Addressing the Peak Power Problem Through Thermal Energy Storage

Abstract

In the United States, the electrical power grid is divided into three primary regions: the Western Interconnection, the Eastern Interconnection, and the Texas Interconnection. Each of these regions struggles with peak power issues, but this case study will focus on the Texas Interconnection, which is operated by the Electricity Reliability Council of Texas (ERCOT).This chapter discusses the opportunity to shift one of the largest electricity loads (air-conditioning) from the expensive aftrenoon peak to the cheaper nighttime hours using Thermal Energy Storage (TES), which is used for storing “cooling” in the form of chilled wate, and outlines a model for finding an optimal design for it.

Full text

Chapter 14
Addressing the Peak Power Problem through Thermal
Energy Storage
Wesley Cole, JongSuk Kim, Kriti Kapoor, Thomas Edgar
McKetta Department of Chemical Engineering
University of Texas at Austin
78712
[wcole or jongkim or kriti or edgar]@che.utexas.edu
Introduction
In the United States, the electrical power grid is divided into
three primary regions: the Western Interconnection, the Eastern In-
terconnection, and the Texas Interconnection (see Figure 1). Each of
these regions struggles with peak power issues, but this case study
will focus on the Texas Interconnection, which is operated by the
Electricity Reliability Council of Texas (ERCOT).

2
Figure 1: Map of the North American electricity interconnections.
Image is from
http://www.eia.gov/todayinenergy/detail.cfm?id=5070.
Electricity demand fluctuates throughout the day. During the
summer months in ERCOT, peak demand (sometimes just called
“the peak”) occurs during the late afternoon or early evening (see
Figure 2). One of the primary drivers of the peak is air conditioning,
both from residential and commercial buildings.

3
Figure 2: Electricity power demand in the ERCOT system on June
25, 2012. The peak demand occurs during late afternoon, and its
magnitude is nearly twice the demand at 5:00 am.
The electrical grid has to be built to meet the peak demand.
For ERCOT in 2012, that means that they needed more than 65 GW
(gigawatts) of capacity. However, for most of the year, much of this
capacity sits idle. For example, during a mild day in March, the peak
may only reach 35 GW, meaning that more than 30 GW of generat-
ing capacity is sitting unused that day. Even on peak days (like in
Figure 2) there are several hours with significant untapped capacity.
This leads to low capital utilization and increases the overall cost of
electricity. Also, because peak power plants are only used during
peak times, they tend to be single-cycle gas turbines which have
lower capital costs, but also lower efficiency and higher emissions
than their combined-cycle counterparts. Combined-cycle power
plants utilize the waste heat from the gas turbine to produce addi-
tional electricity.
Peak power creates disparity in the electricity prices
throughout the day. In ERCOT, electricity is bought and sold in the
wholesale market. For example, the utility that supplies electricity to
homes to Austin may either produce that power themselves or pur-
chase that power from another generator in the ERCOT market. At
the same time, the local utility has the option of producing extra

4
power and selling it in the ERCOT market. There are a variety of
methods for buying and selling power in ERCOT, including real-
time markets (electricity is purchased at the time of use), day-ahead
markets (electricity is purchased the day before it is used), and nego-
tiated contracts (the parties involved set the individual terms of the
contract). Prices are specified at five-minute intervals in the real-
time market and at one-hour intervals in the day-ahead market. In
both markets, however, the price of the electricity corresponds to the
scarcity of electricity during that time.
For example, during the morning when demand is lowest, the
most efficient, cheapest power plants are operating. As demand in-
creases throughout the day, less-efficient and more expensive plants
are brought online, increasing the market electricity prices. This is
shown in Figure 3, which corresponds to the same day as Figure 2.
Because of these fluctuations in prices during the day, the ability to
have flexibility in when electricity is consumed can be very valua-
ble. In this case study we consider adding that flexibility through the
use of thermal energy storage.
Figure 3: Day-ahead wholesale market prices in the Austin load
zone of ERCOT on June 25, 2012. Prices increase dramatically dur-
ing the peak period.

