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Equations: Artificial lift

The equations of the “Artificial lift” lessons in “Oil and gas production”, with their conditions and units, and what each symbol means in them.

Depth of the static fluid level
Hst=H−pˉR−pannρgH_{\text{st}} = H - \frac{\bar p_R - p_{\text{ann}}}{\rho g}

SI; pressures are gauge (above atmospheric)

Holds when: The well is shut in and the bottomhole pressure has equalised with the reservoir pressure. Below the level there is liquid of constant density with no gas. The weight of the gas column above the level is small. Depths are true vertical depths. In a deviated well, a level measured along the hole is converted using the directional survey (the record of the well's path).

Artificial lift: when it is needed and how to choose the method →
Pressure rise produced by the pump
Δppump=pout−pin,pout=pwh+ρgHpump+Δpfr,pin=pwf−ρg (H−Hpump)\begin{aligned} \Delta p_{\text{pump}} &= p_{\text{out}} - p_{\text{in}},\\ p_{\text{out}} &= p_{wh} + \rho g H_{\text{pump}} + \Delta p_{\text{fr}},\\ p_{\text{in}} &= p_{wf} - \rho g\,(H - H_{\text{pump}}) \end{aligned}

SI (any consistent system of units)

Holds when: The liquid contains no free gas and has the same density in the tubing and below the pump. If there is gas in the tubing, the column is lighter, and its weight is calculated with multiphase flow correlations. Depths are true vertical depths. Friction is calculated along the length of the tubing. The formula also holds for a pump set below the perforated interval. Then H−Hpump<0H - H_{\text{pump}} < 0, and the intake pressure is higher than the bottomhole pressure.

Artificial lift: when it is needed and how to choose the method →
Hydraulic power and power consumption
N=q Δppump,Nel=NηN = q\,\Delta p_{\text{pump}},\qquad N_{\text{el}} = \frac{N}{\eta}

SI

Holds when: The flow is steady. The flow rate is the volume of liquid at intake conditions. Free gas, if it passes through the pump, is compressed, and for it the formula gives only an upper bound (a value that is too high). The system efficiency includes the losses in the motor, the transmission (rods, cable), the pump and the tubing.

Artificial lift: when it is needed and how to choose the method →
Free gas fraction at the pump intake
φg=qo (R−Rs) Bgqo (R−Rs) Bg+qoBo+qwBw\varphi_{\text{g}} = \frac{q_o\,(R - R_s)\,B_g}{q_o\,(R - R_s)\,B_g + q_o B_o + q_w B_w}

any consistent units; pressures in BgB_g are absolute, temperatures are in kelvins

Holds when: All the gas that is not dissolved in the oil at the intake pressure goes to the intake together with the liquid. If part of it is vented into the annulus, put only the remaining part into the formula. We ignore the solubility of gas in water.

Artificial lift: when it is needed and how to choose the method →
Bottomhole pressure from the dynamic fluid level
pwf≈pann+ρliq g (H−Hdyn)p_{wf} \approx p_{\text{ann}} + \rho_{\text{liq}}\,g\,(H - H_{\text{dyn}})

SI

Holds when: The weight of the gas column above the level is small. At a high annulus pressure it is already a few per cent, and then it is added. The liquid in the annulus contains no gas and has a known density.

Artificial lift: when it is needed and how to choose the method →
The limiting number of strokes set by the free fall of the rods
nlim=30π2gS≈42.3Sn_{\text{lim}} = \frac{30}{\pi}\sqrt{\frac{2g}{S}} \approx \frac{42.3}{\sqrt{S}}

SS in metres, gg in m/s², nn in strokes per minute

Holds when: the polished rod moves in a harmonic (sine-wave) motion: the greatest acceleration is ω2S/2\omega^2 S/2, where ω=πn/30\omega = \pi n/30 is the angular velocity of the crank in radians per second the rods fall with the acceleration gg, with no liquid and no friction. In liquid they fall more slowly, because buoyancy reduces their weight and friction resists the motion. So the formula gives an upper limit, not a working value.

Sucker-rod pumping →
Theoretical displacement of a sucker rod pump
qth=1440 Apl S n,Apl=πD24q_{\text{th}} = 1440\,A_{\text{pl}}\,S\,n, \qquad A_{\text{pl}} = \frac{\pi D^2}{4}

metres, square metres, strokes per minute; the result is in m³/d

Holds when: the plunger stroke is equal to the stroke of the polished rod, the barrel fills completely with liquid, and there are no leaks; the volumetric efficiency allows for the departures from this in the practice of the CIS countries, the stroke of the polished rod is put into the formula. In the API method the volume is calculated from the plunger stroke, and the loss of stroke is not part of the volumetric efficiency. When comparing coefficients from different sources, this must be checked.

