P
•
A
•
R
•
T2
PLANT AND
FACILITIES
ENGINEERING
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
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PLANT AND FACILITIES ENGINEERING
7.3
SECTION 7
PUMPS AND
PUMPING SYSTEMS
PUMP OPERATING MODES AND
CRITICALITY
7.3
Series Pump Installation Analysis
7.3
Parallel Pumping Economics
7.5
Using Centrifugal Pump Specific
Speed to Select Driver Speed

Liquid Pumps
7.41
Condensate Pump Selection for a
Steam Power Plant
7.43
Minimum Safe Flow for a Centrifugal
Pump
7.46
Selecting a Centrifugal Pump to
Handle a Viscous Liquid
7.47
Pump Shaft Deflection and Critical
Speed
7.49
Effect of Liquid Viscosity on
Regenerative-Pump Performance
7.51
Effect of Liquid Viscosity on
Reciprocating-Pump Performance
7.52
Effect of Viscosity and Dissolved Gas
on Rotary Pumps
7.53
Selection of Materials for Pump Parts
7.56
Sizing a Hydropneumatic Storage
Tank
7.56
Using Centrifugal Pumps as Hydraulic
Turbines

kW) are being considered. Can energy be conserved, and how much, with some
other pumping arrangement?
Calculation Procedure:
1. Plot the characteristic curves for the pumps being considered
Figure 2 shows the characteristic curves for the proposed pumps. Point 1 in Fig. 1
is the proposed operating head and flow rate. An alternative pump choice is shown
at Point 2 in Fig. 1. If two of the smaller pumps requiring only 0.25 hp (0.19 kW)
each are placed in series, they can generate the required 26-ft (7.9-m) head.
2. Analyze the proposed pumps
To analyze properly the proposal, a new set of curves, Fig. 2, is required. For the
proposed series pumping application, it is necessary to establish a seriesed pump
curve. This is a plot of the head and flow rate (capacity) which exists when both
pumps are running in series. To construct this curve, double the single-pump head
values at any given flow rate.
Next, to determine accurately the flow a single pump can deliver, plot the
system-head curve using the same method fully described in the previous calcula-
tion procedure. This curve is also plotted on Fig. 2.
Plot the point of operation for each pump on the seriesed curve, Fig. 2. The
point of operation of each pump is on the single-pump curve when both pumps are
operating. Each pump supplies half the total required head.
When a single pump is running, the point of operation will be at the intersection
of the system-head curve and the single-pump characteristic curve, Fig. 2. At this
point both the flow and the hp (kW) input of the single pump decrease. Series
pumping, Fig. 2, requires the input motor hp (kW) for both pumps; this is the point
of maximum power input.
3. Compute the possible savings
If the system requires a constant flow of 45 gal/min (2.84 L/ s) at 26-ft (7.9-m)
head the two-pump series installation saves (0.75 hp
Ϫ
2

15
10
5
0
0 1020304050607080
0
2.5
5.0
7.5
10.0
12345
3/4 HP PUMP
(0.56 kW)
1/2 HP PUMP
(0.37 kW)
1/4 HP PUMP
(0.19 kW)
1/6 HP PUMP
(0.12 kW)
1
2
GPM
L/s
HEAD - FEET
Head, m
FIGURE 1 Pump characteristic curves for use in series installation.
PARALLEL PUMPING ECONOMICS
A system proposed for heating a 20,000-ft
2
(1858-m

