Products Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 12.1

Consider the cooling load profile for an office building shown in Figure 12.8. The cooling is provided by a chiller having a capacity of 1,000 kW and with an average seasonal COP of 3.5. The noncooling profile experienced by the same office building is illustrated in Figure 12.9. It is proposed to install an ice storage system. When making ice the chiller has an average COP of 3.0. Determine the simple payback period of installing an ice storage system if the cost of electricity is as follows:
Energy Charges: $0.07/kWh for on-peak hours (between 10:00 • and 15:00 during weekdays) and only$ 0.02/Kwh during other hours.
• Demand charges: $15/kW during on-peak hours and$ 0/kW during off-peak hours. The
demand charges are assessed on a monthly basis.

The installed cost of the TES system is $100/kW. Assume that the number of typical cooling days during the entire year both before and after the installation of the TES system is 250 days. The TES system is operated with demand-leveling control so that the power demand during onpeak hours never exceeds 500 kW (which is the maximum noncooling load). ## Step-by-Step ## Report Solution ## Verified Solution In this example, the entire cooling load during on-peak has to be shifted and thus in Eq. (12.2) the fraction X = 1. Therefore, the savings in the electric power demand can be calculated using $SEER_{CHW}$ = 3.5; and $Q_{C}$ = 1,000 kW: $ΔkW_{TES}=.(\frac {\dot{Q}_{c}}{SEER_{CHW}})_{e}-.(\frac {(1-X).\dot{Q}_{c}}{SEER_{CHW}})_{r}$ (12.2) $ΔkW_{TES}=1000kW*(\frac {1}{3.5})=286 kW$ Thus, the cost savings due to demand charges are: 286 kW*$15/kW*12 months/yr = \$51, 480/yr. The energy cost savings of using the TES can be calculated using Eq. (12.3) with the assumption that there are 250 typical cooling days per year:

$ΔEC_{TES}=\dot{Q}_{c}.N^{TES}_{h,c}.(\frac {C_{on,pk}}{SEER_{CHW}}-\frac {C_{off,pk}}{SEER_{ICE}})$                  (12.3)

$ΔEC_{TES}=1000kW*5 hrs/day * 250 days/yr(\frac {\0.07/k Wh}{3.5}-\frac {\0.02/k Wh}{3.0})=\16.667/yr$

The size of the ice storage tank is such that it can hold the cooling energy required during on-peak period; that is, 1,000 kW * 5 hr = 5,000 kWh. Therefore, the simple payback period for installing TES is estimated as follows:

$SPB=\frac {\100/k Wh* 5000k Wh}{(\51480-\16667)}=7.3 years$

A life-cycle cost analysis may be required to determine if the investment in a TES system is really warranted.
It should be noted that more savings can be obtained if the chiller has to be replaced. Indeed,when the TES is installed the chiller capacity can be reduced. For the case of this example, a chiller with a 500-kW cooling capacity is sufficient (instead of a 1,000-kW chiller) to charge the TES system and cool the office building during off-peak periods.