Sucker-rod pumping →
Pump volumetric efficiency and its components
αp=qqth=αs αf αleak αsh\alpha_{\text{p}} = \frac{q}{q_{\text{th}}} = \alpha_{\text{s}}\,\alpha_{\text{f}}\,\alpha_{\text{leak}}\,\alpha_{\text{sh}}

dimensionless fractions

Holds when: the factors are treated as independent; this is a convenient approximation for finding the main cause of the losses, not an exact law the delivery was measured in steady operation, not just after a start-up

Sucker-rod pumping →
Static loads on the polished rod
P↑=Prod′+Pliq,P↓=Prod′,Prod′=mrod g L kb,Pliq=(pout−pin) Apl\begin{aligned} P_{\uparrow} &= P'_{\text{rod}} + P_{\text{liq}}, \qquad P_{\downarrow} = P'_{\text{rod}},\\ P'_{\text{rod}} &= m_{\text{rod}}\,g\,L\,k_{\text{b}}, \qquad P_{\text{liq}} = (p_{\text{out}} - p_{\text{in}})\,A_{\text{pl}} \end{aligned}

SI; both pressures gauge or both absolute, because only their difference enters the formula

Holds when: very slow motion: no inertia, friction or vibration the well is vertical; in a deviated well the weight of the rods is calculated from the vertical setting depth, not from the length of the string, and the friction of the rods against the tubing there increases the load on the upstroke and reduces it on the downstroke the rod string has one diameter; for a tapered string the weights of the tapers are added the fluid load is calculated on the full plunger area precisely because the weight of the rods is taken in liquid. If the weight of the rods is taken in air, the discharge pressure is multiplied by the plunger area less the cross-section of the rods, and the intake pressure still by the full area. The result is practically the same (the derivation is in the advanced section) the force with which the wellhead pressure pushes the rod out of the stuffing box is not included; it is usually a fraction of a kilonewton

Sucker-rod pumping →
Acceleration factor and loads with inertia
Ka=ω2S2g=S n21790,Pmax⁡≈Prod′+Pliq+Ka Prod,Pmin⁡≈Prod′−Ka Prod\begin{aligned} K_{\text{a}} &= \frac{\omega^2 S}{2g} = \frac{S\,n^2}{1790},\\ P_{\max} &\approx P'_{\text{rod}} + P_{\text{liq}} + K_{\text{a}}\,P_{\text{rod}}, \qquad P_{\min} \approx P'_{\text{rod}} - K_{\text{a}}\,P_{\text{rod}} \end{aligned}

SS in metres, nn in strokes per minute; the loads in any identical units

Holds when: harmonic motion of the polished rod; in an ordinary pumping unit the acceleration at the bottom is larger than the harmonic one, and at the top it is smaller the rods are rigid and move as one piece; elastic vibrations, the inertia of the liquid and friction are not included the estimate is good for small depths and small numbers of strokes. For design, engineers use methods based on solving the wave equation for the rod string (for example API RP 11L, now designated TR 11L)

Sucker-rod pumping →
Stretch of the rods and tubing, and the plunger stroke
λrod=Pliq LE Arod,λtbg=Pliq LE Atbg,Spl=S−λrod−λtbg\lambda_{\text{rod}} = \frac{P_{\text{liq}}\,L}{E\,A_{\text{rod}}}, \qquad \lambda_{\text{tbg}} = \frac{P_{\text{liq}}\,L}{E\,A_{\text{tbg}}}, \qquad S_{\text{pl}} = S - \lambda_{\text{rod}} - \lambda_{\text{tbg}}

SI

Holds when: the rod string has one diameter; for a tapered string the stretches of the tapers are added: λrod=(Pliq/E)∑Li/Ai\lambda_{\text{rod}} = (P_{\text{liq}}/E)\sum L_i/A_i the tubing is not anchored; if the bottom of the tubing is fixed with an anchor, λtbg=0\lambda_{\text{tbg}} = 0 statics: at large depths and numbers of strokes the inertia of the rods adds to the plunger stroke (advanced section) the rods are steel; for fibreglass rods the elastic modulus is several times smaller, and the stretch is correspondingly larger

