How a reciprocating pump works

A reciprocating pump is a positive displacement mechanism, liquid discharge pressure being limited only by the strength of the structural parts. Liquid volume or capacity delivered is constant regardless of pressure, and is varied only by driver speed, speed reduction, and/or plunger/piston size changes.

Reciprocating motion is imparted to a plunger/piston by a slider crank linkage which results in a piston motion closely approximating simple harmonic motion, as shown in Appendix B. This reciprocating motion alternately lowers the pressure in front of the plunger/piston when filling the pump, and increases the pressure when emptying the pump. The incoming liquid opens the suction/inlet valve. At the same time the discharge valve is held closed by the downstream line pressure. Outgoing liquid closes the inlet valve and opens the discharge valve. This simple mechanism provides high volumetric efficiency, approximately 95 percent, for most incompressible liquids.

Characteristics of a WGI reciprocating pump are;

  • positive liquid displacement
  • high pulsations caused by the sinusoidal motion of the plunger/piston
  • high volumetric efficiency
  • high mechanical efficiency
  • low pump maintenance cost

 

Plunger or Piston Rod Load

Plunger or piston rod load is an important power end “design” consideration for WGI reciprocating pumps. Rod load is the force caused by the liquid pressure acting on the face of the plunger/piston. This load is transmitted through the adapter or piston rod to the crosshead, then to the crosshead pin, wrist pin bushing, connecting rod, crank shaft, main bearings and power frame. This load is directly proportional to the discharge gauge pressure and the square of the plunger/piston diameter.

Occasionally, allowable liquid end pressures limit the “allowable” rod load to a value below the “design” rod load.

It is important that liquid end pressure DOES NOT exceed WGI’s latest published recommendations.

Rod loads are not generally used when “applying” the pump. They are used to establish the power frame design – not to determine the allowable pressures, in most cases, for each size plunger/piston.. Also see the Sizing pumps in high inlet pressure conditions section.

 

Calculations of Volumetric Efficiency

Volumetric Efficiency (EV) is defined as the ratio of actual pump capacity to ideal pump displacement. The EV calculation depends upon the internal configuration of each individual liquid end cylinder, plunger/piston size, and the liquid being pumped. Given full details regarding differential pressure, pumped liquid mixture, and expected temperature rise on the discharge stroke – WGI can calculate this efficiency.

  1. Calculating EV for WaterSee TABLE 4 Page A31 for the water compressibility chart. An example of volumetric efficiency calculation for water is also shown.
  2. Calculating EV for HydrocarbonsSee TABLE 6 Page A34 for the physical properties of hydrocarbons. For compressible liquids such as these, horsepower calculations are slightly more complex than for incompressible liquids. However, the magnitude of the horsepower required will be slightly less than if calculated for the full displacement.
  3. Displacement TableTable 1 can be used to determine pump displacement and whether or not the proper efficiency is being obtained.

Example 1:

Find the capacity (Q), in Barrels Per Hour (BPH), for a single acting 3-1/2″ x 4″ triplex plunger pump operating at 95% volumetric efficiency (EV ) and a speed of 350 rpm.

From TABLE 1 a 3-1/2″ plunger with a 4″ stroke will displace 0.167 gallons per stroke. This type of pump will displace liquid at a rate of 3 forward strokes only per revolution. Therefore, the displacement per revolution is 0.167 gal./stroke x 3 strokes/rev. = 0.501 gal./rev. (gpr). At 350 rpm, the total displacement is 175.35 gallons per minute (GPM). Dividing by 0.7 gives 250.5 BPH displacement (D).  Therefore, the pump capacity at a 95% volumetric efficiency is Q = D Ev = 250.5 (95) (100) (100) Q = 237.9 BPH


Example 2: 

Find the rpm (n) required for a single acting 3″ x 5″ triplex plunger pump operating at EV = 85% and capacity (Q) = 200 BPH. D = displacement = Q (100)  Ev D = 200 = 235.29 BPH 0.85 gpr = (0.153 gal./stroke) (3 strokes/rev.) gpr = 0.459 GPM = BPH x 0.7 = 235.29 x 0.7 GPM = 164.7 At the given displacement and volumetric efficiency, the pump speed (n) is n = GPM = 164.7 rpm gpr 0.459 n = 358.8 rpm


Example 3:

Find the displacement (D), in gallons per minute (GPM), for a double acting 4″ x 10″ duplex piston pump running at a crankshaft speed of 60 rpm. Pump has 1-1/2″ diameter piston rods. This type pump will displace liquid at a rate of; 2 forward strokes/rev. and 2 rearward strokes/rev. (the 2 rearward strokes/rev. will displace less liquid than 2 forward strokes due to the amount of volume taken up by the piston rod). Assume the pump  EV = 95%. gpr = gprf + gprr Where, gprf = gal./rev. for the 2 forward strokes  gprf = 0.544 x 2 = 1.088 gprr = gal./rev. for the 2 rearward strokes gprr = [0.544 – (0.544 – 0.077){piston rod}] x 2  gprr = 0.934 gpr = (1.088 + 0.934) = 2.022 gpr D = Q (100) = 2.022 (60)(95) = 115.3 GPM Ev 100

 

Note 1: One barrel equals 42 U.S. gallons. Gallons per minute (GPM) divided by 0.7 equals barrels per hour (BPH). Displacement is the ideal volume swept by the plunger or piston on the discharge stroke during any selected time period.

Note 2: Under the conditions and definitions given above, the pump must run at a higher rpm in order to displace enough volume to deliver the required amount of liquid.

 

Horsepower Calculations

(see Notes 3, 4, 5, and 6 at the end of this section)

The required horsepower (hp) for driving a single acting reciprocating pump is calculated by the following equation:

hp = Pd Q (100) – Pi Q (Em-5) 1714 Em 1714 (100) Where: D = pump displacement, U.S. gallons per minute (GPM) Q = pump capacity, (GPM). Where Q = D x Ev Pd = liquid pressure at pump discharge, lbs/in2 gauge (PSIG). Pi = liquid pressure at pump inlet, lbs/in2 gauge (PSIG).  If Pi < 50 PSIG, then Pi Q (Em – 5) = 0 1714 x (100) Em = mechanical efficiency of pump, percent (for the above Pd and Pi < 50 PSIG): a. 90% for pumps without built-in reducer (power input from crankshaft). b. 85% for pumps with built-in or bolted-on gear reduction (power input from pinion shaft). c. Reduce Em by the mechanical efficiency losses of any intermediate speed reduction between drive and pump.G  eneral Em values:: V-belt drive 5% HTD drive 5% Parallel shaft gear reducer 5% Etc… Em – 5 = efficiency of power recovery due to liquid pressure at pump inlet. Ev = volumetric efficiency, where Ev = Q


Example 4:

Find the required horsepower (hp) and speed (rpm) for driving a single acting 2-3/4″ x 5″ triplex plunger pump under the following operating conditions:

Pump displacement = D = 138.9 GPM Pressure at pump inlet = Pi = 200 psig Pressure at pump discharge = Pd = 2020 psig Mechanical efficiency of pump = Em = 75% Volumetric efficiency of pump = Ev = 80% Q = D x Ev = 138.9 x 0.8 = 111.12 GPM hp = (Q x Pd) (100) – (Q x Pi ) ( Em – 5) 1714 x (Em) 1714 x (100) hp = (111.12 x 2020) (100) – (111.12 x 200) (75-5) 1714 x (75) 1714 x (100) hp = 165.53 gpr = 0.129 gal./plunger x 3 plungers/rev.  gpr = 0.387 gal./rev. (see Table 1) n = GPM = 138.9 = 358.9 rpm gpr 0.387 2. The required horsepower (hp) for driving a double acting duplex piston pump is calculated by the following equation: hp = (Q) (Pd – Pi) (100) 1714 (Em) Where, D = pump displacement, gallons per minute (GPM) Q = pump capacity, GPM where (Q = D / Ev ) Pd = liquid pressure at pump discharge, lbf/in2 gauge (psig). Pi = liquid pressure at pump inlet, lbf/in2 gauge (psig). Em = pump mechanical efficiency, percent (for above Pd and Pi). a. 90% for pumps without built-in reducer  (with power input from crankshaft) b. 85% for pumps with built-in reducer (with power input from pinion shaft) 3. Reduce Em by the mechanical efficiency losses of any intermediate speed reduction between driver and pumpG  eneral Em values: V-Belt drive 5% HTD drive 5% Parallel shaft gear reducer 5% Etc… EV = pump volumetric efficiency, % = Q (100) D


 

Example 5:

Find the required horsepower (hp) and pump speed (n) for driving a double acting 5″ x 10″ duplex piston pump with the following conditions.

