has eight 1,350 -square-foot units, each served with a $120 / 240$-volt THWN copper conductor feeder, with the following in each unit: $-3.5 \mathrm{~kW}, 240 \mathrm{~V}$ water heater - $12 \mathrm{~kW}, 120 / 240 \mathrm{~V}$ range $-1,750 \mathrm{~W}, 120 \mathrm{~V}$ dishwasher - $4 \mathrm{~kW}, 120 / 240 \mathrm{~V}$ dryer - $120 \mathrm{~V},{ }^{1} /{ }_{3}$ hp garbage disposal $\cdot 7^{1} / 2^{\mathrm{hp}}, 240 \mathrm{~V}$ air-conditioning unit Calculate the feeder demand load for each unit. Round the answer to the nearest whole number.
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Solution

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Solution Steps

Step 1: Calculate the Demand Load for the Water Heater

The water heater has a power rating of 3.5 kW. Since it is a fixed appliance, we use its full rating for the demand load.

\[ \text{Water Heater Demand Load} = 3.5 \, \text{kW} \]

Step 2: Calculate the Demand Load for the Range

The range has a power rating of 12 kW. According to the National Electrical Code (NEC) demand factors for household cooking appliances, for one range, the demand load is 8 kW.

\[ \text{Range Demand Load} = 8 \, \text{kW} \]

Step 3: Calculate the Demand Load for the Dishwasher

The dishwasher has a power rating of 1,750 W, which is equivalent to 1.75 kW. Since it is a fixed appliance, we use its full rating for the demand load.

\[ \text{Dishwasher Demand Load} = 1.75 \, \text{kW} \]

Step 4: Calculate the Demand Load for the Dryer

The dryer has a power rating of 4 kW. According to the NEC, the demand load for one dryer is 5 kW.

\[ \text{Dryer Demand Load} = 5 \, \text{kW} \]

Step 5: Calculate the Demand Load for the Garbage Disposal

The garbage disposal is rated at \( \frac{1}{3} \) hp. Converting horsepower to kilowatts:

\[ \frac{1}{3} \, \text{hp} = \frac{1}{3} \times 0.746 \, \text{kW/hp} = 0.2487 \, \text{kW} \]

Since it is a motor load, we use its full rating for the demand load.

\[ \text{Garbage Disposal Demand Load} = 0.2487 \, \text{kW} \]

Step 6: Calculate the Demand Load for the Air-Conditioning Unit

The air-conditioning unit is rated at \( 7 \frac{1}{2} \) hp. Converting horsepower to kilowatts:

\[ 7.5 \, \text{hp} = 7.5 \times 0.746 \, \text{kW/hp} = 5.595 \, \text{kW} \]

Since it is a motor load, we use its full rating for the demand load.

\[ \text{Air-Conditioning Unit Demand Load} = 5.595 \, \text{kW} \]

Step 7: Calculate the Total Demand Load for Each Unit

Add up all the individual demand loads to find the total demand load for each unit:

\[ \begin{align_} \text{Total Demand Load} &= \text{Water Heater} + \text{Range} + \text{Dishwasher} + \text{Dryer} + \text{Garbage Disposal} + \text{Air-Conditioning Unit} \\ &= 3.5 + 8 + 1.75 + 5 + 0.2487 + 5.595 \\ &= 24.0937 \, \text{kW} \end{align_} \]

Rounding to the nearest whole number:

\[ \text{Total Demand Load} = 24 \, \text{kW} \]

Final Answer

\[ \boxed{24 \, \text{kW}} \]

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