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Credit FAQ: From Coal To Hydrogen: An Energy Math Primer

A camera with a shutter speed of more than a second does not have much use in high-speed photography, such as taking the picture of an Olympic sprinter. The picture would be too blurred. So too is the problem in studying the nature of electrons within atoms. They whizz around so fast that the pulse required to "see" them has to be extremely tiny.

The Nobel prize in physics for 2023 went to researchers who figured out a way to produce attosecond pulses. By providing shorter snapshots of atoms and molecules, attosecond spectroscopy helps researchers understand electron behavior in single molecules, such as how electron charge migrates or how chemical bonds between atoms break.

But what is really interesting here is the timescale--atto is Greek for 18, or one quintillionth, and in this case 1 times 10 to the power of minus 18 seconds (i.e., 10^(-18)). To appreciate this tiny scale, we note that there are more than twice the attoseconds in a second than the number of seconds that have passed since the Big Bang. We will repeat that because you think we made a mistake: To be precise, an attosecond to a second is what a second is to about 31.7 billion years.

We mention this only because humans intuitively cannot comprehend very large, or very small, numbers. It is relevant to underscore this point, because in this report you will encounter very large (and also very small) exponential powers.

Investors have often asked us for an energy primer that spans the many measures, statistics, and conversion rates along the energy spectrum. We not only provide a list of measures and conversions here, but also show how some of these conversions between fuel sources and the energy they produce work.

Frequently Asked Questions

How is $/kilowatt (KW)-month converted to $/megawatt-day?

We'll start with a relatively easy one. For legacy reasons, regional transmission organizations (RTO) use different units to quote prices set in their capacity markets. Historically, this reservation charge was usually quoted in $/KW-month (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 to subsume the capacity charge into a full requirements $/MW-hour energy price, likely because it is easier to convert that to $/MW-hour (table 1).

Table 1

Converting $/megawatt-day to $/kilowatt-month
Value Formula
MW to KW 1,000 a
Days in a month 30 b
$1.0/KW-month to $/MW-day 33.3 1/((1/a)*b)
Rule of thumb
$/KW-month $/MW-day
3 100
6 200
MW--Megawatt. KW--Kilowatt.
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 different units, the U.S. Energy Information Administration (EIA) uses the 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 1 degree Fahrenheit at the temperature at which water has its greatest density, which is about 39 degrees Fahrenheit. Using the conversion factors of 2.2 pounds per kilogram (kg) and 1.8 degrees Fahrenheit/Celsius, and the Joules equivalent (table 12), 1 Btu equals 252 calories and 1,055 degrees J. It is approximately the amount of heat released by burning a matchstick.

Understand that and we can introduce two of the most basic formulas for the power sector. The first:

  • 1 million Btu equals 1,000 thousand Btus, or 1 MMBtu.

Here, M is the Roman numeral for 1,000. So, 1 MMBtu really means 1 million Btu.

Coincidentally though, 1 MMBtu is nearly the heat content in about 1,030 cubic feet (cf) of natural gas, leading us to the second basic formula:

  • 1 MMBtu is approximately 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% (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 hydroelectric, solar, and wind energy generators.

Table 2

The coal-to-gas equivalency
At 100% thermal efficiency
Btu
KW-hour 1 3,413 (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-cycle gas turbine 50% c 6,826 a/c
Combustion turbine (peaker) 28% d 12,189 a/d
KW--Kilowatt.

Operators of a number of coal-fired generation units consider converting to gas-fired units before decommissioning. This means weighing the economic option of a coal-to-gas fired conversion. Eastern coal (produced in Central and Northern Appalachia) has a heat content of 12,000-13,000 Btu per pound, while Powder River Basin (PRB) coal has a heat content of 8,400-8,800 per pound. To convert the price of coal from $/ton to $/MMBtu, we perform the following calculations (table 3).

The variables to consider 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/pound, Central Appalachian coal is 12,500, while PRB coal's heat content can vary between 8,400 and 8,800 Btu/pound.

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 being equal, coal prices need to be 30% lower than natural gas prices for coal to compete with gas. We note that even without coal-fired generation's environmental considerations, natural gas prices must be north of $3 for coal to be economically competitive.

