When calculating maximum speed, power pump valve springs are an under-valued factor.
History tells us that a fellow named Ctesbius built a reciprocating power pump in about 150 B.C. on the Nile1. It sat near the surface of the river, was driven by an undershoot waterwheel and pushed river water into an irrigation system. During the more than 2,000 years since, it seems we have struggled to analyze the valve dynamics of this basic machine.
I have often wondered why reciprocating power pumps cannot run as fast as reciprocating compressors. The power ends look the same and are similar to internal combustion engines, which also run much faster. It, therefore, must be a fluid end limitation.
Could it be only the difference in the densities of the different fluids, and if so, what is the mathematical relationship between the fluid density and the maximum acceptable running speed?
Could it be only the difference in the densities of the different fluids, and if so, what is the mathematical relationship between the fluid density and the maximum acceptable running speed?
Figure 1. (left) A three-inch stroke, triplex power pump showing the springs on both valves.Early in my exposure to power pumps, I learned that some experienced users had established maximum allowable speeds for power pumps, based on stroke length. After a few years of wrestling field problems with power pumps running at high speeds, I established my own set of maximum speeds, based strictly on field observations. My standards were also based on stroke length.
These standards were published in References 10 and 11. I’ve never been totally comfortable with these standards, wondering if there could be an engineering basis for establishing maximum speeds.
The Role of Valve Springs
After observing a number of pumps, in operation on test stands and in the field, I concluded that the seemingly-innocuous valve spring plays a major role in establishing the maximum speed and is, in fact, the most important part of a power pump. Strong springs push the valves onto the seats near the ends of both the suction and discharge strokes, resulting in a smooth, quiet pump which produces acceptable pulsations.
On the other hand, weak valve springs allow the valves to be off the seats so far that, as the plunger reverses, the valves are slammed, by the reverse flow of pumpage, onto the seats. This results in a noisy, excessively-pulsing pump. The resulting hydraulic shocks shake pipe, damage instruments, damage drive components (such as gears and couplings) and damage pump components (such as valves and packing). The excessive pulsations can also cause cavitation. I have solved a number of field problems caused by excessive pulsations by installing semi-arbitrarilly-selected stronger valve springs in the pumps.
For years, I searched, numerous technical documents for equations that would provide the requirements for valve springs that would result in a smooth-running pump. The Worthington group (References 2, 3, and 4) provided significant information on valve characteristics, but their equations yielded inconsistent results.
Finally, after searching for more than 40 years and after many calculations and analyses, I discovered the mathematical models (equations) which seem to accurately describe power pump valve dynamics. Those models were presented in a technical paper at the 2009 International Pump Users Symposium (Reference 13) and in an article that appeared in the COMPRESSOR Tech Two magazine in July 2010 (reference 14). The models are verified by test results, as shown by the examples in the article.
Figure 3. (left) A bevel-face, wing guided valveThe above documents explain how to calculate the required spring force for a particular set of physical and operating parameters or how to calculate the actual valve lift for a valve assembly in a particular application.
After writing these documents, I began to wonder…rather than calculate the spring force required (which is based partially on pump speed), can I use the equations to calculate the maximum speed of a power pump, based partially on the maximum available spring force? The maximum available spring force is a function of the maximum allowable shear stress in the spring and the spring dimensions. Equation 6 (detailed later in the article), is the resultant which ties it all together.
Equation 6 combines the equation that calculates the spring force required to obtain smooth pump operation with the equation that relates spring force to spring stress, wire diameter and spring mean diameter. Using a number of half-reasonable approximations for spring dimensions and stress, an equation was also obtained which enables the calculation of the maximum speed for a “typical” power pump. Examples are shown.
Types of Valves
The equations in this article apply to the more common pump valves as shown in Figures 2 & 3. The pumpage flows radially outward between the valve (disc) and seat seating surfaces. Both the flat face disc valve and the bevel face, wing guided valve are covered. Different equations are required for a valve if the flow is radially inward, or if the valve is double ported.
