16.3.1 Conventional Pumping Unit

For this type of unit, the acceleration at the bottom of the stroke is somewhat greater than true simple harmonic acceleration. At the top of the stroke, it is less. This is a major drawback for the conventional unit. Just at the time the TV is closing and the fluid load is being transferred to the rods, the acceleration for the rods is at its maximum. These two factors combine to create a maximum stress on the rods that becomes one of the limiting factors in designing an installation. Table 16.1 shows dimensions of some API conventional pumping units. Parameters are defined in Fig. 16.6.

16.3.2 Air-Balanced Pumping Unit

For this type of unit, the maximum acceleration occurs at the top of the stroke (the acceleration at the bottom of the stroke is less than simple harmonic motion). Thus, a lower maximum stress is set up in the rod system during transfer of the fluid load to the rods.

The following analyses of polished rod motion apply to conventional units. Fig. 16.7 illustrates an approximate motion of the connection point between pitman arm and walking beam.

If x denotes the distance of B below its top position C and is measured from the instant at which the crank arm and pitman arm are in the vertical position with the crank arm vertically upward, the law of cosine gives

${\left(AB\right)}^2={\left(OA\right)}^2+{\left(OB\right)}^2-2\left(OA\right)\left(OB\right)\cos AOB,$

that is,

$h^2=c^2+{\left(h+c-x\right)}^2-2c\left(h+c-x\right)\cos \omega t,$

where ω is the angular velocity of the crank. The equation reduces to

$x^2-2x\left[h+c\left(1-\cos \omega t\right)\right]+2c\left(h+c\right)\left(1-\cos \omega t\right)=0$

so that

$x=h+c\left(1-\cos \omega t\right)\pm \sqrt{c^2{\cos }^2\omega t+\left(h^2-c^2\right).}$

When ωt is zero, x is also zero, which means that the negative root sign must be taken. Therefore,

$x=h+c\left(1-\cos \omega t\right)-\sqrt{c^2{\cos }^2\omega t+\left(h^2+c^2\right)}.$

Acceleration is

$a=\frac{d^{2}x}{d{t}^2}.$

Carrying out the differentiation for acceleration, it is found that the maximum acceleration occurs when ωt is equal to zero (or an even multiple of p radians) and that this maximum value is

$a_{\max }={\omega }^{2}c(1+\frac{c}{h}).$ (16.1)

(16.1)

It also appears that the minimum value of acceleration is

$a_{\min }={\omega }^{2}c(1-\frac{c}{h}).$ (16.2)

(16.2)

If N is the number of pumping strokes per minute, then

$\omega =\frac{2\pi N}{60}(\mathrm{rad}/\text{sec}).$ (16.3)

(16.3)

The maximum downward acceleration of point B (which occurs when the crank arm is vertically upward) is

$a_{\max }=\frac{c{N}^2}{91.2}\left(1+\frac{c}{h}\right)({\text{ft}/\text{sec}}^2)$ (16.4)

(16.4)

or

$a_{\max }=\frac{c{N}^{2}g}{2936.3}\left(1+\frac{c}{h}\right)({\text{ft}/\text{sec}}^2).$ (16.5)

(16.5)

Likewise the minimum upward (a_min) acceleration of point B (which occurs when the crank arm is vertically downward) is

$a_{\max }=\frac{c{N}^{2}g}{2936.3}\left(1-\frac{c}{h}\right)({\text{ft}/\text{sec}}^2).$ (16.6)

(16.6)

It follows that in a conventional pumping unit, the maximum upward acceleration of the horse's head occurs at the bottom of the stroke (polished rod) and is equal to

$a_{\max }=\frac{d_1}{d_2}\frac{c{N}^{2}g}{2936.3}\left(1+\frac{c}{h}\right)({\text{ft}/\text{sec}}^2),$ (16.7)

(16.7)

where d₁ and d₂ are shown in Fig. 16.5. However,

$\frac{2c{d}_2}{d_1}=S,$

where S is the polished rod stroke length. So if S is measured in inches, then

$\frac{2c{d}_2}{d_1}=\frac{S}{12}$

or

$\frac{c{d}_2}{d_1}=\frac{S}{24}.$ (16.8)