5.0
7.5
10.0
12 3 4 5
GPM
L/s
HEAD - FEET
Head, m
OPERATING POINT
OF EACH PUMP WHEN
BOTH ARE RUNNING
SINGLE PUMP
OPERATING POINT
SINGLE PUMP
CURVE
SYSTEM CURVE
SERIESED
PUMP CURVE
DESIGN
OPERATING
CONDITION
FIGURE 2 Seriesed-pump characteristic and system-head curves.
required system flow can be handled by two pumps, one an operating unit and one
a spare unit. Each pump will have an 0.5-hp (0.37-kW) drive motor. Could there
be any appreciable energy saving using some other arrangement? The system re-
quires 50 hours of constant pump operation and 40 hours of partial pump operation
per week.
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Figure 3 shows the proposed hot-water heating-pump selection for this industrial
building. Looking at the values of the pump head and capacity in Fig. 3, it can be
seen that if the peak load of 80 gal /min (7.4 L /s) were carried by two pumps, then
each would have to pump only 40 gal/min (3.7 L/s) in a parallel arrangement.
2. Plot a characteristic curve for the pumps in parallel
Construct the paralleled-pump curve by doubling the flow of a single pump at any
given head, using data from the pump manufacturer. At 13-ft head (3.96-m) one
pump produces 40 gal /min (3.7 L /s); two pumps 80 gal/min (7.4 L /s). The re-
sulting curve is shown in Fig. 4.
The load for this system could be divided among three, four, or more pumps, if
desired. To achieve the best results, the number of pumps chosen should be based
on achieving the proper head and capacity requirements in the system.
3. Construct a system-head curve
Based on the known flow rate, 80 gal/ min (7.4 L /s) at 13-ft (3.96-m) head, a
system-head curve can be constructed using the fact that pumping head varies as
the square of the change in flow, or Q
2
/Q
1
ϭ
H
2
/H
1
, where Q
1
ϭ
known design
flow, gal /min (L/s); Q
2

4.5
3.0
1.5
0
Head, m
0 20 40 60 80 100 120 140 160
GALLONS PER MINUTE
ONE PUMP TWO PUMPS
Paralleled
FIGURE 4 Single- and dual-parallel pump characteristic curves.
25
20
15
10
5
0
9.0
7.5
6.0
4.5
3.0
1.5
0
5010
L/s
FEET OF HEAD
Head, m
0 20 40 60 80 100 120 140 160
TWO PUMPS
SINGLE PUMP

curves are available free of charge from the pump manufacturer.
The pump operating point is at the intersection of the pump characteristic curve
and the system-head curve in conformance with the first law of thermodynamics,
which states that the energy put into the system must exactly match the energy
used by the system. The intersection of the pump characteristic curve and the
system-head curve is the only point that fulfills this basic law.
There is no practical limit for pumps in parallel. Careful analysis of the system-
head curve versus the pump characteristic curves provided by the pump manufac-
turer will frequently reveal cases where the system load point may be beyond the
desired pump curve. The first cost of two or three smaller pumps is frequently no
greater than for one large pump. Hence, smaller pumps in parallel may be more
desirable than a single large pump, from both the economic and reliability stand-
points.
One frequently overlooked design consideration in piping for pumps is shown
in Fig. 6. This is the location of the check valve to prevent reverse-flow pumping.
Figure 6 shows the proper location for this simple valve.
5. Compute the energy saving possible
Since one pump can carry the fluid flow load about 90 percent of the time, and
this same percentage holds for the design conditions, the saving in energy is
0.9
ϫ
(0.5 kW
Ϫ
.25 kW)
ϫ
90 h per week
ϭ
20.25 kWh / week. (In this com-
putation we used the assumption that 1 hp
ϭ

SELECT DRIVER SPEED
A double-suction condenser circulator handling 20,000 gal /min (75,800 L/min) at
a total head of 60 ft (18.3 m) is to have a 15-ft (4.6-m) lift. What should be the
rpm of this pump to meet the capacity and head requirements?
Calculation Procedure:
1. Determine the specific speed of the pump
Use the Hydraulic Institute specific-speed chart, Fig. 7, page 7.11. Entering at 60
ft (18.3 m) head, project to the 15-ft suction lift curve. At the intersection, read the
specific speed of this double-suction pump as 4300.
2. Use the specific-speed equation to determine the pump operating rpm
Solve the specific-speed equation for the pump rpm. Or rpm
ϭ
N
s
ϫ
0.75 0.5
H /Q ,
where N
s
ϭ
specific speed of the pump, rpm, from Fig. 7; H
ϭ
total head on pump,
ft (m); Q
ϭ
pump flow rate, gal/min (L/min). Solving, rpm
ϭ
4300
ϫϭ
655.5 r/min. The next common electric motor rpm

