Pressure

Liquid pressure is defined as the normal component of force per unit area. In common practice and general function, pressures are frequently measured in pounds force per square inch (lbf/in2). Gauge pressure (psig) is the difference between absolute pressure (psia) and the atmospheric pressure (Pa). Appendix A, Table 7 shows the relationship between atmospheric pressure and elevation. Vapor pressure is the absolute pressure exerted by the liquid and its vapor to maintain an equilibrium condition at a given temperature of the liquid.

Example 7:

Find vapor pressures of water at 76°F and 212°F at sea level..

From Appendix E – Table of Vapor Pressure of Water:

The vapor pressure of 76°F water is 0.4443 psia at sea level.

The vapor pressure of 212°F water is 14.696 psia at sea level.

The English unit for measuring head is feet. The equation, expressing pressure (psi) in units of feet, is:

S.G.

Where,

S.G. = Specific Gravity @ pumping temperature.

Example 8:

Find the head in units of feet (ft.) of crude oil, with a S.G. = 0.8 @ pumping temperature, at 20 psi pressure.

Head = 20 x 2.31 = 57.75 ft.

0.8

Example 9:

Find the head in units of feet (ft.) of mercury with a S.G. = 13.6 @ pumping temperature), at 20 psi pressure.

Head = 20 x 2.31 = 3.39 ft.

13.6

Viscosity

Basic metric viscosity units are the poise (absolute/dynamic viscosity) and the stokes (kinematic viscosity). More customary expression of these units are centipoise and centistokes respectively, each equal to 1/100th the of basic metric viscosity unit. The relationship between the English units for medium viscosity liquids, SSU (Saybolt Universal Seconds), and metric absolute viscosity is:

n (absolute viscosity, centistokes) = 0.22 (SSU) – 180

(SSU)

Introducing the mass density of the liquid (r) allows the expression of the relationship between absolute viscosity to Kinematic viscosity as follows:

m (Kinematic viscosity, centipoise) = rn

Example 10:

Find viscosity in centistokes (n) and centipoise (m) of a liquid with S.G. = 0.8 with a viscosity of 500 SSU.

n = 0.22 (500) – 180 = 109.64 centistokes

500

Then,

m = 0.8 x 109.64 = 87.71 centipoise

The basic pump speed, and its relationship with various ranges of liquid viscosity, are discussed in further detail in Appendix A.

Pipe, valves, fittings, hoses, and meters installed in the liquid supply piping system generate resistance to the liquid flow. The friction head is the hydraulic pressure required to overcome frictional resistance of a piping system. The Table in Appendix C shows an equivalent length in feet, of 100 percent opening valves and fittings. Pressure drop in liquid lines versus liquid flow rates is shown in Appendix D.

Reynolds Number

The Reynolds Number (Re) is used in closed conduit/pipe flow, deals with the viscous force in a liquid, and is defined by the following equation:

Re = r1 df n1

m1

Where,

Re = Reynolds Number

r1 = liquid density at flowing temperature, lbm/ft3

df = pipe inside diameter, feet

n1 = liquid flow velocity, ft/sec

m1 = liquid viscosity

(centipoise divided by 1488 or centistokes multiplied by S.G. then divided by divided by 1488)

Customarily; turbulent flow occurs when the Re is greater than 3000, laminar flow occurs when the Re is less than 2000. The transition period is when the Re is between 2000 and 3000.

Example 11:

A 14″ schedule 30 piping system is designed to deliver 18,970 BPD (553.3 GPM) of crude oil with a Kinematic viscosity of 50 centistokes and S.G. = 0.8 @ 100°F.

