Fossil Energy Resources: R/P Ratio, Reserves, Unconventional Gas and Peak Oil

This chapter, the first of the Engineering Sciences (XE) paper's Energy Science material, answers the first topic of Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam: “Fossil energy resources: R/P ratio, estimation of reserves, unconventional fossil resources (coal bed methane, shale gas, gas hydrates, frozen methane), peak oil theory.” It teaches calorific value basics, the R/P ratio with and without demand growth, the classes of reserves and resources and volumetric estimation, the main unconventional resources, and Hubbert's logistic peak-oil model. The numericals are lifetime under growing demand, volumetric reserves, Langmuir gas release, LHV from HHV, and logistic peak production, linearisation and depletion time.

1. Fossil fuels and calorific value: the basics

Fossil fuels (coal, crude oil, natural gas) are the buried, partly decayed remains of ancient organisms, so they are a stock: they form over geological time and are consumed over human time, which makes them non-renewable. Coal is mostly carbon with hydrogen, oxygen, sulphur, moisture and ash; crude oil is a mixture of hydrocarbons; natural gas is mainly methane (CH₄). The ratio H/C rises from coal to oil to gas, which is why gas gives the least CO₂ per unit of heat.

The calorific value is the heat released when unit mass (or volume) of fuel burns completely and the products return to the reference temperature. The higher (gross) calorific value, HHV counts the latent heat recovered when the water formed by burning hydrogen condenses; the lower (net) calorific value, LHV assumes that water leaves as vapour. Each kg of hydrogen forms 9 kg of water (2 H₂ + O₂ → 2 H₂O, 4 g gives 36 g), so for a fuel with hydrogen mass fraction h and moisture fraction w:

LHV = HHV − (9h + w) × h_fg, with h_fg ≈ 2.44 MJ per kg of water at ordinary temperature.

⚠️ Exam trap
The two values differ only through the latent heat of the water. A fuel with no hydrogen and no moisture (pure carbon) has HHV = LHV. A quoted efficiency is meaningless without its basis. Quote which basis (HHV or LHV) any efficiency or emission figure uses; mixing them gives an error of up to about 10 per cent for natural gas.

2. The R/P ratio and its limits

The reserves-to-production ratio is R/P = (remaining proved reserves at the end of a year) ÷ (production in that year), in years. It states how long the reserves would last if production stayed at this year's rate. It is a quick index of scarcity and a good number to compare regions or fuels, but it is not a forecast of exhaustion.

  • Production is not constant: with growing demand the reserves die sooner than R/P says; with a falling field profile the tail is longer.
  • Reserves are not fixed: price rises, better technology and new discoveries move resources into reserves, so R/P often stays roughly constant for decades.
  • Reserve numbers carry reporting uncertainty and depend on the proved or proved-plus-probable definition used.
  • Production is limited by the rate wells and mines can deliver, not only by the amount underground.

Growing consumption. If the first-year production is P₀ and it grows exponentially at fractional rate g, the reserves R last a time T given by R = ∫₀ᵀ P₀ e^(gt) dt = P₀ (e^(gT) − 1)/g. Therefore

T = (1/g) ln(1 + g R/P₀).

Worked example: R/P₀ = 50 years and g = 2 per cent a year give T = ln(1 + 0.02 × 50)/0.02 = ln 2/0.02 = 34.7 years, not 50. As g → 0, T → R/P₀ (check: ln(1+x) ≈ x). If g = 0 the formula reduces to the plain ratio.

⚠️ Exam trap
With growth the lifetime is always shorter than R/P. A common slip is to apply T = R/P to a growing demand, or to forget that the reserve is in the same units as production (barrels with barrels a year).

3. Resources, reserves and their estimation

A resource is the whole accumulation that is believed to exist, including undiscovered and uneconomic parts. A reserve is the part that is discovered, technically recoverable and commercially viable under present economics and regulations. Reserves are therefore a subset of resources, and they move with price and technology; resources change only with geology and discovery.