(16.8)

So substituting Eq. (16.8) into Eq. (16.7) yields

$a_{\max }=\frac{S{N}^{2}g}{70471.2}\left(1+\frac{c}{h}\right)({\text{ft}/\text{sec}}^2),$ (16.9)

(16.9)

or we can write Eq. (16.9) as

$a_{\max }=\frac{S{N}^{2}g}{70,471.2}M({\text{ft}/\text{sec}}^2),$ (16.10)

(16.10)

where M is the machinery factor and is defined as

$M=1+\frac{c}{h}.$ (16.11)

(16.11)

Similarly,

$a_{\min }=\frac{S{N}^{2}g}{70471.2}\left(1-\frac{c}{h}\right)({\text{ft}/\text{sec}}^2).$ (16.12)

(16.12)

For air-balanced units, because of the arrangements of the levers, the acceleration defined in Eq. (16.12) occurs at the bottom of the stroke, and the acceleration defined in Eq. (16.9) occurs at the top. With the lever system of an air-balanced unit, the polished rod is at the top of its stroke when the crank arm is vertically upward (Fig. 16.5B).

16.4.1 Maximum PRL

The PRL is the sum of weight of fluid being lifted, weight of plunger, weight of sucker rods string, dynamic load due to acceleration, friction force, and the up-thrust from below on plunger. In practice, no force attributable to fluid acceleration is required, so the acceleration term involves only acceleration of the rods. Also, the friction term and the weight of the plunger are neglected. We ignore the reflective forces, which will tend to underestimate the maximum PRL. To compensate for this, we set the up-thrust force to zero. Also, we assume the TV is closed at the instant at which the acceleration term reaches its maximum. With these assumptions, the PRL_max becomes

$\begin{array}{ll}{\mathrm{PRL}}_{\max }\hfill & =S_f(62.4)D(\frac{A_p-A_r}{144})+\frac{{\gamma }_{s}D{A}_r}{144}\hfill \\ \hfill & +\frac{{\gamma }_{s}D{A}_r}{144}\left(\frac{S{N}^{2}M}{70,471.2}\right)\hfill \end{array},$ (16.13)

(16.13)

where

Note that for the air-balanced unit, M in Eq. (16.13) is replaced by 1-c/h.

Eq. (16.13) can be rewritten as

$\begin{array}{ll}{\mathrm{PRL}}_{\max }\hfill & =S_f(62.4)\frac{D{A}_p}{144}-S_f(62.4)\frac{D{A}_r}{144}+\frac{{\gamma }_{s}D{A}_r}{144}\hfill \\ \hfill & +\frac{{\gamma }_{s}D{A}_r}{144}\left(\frac{S{N}^{2}M}{70,471.2}\right).\hfill \end{array}$ (16.14)

(16.14)

If the weight of the rod string in air is

$W_r=\frac{{\gamma }_{s}D{A}_r}{144},$ (16.15)

(16.15)

which can be solved for A_r, which is

$A_r=\frac{144W_r}{{\gamma }_{s}D}.$ (16.16)

(16.16)

Substituting Eq. (16.16) into Eq. (16.14) yields

$\begin{array}{ll}{\mathrm{PRL}}_{\max }\hfill & =S_f(62.4)\frac{D{A}_p}{144}-S_f(62.4)\frac{W_r}{{\gamma }_s}+W_r\hfill \\ \hfill & +W_r\left(\frac{S{N}^{2}M}{70,471.2}\right).\hfill \end{array}$ (16.17)

(16.17)

The above equation is often further reduced by taking the fluid in the second term (the subtractive term) as a 50 °API with S_f=0.78. Thus, Eq. (16.17) becomes (where γ_s=490)

${\mathrm{PRL}}_{\max }=S_f(62.4)\frac{D{A}_p}{144}-0.1W_r+W_r+W_r\left(\frac{S{N}^{2}M}{70,471.2}\right)$

or

${\mathrm{PRL}}_{\max }=W_f+0.9W_r+W_r\left(\frac{S{N}^{2}M}{70,471.2}\right),$ (16.18)