Figure 2. (right) A flat-faced, single-port, outward flow disc valveThe Preferred Valve Lift at a 90-Degree Crank Angle (L90)
Previous attempts at analyzing power pump valve dynamics, as seen in References 2, 3, and 4, were based on the maximum acceptable valve lift being proportional to the diameter of the valve. (The larger the valve, the higher it could be allowed to lift.) Such a relationship resulted in large valves that slammed onto seats and small valves that were unnecessarily limited in their lifts. My discovery was that the maximum acceptable valve lift, for smooth operation, is entirely a function of the pump speed (rpm of the crankshaft). It is independent of valve size. That simple concept formed the basis for all the analyses that followed.
Nomenclature
As Flow area through valve seat (wings & webs are ignored), ft2 (m2) = (π/4)(D32 – D22)
A2 Valve area acted upon by P2, ft2 (m2) = (π/4)(D32 – D12) = A4 - A3
A3 Seating surface area exposed to P3, ft2 (m2) = (π /4)(D42 – D32)
A4 Area of top of valve (that exposed to P4), ft2 (m2) = (π /4)(D42 – D12)
c Orifice coefficient of valve “escape” area
d Diameter of wire in the helical valve spring, ft (m)
Dm Mean diameter of the helical spring, ft (m)
DP Diameter of plunger or piston, ft (m)
D1 Diameter of hole in center of valve disc, feet (m)
D2 OD of inner seating surface, feet (m) (ID of port opening in seat)
D3 ID of outer seating surface, feet (m) (OD of port opening in seat)
D4 OD of valve outer seating surface, ft (m)
FS90 Spring force on valve when crank angle, θ, = 90 degrees, lb (N)
g Acceleration of gravity, 32.2 ft/sec2 (9.8 m/s2)
K Whal stress correction factor for a helical spring made of round wire
Kc Clinging coefficient
for an OD-flow valve = (( 4/D3)2 + (D3/D4)2 - 2)/8π(sinα)2
for an ID-flow valve = (2 - ( 2/D1)2 - (D1/D2)2)/8π(sinα)2
Ki Coefficient of inertial impact of pumpage on upstream side of valve
Ls Stroke length of plunger, feet (m)
L90 The lift of the valve at about 90 degrees of crank rotation, ft (m)
N Rotative speed of pump crankshaft, rev/min
P1 Stagnation pressure upstream of valve assembly, lb/ft2 (Pa)
P2 Static pressure acting on upstream disc area A2, Ib/ft2 (Pa) = P1 - ρ(V S)2/2
P3 Static pressure in lift flow area between valve and seat, lb/ft2 (Pa)
P4 Stagnation pressure downstream of valve assembly, lb/ft 2 (Pa)
Q1 Flow rate created by plunger at mid-stroke (the peak) = Q90, assuming pure sinusoidal movement of plunger, ft3/sec (m3/s) = π2NLsDP2/240
Q2 Instantaneous flow rate created by the plunger (plunger volume displacement), ft3/sec (m3/s) = (Q1)(sin θ)
SS Shear stress in the wire of the helical spring, lb/ft2 (Pa)
t Time increment, sec = 30θ/πN (for θ in radians) = θ/6N (for θ in degrees)
VS Velocity of pumpage moving through seat, ft/sec (m/s) = (Q2)/(AS)
VV Velocity of valve, ft/sec (m/s)
W1 Weight of valve (in air), lb (N)
W2 Weight of valve if axis is vertical, less the bouyant effect of the pumpage, lb (N) = W1(1 - W3/W4) (if axis is horizontal, W2 = 0)
W3 Specific weight of pumpage, lb/ft3 (N/m3) = ρg
W4 Specific weight of valve, lb/ft3 (N/m3)
α Angle between valve seating surface and valve axis, degrees
θ Angle of crank rotation from start of stroke of plunger, radians = ωt = πNt/30
ρ Density of pumpage, slugs/ft3 (lb-s2/ft4) (kg/m3) = W3/g
ω Angular velocity of pump crankshaft, radians/sec = πN/30
Although the maximum valve lift (except for the case when a late opening valve overshoots upon opening) occurs at angles only near 90 degrees of crankshaft rotation, the symbol L90 is used here to represent this value. For smooth valve action and minimal hydraulic shock, the following has been established, from field experience, and gleaned from References 3, 6, and 7, as the maximum plunger-mid stroke lift of the valve:
Equation 1
L90 = 6/N feet (72/N inches) (1800/N mm)
This equation results from the need for the valve to impact the seat, upon closing, at a velocity low enough to prevent damage to the valve and seat, and to minimize hydraulic shock. This limit applies to both the suction and discharge valves. (The suction valve should not lift higher than the discharge valve.)