Find the Reynolds Number (Re).
Re = r1 df n1 = 49.92 x 1.104 x 1.28
m1 0.02688
Re = 2624.2

Where,
r1 = 62.4 x 0.8 = 49.92 lbm/ft3
df = 13.25 = 1.104 feet
12
n1 = 553.3 = 1.28 ft/sec
2.45 (13.25)2
m1 = 50 x 0.8 = 0.02688
1488

Whenever a column of liquid is either accelerated or decelerated, pressure surges exist. This condition is found on the suction/inlet side, as well as discharge side, of a reciprocating pump. Not only can the surges cause vibration in the inlet line, but they can restrict and impede the flow of liquid and cause incomplete filling of the inlet valve chamber. The magnitude of the surges, and how they will react in the system, is impossible to predict without an extremely complex and costly analysis of the system. Since the behavior of the natural frequencies in the system is not easily predicted, as much of the surge as possible must be eliminated at the source. Proper sizing, installation and charging of a dampening device will absorb a large percentage of the surge before it travels into and through the system and cause trouble. The function of the device is to absorb the peak of the surge and feed it back at the low part of the cycle. The preferred position for the device is in the liquid supply line, as close as physically possible to the reciprocating pump, or alternately attached to the blind side of the pump inlet. In either location, the surges will be significantly dampened and the possibility of harmful vibrations considerably reduced.

The experimental formula for calculating acceleration head is:
ha = L V n C and V = GPM
K g (2.45) (ID)2

Where,
L = length of liquid supply line (ft.)
V = average velocity in liquid supply line (fps)
n = Pump speed (rpm)
C = Constant depending on the type of pump (see page A35)
K = liquid compressibility factor:
K = 2.5 for relatively compressible liquids (ethane, hot oil)
K = 2.0 for most other hydrocarbons
K = 1.5 for amine, glycol and water
K = 1.4 for liquids with almost no compressibility (hot water)
g = standard gravity = 32.2 ft/sec2
ID = inside diameter of pipe (in.)

Example 12:

Find the acceleration head (ha) for a single acting 3-1/2″ x 4″ triplex plunger pump operating at 350 rpm and 175.35GPM capacity. Supply to the pump is through 50 feet of 4″ Schedule 40 pipe fed from an open tank.
Assume the liquid is water @ 80°F.

ha = (50) (175.35) (350) (0.066)
(1.4) (32.2) (4.026)2 (2.45)
ha = 113.13 feet

The acceleration head would be reduced to 49.85 feet if 6″ Schedule 40 pipe were used, but this is still too high. An 8″ schedule 40 pipe would reduce ha to 28.78 feet. Proper application of dampener device could reduce this value to approximately 10 feet of 4″ schedule 40 pipe. This would appreciably help the net positive suction head available (NPSHA) – in addition to the benefits in avoiding harmful vibrations.

NPSHR (Net Positive Suction Head Required)

The NPSHR is the head of liquid, in feet, required at the centerline of the liquid end inlet/suction connection to completely fill each cylinder on the reverse/suction stroke. It is the feet of liquid necessary to;

• overcome the frictional losses through inlet manifold, valves and liquid chamber,
• overcome the valve weight and spring force acting on the valve,
• overcome the valve velocity head losses,
• accelerate the liquid from rest to required velocity.

NPSHR is a function of the liquid and pump characteristics (stroke, plunger size, liquid end design, and operating speed).

NPSHR is usually determined by test or estimated by computation. Figure 55 and 56, on pages A36 and A37 respectively, show the liquid supply system relationships for open and closed supply tanks.

NPSHA (Net Positive Suction Head Available)

NPSHA is the total inlet head from the system at the pump inlet connection minus the vapor pressure of the liquid at the pumping temperature. It is a function of the supply system and the liquid and pump characteristics (stroke, plunger size, liquid end design, operating speed). The pump characteristic is important to both supply system and NPSHR. A liquid supply system designed without consideration of the pump characteristics has not been properly designed.

Pump Cavitation

A WGI reciprocating power pump is a device to move liquid under specified operating conditions. If the liquid is not arriving at the inlet side of the pump promptly, evenly, and with the least amount of resistance; the pump cannot operate efficiently, nor can it move the liquid through the discharge system smoothly. A poor liquid supply system to the inlet side of the pump will create pump cavitation (where liquid moving through the pump vaporizes rapidly wherever the local absolute pressure falls to, or attempts to fall below, the liquid vapor pressure). Cavitation can cause premature failure of the pump valve, piston or plunger packing, pitting of the cylinder walls, and damage to the pump and system by subjecting all parts to undue stresses. Consider the interactions between a piston and the liquid it pumps. The piston must stay in contact with the liquid through its entire stroke.
For reciprocating power pumps, plunger/piston velocity varies sinusoidally with crankshaft position. Maximum velocity occurs approximately at mid-stroke and zero velocity occurs at both ends (full forward position and full reverse position) of the stroke. Under certain inlet conditions, liquid looses contact with the piston and creates a cavity. When the piston slows, and the liquid catches up with the piston, collapsing of the cavity will occur. This cavitation creates shock waves that travel throughout the pump and pumping system, generating noise, vibration and wear.