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/lbs) 12,500 c
Btu to MMbtu 1,000,000 d
MMBtu/ton 25 e=b*c/d
Eastern coal in $/MMBtu 3 a/e
Delivered cost of Powder River Basin coal ($/ton) 40 a
Pounds in a short ton 2,000 b
Heat content per lb. (Btu/lbs) 8,800 c
Btu to MMBtu 1,000,000 d
MMBtu/ton 17.6 e=b*c/d
PRB coal in $/MMBtu 2.27 a/e
MMBtu--Thousand Btu. Lbs--Pounds.

While the coal-fired generation industry is wrestling with advancing environmental considerations, economic factors are causing the ongoing wave of coal-fired closures. We think coal-fired generation retirements in the PJM Interconnection will be a major reason for a significant uplift in capacity prices in the next RPM capacity auction.

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

Mining coal also generates emissions. We next discuss how much mined coal does a coal-fired generation unit actually need.

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

Table 4

Fuel that a 1-gigawatt coal-fired plant requires per year
Value Formula Remarks
Heat content
Central Appalachian (CAPP) bituminous coal 12,500 Btu/lb 25 MMBtu/ton (i)
Powder Rive Basin (PRB) coal 8,800 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 minute 60 e
Hours/year 8,760 f=b*c
Capacity factor 80% g This varies based on the unit's efficiency on the power supply dispatch.
Output (in MW-hour)- (h) 7,008,000 h=a*f*g
Heat content required (MMBtu) 63,072,000 l=h*k We assumed 9 MMBtu/MWh for k.
Results
Tons of CAPP coal required 2,522,880 m=l/i 2.5 million short tons.
Tons of PRB coal required 3,583,636.4 n=l/j 3.6 million short tons.
GW--Gigawatt. MW--Megawatt. MMBtu--Thousand Btu. lb--Pound. j--Joule.

As bookends, the fuel use for an efficient and relatively inefficient coal-burning unit can be calculated. 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 a capacity factor of 57%, it burns 2 million tons. The variables in this exercise are:

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

These are related because an efficient coal-fired unit will dispatch more at lower cost, thus have a higher capacity factor.

Finally, and this is a simplification, a coal unit generates about 1 ton per MWh of carbon dioxide, while a natural gas unit produces about 0.5 tons per MWh. Introducing a cost for carbon will mean that natural gas prices have to be higher still for coal-fired generation to be economic.

How much fuel does a 1 GW nuclear unit require per year?

This problem needs to be set up a bit. As a sidebar, we also compare the energy density of uranium with that of bituminous coal.

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 about 40 or 55 nucleons and are more stable. An example is iron 56. As uranium splits, each of the 235 nucleons gain stability and release 0.8 megaelectron volts (MeV) of energy in the process. They combine to form the 200 MeV of kinetic energy carried away by the fission products.

For the terribly curious, these are the details. Over many fission reactions, the total liberated energy is about 200 MeV 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
  • Implied energy of the invisible neutrinos: 13

These calculations show that to put out a gigawatt of energy each year, a nuclear plant requires just under seven tons of uranium (table 5).

Table 5

Annual fuel usage of a 1-gigawatt nuclear unit
Energy produced in one year (Joules/year)
Value Formula
One GW to KW 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 90.00% f
KW-hour 7.88E+09 g=a*b*c*f
KW-hour to Joules 3.60E+06 h
Joules/year 2.84E+16 i=g*h; h =KW-hour to joules conversion. See table 12
One eV to Joules 1.60E-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)*1000;inverse of l in kg
Atom/mole 6.02E+23 n, Avagadro's number
Energy in one kg of U235 8.20E+13 o=j*k*m*n*E06;convert Mev to eV
Energy density of U235 8.20E+13 About 80 trillion joules/kg
Enrichment 5% p
Converted to electricity 4.10E+12 q=o*p
Mass of uranium needed (kgs) 6.92E+03 i/q
Mass of uranium (tons) 6.92E+00
Energy in one kg of U235 (J/kgs) 8.20E+13
Energy in one kg of coal (J/kgs) 3.10E+07
Energy density relative to coal 2.65E+06 About 2.6 million times
GW--Gigawatt. KW--Kilowatt. MeV--Megaelectron volt. G--Gram. Kg--Kilogram. J--Joule.