The Spring Force Required at 90-Degree Crank Angle
As shown in Reference 13, the following spring force is required, when the valve is at L90:
Equation 2
FS90 = ρA2(Q1/cπD3 L90sinα)2/2 – ρ(Q1/AS)2A2/2 – KCρQ12/ L902 + KiρQ12/AS – W2 + W1L90(πN/30)2/g
On the right side of the equation, the first term represents the pressure drop through the lift area and is the predominant term at higher pump speeds. The second term represents the pressure drop within the valve seat caused by the velocity head. The third term is the “clinging” effect. The fourth is the pumpage impact effect. The fifth is valve weight (if the valve axis is vertical), and the sixth is the valve acceleration (inertial) effect.
In the case of a wing guided valve or a disc valve guided by wings on the O.D., D1 = 0 and D2 = 0, the above equation reduces to:
Equation 3
FS90 = ρ(Q1/cL90sinα)2/8π + ρ(Q12/AS)(Ki – ½) – KCρQ12/ L902 – W2 + W1L90(πN/30)2/g
Surprisingly, the valve size drops from the first term, and as stated above, the first term is the predominant term at higher pump speeds. Therefore at higher speeds, a close approximation of the required FS90 spring force can be calculated with the first term only, and without knowing the size of the valve. To simplify the analysis presented, only the first term will be used to represent the spring force required for smooth operation. With L90 = 6/N, the first term can be rewritten as follows:
Equation 4
FS90 = ρ/8π [ π2N2LSDP2(240)(6)csinα]2
A2 Valve area acted upon by P2, ft2 (m2) = (π/4)(D32 – D12) = A4 - A3
A3 Seating surface area exposed to P3, ft2 (m2) = (π /4)(D42 – D32)
A4 Area of top of valve (that exposed to P4), ft2 (m2) = (π /4)(D42 – D12)
c Orifice coefficient of valve “escape” area
d Diameter of wire in the helical valve spring, ft (m)
Dm Mean diameter of the helical spring, ft (m)
DP Diameter of plunger or piston, ft (m)
D1 Diameter of hole in center of valve disc, feet (m)
D2 OD of inner seating surface, feet (m) (ID of port opening in seat)
D3 ID of outer seating surface, feet (m) (OD of port opening in seat)
D4 OD of valve outer seating surface, ft (m)
FS90 Spring force on valve when crank angle, θ, = 90 degrees, lb (N)
g Acceleration of gravity, 32.2 ft/sec2 (9.8 m/s2)
K Whal stress correction factor for a helical spring made of round wire
Kc Clinging coefficient
for an OD-flow valve = (( 4/D3)2 + (D3/D4)2 - 2)/8π(sinα)2
for an ID-flow valve = (2 - ( 2/D1)2 - (D1/D2)2)/8π(sinα)2
Ki Coefficient of inertial impact of pumpage on upstream side of valve
Ls Stroke length of plunger, feet (m)
L90 The lift of the valve at about 90 degrees of crank rotation, ft (m)
N Rotative speed of pump crankshaft, rev/min
P1 Stagnation pressure upstream of valve assembly, lb/ft2 (Pa)
P2 Static pressure acting on upstream disc area A2, Ib/ft2 (Pa) = P1 - ρ(V S)2/2
P3 Static pressure in lift flow area between valve and seat, lb/ft2 (Pa)
P4 Stagnation pressure downstream of valve assembly, lb/ft 2 (Pa)
Q1 Flow rate created by plunger at mid-stroke (the peak) = Q90, assuming pure sinusoidal movement of plunger, ft3/sec (m3/s) = π2NLsDP2/240
Q2 Instantaneous flow rate created by the plunger (plunger volume displacement), ft3/sec (m3/s) = (Q1)(sin θ)
SS Shear stress in the wire of the helical spring, lb/ft2 (Pa)
t Time increment, sec = 30θ/πN (for θ in radians) = θ/6N (for θ in degrees)
VS Velocity of pumpage moving through seat, ft/sec (m/s) = (Q2)/(AS)
VV Velocity of valve, ft/sec (m/s)
W1 Weight of valve (in air), lb (N)
W2 Weight of valve if axis is vertical, less the bouyant effect of the pumpage, lb (N) = W1(1 - W3/W4) (if axis is horizontal, W2 = 0)
W3 Specific weight of pumpage, lb/ft3 (N/m3) = ρg
W4 Specific weight of valve, lb/ft3 (N/m3)
α Angle between valve seating surface and valve axis, degrees
θ Angle of crank rotation from start of stroke of plunger, radians = ωt = πNt/30
ρ Density of pumpage, slugs/ft3 (lb-s2/ft4) (kg/m3) = W3/g