Even if velocity-matching requirements are met, pressure at the pump inlet must be high enough to prevent gas formation/separation.

Liquid Supply System Relationships

1. Open Supply – Elevated Inlet Situation (also see Page A36).

If atmospheric pressure available at site is greater than the equation (NPSHR + Pv + hf + ha), it is possible to install the pump inlet above the level of the liquid. The maximum distance (Ie(max)) the pump inlet can be placed above the liquid level is determined by the following equation converted to units of feet:

Ie(max) = Pa – (NPSHR + Pv + hf + ha)

Where,

Ie = elevation distance from the center line of the pump inlet connection to the liquid level, feet

Ie(max) = maximum inlet elevation distance, feet

Pa = atmospheric pressure available at site

NPSHR = net positive suction head required (specified by pump manufacturer)

Pv = absolute vapor pressure (@ pumping temperature) plus 7 feet

hf = friction losses through pipe and fittings, feet

To convert psi (pounds per square inch) to feet of head, multiply psi by 2.31 then divide by specific gravity (S.G.).

feet of head = psi x 2.31

S.G.

If it is found that the actual distance which the pump may be installed above the liquid level is less than the maximum calculated distance, then NPSHA is equal to the equation NPSHA – Pa – (Pv + hf + ha + Ie), converted to feet.

In order to find whether the pump can be installed above the liquid level, a comparison of Pa and the evaluation of the equation (NPSHR + Pv + hf + ha) is necessary. If Pa is found to be the greater, the pump may be placed above the liquid level.

Example 13:

A single acting 1-3/4″ x 1-1/2″ triplex plunger pump is operating under the following conditions:

• the pump speed (n) is 200 rpm
• 9.18 GPM pump capacity (Q), barrels per day (BPD) = GPM x 34.3
• the volumetric efficiency (Ev) is 100%
• pumping fresh water @ 120 °F, specific gravity (S.G.) = 1.0
• NPSHR specified by manufacturer is 6 feet
• the entire system installation is at 9000 ft. altitude
• the supply tank is open to atmosphere
• the pump inlet connection is 1-1/2″ diameter
• inlet supply system consists of 10 feet of 2″ schedule pipe and a single 2″ x 1-1/2″ eccentric pipe reducer.

Problem: 1. Can this pump be installed above the liquid level? If yes, find Ie(max) ?

Finding values for the above mentioned equations, we ascertain the following:

Pa = atmospheric pressure = 10.5 psia (from Table 7 Appendix A). To convert to feet; multiply by 2.31 then divide by S.G.
Pa = 24.25 feet
Pv = water vapor pressure = 1.6924 psia (from Appendix E). To convert to feet; multiply by 2.31 then divide by S.G.

= 3.9 feet + 7 feet

Pv = 10.9 feet

hf = friction losses through pipe and fittings

2″ schedule pipe = 0.155 psi /100 feet (from Appendix D)

eccentric reducer = creates a friction loss in an equivalent length of 1 foot of 2″ pipe (see Appendix C)

Therefore,

hf (2) = 0.155 (10 + 1) psi
100
= 0.017 (2.31) feet
hf (2) = 0.039 feet
ha = acceleration head = L V n C
K g

Where,

L = length of liquid supply system piping = 10 feet
V = average liquid velocity through the pipe
= GPM = 9.18 = 0.996 fps
2.45 (D)2 2.45 (1.939)2 n = pump speed = 200 rpm
C = empirical constant for triplex = 0.066
(single acting or double acting)
K = liquid compressibility factor (water) = 1.5
g = standard gravity = 32.2 ft/sec2
Substituting into the equation for acceleration head,
ha = 2.72 feet
Substituting these values into the equation
(NPSHR + Pv + hf + ha) = (6 + 10.9 + 0.039 + 2.72) feet = 19.66 feet

The atmospheric pressure @ 9000 feet altitude is 24.25 feet and is greater than the summation of the equation above. Therefore, the pump can be installed above the liquid level.

