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Credit FAQ: Adventures In Vitalstatistix: An Energy Math Primer

The Adventures of Asterix is a comics series, widely read in Europe and Asia that first appeared in 1959, and follows the adventures of a village of Gauls as they resist Roman occupation circa 50 BC. The protagonist, Asterix, embarks on numerous adventures to defeat Julius Cesar's fierce battalions. He is aided by Getafix, the village druid, whose magic potions give him superhuman strength, and they are led by the village chief: Vitalstatistix. For a village surrounded by fortified garrisons, the authors figured, command of the numbers is, well, vital.

So too, is it, with understanding the energy business. Investors have asked us for a primer that spans the many measures, statistics, and conversion rates along the energy spectrum. We realized that these numbers are vital statistics for understanding the energy business. So we have decided to turn to Vitalstatistix to help lead this effort and protect us, not from the Romans, but from an onslaught of bewilderment that confuses those with financial interest in power plants, utilities, and oil and gas producers.

We've not only provided a list of measures and conversions, but also showed how some of these conversions between fuel sources and the energy they produce work. This primer leans more towards the power sector but we have also include data related to liquefied natural gas (LNG) and refineries. Unfortunately, there is no magic potion to help understand these involved calculations. You will just have to work through some of these examples. Now if we could only figure out the energy content in that cauldron full of Getafix's magic potion.

Frequently Asked Questions

How do you convert $/KW-month to $/MW-day?

We'll start with an easy but annoying one. For legacy reasons, different regional transmission organizations (RTOs) use different units to quote prices set in their capacity markets. Historically, this reservation charge was usually in $/KW-mo. (or $/KW-year) to calculate the revenue requirement of a power generation unit. Over time, one RTO decided to use $/MW-day as the unit, likely because it is easier to convert $/MW-day into $/MW-hour, to subsume the capacity charge into a full-requirements $/MW-hour energy price.

Table 1

Converting $/MW-day to $/KW-month
Value Formula
MW to KW 1000 a
Days in a month 30 b
$/KW-month to $/MW-day 33.3 1/((1/a)*b)
Rule of thumb:
$/KW-month $/MW-day
3 100
6 200
What is the heat content in coal and natural gas, and what does that mean for coal-to-gas equivalency in power generation?

Fuels can be converted from physical units of measure (such as weight or volume) to a common measurement of the energy or heat content of each fuel. While energy content can be calculated using several different units, The U.S. Energy Information Administration (EIA) uses the British thermal unit (Btu) as its standard for a unit of energy content. As a result, the Btu has become the industry standard.

One Btu equals the quantity of heat required to raise the temperature of one pound of liquid water by one degree Fahrenheit at the temperature at which water has its greatest density, which is about 39 degrees Fahrenheit. Understand that, and we can introduce two of the most basic formulas for the power sector. The first:

  • 1,000,000 Btu=1,000 Mbtu= 1MMBtu.

Remember that M is not Mega but is the roman notation for a 1,000. So, 1MMBtu really means 1 million btu. Coincidentally though, 1 MMBtu is nearly the heat content in 1 Mcf (1,000 cubic feet) of natural gas, leading us to the second basic formula:

  • 1 MMBtu ~ 1 Mcf

In the power industry, the typical measure of efficiency is the amount of fuel used to generate 1 KW-hour of net generation. If a unit is 100% efficient, it will produce 1 KW-hour of electricity using 3,413 btu. However, thermal efficiencies of generation units are typically below 50% (see table 2), which gives us the 'unit heat rates', or the efficiency with which these units convert the heat content of the fuel into electricity. If, for instance, a Siemens 8000H series (6,400 btu/KWh) has a lower unit heat rate than the 5000H series (6,800 btu/kwh) it is because the thermal efficiency of the 8000H combined cycle gas turbines (CCGT) has improved with technological evolution.

The EIA, however, does not have similar heat rate estimators for the efficiency of hydro, solar and wind energy generators.