ω Angular velocity of pump crankshaft, radians/sec = πN/30
Although the maximum valve lift (except for the case when a late opening valve overshoots upon opening) occurs at angles only near 90 degrees of crankshaft rotation, the symbol L90 is used here to represent this value. For smooth valve action and minimal hydraulic shock, the following has been established, from field experience, and gleaned from References 3, 6, and 7, as the maximum plunger-mid stroke lift of the valve:
Equation 1
L90 = 6/N feet (72/N inches) (1800/N mm)
This equation results from the need for the valve to impact the seat, upon closing, at a velocity low enough to prevent damage to the valve and seat, and to minimize hydraulic shock. This limit applies to both the suction and discharge valves. (The suction valve should not lift higher than the discharge valve.)
The Spring Force Required at 90-Degree Crank Angle
As shown in Reference 13, the following spring force is required, when the valve is at L90:
Equation 2
FS90 = ρA2(Q1/cπD3 L90sinα)2/2 – ρ(Q1/AS)2A2/2 – KCρQ12/ L902 + KiρQ12/AS – W2 + W1L90(πN/30)2/g
On the right side of the equation, the first term represents the pressure drop through the lift area and is the predominant term at higher pump speeds. The second term represents the pressure drop within the valve seat caused by the velocity head. The third term is the “clinging” effect. The fourth is the pumpage impact effect. The fifth is valve weight (if the valve axis is vertical), and the sixth is the valve acceleration (inertial) effect.
In the case of a wing guided valve or a disc valve guided by wings on the O.D., D1 = 0 and D2 = 0, the above equation reduces to:
Equation 3
FS90 = ρ(Q1/cL90sinα)2/8π + ρ(Q12/AS)(Ki – ½) – KCρQ12/ L902 – W2 + W1L90(πN/30)2/g
Surprisingly, the valve size drops from the first term, and as stated above, the first term is the predominant term at higher pump speeds. Therefore at higher speeds, a close approximation of the required FS90 spring force can be calculated with the first term only, and without knowing the size of the valve. To simplify the analysis presented, only the first term will be used to represent the spring force required for smooth operation. With L90 = 6/N, the first term can be rewritten as follows:
Equation 4
FS90 = ρ/8π [ π2N2LSDP2(240)(6)csinα]2
Spring Stress and Size Parameters
The equation that relates the stress and spring dimensions to the force in a helical spring, made of round wire, is:
Equation 5
F = πd3SS/8KDm
Equating 4 and 5 and solving for pump speed, N, yields the following:
Equation 6
Nmax = 21.4/DP[ csinαLS ]1/2[ d3SSρKDm ]1/4
Equation 5
F = πd3SS/8KDm
Equating 4 and 5 and solving for pump speed, N, yields the following:
Equation 6
Nmax = 21.4/DP[ csinαLS ]1/2[ d3SSρKDm ]1/4
To obtain an approximate solution, we can substitute the approximate value of 0.6 for c sin α, which yields:
Equation 7
Nmax = 16.6/ DP[ 1/LS ]1/2[ d3SSρKDm ]1/4
Equation 7
Nmax = 16.6/ DP[ 1/LS ]1/2[ d3SSρKDm ]1/4
This equation can be used to calculate the maximum speed for a pump with a single-spring, single-ported, outward-flow valve with the optimum valve lift. For the popular 302 stainless steel valve spring material, the approximate maximum shear stress is 40,000 psi (5.76 x 106 lb/ ft2), and with a Whal spring stress correction factor of K = 1.2, the following is obtained:
Equation 8
Nmax = 776/DP[ 1/LS ]1/2[ d3/ρDm ]1/4
Equation 8
Nmax = 776/DP[ 1/LS ]1/2[ d3/ρDm ]1/4
To be able to calculate the approximate maximum speed for a “typical” power pump, we can substitute approximate spring dimension ratios as follows:
Equation 9
Dm = P/2 d = Dm/8 = D P/16
which results in:
Nmax = 116 [1/LsDp ]1/2[ 1/ρ]1/4
For cool water, ρ=1.94 lb-s2/ft4, and the equation reduces to:
Equation 10
Nmax = 98(LSDP)1/2
where the stroke length, LS, and the plunger diameter, D P, are both in feet. To substitute inches directly, change the numerator from 98 to 1180.