As a corollary, we underscore the fact that the energy density of uranium is about 2.6 million times the energy density of coal. Despite the fact that only a small 5% percentage is enriched, the energy density of nuclear fuel is also about 130,000 times higher

A nuclear unit has no carbon emissions. In fact, on Feb. 2, 2022, the European Commission labeled nuclear-fired generation as sustainable fuel under its climate taxonomy. Similarly, the U.S. Inflation Reduction Act directs billions of dollars in federal funding to clean and reliable energy, with the goal of significantly improving resiliency of the grid, increasing onshore dependence, and lowering carbon emissions by the end of this decade. The act provides for funds delivered through a variety of tax incentives, grants, and loan guarantees for the production of clean and firm electricity, including nuclear energy, which has been largely neglected by past initiatives. The act unequivocally recognized nuclear as an integral part of the carbon-free energy taxonomy.

From a ratings perspective, we see the production tax credits (PTC) provisions in the Inflation Reduction Act as a gamechanger for nuclear power. PTCs change the model for unregulated nuclear power generators from a merchant business to a nine-year contracted business, with floor pricing of $40-$44 per MWh.

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, German physicist Albert Betz calculated that the most efficiency that can be extracted from a wind turbine is 59%. To explain this, a turbine that could extract 100% of the kinetic energy in the wind would mean no velocity is left in the wind. If the velocity of air leaving the blades is zero, it couldn't get out of the way of the air coming in; therefore, it produces no power.

To keep the wind moving through the turbine, there must be some velocity in the air after going through the blades. This is why the machine's efficiency cannot be 100%. Betz calculated that a perfect wind turbine could convert about 59% of the power in the wind into mechanical rotating power.

In our example below (table 6), a typical wind turbine blade will operate at an efficiency of about 33% in a 20 mph wind. The variables in this exercise are:

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

Doubling the wind speed would increase the kinetic energy by 8x. Thus, m times v squared implies a factor of 8x because doubling the velocity also means doubling the mass.

From a ratings perspective, this is the reason resource availability is the biggest risk for wind generation. We note that wind resource has underperformed in 2015, 2019 and in 2023.

Table 6

Electric power a wind turbine generates
Value Formula Remarks
Radius (meters) 50 a
Diameter of the blade 100 2a This is a variable.
Area of the circle swept (m2) 7,850 b= pi *a2
Speed of air (m/s) 10 c 20 mph; see conversiion table from meters/sec to mph.
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 f=0.5 *m*v2 Kinetic energy of the wind.
Kinetic energy (in MW) 3.925 g=f/1,000,000 joules/sec = Watts. Converted to MW
Efficiency of a turbine blade 33% h Ths is anoother variable but constrained by maximum efficiency of 59%.
Power (MW) 1.3 i=g*h
KG--Kilogram. MW--Megawatt.
How much solar energy does the Earth get annually, and how much is that relative to human energy needs?

The rate at which this energy is received is referred to as solar flux, representing the power per unit area received at a given location. At the position of the Earth's orbit, this number is about 1,400 watts per meter squared and is referred to as the solar constant. This means that a flat panel of 1 meter squared placed outside the Earth's atmosphere and oriented perpendicular to the sun's rays would receive 1,400 joules per second of solar energy.

The atmosphere absorbs about half of this energy, so that 700 watts per meter squared is about the most that reaches the Earth on a hot summer day in the tropics. Including the effects of latitudes (angle of incidence changes), seasons, cloud covers, and nighttime decreases the total flux by about a factor of about 10-130 watts per meter squared. There is variation though. Tucson, Ariz., may have an annual average solar flux of 250 watts per meter squared, but Philadelphia likely receives only 160.

With that preamble, here's the math (table 7). If we cut a hypothetical slice through the middle of the Earth, that circle is perpendicular to the sun and represents the area that will get sunlight directly overhead. The radius of this slice is about 6,310 kilometers (we simplified that to 6,000 kilometers in our calculations).