Sucker-rod pumping →
The depth of the level from the echometer record
Hdyn=v t2,v=2 Nc ltbgtcH_{\text{dyn}} = \frac{v\,t}{2}, \qquad v = \frac{2\,N_{\text{c}}\,l_{\text{tbg}}}{t_{\text{c}}}

metres, seconds, metres per second

Holds when: the speed of sound is the same over the whole height of the gas column. In fact it depends on the composition, the pressure and the temperature of the gas, and it is found from the upper part of the wellbore if the echoes from the collars cannot be read, the speed is taken from the composition and pressure of the gas, or from the echo of a reference marker (a reflector fixed on the tubing at a known depth) the depth is measured along the wellbore; in a deviated well it is converted to vertical depth with the directional survey

Sucker-rod pumping →
The clearance volume: the delay of the intake and the condition for gas lock
xopen=lcl(poutpin−1),gas lock when1+Spllcl≤poutpinx_{\text{open}} = l_{\text{cl}}\left(\frac{p_{\text{out}}}{p_{\text{in}}} - 1\right), \qquad \text{gas lock when}\quad 1 + \frac{S_{\text{pl}}}{l_{\text{cl}}} \le \frac{p_{\text{out}}}{p_{\text{in}}}

any consistent units; the pressures are absolute

Holds when: the clearance volume is completely filled with gas. This is the worst case: liquid that stays in it reduces the volume of gas and the delay an ideal gas, with isothermal (constant-temperature) compression and expansion. When the gas is compressed fast, it heats up and the pressure grows faster, so the estimate is on the safe side the valves are tight, and the pressure losses in them are not included

Sucker-rod pumping →
The force balance of the rod string with the plunger
P=Prod+pout (Apl−Arod)−pbar AplP = P_{\text{rod}} + p_{\text{out}}\,(A_{\text{pl}} - A_{\text{rod}}) - p_{\text{bar}}\,A_{\text{pl}}

SI; gauge pressures

Holds when: statics; rods of one cross-section; the pressure of the liquid on the side surface of the rods gives no vertical force the pressure losses in the valves are not included

Sucker-rod pumping →
The counterbalance effect and the power at the polished rod
C=Prod′+Pliq2,Nrod=Pliq Spl n60C = P'_{\text{rod}} + \frac{P_{\text{liq}}}{2}, \qquad N_{\text{rod}} = \frac{P_{\text{liq}}\,S_{\text{pl}}\,n}{60}

SI; nn in strokes per minute

Holds when: an ideal static dynamometer card, with no inertia, friction or gas in the barrel. The real power is found from the area of the recorded dynamometer card in practice the balance is checked by comparing the largest motor current on the upstroke with that on the downstroke

Sucker-rod pumping →
An estimate of the added plunger stroke from inertia
λov≈ρsteel g L2E Ka,Spl=S−λrod−λtbg+λov\lambda_{\text{ov}} \approx \frac{\rho_{\text{steel}}\,g\,L^2}{E}\,K_{\text{a}}, \qquad S_{\text{pl}} = S - \lambda_{\text{rod}} - \lambda_{\text{tbg}} + \lambda_{\text{ov}}

SI

Holds when: a steel string of constant cross-section; harmonic motion of the polished rod the acceleration is the same along the whole length of the string, so there are no wave effects. This is an estimate of the order of magnitude, not a design formula a bar of length LL under its own weight stretches by ρsteelgL2/(2E)\rho_{\text{steel}} g L^2/(2E). Inertia adds the fraction KaK_{\text{a}} of this stretch at the bottom of the stroke and takes away the same amount at the top, so the total change is doubled

Sucker-rod pumping →
The pressure rise, the power and the efficiency of a stage
Δpstg=ρgh,Nstg=ρg qin h,η=NstgNshaft,stg\Delta p_{\text{stg}} = \rho g h, \qquad N_{\text{stg}} = \rho g\,q_{\text{in}}\,h, \qquad \eta = \frac{N_{\text{stg}}}{N_{\text{shaft,stg}}}

SI

Holds when: a low-viscosity liquid (its viscosity is close to that of water) with no free gas. The manufacturer measures the curves on water, and viscosity and gas reduce the head and the efficiency the head and the efficiency are taken from the performance curve at the rate and the speed at which the stage works

Electric submersible pumps (ESP) →
The required head of the pump
Hreq=pout−pinρg=Hdyn+pwh−pann+ΔpfrρgH_{\text{req}} = \frac{p_{\text{out}} - p_{\text{in}}}{\rho g} = H_{\text{dyn}} + \frac{p_{wh} - p_{\text{ann}} + \Delta p_{\text{fr}}}{\rho g}

