# XYLENE POWER LTD.- REVISE

## ELECTRICITY RATE DERIVATION

#### By C. Rhodes

INTRODUCTION:
The purpose of this web page is to derive electricity rates that encourage high generator capacity factor and high customer load factor. These rates are applicable to generators and load customers that are not directly controlled by the Independent Electricity System Operator(IESO). These rates are intended to financially enable behind the meter energy storage, load management and energy generation.

Allocation of the revenue from these rates to various parties entitled to receive this revenue is a separate issue. The problem with the present electricity rate structure is that it is a partial composite of certain existing costs, but the rate structure does not convey to either generators or load customers the appropriate signal as to the equipment and operational changes that these parties should adopt to reduce both their own costs and overall electricity system costs.

The message that must be clearly communicated via the rate structure is that in order to minimize transmission/distribution costs generators should operate at very high capacity factors and load customers should operate at very high load factors. As electricity displaces fossil fuels the average load on the electricity system will more than double. Unless the average generator capacity factor and the average load factor also both approximately double the ratepayers of Ontario will face an enormous increase in transmission/distribution costs.

The only practical tool for triggering wide spread increases in generator capacity factor and customer load factor is a suitable electricity rate structure.

ENERGY CONSERVATION:
Let:
T = time
P(T) = power as a function of time
E(T) = cumulative metered energy consumption
E(T) - Eo = Integral from T = To to T = T of P(T) dT
Then the energy consumption Et - Eo in a 24 hour period is given by:
Et - Eo = Integral from T = To to T = To + 24 h of P(T) dT

ENERGY COST TO LOAD CONSUMER:
Let dE be an element of energy
Let dC be an element of energy cost to the load consumer
Assume that the electricity cost increases continuously with increasing energy consumption. Then:
dC = C(E) dE
where:
C(E) = cost / unit energy to the load consumer at cumulative energy E
Let Ct = electricity cost for a 24 hour period
Then:
Ct = Integral from E = Eo to E = Et of C(E) dE

Choose C(E) such that at E = Eo, C(E) = Co.
In order to financially enable energy storage the function C(E) should keep getting smaller over the 24 hour period with increasing energy usage but with no increase in maximum net power drawn from the grid.

Try:
C(E) = Co exp[-K (E - Eo) / (Pm (24 hours))]
Where Pm is the daily maximum power drawn from the grid.
Note that C(E) starts at Co and gradually decreases but never falls below Co exp(-K).

Note that for a load customer the last element of energy used to achieve a 100% load factor is the least expensive.

Thus the daily cost of electricity is given by:
Ct = Integral from E = Eo to E = Et of Co dE exp[-K (E - Eo) / (Pm (24 h))]
= Co [Pm (24 h) /(-K)]exp[-K (Et - Eo) / (Pm (24 h))]
- Co [Pm (24 h) /(-K)]exp[-K (Eo - Eo) / (Pm (24 h))]
= Co [Pm (24 h) /(-K)]exp[-K (Et - Eo) / (Pm (24 h))] - Co [Pm (24 h) /(-K)]
= Co [Pm (24 h) / K][1 - exp[-K (Et - Eo) / (Pm (24 h))]]

In order to financially enable energy storage we need the marginal electricity cost to drop by about a factor of 3 over 12 hours. For a unity load factor customer, after 12 hours:
Et - Eo =Pm (12 h)
or
exp[-K / 2] = 1 / 3
or
exp[K / 2] = 3
or
K = 2 Ln(3) = 2.197 ~ 2.2

The daily Load Factor Lf is:
Lf = (Et - Eo) / (Pm (24 h))

Hence:
Ct = Co [Pm (24 h) / K][1 - exp[-K (Et - Eo) / Pm (24 h)]
= Co [Pm (24 hr) / K][1 - exp(-K Lf)]
= Co [(Et - Eo) / (K Lf)][1 - exp(-K Lf)]
or
Ct = Co (Et - Eo){[1 - exp(-K Lf)] / [K Lf]}

An advantage of this electricity rate is that via the Pm component of Lf it automatically aligns itself to the time of peak energy usage.