Examples
DP = 1 in., LS = 2 in. Nmax = 1180/(2)1/2 = 834 RPM
DP = 2 in., LS = 5 in. Nmax = 1180/(10)1/2 = 373 RPM
DP = 4 in., LS = 6 in. Nmax = 1180/(24)1/2 = 241 RPM
DP = 6 in., LS = 12 in. Nmax = 1180/(72)1/2 = 139 RPM
Figure 4 is a plot of Nmax = 1180/(LS P)1/2 with LS and DP expressed in inches.
Equation 9
Dm = P/2 d = Dm/8 = D P/16
which results in:
Nmax = 116 [1/LsDp ]1/2[ 1/ρ]1/4
For cool water, ρ=1.94 lb-s2/ft4, and the equation reduces to:
Equation 10
Nmax = 98(LSDP)1/2
where the stroke length, LS, and the plunger diameter, D P, are both in feet. To substitute inches directly, change the numerator from 98 to 1180.
Examples
DP = 1 in., LS = 2 in. Nmax = 1180/(2)1/2 = 834 RPM
DP = 2 in., LS = 5 in. Nmax = 1180/(10)1/2 = 373 RPM
DP = 4 in., LS = 6 in. Nmax = 1180/(24)1/2 = 241 RPM
DP = 6 in., LS = 12 in. Nmax = 1180/(72)1/2 = 139 RPM
Figure 4 is a plot of Nmax = 1180/(LS P)1/2 with LS and DP expressed in inches.
The Unexpected Results
Equations 6 and 9 reveal some of the following unexpected results:
The maximum allowable speed varies inversely as the square root of the product of the plunger stroke and diameter.
The density of the pumpage affects the maximum speed only to the one-fourth power, a small amount.
The maximum allowable spring shear stress affects speed only to the one-fourth power, a small amount. Doubling the allowable stress (to 80,000 psi) increases the maximum speed by only 19 percent. Doubling the spring force with two springs in parallel also increases the maximum speed by only 19 percent.
Increasing the spring wire diameter, d, increases the maximum allowable speed to the three-quarter power.
The maximum allowable speed varies inversely as the square root of the product of the plunger stroke and diameter.
The density of the pumpage affects the maximum speed only to the one-fourth power, a small amount.
The maximum allowable spring shear stress affects speed only to the one-fourth power, a small amount. Doubling the allowable stress (to 80,000 psi) increases the maximum speed by only 19 percent. Doubling the spring force with two springs in parallel also increases the maximum speed by only 19 percent.
Increasing the spring wire diameter, d, increases the maximum allowable speed to the three-quarter power.
Increasing the spring mean diameter, Dm, increases the maximum allowable speed to the one-half power (square root).
Figure 4 (above). Plot of the approximate maximum power pump speed, on water, for a single-spring, outward-flow valve.About That Compressor Speed
The density of air at standard conditions is approximately 1/800th that of cool water. Therefore, an air compressor with a single-ported, single-spring valve would be limited to an approximate maximum speed that is about (800)1/4 = 5.3 times that of the same size pump on water, and that looks about right. Because compressors are typically made with multi-port, multi-spring valves, the multiplier is higher than 5.3. That seems to provide a reasonable answer to the question: why can’t reciprocating power pumps run as fast as reciprocating compressors.