Table 7

Annual solar energy reaching Earth
Value Formula Remarks
Energy output by the sun (watts) 4.00E+26 a Sun's luminosity.
Mean distance of Earth from the sun (meters) 1.50E+11 b
Solar flux (W/m2) 1415.4 c=a/(4* pi *b2)
Radius of the Earth (meters) (R) 6.00E+06 r
Area of the earth 1.1.E+14 d= pi *r2
Solar flux (W/m2) 1.4.E+03 c
Seconds in a year 3.1.E+07 e See conversion table appendix.
Energy received per year (joules/year) 4.97.E+24 f=c*d*e Joules=watt*seconds
Annual global energy consumption (2022) 5.00.E+20 g Estimates vary from 425 quintrillion to 575 quintrillion. We cut it down the middle.
Global consumption as a % of solar energy received 1.01E-04 h=g/f Or about 0.01%.

As our calculations show, global energy consumption can be addressed by a very tiny amount (0.01%) of sunlight that hits the Earth. Further, based on 2022 consumption, watts of electric power used in the U.S. was about 4.8 times 10 to the 11th power. Even if we assume that the efficiency of a solar panel module is about 12% (i.e., produces about 15 w/m squared) we would need an area equal to a square with a side measuring just 300 kilometers to address the electric needs of the U.S.

As solar panel costs decline and efficiency improves, there will be cheaper power. The problem will be the intermittent nature of this power (and distance from load that would need transmission).

Renewable power is somewhat destructively disruptive because it is not firm power. While "as-produced" solar power produced at $30/MWh as recently as the first quarter of 2021, it is now about $50/MWh due to a combination of factors that includes more expensive financing, significant increase in labor and equipment cost, and supply chain constraints.

So solar (and wind) need to be supplemented with co-located batteries. But that can be expensive. In fact, firm renewable power in California even tops $100/MWh once resource adequacy payments are subsumed.

Indeed, in the effort to go green, technology to deliver this firm, renewable power has not kept pace with aspirations. Between cheap, clean, and firm power, currently you can only pick any two. For renewables to be truly become a constructive disruption, deployment of batteries is needed.

What are the cost economics of a four-hour stand-alone battery?

Battery costs have declined enough to replace gas-fired peaking generation in certain markets such as Hawaii and California. We showed the economics of a co-located battery in an earlier commentary ("Going With The Flow: The Competitiveness Of Battery Storage Economics In The Power Sector", published Feb. 4, 2021). Below we present the economics of a stand-alone battery. With continually declining capital costs, the economics of a four-hour, grid-level battery storage unit is roughly those of a gas-fired peaker plant. But with no moving parts, the battery bank needs less maintenance (yet, battery asset life is shorter). It also requires no fuel, making its long-term cost of operation more stable and predictable. A battery bank can respond to power demand almost instantly--less than a millisecond rather than several minutes. While a gas turbine is strictly an energy generator, a battery bank can also store surplus energy. Finally, a battery bank is scalable. More units can be added as needed without dramatic cost increases.

First, when we think in terms of capital costs for batteries, units are in kWh of operation. We expect batteries to be duration products for peak shifting (or peak shaving) solutions. Stated differently, since a battery is used to store energy, the capital cost of a battery is expressed in dollars per unit of energy stored, or $/kWh. To be clear, this is different from energy prices, the cost of electric energy expressed in cents/kWh.

Utility scale battery economics are at about $325/kWh. (Installed system costs have fallen 10% in 2023 as rare earth mineral prices have declined compared with prices in 2022.) So a utility scale, four-hour duration battery would cost about $1,425-$1,525/kW (including balance of plant). We think those costs are comparable with the cost of building a natural gas-fired peaker plant in California. We present a simplified calculation (table 8) of the installed cost of a battery pack that could replace a peaking unit.

The cost of a lithium-ion battery system comprises the lithium-ion battery cell, the rest of the direct current (DC) block, and integration and soft (installation and development) costs.

  • The cathode is the most expensive component of a lithium-ion cell, roughly accounting for a third of the cost of a cell using a lithium-iron phosphate (LFP) cathode, compared to about half for a cell using a nickel-based cathode.
  • Lithium-ion battery cell costs are sensitive to a wide variety of materials costs; however, lithium has the biggest impact. For example, a 50% increase in lithium carbonate price would increase the cost of an LFP cell by 20%, compared to a 50% increase in nickel sulphate prices increasing the cost of a nickel manganese cobalt-622 cell by 7%. (These are S&P Global Commodities estimates.)
  • The direct current block (which includes the cell) typically accounts for about 60% ($200/kWh) of the system cost, roughly split evenly between cell costs and the balance of direct current block costs. Integration costs (e.g., transformer and switchgear, precision control systems, etc.) account for about 15%, and soft costs account for the balance 25%.
  • As system durations increase, cell costs account for a larger proportion, increasing sensitivity to raw material prices and supply dynamics.

