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Equations: Reservoir conditions and development tasks

The equations of the “Reservoir conditions and development tasks” lessons in “Reservoir engineering”, with their conditions and units, and what each symbol means in them.

Where the oil is left after waterflooding
NpN=ED EAEV,NorN=EAEV (1−ED),NuN=1−EAEV\frac{N_p}{N} = E_D\,E_AE_V,\qquad \frac{N_{or}}{N} = E_AE_V\,(1 - E_D),\qquad \frac{N_u}{N} = 1 - E_AE_V

fractions; volumes in the same units

Holds when: The displacement and sweep efficiencies are those of the lesson Why water is injected: pressure maintenance and oil displacement. Their product is the recovery factor. The pressure is maintained, the oil formation volume factor hardly changes and there is no free gas. So the shares of the oil in place equal the shares of the oil-saturated pore volume. In the bypassed part the oil saturation is the initial one, and in the swept part it is the residual one. In reality, partly swept zones with an in-between saturation lie between them. So the split is a simplified picture, and it exaggerates the contrast between the parts.

Reservoir engineering: tasks across the life of a field →
Exponential decline
q=qi e−Dt,ΔNp=365 q1−q2Dq = q_i\,e^{-Dt},\qquad \Delta N_p = 365\,\frac{q_1 - q_2}{D}

rate in m³/d, time in years, production in m³

Holds when: The rate falls by the same fraction in equal intervals of time. The operating mode does not change: no new wells, no moving of wells to other layers (recompletion), no changes of injection and no limits set by the facilities. The factor 365 converts the product “m³/d × year” into cubic metres.

Reservoir engineering: tasks across the life of a field →
Hyperbolic decline
q=qi(1+bDit)1/b,Np=365 qi(1−b) Di[1−(qqi)1−b]q = \frac{q_i}{(1 + bD_it)^{1/b}},\qquad N_p = 365\,\frac{q_i}{(1 - b)\,D_i}\left[1 - \left(\frac{q}{q_i}\right)^{1-b}\right]

as in the equation for exponential decline

Holds when: 0<b<10 < b < 1. As b→0b \to 0 the formulas turn into the exponential ones, and at b=1b = 1 the production is 365 (qi/Di)ln⁡(qi/q)365\,(q_i/D_i)\ln(q_i/q). We fit the exponent bb to a long history. Without a long history, the choice of bb is only an assumption.

Reservoir engineering: tasks across the life of a field →
Absolute pressure from a gauge reading
pabs=pgauge+patmp_{\text{abs}} = p_{\text{gauge}} + p_{\text{atm}}

any pressure units, the same for all three quantities

Holds when: For the conversion we usually take the standard atmospheric pressure. The actual pressure depends on altitude and weather. At 1000 m above sea level it is about 0.09 MPa. Pressure differences do not depend on the choice of zero. Examples are drawdown (the pressure drop that pushes fluid into the well), the pressure drop across a choke (a restriction that controls the flow) and a pressure gradient.

Systems of units: SI, oilfield and metric oilfield →
Temperature conversion
T [K]=t [∘C]+273.15,t [∘F]=1.8 t [∘C]+32,T [∘R]=t [∘F]+459.67=1.8 T [K]T\,[\text{K}] = t\,[{}^\circ\text{C}] + 273.15,\qquad t\,[{}^\circ\text{F}] = 1.8\,t\,[{}^\circ\text{C}] + 32,\qquad T\,[{}^\circ\text{R}] = t\,[{}^\circ\text{F}] + 459.67 = 1.8\,T\,[\text{K}]

given in square brackets

Holds when: We convert temperature differences without the shift (the added constant): 1 K = 1 °C, 1 °R = 1 °F, 1 K = 1.8 °R. °R in oilfield formulas is the degree Rankine. In Russian physics reference books it is written °Ra, because there °R is the obsolete degree Réaumur. In older Russian-language literature the name Rankine is also transliterated as “Renkin”. A geothermal gradient (how fast the temperature rises with depth) of 1 °F per 100 feet equals 1.823 °C per 100 m.

Systems of units: SI, oilfield and metric oilfield →
Conversion of a gas volume to other reference conditions
V2=V1 T2T1 p1p2V_2 = V_1\,\frac{T_2}{T_1}\,\frac{p_1}{p_2}

temperatures absolute, pressures absolute and in the same units

Holds when: The amount of gas is the same. At reference conditions we treat the gas as ideal: the gas deviation factor (Z, which shows how far a real gas departs from an ideal one) is close to one, and its small difference almost cancels. The moisture content is the same: under GOST 2939-63 the gas is dry, and water vapour is not included in the volume.