The average cost / kWh is given by:
Ct / (Et - Eo) = Co {[1 - exp(-K Lf)] / [K Lf]}

Now numerically investigate the effect of Lf on average cost / kwh:
For Lf = 1.0:
{[1 - exp(-K Lf)] / [K Lf]} = {[1 - (1 /8.997)] / 2.2} = .4039

For Lf = 0.5:
{[1 - exp(-K Lf)] /[K Lf]} = {[1 - (.33287)] / 1.1} = .6065

For Lf = 0.4:
{[1 - exp(-K Lf)] /[K Lf]} = {[1 - (1 / 2.4108997)] / .88} = .665
Thus there are significant electricity cost advantages in increasing the load factor. However, some of these cost advantages will be offset by storage losses.

GENERATOR COMPENSATION:
Let P(E) be the price per kWh paid to a generator at cumulative energy E. In a 24 hour period the generator earns:
Pt = Integral from Eo to Et of P(E) dE
where:
P(E) = Po [1 - {exp(-K(E - Eo) / Pm (24 hr))}]
where Pm = maximum net power supplied to the grid.
The initial condition is E = Eo at T = To. Note that P(E)= 0 at T = To and P(E) gradually increases but never exceeds:
Po [1 - exp(-K)].
Note that for a generator the last energy supplied to achieve a 100% capacity factor attracts the highest compensation.

The corresponding value of daily generator compensation Pt is:
Pt = Integral from Eo to Et of dP
= Integral from Eo to Et of Po (1 - exp(-K(E - Eo) / Pm (24 h))) dE
= Po (Et - Eo) - Po{[Pm (24 h) / (-K)]exp(-K(Et - Eo) / Pm (24 h))
- [Pm (24 h) / (-K)]exp(-K(Eo - Eo) / Pm (24 h))}
= Po (Et - Eo) - Po{Pm (24 h) / (-K)]exp(-K(Et - Eo) / Pm (24 h))
+ Po[Pm (24 h) / (-K)}
= Po (Et - Eo) - Po[Pm (24 h) / (-K)][exp(-K(Et - Eo) / Pm (24 h)) - 1]
= Po (Et - Eo) + Po[Pm (24 h) /K][exp(-K(Et - Eo) / Pm (24 h)) - 1]

The generator daily capacity factor Cf is given by:
Cf = (Et - Eo) / (Pm (24 h))

Hence Pt is given by:
Pt = Po (Et - Eo) + Po[Pm (24 h) /K][exp(-K(Et - Eo) / Pm (24 h)) - 1]
= Po (Et - Eo) + Po[(Et - Eo) /(K Cf)][exp(-K Cf) - 1]
or
Pt = Po (Et - Eo) {1 - ([1 - exp(-K Cf)] / [K Cf])}

An advantage of this rate is that via the Pm component of Cf it automatically aligns itself to the time of peak energy generation.

The maximum daily generator earnings occur when Cf = 1. Again using K = 2.2 to encourage energy storage, for Cf = 1:
{1 - (1 /(Cf K))[1 - exp(-K Cf)]}
= {1 - (1 /(2.2))[1 - exp(-2.2)]}
= {1 - .454545[1 - .1108]}
= {1 - .4041799}
= .59582

For Cf = 0.3, which is typical of wind generation:
{1 - (1 /(Cf K))[1 - exp(-K Cf)]}
={1 - (1 / .66)[1 - .51685]}
= {1 - .732}
= .2679
Thus increasing the capacity factor increases the value of the generated energy. Some of this increase in energy value is offset by storage losses.

SUMMARY:
The optimum generation compensation formula is:
Pt = Po (Et - Eo) {1 - ([1 - exp(-K Cf)] / [K Cf])}
where Cf is the daily capacity factor given by:
Cf = (Et - Eo) / (Pm (24 h)),
Pm = maximum net output power to the grid during that day
(Et - Eo) = energy generated that day
and K = 2.2
Po = scaled power rate

The optimum load rate formula is:
Ct = Co (Et - Eo){[1 - exp(-K Lf)] /[K Lf]}
where the load factor Lf is given by:
Lf = (Et - Eo) / (Pm (24 h))
where:
(Et - Eo) = daily energy consumption
Pm = maximum net power drawn from the grid that day
K = 2.2
Co = peak energy rate

Note that Po and Co may be subject to seasonal variation to compensate for sesonal changes in Pm values triggered by seasonal changes in river flow, wind and outside air temperature.

Note that Pt and Ct are integrals of time dependent components and hence are in effect averages of time-of-use rates.

This web page last updated February 23, 2009.