Thermodynamic Analysis | Table of Contents | Micro-Cogen Components

Economic, Probability and Environmental Analyses

Economic Analysis

The previous analysis shows that a cogeneration system requires more fuel to produce the same quantity of hot water as a gas-fired water heater. This was shown in equation (2.17). The benefit in using a cogeneration system is the electricity it simultaneously produces, which a conventional water heater does not. It is logical at this point to ask does the value of the electricity produced by the cogenerator exceed the added value of the gas necessary to run it? Whether or not it is worth while to implement the cogeneration system also depends on how much capital has to be invested, and therefore, the length of the simple pay back period15. Defining $SAV as the net periodic savings from operating the cogeneration system, this term is comprised of three components:

(3.1)

where $E,COG represents the value of the electricity produced by the cogenerator, $GAS,COG represents the cost of gas spent running the cogenerator and $GAS,WH represents the cost of gas saved by not using the water heater. These terms are each individually calculated by using the equations derived in the thermodynamic analysis and simply multiplying them by the cost of electricity, CE, and cost of gas, CG:

(3.2a)

(3.2b)

(3.2c)

Substituting equations (2.12), (2.11), and (2.1) respectively into equations (3.2), and assuming hWH = hR, yields the following set of equations:

(3.3a)

(3.3b)

(3.3c)

Substituting equations (3.3) into equation (3.1) yields:

(3.4a)

Equation (3.4a) can be simplified to obtain the following result:

(3.4b)

Note that if VW is water consumed on a daily basis, $SAV is dollars saved on a daily basis, or with unit conversions, on an annual basis. Either way, $SAV is in units of dollars per time period. The net periodic savings can also be described as a function of the load factor. Combining equation (2.15) with equation (3.4b) gives the following relationship:

(3.4c)

This equation shows that if the costs of electricity and gas are fixed, as well as the power rating of the cogenerator, then the savings increases proportionately as the load factor is increased. If the load factor is doubled, the savings is also doubled.

Equation (3.4c) can be used to find a break-even point. This is to say, at what cost of electricity does it become profitable to run the cogenerator? The first step is to set equation (3.4c) greater than zero:

(3.5a)

This criterion can be reduced to:

(3.5b)

Note, however, that the units in equation (3.5c) could be mixed. Knowing that CE is typically available in units of $/kW-hr and CG in units of $/MBtu, the following conversion makes the comparison straightforward:

(3.5c)

The current cost of natural gas for a typical health facility is $3.40/MBtu, which includes delivery. The rate structure for the cost of gas is flat2, i.e., no matter how much or how little natural gas the facility purchases it costs $3.40/MBtu. Applying that value to equation (3.5c) demonstrates that the cost of electricity must be greater than $0.0116/kW-hr in order to profit from running a cogenerator. The fuel recovery cost of electricity7 in the electric rate structure7 is $0.013994/kW-hr (see Appendix C). This does not include the cost of providing electric service, which varies widely depending on several factors, such as the peak electrical load. Therefore, at the rates under which gas and electricity are being purchased by the facility, it is always profitable to run the cogenerator.

The simple pay back period is directly related to the cost of the cogeneration energy system equipment divided by the net periodic savings:

(3.6a)

The simple pay back period can also be described as a function of the load factor. Substituting equation (3.4c) into equation (3.6a) gives the following relationship:

(3.6b)

or

(3.6c)

It can be seen from equations (3.6b) and (3.6c) that the simple pay back period is inversely proportional to the load factor, i.e., if the load factor is doubled, the simple pay back period is halved, just as the savings is doubled. Also, if the cost of electricity is increased, the simple pay back period decreases. Figure 3-1 is a graphical presentation of equation (3.6c).


Figure 3-1: Simple Pay Back Period vs. Electrical Rate

The simple pay back period does not account for the time value of money. In order to account for the time value of money, an equivalent interest rate representing inflation would have to be assumed. The simple pay back also does not account for increases in the costs of natural gas and electricity. Since these factors are both somewhat unpredictable, it is better to use the simple pay back period first and then account for other factors when they are better known.

Probability Analysis

In order to obtain a statistical distribution of a facility’s hot water consumption, data must be taken and then compiled. In this study, it was necessary to use the natural logarithm of the data taken. Hence some variables have the subscript "LN" attached to them. The lognormal distribution8 had to be used because the normal distribution would have said that there were small probabilities that the water consumption of the facility would be negative. In reality, that would be impossible. Contrarily, the natural logarithm of data cannot result in negative values because even when a number approaches negative infinity, the exponent of that number approaches zero. This is acceptable because on certain holidays, a facility may close for a full day and on that day their water usage would be zero.

