Question-and-Answer Resource for the Building Energy Modeling Community
 | 1 | initial version |
If you want to maintain a 1:1 ratio of occupants to computers, then the occupancy schedule and computer room schedule should be the same. When you define the occupant and equipment objects in the model, you set a peak usage value and reference a schedule that is just multiplied with that peak usage value. This can be defined as either a total value (occupants, W, etc.) or as a per-floor-area value (occupants/ft^2, W/ft^2, etc.).
Let's say that the computer room has 20 seats, and each computer uses 100W. When you define the occupant object, the peak total occupant is 20. When you define the equipment object for the computers, the peak electricity use is 2,000 W (20 people * 100 W/person). Both objects should then reference the same occupancy schedule.
If you are defining the peak occupancy with a per-floor-area value, which is most common, then set the computer electricity use per floor area equal to the power draw of one computer [W / person] multiplied by the occupant density [person / ft^2]. Using my example above, let's say the computer room was 1,000 ft^2. This means that the occupant density is 20 / 1,000 = 0.02 people / ft^2, and that the equipment power density is 2,000 / 1,000 = 2 W / ft^2.
| | 2 | No.2 Revision |
If you want to maintain a 1:1 ratio of occupants to computers, then the occupancy schedule and computer room schedule should be the same. When you define the occupant and equipment objects in the model, you set a peak usage value and reference a schedule that is just multiplied with that peak usage value. This can be defined as either a total value (occupants, ($ occupants, W, etc.) etc. $) or as a per-floor-area value (occupants/ft^2, ($occupants/ft^2$, W/ft^2, etc.).
Let's say that the computer room has 20 seats, and each computer uses 100W. When you define the occupant object, the peak total occupant is 20. When you define the equipment object for the computers, the peak electricity use is 2,000 W (20 people * 100 W/person). Both objects should then reference the same occupancy schedule.
If you are defining the peak occupancy with a per-floor-area value, which is most common, then set the computer electricity use per floor area equal to the power draw of one computer [W / person] multiplied by the occupant density [person / ft^2]. Using my example above, let's say the computer room was 1,000 ft^2. This means that the occupant density is 20 / 1,000 = 0.02 people / ft^2, and that the equipment power density is 2,000 / 1,000 = 2 W / ft^2. ft^2.
| | 3 | No.3 Revision |
If you want to maintain a 1:1 ratio of occupants to computers, then the occupancy schedule and computer room schedule should be the same. When you define the occupant and equipment objects in the model, you set a peak usage value and reference a schedule that is just multiplied with that peak usage value. This can be defined as either a total value ($ occupants, ($occupants, W, etc. $) etc.$) or as a per-floor-area value ($occupants/ft^2$, ($occupants/ft^2, W/ft^2, etc.).etc.$).
Let's say that the computer room has 20 seats, and each computer uses 100W. When you define the occupant object, the peak total occupant is 20. When you define the equipment object for the computers, the peak electricity use is 2,000 W (20 ($20 people * 100 W/person). W/person$). Both objects should then reference the same occupancy schedule.
If you are defining the peak occupancy with a per-floor-area value, which is most common, then set the computer electricity use per floor area equal to the power draw of one computer [W / person] [$W / person$] multiplied by the occupant density [person / ft^2]. [$person / ft^2$]. Using my example above, let's say the computer room was 1,000 ft^2. $1,000 ft^2$. This means that the occupant density is 20 $20 / 1,000 = 0.02 people / ft^2, ft^2$, and that the equipment power density is 2,000 $2,000 / 1,000 = 2 W / ft^2.ft^2$. This would be the same as taking $0.02 people / ft^2 * 100 W / person = 2 W / ft^2$.
| | 4 | No.4 Revision |
If you want to maintain a 1:1 ratio of occupants to computers, then the occupancy schedule and computer room schedule should be the same. When you define the occupant and equipment objects in the model, you set a peak usage value and reference a schedule that is just multiplied with that peak usage value. This can be defined as either a total value ($occupants, W, etc.$) or as a per-floor-area value ($occupants/ft^2, W/ft^2, etc.$).etc.$). Using either method, you should set the peak computer electricity use equal to the power draw of one computer [$W / person$] multiplied by the peak occupant use.
Let's say that the computer room is 1,000 $ft^2$, has 20 seats, and each computer uses 100W. When you define the occupant object, the peak total occupant is 20. When you define the equipment object for the computers, the peak electricity use is 2,000 W ($20 people * 100 W/person$). Both objects should then reference the same occupancy schedule.
If you are defining the peak occupancy with a per-floor-area value, which is most common, then set the computer electricity use per floor area equal to the power draw of one computer [$W / person$] multiplied by the occupant density [$person / ft^2$]. Using my example above, let's say the computer room was $1,000 ft^2$. This means that the occupant density is $20 / 1,000 = 0.02 people / ft^2$, and that the equipment power density is $2,000 / 1,000 = 2 W / ft^2$. This would be the same as taking $0.02 people / ft^2 * 100 W / person = 2 W / ft^2$.
| | 5 | No.5 Revision |
If you want to maintain a 1:1 ratio of occupants to computers, then the occupancy schedule and computer room schedule should be the same. When you define the occupant and equipment objects in the model, you set a peak usage value and reference a schedule that is just multiplied with that peak usage value. This can be defined as either a total value ($occupants, W, etc.$) or as a per-floor-area value ($occupants/ft^2, W/ft^2, etc.$). Using either method, you should set the peak computer electricity use equal to the power draw of one computer [$W / person$] multiplied by the peak occupant use.
Let's say that the computer room is 1,000 $ft^2$, has 20 seats, and each computer uses 100W. 100 $W$. When you define the occupant object, the peak total occupant is 20. count is 20 $people$. When you define the equipment object for the computers, the peak electricity use is 2,000 W ($20 people $W$ (20 $people$ * 100 W/person$). $W/person$).
If you are defining the peak occupancy with a per-floor-area value, which is most common, then the occupant density is $20 20 $people$ / 1,000 $ft^2$ = 0.02 people $people / ft^2$, and the equipment power density is $2,000 2,000 $W$ / 1,000 $ft^2$ = 2 W $W / ft^2$. This would be the same as taking $0.02 people / ft^2 0.02 $people / ft^2$ * 100 W / person $W / person$ = 2 W $W / ft^2$.