Table 2

The coal-to-gas equivalency
At 100% Thermal efficiency
BTU
KW-hour 1 3413 (a)
Typical Thermal Efficiencies of power generators Implied Unit Heat rate BTU/KW-hour)
Formula
Coal-fired Unit 34% b 10,038 a/b
Combined Cyle Gas Turbine 50% c 6,826 a/c
Combustion Turbine (Peaker) 28% d 12,189 a/d

Now, let's look at the next example. Eastern coal (produced in Central and Northern Appalachia) has a heat content of 12,000-13,000 Btu pound, while Powder River Basin (PRB) coal has a heat rate of 8,400-8,800 per pound. So to convert the price of coal from $/Ton to $/MMBtu we perform the following calculations (see table 3).

Table 3

Converting The Price Of Coal from $/Short Ton To $/MMbtu
Value Formula
Delivered cost of Eastern coal ($/short ton) 75 a
Pounds in a U.S. short ton 2,000 b
Heat content per Lb (btu/kwh) 12,500 c
Btu to MMbtu 1,000,000 d
MMBtu/ton 25 e=b*c/d
Eastern Coal in $/MMbtu 3.0 a/e
Delivered cost of PRB Coal ($/ton) 38 a
Pounds in a short ton 2,000 b
Heat content per Lb (btu/kwh) 8,800 c
Btu to Mmbtu 1,000,000 d
MMBtu/ton 17.6 e=b*c/d
PRB coal in $/MMbtu 2.16 a/e

The variables to consider in this exercise are

  • The cost of coal per ton (delivered cost should include transportation); and
  • The heat content of the fuel. For example, Northern Appalachian coal tends to be about 13,000 btu/kwh, Central Appalachian coal is 12,500, while PRB coal's heat content can vary between 8,400-8,800 btu/kwh.

Moreover, the variable cost of power generated by a unit is roughly its heat rate times the cost of fuel. On average, a CCGT is about 30% more efficient than a coal-fired unit (a heat rate of 7,000 btu/kwh compared with 10,000 btu/kwh). This means that, all else equal, coal prices need to be 30% lower than natural gas prices for coal to compete with gas. From table 3, we note that while PRB coal is competitive with natural gas ($2.16 per MMBtu compared with natural gas at about $2.75/MMBtu), but eastern coal is not.

While coal-fired generation is wrestling with advancing environmental considerations, it is the economic factors that are causing the ongoing wave of coal-fired closures.

How much fuel does a 1 GW coal-fired plant require per year?

We have to build off the calculations we've just shown to answer this question. Moreover, to answer this question we must also assume that we know the btu content of the specific fuel we're talking about. And we can see, when we look at the calculations in table 4, that it will take nearly 50% more PRB coal than Central Appalachian coal. PRB coal has about 25% moisture by weight (hence the lower heat content compared to eastern coal) but is much cheaper per ton.

Table 4

How Much Fuel Does A 1 GW Coal-fired Plant Require Per Year?
Value Formula
Heat Content:
Central Appalachian (APP) bituminous coal 12500 btu/lb 25 MMbtu/ton (i)
Powder Rive Basin (PRB) coal 8800/btu/lb 17.6 MMbtu/ton (j)
Heat rate of a coal fired unit 8,800-10,000 btu/kwh 8.8-10 MMbtu/MWH (k)
Power generation in one year
1 GW plant to MW 1,000 a
Days in a year 365 b
Hours in a day 24 c
Minutes in an hour 60 d
Seconds in a munite 60 e=a*b
Hours/year 8,760 f=b*c
Capacity factor 80% g
Output (in MW-hour) 7.01E+06 h=a*f*g
MMbtu 63,072,000 l=h*k
Results:
Tons (APP) 2.5+06, or about 2.5 million short tons m=l/i
Tons (PRB) 3.58+06, or about 3.6 million short tons n=l/j

As bookends, we can also calculate the fuel use for an efficient and a relatively inefficient coal-burning unit. At a heat rate of 9,000 btu/kwh and an 80% capacity factor the power generation plant burns 2.5 million tons. On the other hand, at a heat rate of 10,000 btu/kwh and at a capacity factor of 57%, it burns 2.0 million tons. The variables in this exercise are:

  • The capacity factor (utilization rate) of the coal power plant; and
  • The unit heat rate of the coal-fired unit.

These are related because an efficient coal-fired unit will dispatch more and have a higher capacity factor.