The lognormal version of the probability density function8 states:

(3.7)

The variable x is the hot water consumption of a facility for a given day. The variance, mLN, is the mean of the natural logarithm of the daily water consumption data taken at the facility, VW:

(3.8)

where n is the number of data points taken. The standard deviation, sLN, is also defined below:

(3.9)

The function f(x), plotted with the daily hot water consumption along the horizontal axis, results in a sort of bell-shaped curve, the bell becoming elongated to the right (see Figure 3-2). Hypothetical "data" was used for illustration. The hypothetical data had a natural-log-based variance of 2.45 and a natural-log-based standard deviation of 0.274. Notice the difference in the shape of the curve when the standard deviation is doubled and then tripled. Statistical theory8 says the probability that a normal random variable is within one standard deviation of its mean is 68%. The probability that it is within two standard deviations is 95%, and within three standard deviations 99.7%.


Figure 3-2: Sample Probability Density Function

It is desirable to calculate the probability that the facility would use x gallons of hot water per day or less. This is directly related to the probability that a given size cogeneration system would meet 100% of the facility’s hot water requirement on a given day. When this probability is multiplied by the number of days in a year, the curve would represent how many days out of the year that much water, or size cogenerator, would meet the facility’s needs (see Figure 3-3). To normalize the curve, the variable Z is defined:

(3.10)

The cumulative normal distribution8 is then defined:

(3.11)


Figure 3-3: Sample Cumulative Distribution

It then becomes helpful to exchange the axes in Figure 3-3, obtaining Figure 3-4. The integral under the curve of Figure 3-4 shows the amount of hot water the club consumes per year. The curve can be integrated by the trapezoid method9. Then, the maximum amount of hot water that would be produced by the cogenerator is the sum of two terms. The two terms are the integral of x over the interval of days the requirement is satisfied, 0 to Fo, and the product of the corresponding daily hot water usage, xo, with the number of days left in the remainder of the year, (365 - Fo). The symbol xo represents the maximum amount of hot water the cogenerator can provide on a given day. This is clarified by drawing a horizontal line from the coordinate (Fo, xo) through the end of the year to the right in Figure 3-4. An equation for this calculation is expressed below:

(3.12)

where VW,Y is the probable annual hot water demand.


Figure 3-4: Sample Inverted Cumulative Distribution

Global Warming Analysis

Global warming is a controversial issue that many people are concerned about today. It is important to determine whether or not cogeneration technology would adversely affect global warming. It is not desirable to introduce new technology that would increase carbon dioxide emissions.

Basic chemistry dictates that when a hydrocarbon is combusted with air, the products are water and carbon dioxide12. Stoichiometric analyses reveal at what molar ratios the carbon dioxide is produced. From these analyses, it can be calculated how much carbon dioxide a particular fuel emits when burned. The fuel in question is natural gas, which is dominantly methane, CH4. Burned with oxygen, i.e., assuming nitrogen does not partake in the chemical reaction, the following chemical balance describes the combustion of methane:

(3.13)

Data obtained from the Energy Information Administration of the Department of Energy reveals that in 1996, 2.27% of the electric power generated in the United States was fueled by petroleum, 55.74% by coal, and 8.71% by natural gas5. These are the only types of electric power generation that emit any significant amount of carbon. Nuclear power plants do not emit any carbon, and hydroelectric emissions are negligible. In 1996, 32.13 quadrillion Btus of electric power were generated. The Department also reports how much carbon each of those three fuels emit. Petroleum emits 46.6 lb C/MBtu, coal emits 57 lb C/MBtu and natural gas only 32 lb C/MBtu. In 1996, of the 32.13 quadrillion Btus used for electric power generation, 11.05 quadrillion Btus became electricity. These numbers were used to calculate a factor that shows how much carbon is emitted by the average U.S. power plant on the basis of electricity generated for use.

Petroleum: 2.27% (32.13E15 Btu) (46.6 lb C/MBtu) = 0.03399E12 lb C

Coal: 55.74% (32.13E15 Btu) (57 lb C/MBtu) = 1.021E12 lb C

Natural Gas: 8.71% (32.13E15 Btu) (32 lb C/MBtu) = 0.08955E12 lb C

Summing the above three numbers together, the total carbon emissions in 1996 was 1.144E12 lb. Now a carbon emissions factor can be found, per kW-hr of electricity generated:

0.353 lb C/kW-hr of electricity generated

Although a cogenerator burns more natural gas than a water heater, producing more carbon in order to heat water, it simultaneously produces electricity, thereby decreasing much more dramatically the amount of emissions produced by the power plant. The total carbon emissions reduction (CER) due to using a cogenerator, is given by:

(3.14)

where the term hE QGAS,COG, as shown in equation (2.16b), is the increase in gas burned by the cogenerator over the water heater. Again, all of the terms on the right hand side are expressed on a daily basis, so CER is the carbon emissions reduction on a daily, or any desired time period, basis as well. Substituting equation (2.11) into equation (3.14) gives:

(3.15)