Reserve classes are cumulative; the more optimistic the class, the larger the volume and the lower the confidence.
CategoryMeaning (probabilistic convention)Common label
ProvedQuantity at least 90 per cent likely to be recovered or exceeded1P, P90
ProbableProved plus probable at least 50 per cent likely (best estimate)2P, P50
PossibleProved + probable + possible at least 10 per cent likely (optimistic)3P, P10

Volumetric estimation for an oil field: oil initially in place OIIP = A h φ (1 − S_w)/B_o, where A is area, h net pay thickness, φ porosity, S_w water saturation and B_o the formation volume factor (reservoir volume per unit stock-tank volume, B_o > 1 because dissolved gas leaves). Reserves = OIIP × recovery factor RF. Worked example: A = 10 km², h = 20 m, φ = 0.20, S_w = 0.25, B_o = 1.2, RF = 0.30: OIIP = 10⁷ × 20 × 0.20 × 0.75/1.2 = 2.5 × 10⁷ m³ and reserves = 7.5 × 10⁶ m³. For coal the same idea is tonnes = area × seam thickness × density × recovery.

🎯 Why it matters
Other methods (decline-curve analysis, material balance) are used once a field is producing; the volumetric method is used early. Each step cuts the number: the difference between in-place and recoverable is the recovery factor, which is why 'oil in place' is never the reserve.

4. Unconventional fossil resources

Conventional oil and gas sit in porous, permeable reservoir rock under a seal and flow to a well. Unconventional resources are dispersed or locked in rock of very low permeability, or held in a solid form, so they need extra stimulation, more wells and more energy.

ResourceWhat and whereHow it is produced
Coal bed methane (CBM)Methane adsorbed on the internal surface of coal micropores; the seam is both source and reservoir and is usually water-saturatedPump out the water (dewatering) to lower the pressure; methane desorbs and flows
Shale gasGas held free in pores and fractures and adsorbed on organic matter in fine-grained, very low-permeability shaleHorizontal drilling plus hydraulic fracturing to create flow paths
Gas hydratesIce-like crystalline solid in which water molecules form cages around methane; found in permafrost and in cold, deep ocean-floor sedimentsDissociate by depressurising, heating or inhibitor injection; still at research and pilot stage

Frozen methane is the popular name for methane locked in hydrate in permafrost and sea-floor sediment. Hydrates are stable only at low temperature and high pressure, which is why they occur under deep water or cold ground. Warming or a fall in pressure dissociates them and releases methane, a strong greenhouse gas, so they are both a possible fuel and a climate feedback concern.

Langmuir isotherm for gas stored on coal: V = V_L P/(P_L + P), where V_L is the maximum (monolayer) capacity and P_L the pressure at which half of it is held. Worked example: V_L = 10 m³ of gas per tonne, P_L = 2 MPa. At 6 MPa V = 10 × 6/8 = 7.5 m³/t; after dewatering to 2 MPa V = 10 × 2/4 = 5 m³/t; the gas released is 2.5 m³ per tonne. The curve flattens at high P, so most gas comes out only when the pressure is brought low.

ℹ️ Note
A hydrate crystal packs a very large volume of gas, of the order of a hundred and fifty times its own volume at standard conditions, which is why hydrate deposits are counted as enormous resources; the recoverable part is far smaller and uncertain.

5. Peak oil: Hubbert's logistic theory

Peak oil is the idea that the production of a finite resource rises, reaches a maximum and then declines. Hubbert modelled the cumulative production Q(t) of a region as a logistic curve and the production rate P = dQ/dt as its derivative, a symmetric bell-shaped curve. With Q∞ the ultimately recoverable resource and k the growth constant:

  • Q(t) = Q∞ / [1 + e^(−k(t − t_m))], with t_m the time of the peak.
  • P = dQ/dt = k Q (1 − Q/Q∞): production is proportional to the amount already produced and to the fraction still left.
  • The peak occurs at Q = Q∞/2, and P_max = k Q∞/4. The curve is symmetric about t_m.
  • Time to go from 10 per cent to 90 per cent of Q∞: Δt = 2 ln 9/k ≈ 4.39/k.

Hubbert linearisation. Dividing P = kQ(1 − Q/Q∞) by Q gives P/Q = k − (k/Q∞) Q, a straight line when P/Q is plotted against Q. The intercept on the P/Q axis is k and the intercept on the Q axis is Q∞. Worked example: P/Q = 0.04 yr⁻¹ at Q = 500 Gb and 0.02 yr⁻¹ at Q = 1000 Gb. Slope = −0.02/500 = −4 × 10⁻⁵ Gb⁻¹yr⁻¹; k = 0.04 + 500 × 4 × 10⁻⁵ = 0.06 yr⁻¹; Q∞ = k/|slope| = 0.06/(4 × 10⁻⁵) = 1500 Gb. Another check: with Q∞ = 2000 Gb, k = 0.05 yr⁻¹, P_max = 0.05 × 2000/4 = 25 Gb/yr.