But what if we do not know the heat content of the fuel? Below is an exposition of how we would approximate the same problem without knowing the heat content of the coal, or the heat rate of the coal-fired unit. As because the EIA does estimate the efficiency of nuclear, hydro, solar, and wind energy generators, the method we use next can also be applied to calculate fuel usage for a nuclear plant. Much of the data we'll use from here on in is common knowledge but we've included some tables to refer to at the end of this article.

How much fuel does a 1 GW coal-fired plant require per year?

To answer this questions, we've made some simple assumptions in our calculations (see table 5). We assume a C2H4 hydrogen to carbon ratio in coal because the ratio is more than 0 (pure carbon) and less than 4 (pure methane). Bituminous coal contains moisture of approximately 17 %. About 0.5%-2% of the weight of bituminous coal is nitrogen. So its fixed carbon content ranges up to 85%, with ash content up to 12%-13% by weight.

Ultimately, we find that this plant would require a minimum of just over 2.5 million tons of coal each year to generate a gigawatt of power.

Table 5

Annual Coal Usage Of A 1-GW Coal-Fired Unit
Value Formula Remarks
Energy produced in one year (joules/year)
One GW to Watt 1.00E+06 a
Days per year 365 b
Hour per day 24 c
Minutes per hour 60 d
Seconds per minute 60 e
Seconds per year 3.15E+07 b*c*d*e
Capacity factor 80.0% f
KW-hour per year 7.01E+09 g=a*b*c*f
Joules/year 2.52E+16 g*h h is KW-hours to joules. See conversion tables in side bar
One eV 1.6 x 10^-19 joules i
Energy released in each chemical reaction 1.5eV j In a typical chemical reaction, one electron is exchanged between two atoms. The energy of this reaction is about 1.5 electron volts or 1.5eV
One mole of carbon 12 g k
Moles of carbon in one kg 80 i=1/k
Atom/mole 6.023 *10^23 m
Number of moles of carbon in one kg 2.14E+02 n c2h4 has 6 moles in 1 kg
Energy in one kg of coal 3.10E+07 o=i*j*m*n
Energy density of bituminous coal 3.10E+07 o Actually varies between 24-35 Million Joules/kg (so our approximate calculation works)
Thermal Efficiency 32% p Actually ranges from 30%-42%
Converted to electricity 9.91E+06 q=o*p
Mass of coal (kgs) 2.55E+09 q/h This could be over 3 million tons if energy density of coal is 24 million joules (MJ) instead of 31 MJ assumed
Mass of coal (tons) 2.55E+06
Source: S&P Global Ratings.
How much fuel does a 1 GW nuclear unit require per year?

This problem needs to be set-up a bit.

The total energy released in a fission reaction depends on the average binding energies of its protons and neutrons, which is the amount of energy needed to separate the nucleus into its constituent nucleons divided by the number of these nucleons. The higher the binding energy, the more stable the nucleus. Uranium 235, with its 133 neutrons and 92 protons is relatively unstable. The two fragments formed by the fission of uranium each contain around 40 or 55 nucleons and are more stable. For example, Iron 56. As uranium splits, each of the 235 nucleons gains stability, each releasing 0.8 MeV of energy in the process. These 0.8 MeV combine to form the 200 MeV of kinetic energy carried away by the fission products.

Over a large number of fission reactions, the total liberated energy is about 200 MeV's as follows:

  • Kinetic energy of fission products: 165
  • Kinetic energy of gamma photons: 6
  • Kinetic energy of neutrons: 5
  • Beta and gamma radiations of the fission products: 11
  • Energy of the invisible neutrinos: 13

We do those calculation ad we find that to put out a gigawatt of energy each year, a nuclear plant requires just over seven tons of uranium (see table 6).