Systems of units: SI, oilfield and metric oilfield →
Solution gas–oil ratio and producing gas–oil ratio: m³/m³ and scf/STB
R [scf/STB]=5.6146 R [m3/m3]orR [scf/STB]=5.529 R [m3/m3]R\,[\text{scf/STB}] = 5.6146\,R\,[\text{m}^3/\text{m}^3]\qquad\text{or}\qquad R\,[\text{scf/STB}] = 5.529\,R\,[\text{m}^3/\text{m}^3]

gas — standard cubic metres or cubic feet, oil — cubic metres or barrels at standard conditions

Holds when: We use 5.6146 when the gas on both sides is referred to the same conditions. It is simply the number of cubic feet in a barrel. We use 5.529 when the cubic metres of gas are referred to 20 °C and 101.325 kPa, and the cubic feet to 60 °F and 14.696 psia. The volume of oil between 20 °C and 60 °F changes by about 0.35%. We usually neglect this.

Systems of units: SI, oilfield and metric oilfield →
Specific gravity of oil from its density at 20 °C
ρ60=ρ20[1+β (20−15.56)],γo=ρ60999.0\rho_{60} = \rho_{20}\left[1 + \beta\,(20 - 15.56)\right],\qquad \gamma_o = \frac{\rho_{60}}{999.0}

kg/m³; 999.0 kg/m³ is the density of water at 60 °F

Holds when: We assume linear expansion over a narrow temperature range. β≈8⋅10−4\beta \approx 8\cdot10^{-4} 1/°C is a typical value for medium oils, and the lessons of the library use it. For commercial transactions, the conversion is made with standard tables. The density is that of stock-tank oil at atmospheric pressure.

Systems of units: SI, oilfield and metric oilfield →
Barrels per tonne of oil
nbbl/t=6.2898ρ [t/m3]n_{\text{bbl/t}} = \frac{6.2898}{\rho\,[\text{t/m}^3]}

6.2898 is the number of barrels in a cubic metre

Holds when: The oil is stock-tank oil. The density and the volume are at the same temperature.

Systems of units: SI, oilfield and metric oilfield →
The darcy in SI units
1 D=(10−6 m3/s)⋅(10−3 Pa⋅s)⋅(10−2 m)(10−4 m2)⋅(101,325 Pa)=0.9869233⋅10−12 m21\ \text{D} = \frac{(10^{-6}\ \text{m}^3/\text{s})\cdot(10^{-3}\ \text{Pa·s})\cdot(10^{-2}\ \text{m})}{(10^{-4}\ \text{m}^2)\cdot(101{,}325\ \text{Pa})} = 0.9869233\cdot10^{-12}\ \text{m}^2

SI

Holds when: The definition of the darcy: 1 cm³/s of a liquid of viscosity 1 cP through a sample of cross-section 1 cm² and length 1 cm at a pressure difference of 1 atm. The atmosphere here is the standard atmosphere, exactly 101,325 Pa. With the technical atmosphere (1 kgf/cm²) the number would be different.

Systems of units: SI, oilfield and metric oilfield →
Pressure in a water-bearing layer
pw(z)=patm+ρw g z+Cp_w(z) = p_{\text{atm}} + \rho_w\,g\,z + C

SI, absolute pressure

Holds when: The water is at rest and the column is continuous. The water density is constant. (If it changes with depth, we add up the pressure interval by interval.) CC is the same throughout one hydrodynamically connected system (a system in which pressure is passed on freely).

Reservoir pressure and temperature →
Pressure gradient in a continuous phase
 dp dz=ρ g\frac{\dd p}{\dd z} = \rho\,g

SI

Holds when: The fluid is at rest (static equilibrium) and forms a continuous phase between the measurement points. The density is constant with height — a good approximation for water and oil; for gas see below.

Reservoir pressure and temperature →
Excess of the pressure at the top of the accumulation over the water pressure
Δp=(ρw−ρo) g ho+(ρw−ρg) g hg\Delta p = (\rho_w - \rho_o)\,g\,h_o + (\rho_w - \rho_g)\,g\,h_g

SI

Holds when: The accumulation is in hydrostatic equilibrium and the phases are continuous. If there is no gas cap, hg=0h_g = 0; for a purely gas accumulation ho=0h_o = 0.