⚠️ Exam trap
Peak production is at half of the ultimate recovery, not at the end of the resource. The model is symmetric and assumes a fixed Q∞; it is applied to a field or region, and it fails when price, technology, demand shocks or reclassified unconventional resources change Q∞ or distort the curve.

Key takeaways

  • R/P = reserves ÷ annual production; with exponential growth g the lifetime is T = (1/g) ln(1 + g R/P), always below R/P.
  • Resources include undiscovered and uneconomic quantities; reserves are discovered, recoverable and commercial now. Proved, probable and possible correspond to P90, P50 and P10 and are cumulative.
  • Volumetric reserve = A h φ (1 − S_w)/B_o × RF; reserves are always below oil in place.
  • CBM is adsorbed methane freed by dewatering (Langmuir V = V_L P/(P_L + P)); shale gas needs fracturing; hydrates are methane in water cages, stable only at low T and high P.
  • Hubbert: P = kQ(1 − Q/Q∞), peak at Q∞/2 with P_max = kQ∞/4; the plot of P/Q against Q is a straight line with intercepts k and Q∞.
  • LHV = HHV − (9h + w) h_fg; always say which basis a calorific value uses.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The reserves-to-production (R/P) ratio of a fuel is best described as which of the following?

    1. The number of years the remaining proved reserves would last if production stayed at the current annual rate
    2. The number of years until the last barrel is certain to be produced, after allowing for expected growth in the demand for the fuel
    3. The fraction of the total resource that has already been produced up to the present year
    4. The ratio of the proved reserve to the total undiscovered resource of the region
    Show answer

    Answer: A — The number of years the remaining proved reserves would last if production stayed at the current annual rate

    R/P is reserves divided by this year's production, in years, at constant production. The tempting alternative, a lifetime that allows for demand growth, is exactly what R/P does not give; growth shortens the life.

  2. A country's reserves have R/P = 40 years. Which statement about this number is most defensible?

    1. It is a firm lower bound on the time to exhaustion, because new discoveries and technology can only ever add to reserves
    2. It is a snapshot index: production growth would shorten the life while discoveries and price-driven reclassification can lengthen it
    3. It is an exact forecast of exhaustion, because reserves and production are both measured accurately every year
    4. It equals the time from now at which annual production of the fuel will reach its peak and begin to fall
    Show answer

    Answer: B — It is a snapshot index: production growth would shorten the life while discoveries and price-driven reclassification can lengthen it

    R/P freezes both production and reserves at today's values. Growth in demand shortens the life and additions to reserves extend it, so it is neither a lower bound nor exact, and it says nothing about when the peak occurs.

  3. Which statement correctly relates fossil resources to reserves?

    1. Resources are the part of the reserves that is economic at the present price and can be recovered with current technology
    2. Reserves and resources are the same quantity, quoted in different units by different agencies and countries
    3. Reserves are the discovered, technically recoverable and commercially viable part of the resource, so they are a subset of it
    4. Reserves include all undiscovered accumulations that geologists expect to exist anywhere in the sedimentary basins
    Show answer

    Answer: C — Reserves are the discovered, technically recoverable and commercially viable part of the resource, so they are a subset of it

    Reserves require discovery, technical recoverability and commercial viability, so they are a subset of resources. Undiscovered accumulations are resources, not reserves, which is why the alternative that counts them is wrong.

  4. In the convention where the proved, proved-plus-probable and proved-plus-probable-plus-possible volumes are the quantities with at least 90, 50 and 10 per cent probability of being recovered or exceeded, a field has P90 = 40, P50 = 60 and P10 = 90 million barrels. What is the proved-plus-probable (2P) reserve?

    1. 40 million barrels
    2. 100 million barrels
    3. 90 million barrels
    4. 60 million barrels
    Show answer

    Answer: D — 60 million barrels

    The classes are cumulative and already include the lower ones, so 2P is the P50 value, 60 million barrels. Adding 40 + 60 = 100 double counts, and 90 is the 3P (P10) figure.

  5. A coal block has area 4 km², a seam thickness of 5 m and coal density 1.4 t/m³. Mining is expected to recover 60 per cent of the coal in place. What is the recoverable coal reserve?

    1. 28.0 million tonnes
    2. 11.2 million tonnes
    3. 16.8 million tonnes
    4. 12.0 million tonnes
    Show answer

    Answer: C — 16.8 million tonnes

    In place = 4 × 10⁶ m² × 5 m × 1.4 t/m³ = 2.8 × 10⁷ t = 28 Mt. Recoverable = 0.60 × 28 = 16.8 Mt. Quoting 28 Mt confuses in-place with reserve.

