## 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 + P_{v} + h_{f} + h_{a}), it is possible to install the pump inlet above the level of
the liquid. The maximum distance (I_{e(max)}) the pump inlet
can be placed above the liquid level is determined by the following equation

converted to units of feet:

I_{e(max)} = P_{a}
- (NPSHR + P_{v} + h_{f} + h_{a})

Where,

I_{e} = elevation distance from the center line of the
pump inlet

connection to the liquid level, feet

I_{e(max)} = maximum inlet elevation distance, feet

P_{a} = atmospheric pressure available at site

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

P_{v} = absolute vapor pressure (@ pumping temperature) **plus**

**7 feet**

h_{f} = friction losses through pipe and fittings, feet

h_{a} = acceleration head, 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 - P_{a} - (P_{v}
+ h_{f} + h_{a} + I_{e}), converted to feet.

In order to find whether the pump can be installed above the liquid
level, a comparison of P_{a} and the evaluation of the
equation (NPSHR + P_{v} + h_{f}
+ h_{a}) is necessary. If P_{a}
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 (E_{v}) 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 I_{e(max)} ?

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

P_{a} = atmospheric pressure = 10.5 psia (from Table 7
Appendix A). To convert to feet; multiply by 2.31 then divide by S.G.

P_{a} = 24.25 feet

P_{v} = 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 **

**P _{v} = 10.9 feet**

**h _{f} = 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, **

**h _{f }(2) = 0.155 (10 + 1) psi**

100

= 0.017 (2.31) feet

h_{f }(2) = 0.039 feet

h_{a} = 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/sec

^{2}Substituting into the equation for acceleration head, h

_{a}= 2.72 feet Substituting these values into the equation (NPSHR + P

_{v}+ h

_{f}+ h

_{a}) = (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 I _{e(max)}
I_{e(max)} = P_{a} - (NPSHR + P_{v} + h_{f}
+ h_{a}) = 24.25 - 19.66
I_{e(max)} = 4.59 feet**

_{}

_{}*Example 14:*

**Problem: Calculate NPSHA for the above conditions, if I _{e}
= 2 feet.
Solution: The equation for Net Positive Suction Head Required is**

**NPSHA = P _{a} - (P_{v} + h_{f} + h_{a} + I_{e})
= 24.25 - (13.66 + 2)**

**NPSHA = 8.59 feet**

**2. Open Supply - Submerged Inlet Situation (also see Page A36)**

**If atmospheric pressure (P _{a}) is less than the equation
(NPSHR + P_{v} + h_{f }+ h_{a}), 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 (H _{i(min)} ) is calculated as follows:**

**H _{i(min)} = (NPSHR + P_{v}
+ h_{f} + h_{a}) - P_{a}**

_{}**Where,**

**H _{i(min)} = minimum static head required**

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

**P _{a} = atmospheric pressure available at site elevation**

**P _{v} = absolute vapor pressure plus 7 feet (at pumping**

**temperature)**

**h _{f} = friction losses through pipe, fittings, etc.,
feet**

**h _{a} = acceleration head, 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 (E _{v}) = 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, H _{i(min)} =
(NPSHR + P_{v} + h_{f} + h_{a}) - P_{a}, are found as follows;**

**P _{a} = atmospheric pressure (12.2 psia, Table 7,
Appendix A)**

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

**P _{a} = 28.18 feet**

**P _{v} = 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**

**P _{v} = 7.83 feet**

**NPSHR = 15 feet (specified by WGI)**

**h _{f} = h_{f}(8) + h_{f}(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**

**h _{f}(8) = 0.05 x 34 = 0.017 psi**

**100**

**To convert h _{f}(8) to feet, multiply by 2.31 then divide
by S.G.**

**h _{f}(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**

**h _{f}(6) = 0.17 x 19 = 0.032 psi**

**100**

**To convert h _{f}(6) to feet, multiply by 2.31 then divide
by S.G.:**

**h _{f}(6) = 0.032 x 2.31 = 0.074 feet**

**1**

**Therefore, **

**h _{f} = h_{f}(8) + h_{f}(6) = (0.039 + 0.074) feet**

**h _{f} = 0.133 feet**

**h _{a} = acceleration head = L V n C**

**K g **

**In this case, we have two different h _{a} values; h_{a}
(8) for 8" pipe and h_{a} (6) for 6" pipe.**

**Where:**

**g = 32.2 ft./sec ^{2}**

^{}**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 h _{a} (8),**

**V(8) = 229.5 = 1.47 fps**

**(7.981) ^{2} 2.45 **

**h _{a}(8) = L(8) V(8) n C**

**K g**

**h _{a}(8) = (10) x (1.47) x (300) x (0.04) = 3.65 feet**

**(1.5) x (32.2)**

**Find h _{a} (6),**

**V(6) = 229.5 = 2.55 fps (6.065) ^{2} 2.45**

**h _{a}(6) = L(6) V(6) n C**

**K g**

**h _{a}(6) = (15) x (2.55) x (300) x (0.04) = 9.50 feet**

**(1.5) x (32.2)**

**Therefore,**

**h _{a} = h_{a}(8) + h_{a}(6) = 13.15 feet**

**Using these values, we can now solve the equation**

**(NPSHR + P _{v} + h_{f} + h_{a}) = (15 + 7.83 + 0.11 + 13.15) = 36.09**

**Solution: 1. Since P _{a} (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**

**H _{i(min)} = 36.09 - 28.18 = 7.91 feet**

**If actual inlet static head (H _{i(act)})
is greater than the minimum inlet static head required (H_{i(min)}),
the formula used for calculating NPSHA is**