SI; the pressures are all gauge or all absolute

Holds when: the liquid has no free gas and the same density in the tubing and in the annulus below the level. Gas in the tubing makes the column lighter. Then the discharge pressure is found from multiphase flow, and the required head is smaller the depths are vertical, and the friction is found from the length of the tubing: Δpfr=λ Ld ρv22\Delta p_{\text{fr}} = \lambda\,\dfrac{L}{d}\,\dfrac{\rho v^2}{2}, as in the lesson on nodal analysis the weight of the gas column in the annulus above the level is small

Electric submersible pumps (ESP) →
The number of stages and the shaft power of the pump
zstg=Hreqh(qin),Nshaft=ρg qin Hpumpη(qin)z_{\text{stg}} = \frac{H_{\text{req}}}{h(q_{\text{in}})}, \qquad N_{\text{shaft}} = \frac{\rho g\,q_{\text{in}}\,H_{\text{pump}}}{\eta(q_{\text{in}})}

SI

Holds when: all stages are identical, the liquid has a low viscosity, and the share of gas is small. We allow for viscosity and gas with corrections or with extra stages to the power of the pump we add what the gas separator and the protector use, usually a few per cent

Electric submersible pumps (ESP) →
The affinity laws of a centrifugal pump when the speed changes
q2q1=n2n1,h2h1=(n2n1)2,N2N1=(n2n1)3,η2=η1\frac{q_2}{q_1} = \frac{n_2}{n_1}, \qquad \frac{h_2}{h_1} = \left(\frac{n_2}{n_1}\right)^{2}, \qquad \frac{N_2}{N_1} = \left(\frac{n_2}{n_1}\right)^{3}, \qquad \eta_2 = \eta_1

the ratios are dimensionless; the quantities with the subscripts 1 and 2 are in the same units

Holds when: one and the same stage, a low-viscosity liquid with no gas. When the frequency is lowered a lot (below about 30–35 Hz, when the rated frequency is 50 Hz), the efficiency falls and the laws hold less well the laws link the corresponding points of two curves, not the operating points in the well. We find the operating point at a new frequency as the crossing of the new pump curve with the well curve

Electric submersible pumps (ESP) →
The share of free gas at the pump inlet, allowing for separation
φg=(1−Esep) qo (R−Rs) Bg(1−Esep) qo (R−Rs) Bg+qoBo+qwBw\varphi_{\text{g}} = \frac{(1 - E_{\text{sep}})\,q_o\,(R - R_s)\,B_g}{(1 - E_{\text{sep}})\,q_o\,(R - R_s)\,B_g + q_o B_o + q_w B_w}

any consistent units; the pressures in BgB_g are absolute, and the temperatures are in kelvins

Holds when: the gas that is not dissolved in the oil at the intake pressure comes to the intake together with the liquid; the solubility of gas in water is not included the share EsepE_{\text{sep}} is an estimate: it depends on the rate, the size of the bubbles, the viscosity and the design of the separator. At Esep=0E_{\text{sep}} = 0 the formula is the same as the formula of the lesson on choosing a method

Electric submersible pumps (ESP) →
The velocity of the liquid along the submersible motor
vcool=4 qinπ(Dcas2−Dmot2)v_{\text{cool}} = \frac{4\,q_{\text{in}}}{\pi\left(D_{\text{cas}}^2 - D_{\text{mot}}^2\right)}

SI

Holds when: all the liquid that goes to the intake passes the motor: the motor is below the intake, and the inflow comes from below free gas cools badly, so the velocity is estimated for the liquid

Electric submersible pumps (ESP) →
The motor current and the losses in the cable
I=Nshaft3 U ηmotcos⁡φ,ΔNcab=3I2Rcond,Rcond=ρel LcabS [1+α (t−20)]I = \frac{N_{\text{shaft}}}{\sqrt{3}\,U\,\eta_{\text{mot}}\cos\varphi}, \qquad \Delta N_{\text{cab}} = 3 I^2 R_{\text{cond}}, \qquad R_{\text{cond}} = \rho_{\text{el}}\,\frac{L_{\text{cab}}}{S}\,\bigl[1 + \alpha\,(t - 20)\bigr]