Find Ie(max)
Ie(max) = Pa – (NPSHR + Pv + hf + ha) = 24.25 – 19.66
Ie(max) = 4.59 feet

Example 14:

Problem: Calculate NPSHA for the above conditions, if Ie = 2 feet.
Solution: The equation for Net Positive Suction Head Required is

NPSHA = Pa – (Pv + hf + ha + Ie) = 24.25 – (13.66 + 2)

NPSHA = 8.59 feet

Open Supply – Submerged Inlet Situation (also see Page A36)

If atmospheric pressure (Pa) is less than the equation (NPSHR + Pv + hf + ha), the pump can not be installed above the liquid level. With this situation, a positive static inlet head or a properly sized charging pump is necessary. The charging pump must have more capacity than the reciprocating pump to avoid cavitation from insufficient liquid available. If a charging pump is the only solution, it is usually best to have WGI specify what size and type of charging pump to use.

Keep in mind that, as with any pumping issue, in order for WGI to give you the proper information; we need all details of your specific pumping application.

If a pump inlet static head is possible, the minimum static head required (Hi(min) ) is calculated as follows:

Hi(min) = (NPSHR + Pv + hf + ha) – Pa

Where,

Hi(min) = minimum static head required

NPSHR = net positive suction head required (specified by pump manufacturer)

Pa = atmospheric pressure available at site elevation

Pv = absolute vapor pressure plus 7 feet (at pumping

temperature)

hf = friction losses through pipe, fittings, etc., feet

Example 15:

A single acting 3″ x 5″ quintuplex plunger pump operating under the following conditions:

• pump speed (n) = 300 rpm
• 229.5 GPM pump capacity (Q) = 229.5 GPM, BPD = 7,872
• pump volumetric efficiency (Ev) = 100%
• pumping fresh water @ 70°F with S.G. = 1
• NPSHR specified by pump manufacturer is 15 feet
• the entire system is at 5000 ft. altitude
• the supply tank is open to atmosphere
• the pump inlet connection is 6″ diameter
• liquid supply piping system is composed of:
• 10 ft. 8″ schedule 40 pipe
• 1 ea. 8″ gate valve
• 2 ea. 8″ long radius ells
• 15 ft. 6″ schedule 40 pipe
• 1 ea. 8″ x 6″ eccentric reducer

Problem: 1. Can this pump be installed above the liquid level? If not, find minimum static head required.

Values for the equation, Hi(min) = (NPSHR + Pv + hf + ha) – Pa, are found as follows;

Pa = atmospheric pressure (12.2 psia, Table 7, Appendix A)

To convert to feet, multiply by 2.31 then divide by S.G.

Pa = 28.18 feet

Pv = water vapor pressure (0.3631 psia) plus 7 feet

= to convert to feet, multiply by 2.31 then divide by S.G. +

7 feet = (0.83 + 7) feet

Pv = 7.83 feet

NPSHR = 15 feet (specified by WGI)

hf = hf(8) + hf(6)

equivalent length

description (Appendix C)

8″ gate valve 6 feet

8″ long radius ell 18 feet (9 ft./ea. x 2 ea.)

8″ pipe 10 feet

total equivalent feet 34 feet

The pressure drop through 8″ pipe is 0.05 psi/100 feet (see Appendix D) and for 34 equivalent feet is

hf(8) = 0.05 x 34 = 0.017 psi

100

To convert hf(8) to feet, multiply by 2.31 then divide by S.G.

hf(8) = 0.017 x 2.31 = 0.039 feet

1

equivalent length

description (Appendix C)

6″ gate valve 15 feet

8″ x 6″ eccentric reducer 4 feet

total equivalent feet 19 feet

The pressure drop through 6″ pipe is 0.17 psi/100 feet (see Appendix D) and for 19 equivalent feet is

hf(6) = 0.17 x 19 = 0.032 psi

100

To convert hf(6) to feet, multiply by 2.31 then divide by S.G.:

hf(6) = 0.032 x 2.31 = 0.074 feet

1

Therefore,

hf = hf(8) + hf(6) = (0.039 + 0.074) feet

hf = 0.133 feet

ha = acceleration head = L V n C

K g

In this case, we have two different ha values; ha (8) for 8″ pipe and ha (6) for 6″ pipe.