**NPSHA = (H _{i(act)} + P_{a}) - (P_{v}
+ h_{f} + h_{a})**

**Problem: For above conditions, if H _{i(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 (P _{source} = P_{v}), the minimum
inlet static head required (H_{i(min)} ) must equal or
exceed the sum or all the losses per the following equation: **

**H _{i(min)} = h_{f} + h_{a} + 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 P _{source}
= P_{v}**

_{}**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, H _{i(min)}
,**

**H = h _{f} + h_{a} + NPSHR**

**Where,**

**NPSHR = Net Positive Suction Head Required (specified by**

**manufacturer) is 12 feet**

**h _{f} = friction losses through liquid
supply system pipe**

**and fittings (for this condition there are two**

**different h _{f} values; h_{f
}(14) for 14" pipe and h_{f}(6) for 6"
pipe.**

**Therefore, h _{f} = h_{f}(14)
+ h_{f}(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**

**h _{f}(14) = 0.02 x 113 = 0.023 psi**

**100**

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

**h _{f}(14) = 0.023 x 3.5 = 0.08 psi**

**To convert h _{f} (14) to feet, multiply by 2.31 then divide by
S.G:**

**h _{f}(14) = 0.08 x 2.31 = 0.23 feet**

**0.8**

**h _{f}(6) = equivalent length of the
14" x 6" eccentric reducer in 6" pipe (see Appendix C) is **

**h _{f}(6) = 7 feet**

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

**h _{f}(6) = 0.83 x 7 = 0.058 psi**

**100**

**To convert h _{f} (6) to feet, multiply by 2.31 then divide by
S.G:**

**h _{f}(6) = 0.20 x 2.31 = 0.57 feet**

**0.8**

**Substituting in for h _{f}(14) and h_{f}(6**

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

**h _{f}(6) = 0.058 x 3.5 = 0.20 psi**

**)**

**h _{f} = h_{f}(14)
+ h_{f}(6) = (0.23 + 0.57) feet**

**h _{f} = 0.80 feet**

**V = GPM = 53.4 **

**(ID) ^{2} 2.45 (13.25)^{2 }2.45**

**V = 1.29 fps**

**h _{a} = acceleration head = L V n C**

**K g**

**Where,**

**g = standard gravity = 32.2 ft/sec ^{2}**

^{}**K = liquid compressibility factor (oil) = 2**

**C = empirical constant for duplex = 0.115**

**n = pump speed = 85 rpm**

**L = actual pipe length = 55 feet**

**h _{a} = (55) (1.29) (85) (0.115) = 10.77
feet**

**(2) (32.2)**

**Substituting these values into the equation for minimum static head**

**H _{i(min)} = h_{f}
+ h_{a} + NPSHR = 0.8 + 10.77 + 12**

**H _{i(min)} = 23.57 feet**

**At the above conditions, with actual inlet static head ( H _{i(act)} ) of 25 feet, find the Net Positive Suction Head
Available**

**NPSHA = H _{i(act)} - (h_{f}
+ h_{a}) = 25 - (0.8 + 10.77)**

**NPSHA = 13.43 feet**

**Open Supply - Elevated Inlet System**

P_{a} = atmospheric pressure available at
site

h_{f} = supply system frictional losses
(pipe and fittings)

h_{a} = supply system acceleration head

P_{v} = absolute vapor pressure **plus 7
feet** (at pumping temperature)

h_{v} = velocity head through valve

h_{l} = 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)

I_{e} = inlet elevation

1. If P_{a} > (NPSHR + P_{v}
+ h_{a} + h_{f }) then, the
pump can be installed above the liquid level and maximum elevated inlet distance I_{e(max)} = P_{a} - (NPSHR + P_{v} + h_{f} + h_{a})

2. If actual elevated inlet distance I_{e}
< I_{e(max)} , then the available net positive inlet head
is NPSHA = P_{a} - ( P_{v} + h_{f} + h_{a} + I_{e
})

**Open Supply - Submerged Inlet System**

P_{a} = atmospheric pressure available at
site.

h_{f} = supply system frictional losses
(pipe and fittings)

h_{a} = supply system acceleration head

P_{v} = absolute vapor pressure **plus 7
feet** (at pumping temperature)

h_{v} = velocity head (through valve)

h_{l} = 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)

H_{i} = static inlet head required

1. If P_{a} < (NPSHR + P_{v}
+ h_{a} + h_{f}) then,
the pump can be installed above the liquid level and the static inlet head required is H_{i} = ( NPSHR + P_{v} + h_{f} + h_{a }) - P_{a}

_{}

2. If actual static inlet head exceeds minimum required static inlet
head (H_{i(act)} > H_{i(min)}
) then NPSHA = ( H_{i(act)} + P_{a}
) - ( P_{v} + h_{f} + h_{a })

**Closed Supply System**

P_{ source} = pressure of closed vessel
from liquid source

H_{i(min)} = minimum static head required

h_{a} = supply system acceleration head

P_{v} = absolute vapor pressure **plus 7
feet** (at pumping temperature)

h_{v} = velocity head through valve

h_{l} = 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

P_{source} = P_{v}

_{}

then minimum static inlet head must equal, or exceed, the sum of losses
H_{i(min)} = h_{f} + h_{a }+ NPSHR

2. If actual static inlet head exceeds the minimum inlet head H_{i(act)} > H_{i(min)} then NPSHA
= (H_{i(act)} - ( h_{f} + h_{a })