SI; the conductor cross-section in mm², the resistivity in Ω·mm²/m

Holds when: a balanced three-phase load; the motor works near its rated mode, and its efficiency and power factor are the nameplate values only the resistance of the conductors is included. The drop of the line voltage is estimated as ΔU≈3 IRcondcos⁡φ\Delta U \approx \sqrt{3}\,I R_{\text{cond}}\cos\varphi, and the inductive reactance of the cable increases it a little

Electric submersible pumps (ESP) →
The theoretical head of an impeller (the Euler equation)
Hth=u2 cu2−u1 cu1gH_{\text{th}} = \frac{u_2\,c_{u2} - u_1\,c_{u1}}{g}

SI

Holds when: an ideal impeller: the flow follows the vanes exactly, and there are no losses. The real head is lower because of the finite number of vanes, friction and leaks with an inlet without swirl (cu1=0c_{u1} = 0), what is left is Hth=u2cu2/gH_{\text{th}} = u_2 c_{u2}/g

Electric submersible pumps (ESP) →
The pressure balance of a hydraulic piston pump
ppf−psp=(pout−pin) PE+Fp_{\text{pf}} - p_{\text{sp}} = (p_{\text{out}} - p_{\text{in}})\,\frac{P}{E} + F

any consistent units of pressure

Holds when: work per cycle: the engine receives (ppf−psp)Ve(p_{\text{pf}} - p_{\text{sp}})V_{\text{e}}, the pump end gives (pout−pin)Vp(p_{\text{out}} - p_{\text{in}})V_{\text{p}}, and the difference is friction. P/E=Vp/VeP/E = V_{\text{p}}/V_{\text{e}} (the derivation is in the advanced section) leaks are included separately, in the flow rates; the inertia of the masses and the pressure oscillations during the stroke are neglected FF grows with the number of strokes and with the viscosity of the liquid; it is taken from the manufacturer's data

Hydraulic pumps: piston and jet →
The delivery, the number of strokes and the flow rate of the power fluid
q=αp 1440 Vp n,qpf=1440 Ve nαe=qαp αe (P/E)q = \alpha_{\text{p}}\,1440\,V_{\text{p}}\,n, \qquad q_{\text{pf}} = \frac{1440\,V_{\text{e}}\,n}{\alpha_{\text{e}}} = \frac{q}{\alpha_{\text{p}}\,\alpha_{\text{e}}\,(P/E)}

metric oilfield units; 1440 is the number of minutes in a day

Holds when: a double-acting unit: VpV_{\text{p}} and VeV_{\text{e}} are the volumes over a full cycle, up and down there is no free gas (gas that is not dissolved in the liquid) at the intake; with gas the volumetric efficiency falls, as with a sucker rod pump (Sucker-rod pumping)

Hydraulic pumps: piston and jet →
The dimensionless characteristic of a jet pump
H=B−S1+Kn−B,B=2R+2M2R21−R−(1+Ktd)(1+M)2R2,S=(1+Ks) M2R2(1−R)2\begin{aligned} H &= \frac{B - S}{1 + K_{\text{n}} - B},\\ B &= 2R + \frac{2M^2R^2}{1 - R} - (1 + K_{\text{td}})(1 + M)^2R^2,\\ S &= (1 + K_{\text{s}})\,\frac{M^2R^2}{(1 - R)^2} \end{aligned}

dimensionless quantities

Holds when: one-dimensional flow of an incompressible liquid; the densities of the power fluid and of the production are the same the mixing chamber is cylindrical, and the mixing in it is complete; the velocities at the intake and at the exit of the diffuser are small there is no free gas at the intake; if there is gas, its volume is added to the production rate, and the curve is only approximate the loss coefficients are found by tests; here the jet leaves the nozzle at the pressure at the inlet of the mixing chamber. In common field methods, the pressure drop in the nozzle is counted from the intake pressure. Then the denominator contains B−SB - S, and at the same MM the pressure ratio comes out lower (see the advanced section).