Reservoir pressure and temperature →
Bringing pressure to a datum
pdatum=pmeas+ρ g (zdatum−zmeas)p_{\text{datum}} = p_{\text{meas}} + \rho\,g\,(z_{\text{datum}} - z_{\text{meas}})

SI

Holds when: The reservoir between the two depths is hydrodynamically connected and the fluid is at rest. If a contact lies between the depths, we make the conversion section by section, each with its own density.

Reservoir pressure and temperature →
Temperature at depth from the geothermal gradient
T(z)=T0+GT zT(z) = T_0 + G_T\,z

°C and m (the gradient is often given in °C/100 m)

Holds when: We take the gradient as constant through the section. This is a rough approximation, because the gradient depends on the thermal conductivity of the rocks (how well they conduct heat) (advanced section). The temperature is the stabilised one, undisturbed by drilling and development.

Reservoir pressure and temperature →
Extrapolation of the bottomhole temperature to the stabilised temperature (Horner method)
Tbh(Δt)=Tres−m lg⁡tc+ΔtΔtT_{\text{bh}}(\Delta t) = T_{\text{res}} - m\,\lg\frac{t_c + \Delta t}{\Delta t}

time in the same units (hours); the logarithm is decimal (lg)

Holds when: This is an empirical method, by analogy with the pressure build-up curve. It describes the recovery of temperature around the wellbore with a line heat source model (heat released along a line, the wellbore). We know the duration of circulation only roughly. We need at least three measurements, preferably from different runs to the same depth.

Reservoir pressure and temperature →
Geothermal gradient from heat flow
GT=qλG_T = \frac{q}{\lambda}

SI

Holds when: Steady one-dimensional vertical heat flow. The layers are horizontal. Above and below salt domes, the heat flow shifts sideways and the picture is more complicated.

Reservoir pressure and temperature →
Pressure in a static column of gas
pbh=pwh exp⁡(M g HZˉ R Tˉ)p_{\text{bh}} = p_{\text{wh}}\,\exp\left(\frac{M\,g\,H}{\bar Z\,R\,\bar T}\right)

SI (g=9.81g = 9.81 m/s²)

Holds when: The gas is at rest and there is no liquid in the wellbore. We replace ZZ and TT by their averages over the wellbore. We find Zˉ\bar Z at the average pressure, so we repeat the calculation once or twice.

Reservoir pressure and temperature →

Symbols

SymbolMeaning and context
Δp\Delta p
Zˉ, Tˉ\bar Z,\ \bar T
 dp/ dz\dd p/\dd z
10−2 m; 10−4 m210^{-2}\ \text{m};\ 10^{-4}\ \text{m}^2
10−3 Pa⋅s10^{-3}\ \text{Pa·s}
10−6 m3/s10^{-6}\ \text{m}^3/\text{s}
101,325 Pa101{,}325\ \text{Pa}
bb
CC
DD
DiD_i
EAEVE_AE_V
EDE_D
gg
GTG_T
HH
ho, hgh_o,\ h_g
mm
MM
NN
nbbl/tn_{\text{bbl/t}}
NorN_{or}
NpN_p
NuN_u
ΔNp\Delta N_p
p1, p2p_1,\ p_2
pabsp_{\text{abs}}
patmp_{\text{atm}}
pbh, pwhp_{\text{bh}},\ p_{\text{wh}}
pdatump_{\text{datum}}
pgaugep_{\text{gauge}}
pmeasp_{\text{meas}}
pwp_w
qq
q, qiq,\ q_i
RR
tt
TT
T0T_0
T1, T2T_1,\ T_2
Tbh(Δt)T_{\text{bh}}(\Delta t)
tct_c
TresT_{\text{res}}
  • stabilised reservoir temperature — the value at lg⁡(…)=0\lg(\ldots) = 0, that is, as Δt→∞\Delta t \to \infty, °CReservoir pressure and temperature
T(z)T(z)
V1, V2V_1,\ V_2
zz
zdatum, zmeasz_{\text{datum}},\ z_{\text{meas}}
β\beta
γo\gamma_o
λ\lambda
ρ\rho
ρ20, ρ60\rho_{20},\ \rho_{60}
ρw\rho_w
ρw, ρo, ρg\rho_w,\ \rho_o,\ \rho_g