5
Thermal Energy Storage
Thermal energy storage (TES) is the storage of thermal ener-
gy in some medium. The energy can be stored as latent heat, sensi-
ble heat, or chemical heat, though latent and sensible heat are the
most common. Latent heat is the heat released or absorbed by a body
during a constant temperature process, such as a phase change (e.g.,
boiling water). Sensible heat refers to heat transfer that results in a
temperature change, but no phase change occurs (e.g., heating up a
piece of metal). Chemical heat is heat stored in chemical bonds, such
as H2, which can be reacted to release heat.
In warmer climates such as in the ERCOT region, TES is of-
ten used for storing “cooling” in the form of chilled water or ice. For
example, the University of Texas at Austin recently installed a four
million gallon (15,140 m3) chilled water TES tank. This insulated
tank is used to store cold water when the cooling and power de-
mands are low (e.g., at night time), and then the water can be dis-
charged in the afternoon when the cooling and power demands are
high. Without the TES, the campus buildings are cooled directly by
the electric chillers, resulting in chilled water being used at the same
time it is made. When using the water in the TES to provide cooling,
the electric chillers can be turned down or off, thus reducing the
electricity required by the campus while the TES is discharging.
Though not as versatile as battery storage (because TES can only
meet thermal loads), TES is much cheaper with costs ranging from
$6-43/kWh [1]–[4]. Battery costs range from $74-1484/kWh [5]. For
more information on TES see [6], [7].
Thermal storage provides an opportunity to shift one of the
largest electricity loads (air conditioning) from the expensive after-
noon peak to the cheaper nighttime hours. It also provides an oppor-
tunity for the air conditioning equipment to operate more efficiently.
The efficiency of large chillers, which make the chilled water for
cooling the air in buildings, is a function of the amount of cooling
energy (cooling load) provided by the chillers and the temperature of
the outdoor air. TES gives the chillers more flexibility to operate in

6
the regions where they are most efficient (cooler outdoor tempera-
tures with moderate to high cooling loads).
The thermal storage considered in this problem is chilled wa-
ter thermal storage using a single stratified water tank. This type of
TES tank takes advantage of the fact that colder water is more dense
that warmer water. When cold water is needed it is removed from
the bottom of the tank and sent to a heat source, such as a building,
to provide cooling (see Figure 4). The warmer water is returned to
the top of the tank using carefully designed diffusers that ensure the
flow is not turbulent. The difference in density keeps the warm wa-
ter on top from mixing with the cold water on the bottom. This also
causes the tank to only have two temperatures—the warm water
temperature and the cold water temperature. To recharge the tank,
warm water is taken from the top, cooled in the heat sink (i.e., the
chillers), and returned to the bottom of the tank. The advantage of
the stratified chilled water tank is that capital costs are reduced be-
cause only a single tank is required to store both the hot and the cold
fluids.
Figure 4: Schematic of a stratified chilled water TES tank. The dif-
ferences in density between the warmer and colder water in the tank
ensure that the two do not mix. The heat source on the right could
be, for example, a building that uses cold water to cool air for air
conditioning, while the heat sink could be an electric chiller, which
removes heat from the water. Image from [8].

7
Problem Description
Two variations of a single system are analyzed in this study.
The first is a traditional chiller/cooling tower configuration (see Fig-
ure 5a) that does not have TES. This provides a base case scenario
for comparing the benefits of TES. The second is identical to the
first except that it has a chilled-water TES unit running in parallel
with the chillers that supplies cold water to the buildings during peak
hours (see Figure 5b). The TES is recharged by the chillers during
off-peak hours. Although Figure 5 only shows one chiller, each
chiller system has two identical chillers connected in parallel.

8
Figure 5: Schematic of the system with [Figure (b)] and without
[Figure (a)] thermal energy storage [9].
A section of the University of Texas at Austin campus was
used to produce realistic cooling demands for the chiller-TES con-
figurations. This section consists of five buildings: two mixed-use
(classroom and office) buildings, an office administration building, a
performing arts building, and a lab building with the total floor area
of 65,500 m2. Dynamic building cooing load profiles were calculat-
(a)
(b)

9
ed using DOE 2.2 building modeling software [10] and are provided
with the attached data.
In both cases the chillers, cooling tower, and campus chilled
water loop are identical. The cooling system is sized to meet a max-
imum cooling demand of 6050 kW by using two identical 3025 kW
centrifugal chillers. Two chillers are selected because during times
of low or medium load, a single, large chiller would be operating far
below its design capacity. The minimum thermal operating capacity
of a chiller is 30 kW. The temperatures of the chilled water supply
(TCWS, see Figure 5) and chilled water return (TCWR, see Figure 5) are
fixed at 40°F and 53°F, respectively.
Models
The chillers’ power consumption (Pchiller) is a function of the
chilled water supply temperature (TCWS), the cooling tower supply
temperature (TCTS), and the ratio of the actual cooling load on the
chiller to the nominal cooling load. The chilled water supply tem-
perature is the temperature of the water leaving the chiller’s evapo-
rator and the cooling tower supply temperature is the temperature of
the water leaving the cooling tower and entering the condenser (see
Figure 5). The actual load on the chiller is the cooling load supplied
by the chiller at a given time and the nominal load is the rated capac-
ity of the chiller (in this case, 3025 kW). The following set of equa-
tions predict the chillers’ power consumption [11]:
2
1 1 1 1
2
11
CWS CWS CTS
CTS CWS CTS
CAPFT a b T c T d T
e T f T T
      