Hydraulic pumps: piston and jet →
The flow rate of the power fluid through the nozzle
qpf=An2 (ppf−pin)ρ (1+Kn−S)q_{\text{pf}} = A_{\text{n}}\sqrt{\frac{2\,(p_{\text{pf}} - p_{\text{in}})}{\rho\,(1 + K_{\text{n}} - S)}}

SI

Holds when: the velocity before the nozzle is small compared with the velocity of the jet SS is usually a few hundredths, and in estimates it is dropped. The flow rate then comes out slightly smaller

Hydraulic pumps: piston and jet →
The largest production rate of a jet pump without cavitation
qcav=As2 (pin−pv)ρ (1+Ks),As=At−Anq_{\text{cav}} = A_{\text{s}}\sqrt{\frac{2\,(p_{\text{in}} - p_{\text{v}})}{\rho\,(1 + K_{\text{s}})}}, \qquad A_{\text{s}} = A_{\text{t}} - A_{\text{n}}

SI; the pressures are absolute

Holds when: a one-dimensional estimate. In fact cavitation begins earlier, because in the boundary layer of the jet the pressure is lower than the average. So we leave a margin in the production rate there is no free gas. If the production contains gas, it expands at the inlet of the chamber and limits the production rate in a similar way

Hydraulic pumps: piston and jet →
The displacement per revolution and the theoretical delivery of a progressing cavity pump
V=4 e d ts,qth=1440 VnV = 4\,e\,d\,t_{\text{s}}, \qquad q_{\text{th}} = 1440\,V n

metres and cubic metres; nn in rpm, qthq_{\text{th}} in m³/d (1440 is the number of minutes in a day)

Holds when: a single-lobe rotor in a two-lobe stator is the most common geometry; for pumps with more lobes the displacement is found in a different way the cavities are completely filled with liquid, and there is no slippage; the volumetric efficiency allows for the departures from this

Progressing cavity pumps and hybrid systems →
The actual delivery and the volumetric efficiency of a progressing cavity pump
q=qth−s Δppump,ηv=qqth=1−s Δppump1440 Vnq = q_{\text{th}} - s\,\Delta p_{\text{pump}}, \qquad \eta_{\text{v}} = \frac{q}{q_{\text{th}}} = 1 - \frac{s\,\Delta p_{\text{pump}}}{1440\,V n}

m³/d, MPa, rpm

Holds when: the slippage depends linearly on the pressure drop. This is an approximation within the rated head. Near the limiting pressure drop, and in a worn pump, the slippage grows faster the coefficient ss is found on a test stand for a given liquid, temperature and interference. It falls as the viscosity and the interference grow there is no free gas in the liquid: gas takes up part of the cavities and reduces the delivery of liquid still more

Progressing cavity pumps and hybrid systems →
Laminar flow in the ring-shaped gap between the rods and the tubing
ΔpfrL=8 μ Qπ[r24−r14−(r22−r12)2ln⁡(r2/r1)]\frac{\Delta p_{\text{fr}}}{L} = \frac{8\,\mu\,Q}{\pi\left[r_2^4 - r_1^4 - \dfrac{(r_2^2 - r_1^2)^2}{\ln(r_2/r_1)}\right]}

SI

Holds when: laminar flow of a Newtonian liquid (one whose viscosity does not depend on the flow rate): the Reynolds number Re=ρv (d2−d1)/μRe = \rho v\,(d_2 - d_1)/\mu (the number that tells smooth flow from turbulent flow) is less than about 2000 the rods are coaxial with the tubing. For the axial flow of a Newtonian liquid their rotation has no effect. Rods pressed against the tubing wall reduce the losses, and couplings and rod guides increase them, so the formula gives an estimate the viscosity is constant along the length; in fact the oil cools towards the wellhead and thickens

Progressing cavity pumps and hybrid systems →
The torque and the power at the polished rod
M=Mh+Mfr+Mrod,Mh=V Δppump2π,N=M ω,ω=πn30M = M_{\text{h}} + M_{\text{fr}} + M_{\text{rod}}, \qquad M_{\text{h}} = \frac{V\,\Delta p_{\text{pump}}}{2\pi}, \qquad N = M\,\omega, \qquad \omega = \frac{\pi n}{30}

SI

Holds when: the hydraulic torque is found from the theoretical displacement: the pressure drop acts on the rotor whatever amount of liquid slips back the viscous torque of a smooth rod, coaxial with the tubing, in a Newtonian liquid: Mrod=4πμω r12r22r22−r12 LM_{\text{rod}} = \dfrac{4\pi\mu\omega\,r_1^2 r_2^2}{r_2^2 - r_1^2}\,L (the symbols are as in the formula for friction in the ring-shaped gap). Couplings, rod guides and the pressing of the rods against the tubing in a curved wellbore increase it the friction of the rods against the tubing in deviated sections is not included

Progressing cavity pumps and hybrid systems →
The twist of the rod string and the stored energy
θ=MLGJp,Jp=πdrod432,W=M θ2\theta = \frac{M L}{G J_p}, \qquad J_p = \frac{\pi d_{\text{rod}}^4}{32}, \qquad W = \frac{M\,\theta}{2}

