A 40-unit motel (300 sq ft in each unit) has 3 kVA of air-conditioning and 4 kW of heat in each unit, and every two units share one 1.50 kW water heat

[Electricallearner]

A 40-unit motel (300 sq ft in each unit) has 3 kVA of air-conditioning and 4 kW of heat in each unit, and every two units share one 1.50 kW water heater. What is the calculated load for the motel?

My Calculation - Which maybe wrong..lol (I am wondering what is the use of the 300sq ft in each unit?)

- 4KW (Heater) x 40 units = 160KW
1.5kW/2 units water heater = 1.5KW (20) = 30KW

Water heater = fixed appliance .. so - 30KW times .75 demand factor = 22.5KW

Total = 182.5KW

But correct answer is 202KW

Any help? Ty!

 

[LarryFine]

No room lighting (this is why sq. ft. matters)? No site lighting? No laundry? No office/house loads?

 

[Paul_Briganti]

I don't have my 2020 this is based on the 2023 code. Check the VA / Sq. ft. to be sure it is 1.7Va for a motel in your code year.

MLL 300Sq. Ft * 1.7 * 40 = 20400Va
Heat 4000W * 40 = 160000Va
Water heater 1500W / 2 = 750W * .75 * 40 = 22500W

Grand total 202,900Va, if this was to be rounded in Kw it would be 203Kw

This is based on the Standard Method

 

[david luchini]

But correct answer is 202KW

Lighting/receptacle load 12,200VA
Heat load............................160,000VA
Water Heater load.............30,000VA
Total load............................202,200VA