(16.18)

where $W_f=S_f(62.4)\frac{D{A}_p}{144}$ and is called the fluid load (not to be confused with the actual fluid weight on the rod string). Thus, Eq. (16.18) can be rewritten as

${\mathrm{PRL}}_{\max }=W_f+(0.9+F_1)W_r,$ (16.19)

(16.19)

where for conventional units

$F_1=\frac{S{N}^2\left(1+\frac{c}{h}\right)}{70,471.2}$ (16.20)

(16.20)

and for air-balanced units

$F_1=\frac{S{N}^2\left(1-\frac{c}{h}\right)}{70,471.2}.$ (16.21)

(16.21)

16.4.2 Minimum PRL

The minimum PRL occurs while the TV is open so that the fluid column weight is carried by the tubing and not the rods. The minimum load is at or near the top of the stroke. Neglecting the weight of the plunger and friction term, the minimum PRL is

${\mathrm{PRL}}_{\min }=-S_f(62.4)\frac{W_r}{{\gamma }_s}+W_r-W_r{F}_2,$

which, for 50 °API oil, reduces to

${\mathrm{PRL}}_{\min }=0.9W_r-F_2{W}_r=(0.9-F_2)W_r,$ (16.22)

(16.22)

where for the conventional units

$F_2=\frac{S{N}^2\left(1-{\displaystyle \frac{c}{h}}\right)}{70,471.2}$ (16.23)

(16.23)

and for air-balanced units

$F_2=\frac{S{N}^2\left(1+{\displaystyle \frac{c}{h}}\right)}{70,471.2}.$ (16.24)

(16.24)

16.4.3 Counterweights

To reduce the power requirements for the prime mover, a counterbalance load is used on the walking beam (small units) or the rotary crank. The ideal counterbalance load C is the average PRL. Therefore,

$C=\frac{1}{2}\left({\mathrm{PRL}}_{\max }+{\mathrm{PRL}}_{\min }\right).$

Using Eqs. (16.19) and (16.22) in the above, we get

$C=\frac{1}{2}{W}_f+0.9W_r+\frac{1}{2}\left(F_1-F_2\right)W_r$ (16.25)

(16.25)

or for conventional units

$C=\frac{1}{2}{W}_f+W_r\left(0.9+\frac{S{N}^2}{70,471.2}\frac{c}{h}\right)$ (16.26)

(16.26)

and for air-balanced units

$C=\frac{1}{2}{W}_f+W_r\left(0.9-\frac{S{N}^2}{70,471.2}\frac{c}{h}\right).$ (16.27)

(16.27)

The counterbalance load should be provided by structure unbalance and counterweights placed at walking beam (small units) or the rotary crank. The counterweights can be selected from manufacturer's catalog based on the calculated C value. The relationship between the counterbalance load C and the total weight of the counterweights is

$C=C_s+W_c\frac{r}{c}\frac{d_1}{d_2},$

where

16.4.4 Peak Torque and Speed Limit

The peak torque exerted is usually calculated on the most severe possible assumption, which is that the peak load (polished rod less counterbalance) occurs when the effective crank length is also a maximum (when the crank arm is horizontal). Thus, peak torque T is (Fig. 16.5)

$T=c\left[C-(0.9-F_2)W_r\right]\frac{d_2}{d_1}.$ (16.28)

(16.28)

Substituting Eq. (16.8) into Eq. (16.28) gives

$T=\frac{1}{2}S\left[C-(0.9-F_2)W_r\right]$ (16.29)

(16.29)

or

$T=\frac{1}{2}S\left[\frac{1}{2}{W}_f+\frac{1}{2}\left(F_1+F_2\right)W_r\right]$

or

$T=\frac{1}{4}S\left(W_f+\frac{2S{N}^2{W}_r}{70,471.2}\right)(\mathrm{in}.-\text{lb}).$ (16.30)

(16.30)

Because the pumping unit itself is usually not perfectly balanced (C_s ≠ 0), the peak torque is also affected by structure unbalance. Torque factors are used for correction:

$T=\frac{{\displaystyle \frac{1}{2}}\left[{\mathrm{PRL}}_{\max }\left(T{F}_1\right)+{\mathrm{PRL}}_{\min }\left(T{F}_2\right)\right]}{0.93},$ (16.31)

(16.31)

where

For symmetrical conventional and air-balanced units, TF=TF₁=TF₂.