pump rpm for centrifugal pumps used in boiler feed, industrial, marine, HVAC, and
similar applications. Note that the latest Hydraulic Institute curves should be used.
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PUMPS AND PUMPING SYSTEMS
7.12
PLANT AND FACILITIES ENGINEERING
RANKING EQUIPMENT CRITICALITY TO COMPLY
WITH SAFETY AND ENVIRONMENTAL
REGULATIONS
Rank the criticality of a boiler feed pump operating at 250
Њ
F (121
Њ
C) and 100 lb /
in
2
(68.9 kPa) if its Mean Time Between Failures (MTBF) is 10 months, and vi-
bration is an important element in its safe operation. Use the National Fire Protec-
tion Association (NFPA) ratings of process chemicals for health, fire, and reactivity
hazards. Show how the criticality of the unit is developed.
Calculation Procedure:
1. Determine the Hazard Criticality Rating
(
HCR) of the equipment
Process industries of various types—chemical, petroleum, food, etc.—are giving
much attention to complying with new process safety regulations. These efforts
center on reducing hazards to people and the environment by ensuring the me-
chanical and electrical integrity of equipment.

ϭ
2.
The rating of Other Hazards, O, Table 1, is O
ϭ
1, because of the high tem-
perature of the water. Thus, the Hazard Criticality Rating, HCR
ϭ
2, found from
the higher numerical value of PCH and O.
2. Determine the Process Criticality Rating, PCR, of the equipment
From Table 2, prepared by the PHA Group using the results of its study of the
equipment in the plant, PCR
ϭ
3. The reason for this is that the boiler feed pump
is critical for plant operation because its failure will result in reduced capacity.
3. Find the Process and Hazard Criticality Rating, PHCR
The alphanumeric PHC value is represented first by the alphabetic character for the
category. For example, Category A is the most critical, while Category D is the
least critical to plant operation. The first numeric portion represents the Hazard
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PUMPS AND PUMPING SYSTEMS
PUMPS AND PUMPING SYSTEMS
7.13
TABLE 1
The Hazard Criticality Rating (HCR) is Determined in Three Steps*
Hazard Criticality Rating
1. Assess the Process Chemical Hazard (PCH) by:
●

Potential for release of regulated chemicals is high
Release causes possible serious health safety or environmental effects
Plant requires steam turbine trip mechanisms, fired-equipment shutdown systems,
or toxic- or combustible-gas detectors†
Failure of pollution control system results in environmental damage†
●
Extremely dangerous (3), if:
Equipment rotates at
Ͼ
5000 r/min
Temperatures
Ͼ
500
Њ
F
Plant requires process venting devices
Potential for release of regulated chemicals is low
Failure of pollution control system may result in environmental damage†
●
Hazardous (2), if:
Temperatures
Ͼ
300
Њ
F;
Extended failure of pollution control system may cause damage†
●
Slightly hazardous (1), if:
Equipment rotates at
Ͼ

as: Highest Priority, Category A; High Priority, Category B; Medium Priority, Cat-
egory C; Low Priority, Category D.
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PUMPS AND PUMPING SYSTEMS
7.14
PLANT AND FACILITIES ENGINEERING
TABLE 2
Process Criticality Rating*
Process Criticality Rating
Essential
(4)
The equipment is essential if failure will result in shutdown of the unit,
unacceptable product quality, or severely reduced process yield
Critical
(3)
The equipment is critical if failure will result in greatly reduced capacity,
poor product quality, or moderately reduced process yield
Helpful
(2)
The equipment is helpful if failure will result in slightly reduced capacity,
product quality or reduced process yield
Not critical
(1)
The equipment is not critical if failure will have little or no process conse-
quences
*Chemical Engineering.
TABLE 3
The Process and Hazard Criticality Rating*