Table 6

Annual Fuel Usage Of A 1-GW Nuclear Unit
Energy produced in one year (Joules/year)
Value Formula
One GW/Watt 1.00E+06 a
Days per year 365 b
Hours per day 24 c
Minutes per second 60 d
Seconds per minute 60 e
Hours per year 3.15E+07 b*c*d*e
Capacity factor 90.0% f
KW-hour 7.88E+09 g=a*b*c*f
Joules/year 2.84E+16 i=g*h; h =KW-hour to joules conversion. See sidebar
One eV to Joules 1.6 x 10^-19 j
Fissioning one atom of U235 (MeV) 200 k
One mole of uranuim (g) 235 l
Moles of U235 in 1 kg 4.26 m=1/l
Atom/mole 6.023 *10^23 n
Energy in one kg of U235 8.04E+13 o=j*k*m*n
Energy density of U235 8.04E+13 About 80 billion joules/kg
Enrichment 5% p
Converted to electricity 4.02E+12 q=o*p
Mass of uranium (kgs) 7.06E+03 q/i
Mass of uranium (tons) 7.06E+00
Energy in one kg of U235 (J/kgs) 8.04E+13
Energy in one kg of coal (J/kgs) 3.10E+07
Energy density relative to coal 2.59E+06 About 2.6 million times
Source: S&P Global Ratings.

As a corollary, we underscore the fact that the energy density of nuclear fuel is about 2.6 million times the energy density of coal.

How much electric power does a wind turbine generate?

The problem requires a bit of a context too, specifically as it relates Betz's law.

In 1919, a German physicist, Albert Betz calculated that the most efficiency that you extract from a wind turbine is 59%. To explain this, imagine a turbine that could extract 100% of the kinetic energy in the wind. No kinetic energy left in the wind means no velocity left in the wind. If the velocity leaving the blades is zero then the air wouldn't be leaving at all. As a result, the air after the blades will not be getting out of the way of the air coming in, and no air flowing through the turbine blades would mean no power.

In order to keep the wind moving through the turbine there has to be some velocity in the air after going through the blades to make way for the new air coming in. This is why the machine's efficiency cannot be 100%. Betz proved that the best achieved by a wind turbine is around 59%. In other words, a perfect wind turbine would be able to convert about 59% of the power in the wind into mechanical rotating power.

In our example below (see chart 7), we see that a typical wind turbine blade will operate at an efficiency of about 33% in a 20 mile per hour wind. The variables in this exercise are:

  • Efficiency of the turbine
  • Diameter of the blade
  • Wind speed

Note that doubling the wind speed would increase the kinetic energy by 8x, i.e. m*v2 implies a factor of 8x because doubling the velocity also means doubling the mass. That is why resource risk is the biggest risk for renewable generation.

Table 7

Electric Power From A Wind Turbine
How much electric power does a wind turbine generate?
Value Formula Remarks
Diameter of the blade 100 2a
Radius (meters) 50 a
Area of the circle swept (m2) 7,850 b=pi * a2
Speed of air (m/s) 10 c 20 miles/hr; see conversiion table from mtr/sec to miles/hr
density of air (kg/m3) 1 d
Mass of air in kg/sec 78,500 e=b*c
Kinetic Energy (joules/sec) 3,925,000 0.5 *m*v2 Kinetic energy of the wind
Kinetic Energy (in Watts) 3.925 MW joules/sec= Watts
Efficiency of a turbine blade 33% Maximum efficiency is 59%
Power 1.30 MW
Source: S&P Global Ratings.

The variables in this exercise are:

  • Efficiency of the turbine
  • Diameter of the blade
  • Wind speed

Note that doubling the wind speed would increase the kinetic energy by 8x—i.e. m*v2 implies a factor of 8x because doubling the velocity also means doubling the mass. That is why resource risk is the biggest risk for renewable generation.

How do you calculate the energy density of gasoline?

Gasoline is a refined product of crude oil and is made up of various hydrocarbons. We approximate gasoline as being made up of only octane, whose chemical formula is C8H18.

The calculated energy density of gasoline is about 45 MJ/kg. Our calculation—-at about 33 MJ/kg--is lower because of approximating the formula for gasoline (see table 8). What is interesting, however, is that we from can look back at our previous calculations and see that the energy density of gasoline is roughly the same as that in coal.