There is a limiting relationship between stroke length and cycles per minute. As given earlier, the maximum value of the downward acceleration (which occurs at the top of the stroke) is equal to

$a_{\max /\min }=\frac{S{N}^{2}g\left(1\pm {\displaystyle \frac{c}{h}}\right)}{70,471.2},$ (16.32)

(16.32)

(the ± refers to conventional units or air-balanced units, see Eqs. (16.9) and (16.12)). If this maximum acceleration divided by g exceeds unity, the downward acceleration of the hanger is greater than the free-fall acceleration of the rods at the top of the stroke. This leads to severe pounding when the polished rod shoulder falls onto the hanger (leading to failure of the rod at the shoulder). Thus, a limit of the above downward acceleration term divided by g is limited to approximately 0.5 (or where L is determined by experience in a particular field). Thus,

$\frac{S{N}^2\left(1\pm {\displaystyle \frac{c}{h}}\right)}{70,471.2}\le L$ (16.33)

(16.33)

or

$N_{\mathrm{limit}}=\sqrt{\frac{70,471.2L}{S\left(1\mp {\displaystyle \frac{c}{h}}\right)}}.$ (16.34)

(16.34)

For L=0.5,

$N_{\mathrm{limit}}=\frac{187.7}{\sqrt{S\left(1\mp {\displaystyle \frac{c}{h}}\right)}}.$ (16.35)

(16.35)

The minus sign is for conventional units and the plus sign for air-balanced units.

16.4.5 Tapered Rod Strings

For deep well applications, it is necessary to use tapered sucker rod strings to reduce the PRL at the surface. The larger diameter rod is placed at the top of the rod string, then the next largest, and then the least largest. Usually these are in sequences up to four different rod sizes. The tapered rod strings are designated by 1/8-in. (in diameter) increments. Tapered rod strings can be identified by their numbers, such as:

Tapered rod strings are designed for static (quasi-static) lads with a sufficient factor of safety to allow for random low-level dynamic loads. Two criteria are used in the design of tapered rod strings:

The reason for the second criterion is that it is preferable that any rod breaks occur near the bottom of the string (otherwise macaroni).

ExampleProblem 16.1 The following geometric dimensions are for the pumping unit C–320D–213–86:

If this unit is used with a 2½-in. plunger and ⅞-in. rods to lift 25 °API gravity crude (formation volume factor 1.2 rb/stb) at depth of 3000 ft, answer the following questions:

Solution The pumping unit C–320D–213–86 has a peak torque of gearbox rating of 320,000 in.-lb, a polished rod rating of 21,300 lb, and a maximum polished rod stroke of 86 in.

1. Based on the configuration for conventional unit shown in Fig. 16.5A and Table 16.1, the polished rod stroke length can be estimated as

$S=2c\frac{d_2}{d_1}=(2)(37)\frac{111}{96.05}=85.52\mathrm{in}.$

The maximum allowable pumping speed is

$N=\sqrt{\frac{70,471.2L}{S\left(1-{\displaystyle \frac{c}{h}}\right)}}=\sqrt{\frac{(70,471.2)(0.4)}{(85.52)(1-0.33)}}=22\mathrm{SPM}.$

2. The maximum PRL can be calculated with Eq. (16.17). The 25 °API gravity has an S_f=0.9042. The area of the 2½-in. plunger is A_p=4.91 in.². The area of the ⅞-in. rod is A_r=0.60 in.². Then

$\begin{array}{ll}{W}_f\hfill & =S_f(62.4)\frac{D{A}_p}{144}=(0.9042)(62.4)\frac{(3000)(4.91)}{144}\hfill \\ \hfill & =5770\mathrm{lbs}\hfill \end{array}$