Here, we calculate the economics of a 100 MW/400 MWh battery as a capacity product (table 8). Assuming a capital cost of $325/kWh and a 15-year life, we calculate that a battery still requires a toll of about $15.50/kW-month and about $11.50/kW-month with investment tax credit.

Table 8

Battery as a capacity product
Cost economics of a battery unit that could substitute for a gas-fired peaker
Value Formula Remarks
Size of the battery system deployed (MW) (A) 100 a Peaking capacity that the battery installation can deliver instantaneously. This is the typical size of the LM6000 installed in California.
Peak-shifting application hours (B) 4 b Typical shift required; however, batteries can now deliver six hours too.
Total battery energy (MWh) (C) 400 c=a*b
2023 battery cost that we have assumed ($/kW-hour)-- Inc.301 tariffs with tax equity (D) 320 d Utility scale battery costs have been declining, but reversed during the COVID-19 pandemic, still at $300-$375/kWh. Tax credits would lower this to about $240/kWh.
AC to grid connection one-way efficiency loss (E) 4.00% e Just an inevitable loss, but getting more efficient. Historically, this was as high as 7% one way.
Two-way aggregate loss (F) 8.00% f=2xe At charging and while discharging.
Maximum depth of discharge (G) 95.00% g Lithium-ion batteries are typically recharged before they fully discharge. However, they have low memory effect (i.e., unlike nickel-cadmium batteries, they do not lose their maximum energy capacity if they are repeatedly recharged after being only partially discharged).
Minimum depth of discharge (H) 5.00% h
Battery capacity use (I) 90.00% i=g-h Only a maximum 90% of the battery can be effectively used between charges.
Combined losses (J) 13.60% j=1-(1-e)*i Combines 4% charging inefficiency loss and 20% of capacity not used between battery charge and recharge. It just means this 400 MWh battery actually needs to be sized to about 465 MWh to deliver 400 MWh (400/(0.9*0.96).
Cost of the battery (K) 148,148,148 k=c*d*1,000/(1-j) Cost over life of the battery.
Cost of the 100 MW battery (Mil. $) (L) 148 l=k/1,000,000 Coverted to Mil. $. This means that in nominal dollars (2022 base), a battery installation is about $1,530/KW
PV of cash flow stream for $1 in future over 15 years (M). $8.56 m Net present value (rate, cash flow stream of $1 over 15 years); i.e., cash flow discounted at 8% required return over 15-year battery life. We've decided, somewhat randomly, that 8% is an appropriate discount rate. Please use your assumption.
How capital costs will be recovered (N) 11.68% n=1/m Inverse of M. This indicates that $1 invested today requires about 8.56 cents in revenue over the next 15 years for full capital recovery of $1, discounted at a hurdle rate of 8%.
Annualized O&M costs through life $1.43 p=PV(o&m) Combines present value of O&M at 1.75% of capital cost every year; includes mainanence capex every four years.
Annual revenue (O) 18.74 o=(l*n)+p
Required capacity price ($ /MW-day) (Q) 513.34 q=o/(ax365)x1,000,000 Merchant markets do not support a capacity price of $590/MW-day. This needs to be supported through an offtake contract.
Converted to $ /kW-Mo.(R) 15.6 r=qx365/(1,000*12) An installation without investment tax credits will be about $15.50/kW-month; including ITC, about $11.50/kW-month.
MW--Megawatt. MWh--Megawatt hour. kW--Kilowatt. O&M--Operations and maintenance.

However, adding a $25/MWh energy arbitrage reduces this tolling amount to about $13.50/kW-month (table 9). Declining battery costs, aided by the stand-alone investment tax credit, would accelerate the viability of battery economics to about $8/kW-month.