Another Unexpected Result
Plugging Equation 9 into Equation 4 produces another surprising result:
Equation 11
FS90 = ρ8π[ π2N2LSDP2/(240)(6)csinα]2= ρ[ N2LSDP2/439]2= ρ[ 13460LSDP2 LSDPρ1/2 439 ]2= 940DP2
This equation says that the approximate spring force (in pounds) required for any power pump—with the general design parameters stated above, with any density pumpage, to achieve the optimum valve lift—can be obtained by the simple calculation of 940 times the plunger diameter (in feet) squared. For the plunger diameter in inches, change the 940 coefficient to 6.52.
For example, a 4-inch diameter plunger in a 5-inch stroke pump—running on cool water, at the maximum allowable speed of 264 rpm, with the optimum valve lift of 0.273 inches—will require a spring force, at that valve lift, of about 104 pounds. This same spring would be selected for a 4-inch diameter plunger in a 10-inch stroke pump—running, on cool water, at the maximum allowable speed of 187 rpm, with the optimum valve lift of 0.386 inches. A more accurate value of FS90 can be obtained with Equation 2 or 3.
Can it be so simple? So it seems.
FS90 = ρ8π[ π2N2LSDP2/(240)(6)csinα]2= ρ[ N2LSDP2/439]2= ρ[ 13460LSDP2 LSDPρ1/2 439 ]2= 940DP2
This equation says that the approximate spring force (in pounds) required for any power pump—with the general design parameters stated above, with any density pumpage, to achieve the optimum valve lift—can be obtained by the simple calculation of 940 times the plunger diameter (in feet) squared. For the plunger diameter in inches, change the 940 coefficient to 6.52.
For example, a 4-inch diameter plunger in a 5-inch stroke pump—running on cool water, at the maximum allowable speed of 264 rpm, with the optimum valve lift of 0.273 inches—will require a spring force, at that valve lift, of about 104 pounds. This same spring would be selected for a 4-inch diameter plunger in a 10-inch stroke pump—running, on cool water, at the maximum allowable speed of 187 rpm, with the optimum valve lift of 0.386 inches. A more accurate value of FS90 can be obtained with Equation 2 or 3.
Can it be so simple? So it seems.
High-Speed Warning
Although the above equations allow us to determine the approximate maximum speed allowed by adequate valve springs, many factors may dictate a lower speed. Some of those conditions are:
Weak valve springs
Valve springs that have a stiffness that is less than calculated (not uncommon)
Special valves (such as balls)
Limited NPSH available from the system
Air or other gas dissolved in the pumpage
Air or other gas entrained in the pumpage
Long, and/or undersized, and/or complex, suction pipe
Absent or inadequate pulsation-dampening equipment
High temperature of pumpage
Solids in pumpage (slurry)
Compressible pumpage
Viscous pumpage
High ambient temperature
Desire for less pulsation in suction and/or discharge piping
A high spring force, as required by large pumps, which may be difficult or unsafe to assemble or disassemble
Desire for extended life of packing, plungers, valves, springs and other pump and system components (reduced maintenance requirements)
Therefore, the above equations simply indicate the approximate maximum speed as determined by the valve spring, which usually must be reduced to meet specific application conditions.
Did it really take us 2,000 years to establish the more significant characteristics of the power pump? So it seems.
Reference:
1. Ingenious Mechanisms, New York: Ingersoll-Rand, Hydraulic Division, 1960
2. Worthington Pump and Machinery Corp., formulas and derivations for reciprocating pump valve lift and valve springs, 1949 1953
3. Wright, Elliott F., “New Developments in Reciprocating Power Pumps,” A paper presented to the 6th meeting of the National Conference on Industrial Hydraulics, about 1955
4. Wright, E. F., Marks’ Mechanical Engineers’ Handbook, 6th Edition, Section 14 - Reciprocating Power Pumps, 1958
5. Collier, Samuel L., “Knocking From Valve Hammer in Triplex Pumps,” ASME Paper 83 PET 29, 1983
6. Collier, Samuel L., Mud Pump Handbook, Gulf Publishing Co., 1983
7. Worster, A. R., “Importance of Maintaining Proper Valve Lift on Compressor Valves,” a paper presented to the N.G.A.A. regional meeting in Odessa, Texas, February 26, 1954.