$W_r=\frac{{\gamma }_{s}D{A}_r}{144}=\frac{(490)(3000)(0.60)}{144}=6138\mathrm{lbs}$

$F_1=\frac{S{N}^2\left(1+{\displaystyle \frac{c}{h}}\right)}{70,471.2}=\frac{(85.52){(22)}^2(1+0.33)}{70,471.2}=0.7940.$

Then the expected maximum PRL is

$\begin{array}{ll}{\mathrm{PRL}}_{\max }\hfill & =W_f-S_f(62.4)\frac{W_r}{{\gamma }_s}+W_r+W_r{F}_1\hfill \\ \hfill & =5770-(0.9042)(62.4)(6138)/(490).\hfill \\ \hfill & +6138+(6138)(0.794)\hfill \\ \hfill & =16,076\mathrm{lbs}<21,300\mathrm{lb}\hfill \end{array}$

3. The peak torque is calculated by Eq. (16.30):

$\begin{array}{ll}T\hfill & =\frac{1}{4}S\left(W_f+\frac{2S{N}^2{W}_r}{70,471.2}\right)\hfill \\ \hfill & =\frac{1}{4}(85.52)\left(5770+\frac{2(85.52){(22)}^2(6138)}{70,471.2}\right)\hfill \\ \hfill & =280,056\mathrm{lb}-\mathrm{in}.<320,000\mathrm{lb}-\mathrm{in}.\hfill \end{array}$

4. Accurate calculation of counterbalance load requires the minimum PRL:

$\begin{array}{ll}{F}_2\hfill & =\frac{S{N}^2\left(1-\frac{c}{h}\right)}{70,471.2}=\frac{(85.52){(22)}^2(1-0.33)}{70,471.2}=0.4\hfill \\ {\mathrm{PRL}}_{\min }\hfill & =-S_f(62.4)\frac{W_r}{{\gamma }_s}+W_r-W_r{F}_2\hfill \\ \hfill & =-(0.9042)(62.4)\frac{6138}{490}+6138-(6138)(0.4)\hfill \\ \hfill & =2976\mathrm{lb}\hfill \end{array}$

$C=\frac{1}{2}\left({\mathrm{PRL}}_{\max }+{\mathrm{PRL}}_{\min }\right)=\frac{1}{2}(16,076+2976)=9526\mathrm{lb}.$

A product catalog of LUFKIN Industries indicates that the structure unbalance is 450 lb and 4 No. 5ARO counterweights placed at the maximum position (c in this case) on the crank will produce an effective counterbalance load of 10,160 lb, that is,

$W_c\frac{(37)}{(37)}\frac{(96.05)}{(111)}+450=10,160,$

which gives W_c=11,221 lb. To generate the ideal counterbalance load of C=9526 lb, the counterweights should be placed on the crank at

$r=\frac{(9526)(111)}{(11,221)(96.05)}(37)=36.30\mathrm{in}.$

The computer program SuckerRodPumpingLoad.xls can be used for quickly seeking solutions to similar problems. It is available from the publisher with this book. The solution is shown in Table 16.2.