Table 9

Battery as an energy arbitrage product
Value Formula Remarks
Assuming additional energy margins at $25/MWh The battery can glean a $25/MWh charge spread, arbitraging between on and off peak power.
Battery output in MWh (100*4) 400 c=a*b From C in the previous calculation above.
Battery input in MWh (S) =C/((1-F)*I) 483.09 s=c/((1-f)*i) This time we factor a two-way efficiency loss (i.e., 8%) as the battery will be actively called and not merely charged and on stand-by.
Power price at on-peak delivery ($/MWh) (T) 45.00 t The net spread is $25/MWh, which is the arbitrage opportunity.
Power price at off-peak charging (U) 20.00 u
Charge' spread ($) (V) 8338.16 v=(c*t)-(s*u)
# of charging cycles per year (W) 350.00 w
Aggegate annual charge spread (Mil. $) (X) 2.92 x=((v*w)/1,000,000
Charge spread offset on total revenue (Y) 15.82 y=(o-x)
Required capacity price ($/MW-day) (Z) 433.39 z=y/(ax365)x1,000,000 Still not competitive in merchant markets but competitive when battery costs approach $200/kWh. S&P Global Platts estimates that by 2027.
Converted to $/kW-month 13.18 At a battery cost of $200/kWh with investment tax credit, about $8/kW-month.
MWh--Megawatt hour. kW--Kilowatt.

From market sources, we understand that a $7.50-$10/kW-month fixed resource adequacy (RA) payment is roughly in line with contracts with stand-alone battery projects. An "RA-only" means the project retains the right to all other attributes and can, for example, participate in energy and ancillary service markets and keep the associated revenue. Multiple streams of income allow the project to offer this capacity attributes at a price below its annual revenue requirement.

We think batteries as capacity and energy resources are more economically viable in a merchant market. However, as a natural transition until they become competitive as peaking applications, such investments are increasingly being considered as part of a utility's rate base or supported by a contract in fulfillment of state initiatives.

From a ratings perspective, the challenge for a stand-alone battery project is that forecast arbitrage revenue can account for between 20% and 40% of a four-hour battery energy storage system's levelized costs. However, we see increasing prospects of co-located batteries in project-financed wind and solar projects. As the cost economics of batteries decline, we expect to rate stand-alone battery transactions as well as those co-located with renewable projects in 2024.

Also, we are often asked whether we can rate contracted renewables with merchant tails investment-grade. We think that with greater resource predictability, adequate operations and maintenance, and derisked capital costs, this is possible for solar projects if their capital structures are sized to meet debt service coverage ratio thresholds. Resource predictability of wind generation is a challenge (e.g., 2015, 2019, and 2023 have been weak wind years in the U.S.), so a sponsor would have to specifically address through mitigants that include conservative assumptions for wind resource and liquidity buffers.

What is the cost of producing hydrogen ?

On Oct. 13, 2023, the Biden Administration announced seven regional clean hydrogen hubs that will receive $7 billion in infrastructure funding to accelerate the domestic market for clean hydrogen. Collectively, the hubs aim to produce more than 3 million metric tons of clean hydrogen per year, nearly one-third of the 2030 U.S. clean hydrogen production goal. Hydrogen is already produced at scale for use in making fertilizer and in the petrochemical industry. But more recently, hydrogen is being seen as a way to decarbonize hard-to-electrify sectors like maritime shipping, long-haul trucking, steel-making, industrial heating, and aerospace, albeit at a cost. The inflation reduction act's $3/kg hydrogen production tax credit, along with other policy initiatives, changes the calculation to where hydrogen serves to decarbonize.

We note that our cost economics ignores guidance awaited relating to additionality, regional deliverability and hourly matching to ensure that hydrogen is actually produced in a clean way.

We see both proton exchange membrane (PEM) electrolyzers and solid oxide electrolyzers as having a similar cost of production. The latter's efficiency advantage is offset by its higher capital cost relative to PEM.

In terms of modeling, we see four variables as key to hydrogen economics:

  • Input power price.
  • Efficiency of the process.
  • Capital cost.
  • Capacity factor.

Of these, the most important driver is the price of input electricity. The second is efficiency. More efficiency means less electricity needs.

In terms of energy equivalency, a kg of hydrogen has the same high heating value as 39 kWh of energy units. A PEM electrolyzer takes about 55 kWh of electricity to produce a kg of hydrogen, which makes it about 70% efficient. However, solid oxide is more efficient and reaches efficiency of about 75%, and as high as 85% if there is external high temperature steam.