Table 8

Energy Density Of Gasoline
Value Formula Remarks
Energy from each chemical reaction 1.5eV a
We will assume gasoline to be c8H18
The molecular mass of gasoline (gms) 114 b c*8=96, H*18=18 (c=12 and H=1)
One kg of Gasoline contains (moles) 8.77 c=1/(b*10-2)
Total number of eV released 26 d The 8 carbon atoms will oxidize to form co2 and the 18 hydrogen atoms will oxidize and for h2o
Atoms in a mole 6.023*10^23 e
One eV to Joules 1.6x 10^-19 f
Density of gasoline (to water) 0.9 g
Total energy released (joules/kg) 3.30E+07
Total energy released (joules/ltr) 2.97E+07 h=a*c*d*e*f*g
Source: S&P Global Ratings.
For liquefied natural gas (LNG), how do you convert million tonnes per annum (mtpa) into billion cubic feet per day (BCF/day)?

First, unlike the short ton of 2,000 pounds that we use in the U.S. coal industry, the LNG industry use the metric tonne, which is 2,204 pounds.

These different units arise in the LNG industry because LNG plant capacity is usually specified in mtpa while LNG ship and storage tank capacity is usually measured in cubic meters of liquid.

The typical rate that we use is approximately 150 MMcf/d of gas into a liquefaction plant produces 1 mtpa of LNG. However, 1 mtpa of LNG will create only approximately 130 MMcf/d of gas. The difference between the 150 MMcf/d and 130 MMcf/d of gas would be natural gas losses involved in the liquefaction process.

Thus, using these rates (see table 9), we see that a tonne of LNG is the energy equivalent of 48.53 million btu of natural gas.

Table 9

LNG Industry Units: How To Convert Million Tons Per Annum (mtpa) Into Billion Cubic Feet Per Day (BCF/d)
Value Formula Remarks
Density of LNG in lb per M3 1020 a
Pounds in a MT 2204 b
MT LNG to M3 LNG 2.16 c=b/a
LNG to natural gas (NG) multiple 625 d Cooling natural gas to -259F shrinks its volume by 1/625
MT LNG to M3 NG 1350.5 e=c*d
M3 to cubic feet 35.94 f
MT LNG to cubic feet NG 48,532.6 g=e*f
MT to million MT 1,000,000 h
One million MT LNG to cubic feet NG 4.85E+10 i=g*h
Days in a year 365 j
Cubic feet to BCF 1.00E+09 k
1 MTPA to BCF/d 0.1330 L=i/(k*j) about 130 MMcf/day-150 MMcf/day
BCF/d to 1 MTPA 7.521 m-=i/L
Results:
MT LNG to cubic feet NG 48,532.6 n
MT LNG to MMBTU 48.53 n/1000 1 MMbtu =1 Mcf
How do you calculate the complexity of a refinery?

In 1960, Wilbur Nelson developed the Nelson Complexity Index to evaluate the relative complexity of one refinery to another. The index assigns a complexity factor to each major piece of equipment in a refinery based on its complexity and cost relative to crude distillation, which has an assigned complexity factor of 1.

To calculate the Nelson Complexity of a refinery, the throughput capacity of each piece of refining equipment is multiplied by its assigned complexity factor. Then, all of the equipment unit's complex barrels are summed together to calculate the total complex barrels of a given refinery. The last step is to divide total complex barrels by the distillation capacity, which yields the Nelson Complexity.

A Nelson Complexity of 10, as calculated in our example (tables 10 and 11), implies a refinery is 10 times more complex than crude distillation for the same amount of throughput. In general, a higher number indicates the ability to process more difficult crude oil (i.e., heavy or sour). In theory, refineries with higher complexity are able to achieve better yields of lighter refined products and use a wider slate of feedstocks.

Table 10

Complexity Factors Of Different Processing Units
Complexity factor
Distillation capacity 1.0
Asphalt 1.5
Vaccum Distillation 2.0
Catalytic Hydrotreating 2.25
Thermal Processes 2.75
Catalytic Hydrorefining 3.0
Catalytic Reforming 5.0
Catalytic Cracking 6.0
Catalytic Hydrocracking 6.0
Coking 6.0
Alkylation/Polymerization 10.0
Oxygenates 10.0
Aromatics/Isomerisation 15.0
Lubes 60.0

Table 11

Nelson Complexity Calculation
Barrels Complexity Complexity Barrels
Atmospheric Distillation 210,000 1 210,000
Vaccum Distillation 70,000 2 140,000
Catalytic Hydrotreating 175,000 2.25 393,750
Catalytic Reforming 78,000 5 390,000
Coking 30,000 6 180,000
Catalytic Cracking 50,000 6 300,000
Catalytic Hydrocracking 26,000 6 156,000
Alkylization/Polymerization 13,500 10 135,000
Aromatics/Isomeristion 13,000 15 195,000
Total Complexity Barrels (a) 2,099,750
Distillation Capacity (b) 210,000
Nelson Complexity (a/b) 10.0
So what are the basic units we need to know across the energy spectrum?