  6. In a water-saturated coal seam, what is the physical mechanism by which coal bed methane is made to flow to the well?

    1. Fracturing the overlying sandstone cap so that free gas trapped beneath it can rise into the well bore
    2. Burning part of the seam underground to crack the coal and free the methane
    3. Removing water to lower the seam pressure so that adsorbed methane desorbs from the coal surface
    4. Warming the seam well above the temperature at which methane hydrate is stable, so that hydrate melts
    Show answer

    Answer: C — Removing water to lower the seam pressure so that adsorbed methane desorbs from the coal surface

    CBM is held adsorbed on coal; lowering the pressure by pumping out the water shifts the adsorption equilibrium and the gas desorbs. Hydrate warming and underground burning describe different resources or processes.

  7. Why do shale gas wells typically need hydraulic fracturing and horizontal drilling to be productive?

    1. The shale permeability is extremely low, so gas will not flow to a conventional well; fractures and a long contact length create flow paths
    2. The gas in shale is held as a solid that has to be melted by the warm injected water before it can flow to the well
    3. Shale is a high-permeability rock whose gas would escape far too quickly unless its pore throats were first sealed off by drilling mud
    4. Fracturing is needed only to reduce the water saturation of the rock, as is done by dewatering in a coal seam
    Show answer

    Answer: A — The shale permeability is extremely low, so gas will not flow to a conventional well; fractures and a long contact length create flow paths

    Shale is a source rock of very low permeability, so the well must be connected to a large rock surface by fractures along a long horizontal section. The gas is not solid (that is hydrate), and shale is not high-permeability.

  8. Gas storage on coal follows V = V_L P/(P_L + P) with V_L = 10 m³ per tonne and P_L = 2 MPa. The seam pressure is lowered from 6 MPa to 2 MPa by dewatering. How much gas is released per tonne of coal?

    1. 2.5 m³
    2. 5.0 m³
    3. 7.5 m³
    4. 4.0 m³
    Show answer

    Answer: A — 2.5 m³

    At 6 MPa V = 10 × 6/8 = 7.5 m³/t; at 2 MPa V = 10 × 2/4 = 5.0 m³/t; released = 7.5 − 5.0 = 2.5 m³/t. The values 7.5 and 5.0 are the stored amounts, not the amount released.

  9. Methane hydrate deposits are found under permafrost and in deep ocean-floor sediment because there the conditions are:

    1. low pressure and high temperature, which together make the water cages rigid and gas-tight
    2. high pressure and low temperature, at which the gas-filled water cages are stable
    3. high temperature and high pressure, because heat drives methane into the water
    4. low temperature and low pressure, because hydrate forms by evaporation of seawater
    Show answer

    Answer: B — high pressure and low temperature, at which the gas-filled water cages are stable

    Hydrates are stable only when the temperature is low and the pressure high; warming or depressurising dissociates them into water and methane.

  10. A fuel has HHV = 45 MJ/kg, hydrogen mass fraction 0.12 and no moisture. Take the latent heat of the water formed as 2.44 MJ per kg of water, and note that each kg of hydrogen burns to 9 kg of water. What is the LHV?

    1. 42.4 MJ/kg
    2. 47.6 MJ/kg
    3. 44.7 MJ/kg
    4. 39.9 MJ/kg
    Show answer

    Answer: A — 42.4 MJ/kg

    Water per kg of fuel = 9 × 0.12 = 1.08 kg; latent heat = 1.08 × 2.44 = 2.64 MJ. LHV = 45 − 2.64 = 42.4 MJ/kg. Adding the latent heat gives 47.6 (wrong direction), and using 0.12 × 2.44 forgets the factor 9.

  11. In Hubbert's logistic model, a plot of P/Q against Q (P = annual production, Q = cumulative production) is a straight line. Its intercept on the Q axis gives:

    1. the growth constant k, in per year, of the logistic curve
    2. the peak annual production rate P_max, in Gb per year
    3. the calendar year in which the production rate reaches its maximum
    4. the ultimately recoverable resource Q∞
    Show answer

    Answer: D — the ultimately recoverable resource Q∞

    P/Q = k − (k/Q∞) Q: the Q-axis intercept is Q∞, while the P/Q-axis intercept is k. Peak rate and peak year are not read directly from the intercepts.