In our quick snapshot of hydrogen economics (table 10), it's interesting to note that costs of production decline most with lower electricity prices. In our example, should electricity prices decline to $25/MWh (as projected), hydrogen production costs decline 50%.

Table 10

Economics of hydrogen
Values Formula Remarks
Energy
Electrolyzer efficiency (kWh/kg) 55 a Varies depending on electrolyzer technology and balance of plant energy needs (this should be all-in including balance of plant). Efficicency improvements to 45 kWh/kg results in hydrogen costs declining by a third, all else being equal.
kg/MWh (B) 18.2 b=10^3/a
Levelized energy costs ($/MWh)(C) 40 c This varies from $35/MWh in ERCOT to $45/MWh in California to $55-$60/MWh in the Northeast based on renewable energy used.
Hydrogen production tax credit (H2 PTC)
H2 PTC $/kg (D) $3.50 d The average of the present value of the subsidy escalated over the 10-year production tax credit period. If a tax-payer, we would take a pre-tax amount that would aggregate to $4.50/kg.
Cost
Energy $/kg (F) $2.20 (c/1,000)*a
Capital (includes storage) $/kg $1.20 g Varies significantly depending on size of facility and capacity factor of electrolyzer (e.g., nuclear versus renewables). These are levelized costs over 10 years. Note that this could be a conservative respresentation if some of these costs are longer-term investments that would increase margins.
O&M $/kg $0.40 h Includes labor, maintenance, insurance, and others.
Interest ($/kg) $0.90 i
Total costs $/kg $4.70 j=f+g+h+i
less H2 PTC ($/kg) -$3.50 d
Net levelized cost (w/o transportation) - $/kg $1.20 k=j-d Levelized cost net of H2 PTC.
Natural gas equivalency @7.7x ($/MMBtu) $9.20 k*7.7 130 MMBtu of natural gas have the same energy equivalency as 1,000 kg of hydrogen.
kWh--Kilowatt hour. kg--Kilogram. MWh--Megawatt hour. O&M--Operations and maintenance. MMBtu--Million Btu.

We make the following observations:

  • The choice of the electrolyzer comes down to the end use. A PEM electrolyzer is highly flexible to cycle/change load and operates at a lower temperature than solid oxide. However, if there is a source of external high temperature steam available, the solid oxide electrolyzer is more efficient. Therefore, a PEM electrolyzer is better matched with renewable generation, whereas a solid oxide electrolyzer will likely operate better in an industrial setup that can provide external heat.
  • In energy equivalency terms, about 7.7 kg of hydrogen provides the same energy as 1 MMBtu of natural gas, making our hydrogen calculation equal to about $9/MMBtu of natural gas. However, as hydrogen economics declines below $1/kg, its industrial uses as a substitute for natural gas are possible, such as steelmaking (via direct reduced iron), cement production, chemicals, and other industrial heat applications.
  • A kg of hydrogen has roughly the energy equivalency of a gallon of diesel and gasoline. This should make a fuel-cell electric car economical. However, the huge constraint is the higher cost of a fuel-cell electric vehicle and the cost of duplicating a hydrogen fueling infrastructure as a battery electric vehicle infrastructure is developed.
  • We do not see near-term use of hydrogen as a substitute fuel for natural gas in the power sector given its higher cost in MMBtu equivalency terms. In addition, blending with natural gas and embrittlement issues (corrosion and cracking caused by hydrogen in pipeline metal) are other constraints. However, the installation of electrolyzers at nuclear plants allows the nuclear unit to produce pink hydrogen for industrial usage instead of selling power into a low off-peak power market. Excess renewable generation can similarly be used for producing green hydrogen.
How are electric vehicles affecting power demand?

While RTOs such as the PJM Interconnection and Electric Reliability Council of Texas have issued estimates of power demand based on scenarios for battery electric vehicle adoption, we think it's still early to build a bottoms-up model for estimating electrification (see "Plugged In: How EVs Supercharge Growth For North America’s Investor-Owned Electric Regulated Utilities", published Oct. 31, 2023). However, we're using the following guidance from industry experts in our early modeling.