When preparing for a PHCR evaluation, a spreadsheet, Table 6, listing critical
equipment, should be prepared. Then, as the various rankings are determined, they
can be entered in the spreadsheet where they are available for easy reference.
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PUMPS AND PUMPING SYSTEMS
PUMPS AND PUMPING SYSTEMS
7.15
TABLE 4
The Criticality and Repetitive
Equipment Values*
CRE Values
PHCR
Mean time between
failures, months
0–6 6–12 12–24
Ͼ
24
Aa1a2a3a4
Ba2b1b2b3
Ca3b2c1c2
Da4b3c2d1
*Chemical Engineering.
TABLE 5
Predictive Maintenance
Frequencies for Rotating Equipment
Based on Their CRE Values*
Maintenance cycles
CRE

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PUMPS AND PUMPING SYSTEMS
7.16
PLANT AND FACILITIES ENGINEERING
then be used to ensure the mechanical integrity of critical equipment by prioritizing
predictive and preventive maintenance programs, inventories of critical spare parts,
and maintenance work orders in case of plant upsets.
In any plant, the hazards posed by different operating units are first ranked and
prioritized based on a PHA. These rankings are then used to determine the order
in which the hazards need to be addressed. When the PHAs approach completion,
team members evaluate the equipment in each operating unit using the PHCR sys-
tem.
The procedure presented here can be used in any plant concerned with human
and environmental safety. Today, this represents every plant, whether conventional
or automated. Industries in which this procedure finds active use include chemical,
petroleum, textile, food, power, automobile, aircraft, military, and general manu-
facturing.
This procedure is the work of V. Anthony Ciliberti, Maintenance Engineer, The
Lubrizol Corp., as reported in Chemical Engineering magazine.
Pump Affinity Laws, Operating Speed,
and Head
SIMILARITY OR AFFINITY LAWS FOR
CENTRIFUGAL PUMPS
A centrifugal pump designed for a 1800-r /min operation and a head of 200 ft (60.9
m) has a capacity of 3000 gal/min (189.3 L/ s) with a power input of 175 hp (130.6
kW). What effect will a speed reduction to 1200 r /min have on the head, capacity,
and power input of the pump? What will be the change in these variables if the
impeller diameter is reduced from 12 to 10 in (304.8 to 254 mm) while the speed
is held constant at 1800 r/min?
Calculation Procedure:

1
/N
2
)
3
. For a constant speed, Q
1
/
Q
2
ϭ
D
1
/D
2
; H
1
/H
2
ϭ
(D
1
/D
2
)
2
; P
1
/P
2

; 3000 /
Q
2
ϭ
1800/1200; Q
2
ϭ
2000 gal / min (126.2 L /s). And, H
1
/H
2
ϭ
(N
1
/
N
2
)
2
ϭ
200/ H
2
ϭ
(1800/1200)
2
; H
2
ϭ
88.9 ft (27.1 m). Also, P
1

Q
2
ϭ
D
1
/D
2
; 3000/ Q
2
ϭ
12
⁄
10
; Q
2
ϭ
2500 gal/ min (157.7 L /s). And H
1
/
H
2
ϭ
(D
1
/D
2
)
2
; 200/H
2