8. White, K. H., “Prediction and Measurement of Compressor Valve Losses,” ASME Paper 72 Pet 4, 1972
9. Vetter, Gerhard and Schweinfurter, Friedrich, “Pressure Pulsations in the Piping of Reciprocating Pumps,” Chem. Eng. Technol. 10, 1987
10. Henshaw, Terry L., Reciprocating Pumps, Van Nostrand Reinhold Co. Inc., New York, 1987
11. Positive Displacement Pumps – Reciprocating, API STANDARD 674, American Petroleum Institute, Washington. D.C., June 1995
12. Price, Stephen M., Smith, Donald R., and Tison, James D., “The Effects of Valve Dynamics on Reciprocating Pump Reliability,” a tutorial presented at the 12th International Pump Users Symposium, Houston, Texas, 1995
13. Henshaw, Terry, “Power Pump Valve Dynamics – A Study of the Velocity and Pressure Distribution in Outward-Flow Bevel-Face and Flat-Face Power Pump Valves,” a technical paper presented at the 25th International Pump Users Symposium, Houston, Texas, 2009
14. Henshaw, Terry, “Improve Power Pump Performance With Stronger Valve Springs,” COMPRESSOR Tech Two, July 2010
Weak valve springs
Valve springs that have a stiffness that is less than calculated (not uncommon)
Special valves (such as balls)
Limited NPSH available from the system
Air or other gas dissolved in the pumpage
Air or other gas entrained in the pumpage
Long, and/or undersized, and/or complex, suction pipe
Absent or inadequate pulsation-dampening equipment
High temperature of pumpage
Solids in pumpage (slurry)
Compressible pumpage
Viscous pumpage
High ambient temperature
Desire for less pulsation in suction and/or discharge piping
A high spring force, as required by large pumps, which may be difficult or unsafe to assemble or disassemble
Desire for extended life of packing, plungers, valves, springs and other pump and system components (reduced maintenance requirements)
Therefore, the above equations simply indicate the approximate maximum speed as determined by the valve spring, which usually must be reduced to meet specific application conditions.
Did it really take us 2,000 years to establish the more significant characteristics of the power pump? So it seems.
Reference:
1. Ingenious Mechanisms, New York: Ingersoll-Rand, Hydraulic Division, 1960
2. Worthington Pump and Machinery Corp., formulas and derivations for reciprocating pump valve lift and valve springs, 1949 1953
3. Wright, Elliott F., “New Developments in Reciprocating Power Pumps,” A paper presented to the 6th meeting of the National Conference on Industrial Hydraulics, about 1955
4. Wright, E. F., Marks’ Mechanical Engineers’ Handbook, 6th Edition, Section 14 - Reciprocating Power Pumps, 1958
5. Collier, Samuel L., “Knocking From Valve Hammer in Triplex Pumps,” ASME Paper 83 PET 29, 1983
6. Collier, Samuel L., Mud Pump Handbook, Gulf Publishing Co., 1983
7. Worster, A. R., “Importance of Maintaining Proper Valve Lift on Compressor Valves,” a paper presented to the N.G.A.A. regional meeting in Odessa, Texas, February 26, 1954.
8. White, K. H., “Prediction and Measurement of Compressor Valve Losses,” ASME Paper 72 Pet 4, 1972
9. Vetter, Gerhard and Schweinfurter, Friedrich, “Pressure Pulsations in the Piping of Reciprocating Pumps,” Chem. Eng. Technol. 10, 1987
10. Henshaw, Terry L., Reciprocating Pumps, Van Nostrand Reinhold Co. Inc., New York, 1987
11. Positive Displacement Pumps – Reciprocating, API STANDARD 674, American Petroleum Institute, Washington. D.C., June 1995
12. Price, Stephen M., Smith, Donald R., and Tison, James D., “The Effects of Valve Dynamics on Reciprocating Pump Reliability,” a tutorial presented at the 12th International Pump Users Symposium, Houston, Texas, 1995
13. Henshaw, Terry, “Power Pump Valve Dynamics – A Study of the Velocity and Pressure Distribution in Outward-Flow Bevel-Face and Flat-Face Power Pump Valves,” a technical paper presented at the 25th International Pump Users Symposium, Houston, Texas, 2009
14. Henshaw, Terry, “Improve Power Pump Performance With Stronger Valve Springs,” COMPRESSOR Tech Two, July 2010
Written by Terry Henshaw
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