Where:

g = 32.2 ft./sec2

L = actual length of liquid supply system piping, feet

L(6) = 15 feet

L(8) = 10 feet

V = avg. fluid velocity through pipe, fps = GPM

(ID)2 x 2.45

n = pump speed = 300 rpm

C = empirical constant for quintuplex = 0.040

K = liquid compressibility factor (water) = 1.5

Find ha (8),

V(8) = 229.5 = 1.47 fps

(7.981)2 2.45

ha(8) = L(8) V(8) n C

K g

ha(8) = (10) x (1.47) x (300) x (0.04) = 3.65 feet

(1.5) x (32.2)

Find ha (6),

V(6) = 229.5 = 2.55 fps (6.065)2 2.45

ha(6) = L(6) V(6) n C

K g

ha(6) = (15) x (2.55) x (300) x (0.04) = 9.50 feet

(1.5) x (32.2)

Therefore,

ha = ha(8) + ha(6) = 13.15 feet

Using these values, we can now solve the equation

(NPSHR + Pv + hf + ha) = (15 + 7.83 + 0.11 + 13.15) = 36.09

Solution: 1. Since Pa (28.18 feet) is less than the above equation, the pump cannot be installed above the liquid level.

Solution: 2. The minimum positive static inlet head required is:

Hi(min) = 36.09 – 28.18 = 7.91 feet

If actual inlet static head (Hi(act)) is greater than the minimum inlet static head required (Hi(min)), the formula used for calculating NPSHA is

NPSHA = (Hi(act) + Pa) – (Pv + hf + ha)

Problem: For above conditions, if Hi(act) is 30 feet, find NPSHA

Solution: NPSHA = (30 + 28.18) – (7.83 + 0.11 + 13.15)

NPSHA = 37.09 feet

Closed Supply Vessel (also see Page A37)

When absolute pressure at source is equal to the absolute liquid vapor pressure (Psource = Pv), the minimum inlet static head required (Hi(min) ) must equal or exceed the sum or all the losses per the following equation:

Hi(min) = hf + ha + NPSHR

Example 16:

A double acting 7″ x 10″ duplex piston pump is operating under the following conditions:

• pump crankshaft speed (n) = 95 rpm
• pump capacity (Q) = 553.4 GPM (BPD = 18,980)
• pump volumetric efficiency (%)
• pumping crude oil, S.G. = 0.8 @ 100 centistokes viscosity
• NPSHR specified by the pump manufacturer is 12 feet of crude oil
• the entire system is at sea level
• the supply tank is a closed supply vessel with Psource = Pv
• pump inlet connection is 6″
• the liquid supply system piping consists of:
• 55 ft. 14″ schedule 30 pipe
• 1 ea. 14″ gate valve
• 3 ea. 14″ long radius ells
• 1 ea. 14″ x 6″ eccentric reducer

Problem: Find the minimum inlet static head required, Hi(min) ,

H = hf + ha + NPSHR

Where,

NPSHR = Net Positive Suction Head Required (specified by

manufacturer) is 12 feet

hf = friction losses through liquid supply system pipe

and fittings (for this condition there are two

different hf values; hf (14) for 14″ pipe and hf(6) for 6″ pipe.

Therefore, hf = hf(14) + hf(6)

equivalent feet

description (Appendix C)