3
; P
2
ϭ
101.2 bhp (75.5 kW).
Related Calculations. Use the similarity laws to extend or change the data
obtained from centrifugal pump characteristic curves. These laws are also useful in
field calculations when the pump head, capacity, speed, or impeller diameter is
changed.
The similarity laws are most accurate when the efficiency of the pump remains
nearly constant. Results obtained when the laws are applied to a pump having a
constant impeller diameter are somewhat more accurate than for a pump at constant
speed with a changed impeller diameter. The latter laws are more accurate when
applied to pumps having a low specific speed.
If the similarity laws are applied to a pump whose impeller diameter is increased,
be certain to consider the effect of the higher velocity in the pump suction line.
Use the similarity laws for any liquid whose viscosity remains constant during
passage through the pump. However, the accuracy of the similarity laws decreases
as the liquid viscosity increases.
SIMILARITY OR AFFINITY LAWS IN
CENTRIFUGAL PUMP SELECTION
A test-model pump delivers, at its best efficiency point, 500 gal /min (31.6 L /s) at
a 350-ft (106.7-m) head with a required net positive suction head (NPSH) of 10 ft
(3 m) a power input of 55 hp (41 kW) at 3500 r /min, when a 10.5-in (266.7-mm)
diameter impeller is used. Determine the performance of the model at 1750 r/min.
What is the performance of a full-scale prototype pump with a 20-in (50.4-cm)
impeller operating at 1170 r/min? What are the specific speeds and the suction
specific speeds of the test-model and prototype pumps?
Calculation Procedure:
1. Compute the pump performance at the new speed

(prototype speed, r/min)/(model speed, r/min). Other symbols are the same
as in the previous calculation procedure.
When the model speed is reduced from 3500 to 1750 r/ min, the pump dimen-
sions remain the same and K
d
ϭ
1.0; K
n
ϭ
1750/3500
ϭ
0.5. Then Q
ϭ
(1.0)(0.5)(500)
ϭ
250 r / min; H
ϭ
(1.0)
2
(0.5)
2
(350)
ϭ
87.5 ft (26.7 m); NPSH
ϭ
(1.0)
2
(0.5)
2
(10)

ϭ
1170/3500
ϭ
0.335.
Then Q
p
ϭ
(1.905)
3
(0.335)(500)
ϭ
1158 gal/min (73.1 L/ s); H
p
ϭ
(1.905)
2
(0.335)
2
(350)
ϭ
142.5 ft (43.4 m); NPSH
p
ϭ
(1.905)
2
(0.335)
2
(10)
ϭ
4.06

13,900. For the prototype, N
s
ϭϭ
965;
0.75 0.5 0.75
(10) 1170(1158) /(142.5)
S
ϭϭ
13,900. The specific speed and suction specific
0.5 0.75
1170(1156) /(4.06)
speed of the model and prototype are equal because these units are geometrically
similar or homologous pumps and both speeds are mathematically derived from the
similarity laws.
Related Calculations. Use the procedure given here for any type of centrifugal
pump where the similarity laws apply. When the term model is used, it can apply
to a production test pump or to a standard unit ready for installation. The procedure
presented here is the work of R. P. Horwitz, as reported in Power magazine.
1
SPECIFIC SPEED CONSIDERATIONS IN
CENTRIFUGAL PUMP SELECTION
What is the upper limit of specific speed and capacity of a 1750-r / min single-stage
double-suction centrifugal pump having a shaft that passes through the impeller
eye if it handles clear water at 85
Њ
F (29.4
Њ
C) at sea level at a total head of 280 ft
(85.3 m) with a 10-ft (3-m) suction lift? What is the efficiency of the pump and
its approximate impeller shape?

6040 gal /min (381.1
0.75 0.75
(NH (2000
ϫ
280
St
L/s).
1
R. P. Horwitz, ‘‘Affinity Laws and Specific Speed Can Simplify Centrifugal Pump Selection,’’ Power,
November 1964.
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PUMPS AND PUMPING SYSTEMS
PUMPS AND PUMPING SYSTEMS
7.19
FIGURE 8 Upper limits of specific speeds of single-suction mixed-flow and axial-flow pumps.
(Hydraulic Institute.)
3. Determine the pump efficiency and impeller shape
Figure 9 shows the general relation between impeller shape, specific speed, pump
capacity, efficiency, and characteristic curves. At N
S
ϭ
2000, efficiency
ϭ
87 per-
cent. The impeller, as shown in Fig. 9, is moderately short and has a relatively
large discharge area. A cross section of the impeller appears directly under the
N
S

