## V. SUPPLY SYSTEM CONSIDERATIONS

### J. 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:
a. the pump speed (n) is 200 rpm
b. 9.18 GPM pump capacity (Q), barrels per day (BPD) = GPM x 34.3
c. the volumetric efficiency (Ev) is 100%
d. pumping fresh water @ 120 °F, specific gravity (S.G.) = 1.0
e. NPSHR specified by manufacturer is 6 feet
f. the entire system installation is at 9000 ft. altitude
g. the supply tank is open to atmosphere
h. the pump inlet connection is 1-1/2" diameter
i. 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?

2. 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

2. 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:

a. pump speed (n) = 300 rpm

b. 229.5 GPM pump capacity (Q) = 229.5 GPM, BPD = 7,872

c. pump volumetric efficiency (Ev) = 100%

d. pumping fresh water @ 70°F with S.G. = 1

e. NPSHR specified by pump manufacturer is 15 feet

f. the entire system is at 5000 ft. altitude

g. the supply tank is open to atmosphere

h. the pump inlet connection is 6" diameter

i. 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?

2. 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

3. 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:

a. pump crankshaft speed (n) = 95 rpm

b. pump capacity (Q) = 553.4 GPM (BPD = 18,980)

c. pump volumetric efficiency (%)

d. pumping crude oil, S.G. = 0.8 @ 100 centistokes viscosity

e. NPSHR specified by the pump manufacturer is 12 feet of crude oil

f. the entire system is at sea level

g. the supply tank is a closed supply vessel with Psource = Pv

h. pump inlet connection is 6"

i. 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 )