SI

Holds when: solid rods of one diameter; couplings and upset ends (the thickened ends of the rods) reduce the twist a little the torsion is elastic: the shear stresses in the rods are below the yield strength (the stress at which steel starts to deform permanently)

Progressing cavity pumps and hybrid systems →

Symbols

SymbolMeaning and context
1+Spl/lcl1 + S_{\text{pl}}/l_{\text{cl}}
  • compression ratio of the pump: the ratio of the largest volume under the plunger to the smallest, —Sucker-rod pumping
AnA_{\text{n}}
AplA_{\text{pl}}
Apl, ArodA_{\text{pl}},\ A_{\text{rod}}
ArodA_{\text{rod}}
AsA_{\text{s}}
AtbgA_{\text{tbg}}
BB
  • pressure rise from the inlet of the mixing chamber to the discharge, in fractions of the velocity head of the jet ρvn2/2\rho v_{\text{n}}^2/2, —Hydraulic pumps: piston and jet
BgB_g
  • gas formation volume factor at intake conditions: Bg=pscp⋅TTsc ZB_g = \dfrac{p_{\text{sc}}}{p}\cdot\dfrac{T}{T_{\text{sc}}}\,Z, m³/m³Artificial lift: when it is needed and how to choose the method
  • formation volume factor of the gas at the intake conditions: Bg=pscp⋅TTsc ZB_g = \dfrac{p_{\text{sc}}}{p}\cdot\dfrac{T}{T_{\text{sc}}}\,Z (the pressures are absolute, the temperatures are in kelvins, and Z is the gas compressibility factor, a correction for a gas that is not ideal), m³/m³Electric submersible pumps (ESP)
Bo, BwB_o,\ B_w
CC
  • counterbalance effect: the force of the counterweights, converted into a force at the polished rod, NSucker-rod pumping
cu1, cu2c_{u1},\ c_{u2}
ΔNcab\Delta N_{\text{cab}}
dd
DD
DcasD_{\text{cas}}
DmotD_{\text{mot}}
drodd_{\text{rod}}
ee
EE
EsepE_{\text{sep}}
FF
Δpfr\Delta p_{\text{fr}}
gg
GG
hh
HH
h1, h2h_1,\ h_2
HdynH_{\text{dyn}}
HpumpH_{\text{pump}}
Hpump, HH_{\text{pump}},\ H
HreqH_{\text{req}}
HstH_{\text{st}}
HthH_{\text{th}}
H, HdynH,\ H_{\text{dyn}}
h(qin)h(q_{\text{in}})
II
JpJ_p
KaK_{\text{a}}
  • acceleration factor: the greatest acceleration of the polished rod as a fraction of gg, —Sucker-rod pumping
  • acceleration factor, Sn2/1790S n^2/1790, —Sucker-rod pumping
kbk_{\text{b}}
  • buoyancy factor, 1−ρ/ρsteel1 - \rho/\rho_{\text{steel}}; ρ\rho is the density of the liquid in the tubing, and ρsteel=7850\rho_{\text{steel}} = 7850 kg/m³ is the density of steel, —Sucker-rod pumping
Kn, SK_{\text{n}},\ S
  • loss coefficient in the nozzle, and the fall of the pressure of the production at the inlet of the chamber (from the characteristic), —Hydraulic pumps: piston and jet
Kn, Ks, KtdK_{\text{n}},\ K_{\text{s}},\ K_{\text{td}}
  • loss coefficients in the nozzle, at the suction inlet of the production, and in the mixing chamber with the diffuser (found by experiment; in the lesson 0.03, 0 and 0.30), —Hydraulic pumps: piston and jet
LL
Lcab, SL_{\text{cab}},\ S
lcll_{\text{cl}}
  • height of the clearance volume: its volume divided by the plunger area, mSucker-rod pumping
ltbgl_{\text{tbg}}
ln⁡\ln
MM
MfrM_{\text{fr}}
MhM_{\text{h}}
mrodm_{\text{rod}}
MrodM_{\text{rod}}
nn
NN
n1, n2n_1,\ n_2
  • rotational speeds. For a motor fed by a variable speed drive they are almost proportional to the frequency ff of the current, rpmElectric submersible pumps (ESP)
N1, N2N_1,\ N_2