    
(1)
2
2 2 2 2
2
22
CWS CWS CTS
CTS CWS CTS
EIRFT a b T c T d T
e T f T T
      
    
(2)
/( )
chiller nominal
PLR L Q CAPFT
(3)
23
3 3 3
4 5 6
3 3 3
EIRFPLR a PLR b PLR c PLR
d PLR e PLR f PLR
  
  
(4)

10
chiller nominal
P P CAPFT EIRFT EIRFPLR   
(5)
where CAPFT is capacity as a function of temperature (defined by
(1)), EIRFT is energy input ratio as a function of temperature (de-
fined by (2)), PLR is part load ratio (defined by (3)), EIRFPLR is
energy input ratio as a function of part load ratio (defined by (4)),
Lchiller is the cooling load (in kW) being provided by the chiller,
Qnominal is the nominal rating of the chiller (in this case 3025 kW),
Pnominal is the nominal power usage of the chiller which in this case
is 540 kW, and ai through fi are constants for centrifugal chillers as
given in Table 1. Temperatures for all equations must be in degrees
Fahrenheit. The power consumption of the chillers, given by (5), is a
function of CAPFT, EIRFT, PLR, and EIRFPLR for the chillers.
CAPFT, EIRFT, PLR, and EIRFPLR are all unitless and are further
explained in Table 2.
Table 1: Constants used in (1), (2), and (4) for centrifugal chillers.
a
b
c
d
e
f
CAPFT
-0.299
0.0300
-8.00∙10-4
0.0174
-3.3∙10-4
6.31∙10-4
EIRFT
0.518
-4.00∙10-3
2.03∙10-5
6.99∙10-3
8.29∙10-5
-1.60∙10-4
EIRFPLR
2.54
-7.75
15.5
-15.5
7.69
-1.50
Table 2: Descriptions of CAPFT, EIRFT, PLR, and EIRFPLR.
Acronym
Description
CAPFT
Available chiller capacity as a function of TCWS and TCTS. For
example, a value of 0.9 for CAPFT indicates that 90% of
the chiller’s capacity is available at that TCWS and TCTS.
EIRFT
Full load efficiency of the chiller as a function of TCWS and
TCTS. Full loads indicates that the chiller is operating at
maximum load (i.e., Lchiller = Qnominal∙CAPFT).
PLR
The ratio of the chiller’s current operating load to the maximum
possible operating load.
EIRFPLR
Chiller efficiency as a function of the part load ratio (PLR).

11
The cooling tower supply temperature (TCTS) is a function of
the wet bulb temperature (Tw) and the temperature difference be-
tween the inlet and outlet of the chiller’s condenser (ΔTCOND):
 
 
 
2
52
2
5 6 2
9.14 0.771 0.00149
2.431 0.0196 3.51 10
0.0209 3.91 10 2.71 10
CTS w w
w w COND
w w COND
T T T
T T T
T T T


  
    
      
(6)
The value of ΔTCOND is controlled (i.e., held constant) at 10°F. A
variable speed pump is included on the cooling tower loop to keep
TCTS above 53°F, which means that if (6) predicts TCTS is lower than
53°F, then set TCTS to 53°F.
The cooling load required by the buildings is met by some
combination of the chiller and the TES:
,,building i chiller i i
L L TES
(7)
where Lbuilding,i is the cooling load required by the buildings at time i,
Lchiller,i is the cooling load provided by the chillers at time i, and TESi
is the amount of cooling provided by the TES (when it is negative it
means that the TES is recharging, i.e., requiring rather than provid-
ing cooling). The power consumed by each chiller is given by equa-
tion (5). The TES is limited both by how quickly it can discharge
and by how much energy it can hold:
, , 1TES i TES i i
C C TES


(8)
11i
ub TES ub  
(9)
,2
0TES i
C ub
(10)
where CTES,i is the charge (amount of cooling energy) in the TES at
time i, ub1 and ub2 are the upper bounds of the TES discharge rate
(TES) and the TES storage capacity (CTES), respectively.