Pump capacity = Q = 281.7 GPM Liquid pressure at pump inlet = Pi = 50 psig Liquid pressure at pump outlet = Pd = 330 psig Mechanical efficiency of pump = Em = 90% Volumetric efficiency of pump = EV = 85% Piston Rod diameter = d = 1- 1/2″ hp = Q (Pd – Pi) 100 = 281.7 (330-50) 100 1714 (Em) 1714 (90) hp = 51.13 gpr = [0.850 x 4] – [ π (1.5)2 ] x 10 x 2 x 1 = 3.24 4 231 D = ( Q x 100 ) / EV = (281.7 x 100) / 85 = 331.41 GPM n = D = 331.4 = 102.3 rpm gpr 3.24 3. Quick calculation of horsepower requirement of a reciprocating pump. Where the above formulas are very accurate, they are somewhat detailed. A quick and easy method of calculating horsepower which is almost as accurate as the other methods, is as follows:

A flow of one barrel per hour (1 BPH) of any specific gravity liquid against one pound per square inch gauge pressure (1 psig) requires 0.00040833 correction factor for theoretical horsepower. If the pump mechanism is approximately 90% mechanically efficient, the conversion factor can be corrected, for convenience, to 0.00045 to reflect this efficiency loss.

Therefore, hp = 0.00045 x Pd x BPH capacity BPH capacity = D Ev = (BPH displacement x EV) 100 100 Where, Pd = liquid pressure at pump discharge, psig EV = pump volumetric efficiency, %


 

Example 6:

Find the required horsepower (hp) to displace 100 gallons per minute (GPM) against a Pd = 1000 psig at a Ev = 95%.

hp = 0.00045 x 1000 x 100 x 95 = 61.1

0.7 x 100

Note 3: One barrel is 42 U.S. gallons. Gallons per minute (GPM) divided by 0.7 equals barrels per hour (BPH). Displacement is the theoretical volume swept by the plunger or piston on the discharge stroke during any selected time period.

Under the conditions and definitions given above, the pump speed must be a higher rpm in order to displace enough volume to deliver the required amount of liquid.

Note 4: Mechanical efficiency (Em) expressed as a percentage of total horsepower requirement can be used only if the pump is to be applied at or near its maximum designed rating. The horsepower required to operate a large pump at a small horsepower will usually be a considerably higher percentage of the total horsepower for the application.

We suggest that if the hydraulic horsepower – calculated without correction for mechanical efficiency – is less than 50% of the maximum design rating for the pump, you contact WGI for our recommendation for the driver horsepower.

Note 5: If the horsepower requirement is less than 15, as calculated by any of the above methods, we recommend that the motor driver be one size larger than the calculated requirement. This is because the horsepower required by a speed reduction device (belt, gear reducer, etc.) is relatively fixed and cannot be factored into the equation as a percentage of the smaller horsepower requirements.

Note 6: These computations are intended as a guide to standardization and may be modified if efficiency ratings for the pump application are lower than Ev = 95% and/or Em = 90%.

 

Devices for liquid pulsation control, Suction/Inlet and discharge.

A good inlet/suction and discharge pipe layout for reciprocating pumps of conventional type frequency require no pulsation control devices to compensate for normal variations in velocity flow in the complete piping system.

Where the suction or discharge lines are of considerable length, or the inlet has sufficient head, or liquid handled is hot – a suction or discharge dampening device of suitable size may sometimes be necessary to ensure smooth, quite operation. Dampening devices should be considered as a part of the piping system, rather than as a pump accessory.

The size and pre-charge of the dampener device will depend upon the type of pump, the liquid and the layout of the piping system. Recommendations as to size and type of device(s) should be obtained from the device manufacturer. Be sure to provide full information on the piping installation. Without complete knowledge of the piping system, it is impossible to determine the size and precharge of the dampeners. For bladder type dampeners, provisions should be made to keep the unit(s) charged with nitrogen or similar inert gas, in accordance with the device manufacturers recommendations. An exhausted device is of no value.

Dampeners, particularly on the suction (which are required more frequently than discharge), should be located as-close-as-possible to the pump and in such a position that they will absorb the impact of the moving liquid column and thus cushion the pulsations in the most efficient manner.

A properly sized, located and charged device may reduce the length of pipe used in the acceleration head equation to a value of 5 to 15 nominal pipe diameters. Figure 52 in Appendix A is a suggested piping system for power pumps.