Table 16.2

Solution Given by Computer Program SuckerRodPumpingLoad.xls

SuckerRodPumpingLoad.xls
Description: This spreadsheet calculates the maximum allowable pumping speed, the maximum PRL, the minimum PRL, peak torque, and counterbalance load.
Instruction: (1) Update parameter values in the Input section; and (2) view result in the Solution section.
Input Data
Pump setting depth (D):	3000 ft
Plunger diameter (dp):	2.5 in.
Rod section 1, diameter (dr1):	1 in.
Length (L1):	0 ft
Rod section 2, diameter (dr2):	0.875 in.
Length (L2):	3000 ft
Rod section 3, diameter (dr3):	0.75 in.
Length (L3):	0 ft
Rod section 4, diameter (dr4):	0.5 in.
Length (L4):	0 ft
Type of pumping unit (1=conventional; −1=Mark II or Air-balanced):	1
Beam dimension 1 (d1)	96.05 in.
Beam dimension 2 (d2)	111 in.
Crank length (c):	37 in.
Crank to pitman ratio (c/h):	0.33
Oil gravity (API):	25 °API
Maximum allowable acceleration factor (L):	0.4
Solution
$S=2c\frac{d_2}{d_1}$	=85.52 in.
$N=\sqrt{\frac{70471.2L}{S\left(1-\frac{c}{h}\right)}}$	=22 spm
$A_p=\frac{\pi d_p^2}{4}$	=4.91 in.2
$A_r=\frac{\pi d_r^2}{4}$	=0.60 in.
$W_f=S_f(62.4){\displaystyle \frac{D{A}_p}{144}}$	=5770 lb
$W_r=\frac{{\gamma }_{s}D{A}_r}{144}$	=6138 lb
$F_1=\frac{S{N}^2(1\pm \frac{c}{h})}{70,471.2}$	=0.7940°
${\mathrm{PRL}}_{\max }=W_f-S_f(62.4)\frac{W_r}{{\gamma }_s}+W_r+W_r{F}_1$	=16,076 lb
$T=\frac{1}{4}S\left(W_f+{\displaystyle \frac{2S{N}^2{W}_r}{70,471.2}}\right)$	=280,056 lb
$F_2=\frac{S{N}^2(1\mp \frac{c}{h})}{70,471.2}$	=0.40
${\mathrm{PRL}}_{\min }=-S_f(62.4)\frac{W_r}{{\gamma }_s}+W_r-W_r{F}_2$	=2976 lb
$C=\frac{1}{2}({\mathrm{PRL}}_{\max }+{\mathrm{PRL}}_{\min })$	=9526 lb

16.5.1 Effective Plunger Stroke Length

The motion of the plunger at the pump setting depth and the motion of the polished rod do not coincide in time and in magnitude because sucker rods and tubing strings are elastic. Plunger motion depends on a number of factors including polished rod motion, sucker rod stretch, and tubing stretch. The theory in this subject has been well established (Nind, 1964).

Two major sources of difference in the motion of the polished rod and the plunger are elastic stretch (elongation) of the rod string and overtravel. Stretch is caused by the periodic transfer of the fluid load from the SV to the TV and back again. The result is a function of the stretch of the rod string and the tubing string. Rod string stretch is caused by the weight of the fluid column in the tubing coming on to the rod string at the bottom of the stroke when the TV closes (this load is removed from the rod string at the top of the stroke when the TV opens). It is apparent that the plunger stroke will be less than the polished rod stroke length S by an amount equal to the rod stretch. The magnitude of the rod stretch is

$\delta l_r=\frac{W_f{D}_r}{A_{r}E},$ (16.36)

(16.36)

where

Tubing stretch can be expressed by a similar equation:

$\delta l_t=\frac{W_f{D}_t}{A_{t}E}$ (16.37)

(16.37)

But because the tubing cross-sectional area A_t is greater than the rod cross-sectional area A_r, the stretch of the tubing is small and is usually neglected. However, the tubing stretch can cause problems with wear on the casing. Thus, for this reason a tubing anchor is almost always used.

Plunger overtravel at the bottom of the stroke is a result of the upward acceleration imposed on the downward-moving sucker rod elastic system. An approximation to the extent of the overtravel may be obtained by considering a sucker rod string being accelerated vertically upward at a rate n times the acceleration of gravity. The vertical force required to supply this acceleration is nW_r. The magnitude of the rod stretch due to this force is

$\delta l_o=n\frac{W_r{D}_r}{A_{r}E}(\mathrm{ft}).$ (16.38)

(16.38)

But the maximum acceleration term n can be written as

$n=\frac{S{N}^2(1\pm {\displaystyle \frac{c}{h}})}{70,471.2}$

so that Eq. (16.38) becomes

$\delta l_o=\frac{W_r{D}_r}{A_{r}E}\frac{S{N}^2(1\pm {\displaystyle \frac{c}{h}})}{70,471.2}(\mathrm{ft}),$ (16.39)

(16.39)

where again the plus sign applies to conventional units and the minus sign to air-balanced units.