The U.S. Department of Energy website shows consumption for all current electric cars between 24 kWh and 36 kWh per 100 miles. We think with continued charging efficiency gains in the next couple of year vehicles will tend toward about 24 kWh for 100 miles. A driving range between 12,000 and 14,000 miles corresponds well with the average of 3.4 MWh per year by 2030 that most market experts predict. The consumption varies at about 2 MWh for plug-in hybrid vehicles to 4.5 MWh for passenger light trucks.

The consumption gets significantly higher for heavy duty regional trucks. If we assume 80,000 average annual driven miles and a charging efficiency of about 1.2 kWh per mile, we think annual electric consumption per truck would be about 100 MWh.

As another reference point, based on cooling degree days, we see average consumption per residential customer of about 13,000 kWh/year in the U.S. South and about 6,800 kWh in the Northeast, resulting in one residential customer equivalent load for every four electric vehicles in the South and every two in the Northeast.

With these parameters set, what remains to be modelled is the adoption rate. We've seen various models, including those forecast by our affiliate mobility team. We'll start incorporating these into our electric growth forecasts as data becomes more reliable. We think it will depend more on the deployment of charging infrastructure.

For liquefied natural gas (LNG), how is million metric tons annually (mtpa) converted into billion cubic feet per day (Bcf/day)?

Finally, we're seeing a number of transactions in the LNG sector and have added one useful conversion. First, unlike the short ton of 2,000 pounds that we use in the U.S. coal industry, the LNG industry uses the metric ton, which is 2,204 pounds. These units differ 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 thousand cubic feet (Mcf) per day of gas into a liquefaction plant produces 1 mtpa of LNG. However, 1 mtpa of LNG will create only approximately 130 Mcf/d of gas. The difference would be from natural gas losses involved in the liquefaction process. Using these rates (table 11), a metric ton of LNG is the energy equivalent of 48.53 million Btu of natural gas.

Table 11

Liquefied natural gas industry units: How to convert million tons annually into billion cubic feet per day
Value Formula Remarks
Density of LNG in lb per M3 1,020 a
Pounds in a Metric ton (MT) 2,204 b
MT LNG to M3 LNG 2.16 c=b/a
LNG to natural gas (NG) multiple 625 d Cooling natural gas to -259 degrees Farenheit shrinks its volume by 1/625.
MT LNG to M3 NG 1,350.50 e=c*d
M3 to cubic feet 35.94 f
MT LNG to cubic feet NG 48,532.60 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 0.133 l=i/(k*j) About 130-150 MMcf/day.
1 BCF/d to MTPA 7.521 m-=i/l
Results
MT LNG to cubic feet NG 48,532.60 n
MT LNG to MMBtu 48.53 n/1,000 1 MMBtu =1 Mcf
BCF/d--Billion cubic feet per day. LNG--Liquefied natural gas. MT--Metric ton. mtpa--Million tons per annum. MMBtu--Million Btu. MMcf--Million cubic feet. Mcf--Thousand cubic feet.
So, what are the basic units that we needed for our calculations?

Some of the basic equivalents we used in preparation of this primer are in table 12. We hope you will find it useful. Based on the feedback and requests we receive, we will update this with additional conversions at a later point.

Table 12

Further conversions
Value Formula
Year to seconds
Days in a year 365 a
Hours in a day 24 b
Minutes in an hour 60 c
Seconds in a minute 60 d
Seconds in a year 3.15E+07 a*b*c*d
Meters/second to mph
Kilometer in a meter 0.001 a
Miles in a kilometer 0.6 b
Seconds in a minute 60 c
Minutes in an hour 60 d
Meters/second to mph 2.16 a*b*c*d
1 eV to Joules
1 1.6x 10-19
Joules = Watt *Sec
Joules/second = Watt
Watt to joules/second (energy)
100 watts
Watt 100 a
Year to seconds 3.15E+07 b
Watt to joules/year
100 3.15E+09 a*b
Kilowatt-hour to joules
Kilowatt to watts 1000 a
Hour to minute 60 b
Minute to seconds 60 c
Kilowatt-hour to joules Joules
1 3.60E+06 a*b*c

Related Research

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:Maya Niu, CFA, Toronto;
maya.niu@spglobal.com

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