1. Determine the specific speed and suction specific speed
Ac motors can operate at a variety of speeds, depending on the number of poles.
Assume that the motor driving this pump might operate at 870, 1160, 1750, or
3500 r/min. Compute the specific speed N
S
ϭ
N
ϭ
0.5 0.75
(Q)/(H)
ϭ
3.14N and the suction specific speed S
ϭ
0.5 0.75 0.5
N(10,000) /(100) N(Q)/
ϭϭ
7.43N for each of the assumed speeds. Tabu-
0.75 0.5 0.75
(NPSH) N(10,000) /(32)
late the results as follows:
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PUMPS AND PUMPING SYSTEMS
PUMPS AND PUMPING SYSTEMS
7.21
TABLE 7
Pump Types Listed by Specific
Speed*
TABLE 8

PLANT AND FACILITIES ENGINEERING
FIGURE 10 Typical pump suction and discharge piping arrangements.
of the source of the liquid supply to the free surface of the liquid in the discharge
receiver, or to the point of free discharge from the discharge pipe. When both the
suction and discharge surfaces are open to the atmosphere, the total static head
equals the vertical difference in elevation. Use the free-surface elevations that cause
the maximum suction lift and discharge head, i.e., the lowest possible level in the
supply tank and the highest possible level in the discharge tank or pipe. When the
supply source is below the pump centerline, the vertical distance is called the static
suction lift; with the supply above the pump centerline, the vertical distance is
called static suction head. With variable static suction head, use the lowest liquid
level in the supply tank when computing total static head. Label the diagrams as
shown in Fig. 10.
2. Compute the total static head on the pump
The total static head H
ts
ft
ϭ
static suction lift, h
sl
ft
ϩ
static discharge head h
sd
ft,
where the pump has a suction lift, s in Fig. 10a, b, and c. In these installations,
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PUMPS AND PUMPING SYSTEMS

ft. In these installations,
H
t
ϭ
100
Ϫ
15
ϭ
85 ft (25.9 m).
The total static head, as computed above, refers to the head on the pump without
liquid flow. To determine the total head on the pump, the friction losses in the
piping system during liquid flow must be also determined.
3. Compute the piping friction losses
Mark the length of each piece of straight pipe on the piping drawing. Thus, in Fig.
10a, the total length of straight pipe L
t
ft
ϭ
8
ϩ
10
ϩ
5
ϩ
102
ϩ
5
ϭ
130 ft (39.6
m), if we start at the suction tank and add each length until the discharge tank is

32.2 ft/s
2
(980.67 cm/s
2
). The exit loss occurs when the liquid
passes through a sudden enlargement, as from a pipe to a tank. Where the area of
the tank is large, causing a final velocity that is zero, h
ex
ϭ v
2
/2g .
The velocity
v
ft/s in a pipe
ϭ
gpm /2.448d
2
. For this pipe,
v ϭ
2000/
[(2.448)(7.98)
2
]
ϭ
12.82 ft/ s (3.91 m/ s). Then h
e
ϭ
0.74(12.82)
2
/[2(32.2)]

H
ƒ
(5.86)(248/100)
ϭ
14.53 ft (4.5 m).
4. Compute the total head on the pump
The total head on the pump H
t
ϭ
H
ts
ϩ
For the pump in Fig. 10a,H .
ƒ
H
t
ϭ
110
ϩ
14.53
ϭ
124.53 ft (37.95 m), say 125 ft (38.1 m). The total head on
the pump in Fig. 10b and c would be the same. Some engineers term the total head
on a pump the total dynamic head to distinguish between static head (no-flow
vertical head) and operating head (rated flow through the pump).
The total head on the pumps in Fig. 10d, c, and ƒ is computed in the same way
as described above, except that the total static head is less because the pump has
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