Let us restrict our discussion to conventional units. Then Eq. (16.39) becomes

$\delta l_o=\frac{W_r{D}_r}{A_{r}E}\frac{S{N}^{2}M}{70,471.2}(\mathrm{ft}).$ (16.40)

(16.40)

Eq. (16.40) can be rewritten to yield δl_o in inches. W_r is

$W_r={\gamma }_s{A}_r{D}_r$

and γ_S=490 lb/ft³ with E=30×10⁶ lb/m². Eq. (16.40) becomes

$\delta l_o=1.93\times {10}^{-11}{D}_r^{2}S{N}^{2}M(\mathrm{in}.),$ (16.41)

(16.41)

which is the familiar Coberly expression for overtravel (Coberly, 1938).

Plunger stroke is approximated using the above expressions as

$S_p=S-\delta l_r-\delta l_t+\delta l_o$

or

$\begin{array}{l}{S}_p=S-\frac{12D}{E}\times \left[W_f\left(\frac{1}{A_r}+\frac{1}{A_t}\right)-\frac{S{N}^{2}M}{70,471.2}\frac{W_r}{A_r}\right](\mathrm{in}.).\hfill \end{array}$ (16.42)

(16.42)

If pumping is carried out at the maximum permissible speed limited by Eq. (16.34), the plunger stroke becomes

$\begin{array}{ll}{S}_p\hfill & =S-\frac{12D}{E}\times \left[W_f\left(\frac{1}{A_r}+\frac{1}{A_t}\right)-\frac{1+\frac{c}{h}L{W}_r}{1-\frac{c}{h}{A}_r}\right](\mathrm{in}.).\hfill \end{array}$ (16.43)

(16.43)

For the air-balanced unit, the term is replaced by its reciprocal.

16.5.2 Volumetric Efficiency

Volumetric efficiency of the plunger mainly depends on the rate of slippage of oil past the pump plunger and the solution–gas ratio under pump condition.

Metal-to-metal plungers are commonly available with plunger-to-barrel clearance on the diameter of −0.001, −0.002, −0.003, −0.004, and −0.005 in. Such fits are referred to as −1, −2, −3, −4, and −5, meaning the plunger outside diameter is 0.001 in. smaller than the barrel inside diameter. In selecting a plunger, one must consider the viscosity of the oil to be pumped. A loose fit may be acceptable for a well with high-viscosity oil (low °API gravity). But such a loose fit in a well with low-viscosity oil may be very inefficient. Guidelines are as follows:

An empirical formula has been developed that can be used to calculate the slippage rate, q_s (bbl/day), through the annulus between the plunger and the barrel:

$q_s=\frac{k_p}{\mu }\frac{{\left(d_b-d_p\right)}^{2.9}\left(d_b+d_p\right)}{d_b^{0.1}}\frac{\mathrm{\Delta }p}{L_p},$ (16.44)

(16.44)

where

The value of k_p is 2.77×10⁶–6.36×10⁶ depending on field conditions. An average value is 4.17×10⁶. The value of Δp may be estimated on the basis of well productivity index and production rate. A reasonable estimate may be a value that is twice the production drawdown.

Volumetric efficiency can decrease significantly due to the presence of free gas below the plunger. As the fluid is elevated and gas breaks out of solution, there is a significant difference between the volumetric displacement of the bottom-hole pump and the volume of the fluid delivered to the surface. This effect is denoted by the shrinkage factor greater than 1.0, indicating that the bottom-hole pump must displace more fluid by some additional percentage than the volume delivered to the surface (Brown, 1980). The effect of gas on volumetric efficiency depends on solution–gas ratio and bottom-hole pressure. Down-hole devices, called “gas anchors,” are usually installed on pumps to separate the gas from the liquid.

In summary, volumetric efficiency is mainly affected by the slippage of oil and free gas volume below plunger. Both effects are difficult to quantify. Pump efficiency can vary over a wide range but are commonly 70%–80%.