Vitalstatistix needs his numbers. And so do investors. Here are some of the basic equivalents we have used in preparation of this primer (see below).

We hope you will find this primer as useful as we had fun working on it. No magic potions here, for sure. If only understanding the vital statistics in the world of energy could be mastered with a few swipes of color, and a swig of magic potion. That is not so, but understanding these things is a necessity for energy investors and we will do what is necessary to aid that understanding. If we receive feedback, we'll be happy to post a sequel on energy conversions in other sectors.

We will add that although Asterix has less of a following in North America, this Franco-Belgian series is among the most popular comics in the world, translated into over 110 languages and dialects. In fact, the popularity of this comics series in Asia and Europe compared to the relative absence of it in North America is so stark that our preamble is likely to bring a smile to Asian and European investors even as North American investors wonder what we're talking about.

Anthea Bell, the person credited with translating many of these works into English died on Oct. 18, 2018. She translated numerous Franco-Belgian comics of the bande dessinée genre into English, including Asterix. This is our way of doffing our hat to her.

Table 12

Select List Of Conversion Factors
1 kilowatt hour of electricity 3,413 Btu
1 cubic foot of natural gas 1,008 to 1,034 Btu
1 therm of natural gas 100,000 Btu
1 gallon of liquefied petroleum gas (LPG) 95,475 Btu
1 gallon of crude oil 138,095 Btu
1 barrel of crude oil 5.8 MMBtu
1 gallon of kerosene or light distillate oil 135,000 Btu
1 gallon of middle distillate or diesel fuel oil 138,690 Btu
1 gallon of residual fuel oil 149,690 Btu
1 gallon of gasoline 125,000 Btu
1 gallon of ethanol 84,400 Btu
1 gallon of methanol 62,800 Btu
1 gallon of gasohol (10% ethanol, 90% gasoline) 120,000 Btu
1 pound of coal 8,100 to 13,000 Btu
1 ton of coal 16.2 to 26.0 MMBtu
1 ton of coke 26.0 MMBtu
1 ton of wood 9.0 to 17.0 MMBtu
1 pound of low pressure steam (recoverable heat) 1,000 Btu
1 Joule per second 1 Watt
1 KW-hour 3.60R-06 Joules
Source: S&P Global Ratings

Table 13

Further Conversions
Value Formula
Year to seconds
Days in a year 365 a
Hours in a day 24 b
Min in an hr 60 c
Seconds in a minute 60 d
Seconds in a year 3.15E+07 a*b*c*d
Meters/sec to Miles/Hour
kms in a meter 0.001 a
miles in a kms 0.6 b
seconds in a minute 60 c
Minutes in an hour 60 d
Mtrs/sec to Miles/hr 2.16 a*b*c*d
1 eV to Joules
1 1.6x 10-19
Joules = Watt *Sec
Joules/second = Watt
Watt to Joules/Sec (Energy)
100 Watt=
Watt 100 a
Year to seconds 3.15E+07 b
Watt to Joules/Year
100 3.15E+09 a*b
KW-hour To Joules
KW to Watt 1000 a
Hour to Min 60 b
Min to Seconds 60 c
KW-hour to Joules Joules
1 3.60E+06 a*b*c
Source: S&P Global Ratings

This report does not constitute a rating action.

Primary Credit Analyst:Aneesh Prabhu, CFA, FRM, New York (1) 212-438-1285;
aneesh.prabhu@spglobal.com
Secondary Contact:Michael Tsahalis, New York + 1 (212) 438 0283;
Michael.Tsahalis@spglobal.com

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