14″ gate valve 10

14″ long radius ell (16 ft/ea x 3 ea) 48

14″ schedule 30 pipe 55

total equivalent length 113

The pressure drop through 14″ pipe is 0.02 psi/100 feet (see Appendix D) and for 113 equivalent feet is

hf(14) = 0.02 x 113 = 0.023 psi

100

Correcting for the 100 centistokes Kinematic viscosity (see Appendix D)

hf(14) = 0.023 x 3.5 = 0.08 psi

To convert hf (14) to feet, multiply by 2.31 then divide by S.G:

hf(14) = 0.08 x 2.31 = 0.23 feet

0.8

hf(6) = equivalent length of the 14″ x 6″ eccentric reducer in 6″ pipe (see Appendix C) is

hf(6) = 7 feet

Pressure drop through 6″ pipe is 0.83 psi / 100 feet (see Appendix D) and for 7 equivalent feet is

hf(6) = 0.83 x 7 = 0.058 psi

100

To convert hf (6) to feet, multiply by 2.31 then divide by S.G:

hf(6) = 0.20 x 2.31 = 0.57 feet

0.8

Substituting in for hf(14) and hf(6

Correcting for the 100 centistokes Kinematic viscosity (see Appendix D)

hf(6) = 0.058 x 3.5 = 0.20 psi

)

hf = hf(14) + hf(6) = (0.23 + 0.57) feet

hf = 0.80 feet

V = GPM = 53.4

(ID)2 2.45 (13.25)2 2.45

V = 1.29 fps

ha = acceleration head = L V n C

K g

Where,

g = standard gravity = 32.2 ft/sec2

K = liquid compressibility factor (oil) = 2

C = empirical constant for duplex = 0.115

n = pump speed = 85 rpm

L = actual pipe length = 55 feet

ha = (55) (1.29) (85) (0.115) = 10.77 feet

(2) (32.2)

Substituting these values into the equation for minimum static head

Hi(min) = hf + ha + NPSHR = 0.8 + 10.77 + 12

Hi(min) = 23.57 feet

At the above conditions, with actual inlet static head ( Hi(act) ) of 25 feet, find the Net Positive Suction Head Available

NPSHA = Hi(act) – (hf + ha) = 25 – (0.8 + 10.77)

NPSHA = 13.43 feet

Open Supply – Elevated Inlet System

Pa = atmospheric pressure available at site

hf = supply system frictional losses (pipe and fittings)

ha = supply system acceleration head

Pv = absolute vapor pressure plus 7 feet (at pumping temperature)

hv = velocity head through valve

hl = head of valve spring, valve weight, liquid friction due to viscosity, elevation from inlet center line to center line of piston or plunger

NPSHR = Net Positive Suction Head Required (specified by pump manufacturer)

Ie = inlet elevation

1. If Pa > (NPSHR + Pv + ha + hf ) then, the pump can be installed above the liquid level and maximum elevated inlet distance Ie(max) = Pa – (NPSHR + Pv + hf + ha)

2. If actual elevated inlet distance Ie < Ie(max) , then the available net positive inlet head is NPSHA = Pa – ( Pv + hf + ha + Ie )

Open Supply – Submerged Inlet System

Pa = atmospheric pressure available at site.

hf = supply system frictional losses (pipe and fittings)

ha = supply system acceleration head

Pv = absolute vapor pressure plus 7 feet (at pumping temperature)

hv = velocity head (through valve)

hl = head of valve spring, valve weight, liquid friction due to viscosity, elevation from inlet center line to center line of piston or plunger

NPSHR = Net Positive Suction Head Required (specified by pump manufacturer)

Hi = static inlet head required

1. If Pa < (NPSHR + Pv + ha + hf) then, the pump can be installed above the liquid level and the static inlet head required is Hi = ( NPSHR + Pv + hf + ha ) – Pa

2. If actual static inlet head exceeds minimum required static inlet head (Hi(act) > Hi(min) ) then NPSHA = ( Hi(act) + Pa ) – ( Pv + hf + ha )

Closed Supply System

P source = pressure of closed vessel from liquid source

Hi(min) = minimum static head required

ha = supply system acceleration head

Pv = absolute vapor pressure plus 7 feet (at pumping temperature)

hv = velocity head through valve

hl = head of valve spring, valve weight, liquid friction due to viscosity, elevation from inlet center line to center line of piston or plunger

NPSHR = Net Positive Suction Head Required (specified by pump manufacturer)

1. When absolute pressure at source equals the absolute liquid vapor pressure

Psource = Pv

then minimum static inlet head must equal, or exceed, the sum of losses Hi(min) = hf + ha + NPSHR

2. If actual static inlet head exceeds the minimum inlet head Hi(act) > Hi(min) then NPSHA = (Hi(act) – ( hf + ha )