NcN_{\text{c}}
NelN_{\text{el}}
nlimn_{\text{lim}}
  • the number of strokes per minute at which the acceleration of the polished rod at the top is equal to the acceleration of free fall, min⁻¹Sucker-rod pumping
NrodN_{\text{rod}}
  • power at the polished rod: the work per cycle (the area of the dynamometer card) multiplied by the number of cycles per second, WSucker-rod pumping
NshaftN_{\text{shaft}}
Nshaft,stgN_{\text{shaft,stg}}
NstgN_{\text{stg}}
PP
P↑, P↓P_{\uparrow},\ P_{\downarrow}
pannp_{\text{ann}}
pbarp_{\text{bar}}
  • pressure in the barrel under the plunger: on the upstroke it is equal to the intake pressure, and on the downstroke it is equal to the discharge pressure, PaSucker-rod pumping
pinp_{\text{in}}
PliqP_{\text{liq}}
Pmax⁡, Pmin⁡P_{\max},\ P_{\min}
  • estimates of the greatest and the smallest load on the polished rod in a cycle, NSucker-rod pumping
poutp_{\text{out}}
pout, pinp_{\text{out}},\ p_{\text{in}}
ppfp_{\text{pf}}
ppf, pinp_{\text{pf}},\ p_{\text{in}}
ProdP_{\text{rod}}
  • weight of the rods in air, mrod g Lm_{\text{rod}}\,g\,L: inertia depends on the mass, and the buoyant force does not affect it, NSucker-rod pumping
  • weight of the rods in air, NSucker-rod pumping
pspp_{\text{sp}}
pvp_{\text{v}}
pwfp_{wf}
pwhp_{wh}
pwh, pwfp_{wh},\ p_{wf}
Prod′P'_{\text{rod}}
Prod′, PliqP'_{\text{rod}},\ P_{\text{liq}}
P/EP/E
Δppump\Delta p_{\text{pump}}
qq
QQ
q1, q2q_1,\ q_2
qcavq_{\text{cav}}
qinq_{\text{in}}
qo, qwq_o,\ q_w
qpfq_{\text{pf}}
qthq_{\text{th}}
pˉR\bar p_R
RR
r1r_1
r2r_2
RcondR_{\text{cond}}
RsR_s
ss
SS
SplS_{\text{pl}}
Δpstg\Delta p_{\text{stg}}
tt
tct_{\text{c}}
tst_{\text{s}}
UU
u1, u2u_1,\ u_2
vv
VV
vcoolv_{\text{cool}}
Vp, VeV_{\text{p}},\ V_{\text{e}}
V, nV,\ n
WW
xopenx_{\text{open}}
  • the distance that the plunger travels from the bottom of the stroke before the standing valve opens, mSucker-rod pumping
zstgz_{\text{stg}}
α\alpha
αe\alpha_{\text{e}}
αf\alpha_{\text{f}}
  • fillage: the share of the volume swept by the plunger that is occupied by liquid; the rest is free gas (gas that is not dissolved in the liquid), —Sucker-rod pumping
αleak\alpha_{\text{leak}}
  • the share that is left after leaks through the plunger clearance, the valves and the tubing threads, —Sucker-rod pumping
αp\alpha_{\text{p}}
αs\alpha_{\text{s}}
  • the share of the stroke of the polished rod that reaches the plunger: Spl/SS_{\text{pl}}/S; the rest is the elastic stretch of the rods and tubing, —Sucker-rod pumping
αsh\alpha_{\text{sh}}
  • shrinkage, 1/Bliq1/B_{\text{liq}}: the liquid at the intake contains dissolved gas and takes up more volume than at the surface; BliqB_{\text{liq}} is the formation volume factor of the liquid at the intake conditions, —Sucker-rod pumping
η\eta
η(qin)\eta(q_{\text{in}})
η1, η2\eta_1,\ \eta_2
ηmot, cos⁡φ\eta_{\text{mot}},\ \cos\varphi
ηv\eta_{\text{v}}
θ\theta
λov\lambda_{\text{ov}}
λrod, λtbg\lambda_{\text{rod}},\ \lambda_{\text{tbg}}
  • elastic stretch of the rod string and of the tubing string under the fluid load, mSucker-rod pumping
μ\mu
ρ\rho
ρ, g\rho,\ g
ρ, Ks\rho,\ K_{\text{s}}
ρel\rho_{\text{el}}
ρliq\rho_{\text{liq}}
ρsteel\rho_{\text{steel}}
φg\varphi_{\text{g}}
ω\omega