12
Energy and Power
Power is the rate at which energy is used or consumed. The
SI unit of energy is the Joule (J), and for power is the Watt (W),
which is a Joule per second (J/s). A common unit of energy, espe-
cially when dealing with electricity, is the kilowatt-hour (kWh). One
kWh is the amount of energy used by a system at a rate of one kW
for one hour. Because the time increment in this problem is one
hour, if the chiller operates at 200 kW for one time step, then it con-
sumes 200 kWh.
Another important note to make is that a chiller essentially
converts electrical energy to thermal energy. These two forms of en-
ergy are differentiated because electricity is a more valuable form of
energy than thermal energy. So while the chillers operate at their
thermal capacity of 3025 kW for a one-hour time step, they may on-
ly consume 1000 kWh of electricity during that time step to provide
the 3025 kWh of cooling. The actual amount of electricity consumed
by the chillers is a function of several variables and can be calculat-
ed using (1)-(5).
The TES tank is sized according to volume (in m3), but be-
cause the chilled water supply (TCWS) and chilled water return (TCWR)
are fixed at 40°F and 53°F, respectively, the volume of the TES tank
has an energy equivalent (given in assumption 5 below). For this
problem, it may be easier to consider the TES as an energy tank,
having zero usable energy when the tank is full of 53°F water and
having a full charge of energy (given by ub2) when the tank is full of
40°F water. Thus the amount of energy in the TES tank (CTES) can
be measured in thermal kWh and range from 0 to ub2.
Data and assumptions
Use the following data and assumptions in this problem:
1. Hourly values for Lbuilding, Tw, and electricity prices (pelec) for
a one year period are provided in the accompanying spread-
sheet.

13
2. Use a one hour time step in solving the problem.
3. The building cooling load (Lbuilding) must be met exactly at
each time step.
4. The cost of installing the TES is given by
 
0.455
5265Cost V 

(11)
where Cost is the cost of the TES in $/m3 and V is the vol-
ume of the TES given in m3 (price adapted from [12]).
5. The thermal capacity of the water in the TES tank is 8.5
kWh/m3.
6. The maximum hourly discharge rate (in kWh/h) of the TES
(ub1) is 25% of the maximum storage capacity (in kWh) of
the TES (ub2), i.e., ub1 = 0.25ub2.
7. The annual interest rate (r) is 6%.
8. The TES loses 0.1% of its usable energy every hour.
9. The TES tank can be installed immediately and will last for
20 years.
10. There are no additional maintenance costs due to the TES.
11. The campus cooling load, weather, and electricity prices will
be identical to the 2011 values for the life of the TES. For
example, the cooling load, weather, and electricity prices for
2016 will be identical to those in 2011.
12. There is no cost for turning on or off a chiller.
13. The TES tank is initially full of water at 53°F. This means
that the tank starts with no usable energy (CTES = 0 kWh at
the initial time).
To reduce the complexity of the problem, you can include a
simple heuristic to divide the load between the two chillers. For ex-
ample, if the desired cooling load is less than the maximum cooling
load of one chiller, then only one chiller operates, but if the desired
load is greater than the maximum of the cooling load of one chiller,

14
then the chillers share the cooling load equally. This will eliminate
the need to include binary variables for the chillers’ operation.
Desired Outputs
Your task is to determine the optimal size of the TES tank
using the net present value (NPV) and the payback period (PBP)—a
description of each is given below. You will get a different optimal
size based on each metric. Compare the optimal size of the TES tank
in each case. Note that the optimal size of the TES is a function of
the TES operation, so the annual TES operation will also have to be
determined. The baseline for comparison is the system without the
TES (Figure 5a).
An optimal solution should include the tank volume in m3
(V), the cooling load supplied by the TES (TES) in kWh for every
hour of the year, and the value of the objective function (in dollars
or years, depending on which metric is used).
Economic Metrics
Net Present Value (NPV) – NPV is the present value (in dol-
lars) of a series of cash flows by a project. In this case the cash flows
are the investment cost (a negative cash flow) and the savings in op-
erating costs by using the TES system (positive cash flows). Because
the operating cost savings are accrued over the life of the system, the
cash flows must be discounted according to the interest rate. A pro-
ject is acceptable only if the NPV is at least zero, and projects with a
higher NPV are considered to be more profitable. In mathematical
terms NPV is given as
     
11 n
r
NPV E V V Cost V
r


  
(12)
where E(V) is the savings in operating costs (in dollars) from using
the TES over one year for a given volume (V), r is the annual inter-
est rate, n is the lifetime of the TES system (given in years), and

15
Cost(V) is the investment cost (in dollars) of the TES given by Equa-
tion (11).
Payback Period (PBP) – PBP is the time (in years) in which
the initial expenditure of an investment is expected to be recovered
from the revenues generated by the investment. For this problem, the
payback period is the time in which the initial costs of installing a
thermal storage tank are expected to be recovered by the energy sav-
ings from using the storage:
 
 
V Cost V
PBP EV


(13)
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