16.5.3 Power Requirements

The prime mover should be properly sized to provide adequate power to lift the production fluid, to overcome friction loss in the pump, in the rod string and polished rod, and in the pumping unit. The power required for lifting fluid is called “hydraulic power.” It is usually expressed in terms of net lift:

$P_h=7.36\times {10}^{-6}q{\gamma }_l{L}_N,$ (16.45)

(16.45)

where

and

$L_N=H+\frac{p_{tf}}{0.433{\gamma }_l},$ (16.46)

(16.46)

where

The power required to overcome friction losses can be empirically estimated as

$P_f=6.31\times {10}^{-7}{W}_{r}SN.$ (16.47)

(16.47)

Thus, the required prime mover power can be expressed as

$P_{pm}=F_S\left(P_h+P_f\right),$ (16.48)

(16.48)

where F_s is a safety factor of 1.25–1.50.

Example Problem 16.2 A well is pumped off (fluid level is the pump depth) with a rod pump described in Example Problem 16.1. A 3-in. tubing string (3.5-in. OD, 2.995 ID) in the well is not anchored. Calculate (1) expected liquid production rate (use pump volumetric efficiency 0.8), and (2) required prime mover power (use safety factor 1.35).

Solution This problem can be quickly solved using the program SuckerRodPumpingFlowrate&Power.xls. The solution is shown in Table 16.3.

Table 16.3

Solution Given by SuckerRodPumpingFlowrate&Power.xls

SuckerRodPumpingFlowRate&Power.xls
Description: This spreadsheet calculates expected deliverability and required prime mover power for a given sucker rod pumping system.
Instruction: (1) Update parameter values in the Input section; and (2) view result in the Solution section.
Input Data
Pump setting depth (D):	4000 ft
Depth to the liquid level in annulus (H):	4000 ft
Flowing tubing head pressure (ptf):	100 psi
Tubing outer diameter (dto):	3.5 in.
Tubing inner diameter (dti):	2.995 in.
Tubing anchor (1=yes; 0=no):	0
Plunger diameter (dp):	2.5 in.
Rod section 1, diameter (dr1):	1 in.
Length (L1):	0 ft
Rod section 2, diameter (dr2):	0.875 in.
Length (L2):	0 ft
Rod section 3, diameter (dr3):	0.75 in.
Length (L3):	4000 ft
Rod section 4, diameter (dr4):	0.5 in.
Length (L4):	0 ft
Type of pumping unit (1=conventional; −1=Mark II or Air-balanced):	1
Polished rod stroke length (S)	86 in.
Pumping speed (N)	22 spm
Crank to pitman ratio (c/h):	0.33°
Oil gravity (API):	25 °API
Fluid formation volume factor (Bo):	1.2 rb/stb
Pump volumetric efficiency (Ev):	0.8
Safety factor to prime mover power (Fs):	1.35
Solution
$A_t=\frac{\pi d_t^2}{4}$	=2.58 in.2
$A_p=\frac{\pi d_p^2}{4}$	=4.91 in.2
$A_r=\frac{\pi d_r^2}{4}$	=0.44 in.
$W_f=S_f(62.4)\frac{D{A}_p}{144}$	=7693 lb
$W_r=\frac{{\gamma }_{s}D{A}_r}{144}$	=6013 lb
$M=1\pm \frac{c}{h}$	=1.33
$S_p=S-\frac{12D}{E}\left[W_f\left(\frac{1}{A_r}+\frac{1}{A_t}\right)-\frac{S{N}^{2}M}{70471.2}\frac{W_r}{A_r}\right]$	=70 in.
$q=0.1484\frac{A_{p}N{S}_p{E}_v}{B_0}$	=753 sbt/day
$L_N=H+\frac{p_{tf}}{{0.433}_{{\gamma }_l}}$	=4255 ft
$P_h=7.36\times {10}^{-6}q{\gamma }_l{L}_N$	=25.58 hp
$P_f=6.31\times {10}^{-7}{W}_{r}SN$	=7.2 hp
$P_{pm}=F_s(P_h+P_f)$	=44.2 hp