  12. For a region whose production follows the logistic Hubbert model with ultimate recovery Q∞ and constant k, which statements are correct?

    1. Production is greatest when half of Q∞ has been produced
    2. The production curve is symmetric in time about the peak
    3. The peak production rate equals k Q∞/2
    4. Doubling Q∞ at fixed k moves the peak rate to twice its earlier value
    Show answer

    Answer: A — Production is greatest when half of Q∞ has been produced; B — The production curve is symmetric in time about the peak; D — Doubling Q∞ at fixed k moves the peak rate to twice its earlier value

    P = kQ(1 − Q/Q∞) is largest at Q = Q∞/2, and the logistic derivative is symmetric. P_max = kQ∞/4, not kQ∞/2, so the statement with k Q∞/2 is wrong; since P_max is proportional to Q∞, doubling Q∞ doubles it.

  13. Which statements about unconventional fossil resources are correct?

    1. Coal bed methane is held mainly adsorbed on the coal surface and is released as seam pressure is lowered
    2. Shale gas is produced from high-permeability sandstone reservoirs without any stimulation
    3. A gas hydrate is a crystalline solid in which water molecules form cages enclosing gas molecules
    4. Methane hydrate becomes more stable as the temperature rises while the pressure is held fixed
    Show answer

    Answer: A — Coal bed methane is held mainly adsorbed on the coal surface and is released as seam pressure is lowered; C — A gas hydrate is a crystalline solid in which water molecules form cages enclosing gas molecules

    CBM is adsorbed gas freed by depressurisation, and hydrates are water-cage solids. Shale gas is from very low-permeability rock and needs stimulation, and heating destabilises hydrate.

  14. Which of the following are genuine limitations of the R/P ratio as a measure of how long a fuel will last?

    1. It assumes production stays at the present annual rate
    2. It ignores that price and technology can convert resources into reserves
    3. It cannot be computed unless the undiscovered resource is first known exactly
    4. It does not depend on how reserves are defined, so it is the same for 1P and 2P
    Show answer

    Answer: A — It assumes production stays at the present annual rate; B — It ignores that price and technology can convert resources into reserves

    R/P freezes production and reserves at today's values. It uses only the reported reserve, not the undiscovered resource, and a larger class such as 2P gives a larger R, so the statements that ignore the reserve definition and demand an exact undiscovered resource are false.

  15. Proved reserves of a fuel are 50 times the present annual production. Production is expected to grow exponentially at 2 per cent per year from its present value, and the reserves stay fixed. After how many years are the reserves exhausted? Give the answer in years to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 34.7

    R = P₀(e^(gT) − 1)/g gives T = ln(1 + gR/P₀)/g = ln(1 + 0.02 × 50)/0.02 = ln 2/0.02 = 34.66, about 34.7 years. Python check: log(2)/0.02 = 34.657.

  16. An oil field has area 10 km², net pay 20 m, porosity 0.20, water saturation 0.25 and formation volume factor 1.2. The recovery factor is 0.30. Find the recoverable reserve in million cubic metres of stock-tank oil, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 7.5

    OIIP = A h φ (1 − S_w)/B_o = 10⁷ × 20 × 0.20 × 0.75/1.2 = 2.5 × 10⁷ m³. Reserve = 0.30 × 2.5 × 10⁷ = 7.5 × 10⁶ m³ = 7.5 million m³.

  17. A region's production follows the logistic Hubbert model. Its P/Q values are 0.040 per year when the cumulative production Q is 500 Gb and 0.020 per year when Q is 1000 Gb. Using the straight-line (linearised) form, find the ultimately recoverable resource in Gb (an integer).

    Numerical answer — type the value.

    Show answer

    Answer: 1500

    Slope = (0.020 − 0.040)/(1000 − 500) = −4 × 10⁻⁵; intercept k = 0.040 + 500 × 4 × 10⁻⁵ = 0.060. The line hits zero at Q∞ = 0.060/(4 × 10⁻⁵) = 1500 Gb. Python check gives 1500.0.

  18. Cumulative production of a region follows a logistic curve with growth constant k = 0.05 per year. How many years does it take for the cumulative production to rise from 10 per cent to 90 per cent of the ultimately recoverable resource? Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 87.9

    Q/Q∞ = 1/(1 + e^(−k(t − t_m))): 0.9 gives k(t − t_m) = ln 9 and 0.1 gives −ln 9. So Δt = 2 ln 9/k = 2 × 2.1972/0.05 = 87.9 years.