Energy-Efficient Buildings
Piping and Pump Systems
Approach to Energy-Efficient Fluid Flow
Pumps and fans move fluids in a wide variety of applications. Careful analysis can improve
the energy efficiency of most pump and fan systems. This chapter discusses fundamentals
of fluid flow systems, with an emphasis on energy-efficient design, retrofit and operation.
This chapter includes:
 Derivation of the fluid work equation
 Methods to calculate pressure loss through pipes and fittings
 Introduction to piping and duct system design
 Key principles for designing and operating low-energy fluid flow systems
Fluid Work Equation
The work required to move a fluid through a pipe or duct can be derived from an energy
balance on the system. Assuming steady state conditions, an energy balance on the system
in Figure 1 gives:
Q
2
1
Wf
Figure 1. Control-volume diagram of pumping system.
Wf + m1(h + V2/2 + gz)1 - m2(h + V 2/2 + gz)2 – Q = 0
Wf = m2(h + V 2/2 + gz)2 – m1(h + V 2/2 + gz)1 + Q
[1]
where Wf is the rate of work transmitted to the fluid, m is the mass flow rate, h is the
specific enthalpy, V is the velocity, g is the acceleration of gravity, z is the height above a
fixed reference and Q is the rate of heat loss from the system. From conservation of mass
and from the definition of enthalpy:
m1 = m2 = m
h = u + Pv.
1
Substituting m1 = m2 = m and h = u + Pv into Equation 1 gives:
Wf = m [ (u + Pv + V 2/2 + gz)2 – (u + Pv + V 2/2 + gz)1 + q ]
Wf = m [ (Pv + V 2/2 + gz)2 – (Pv + V 2/2 + gz)1 + (q + u2 – u1) ]
Wf = m [ (Pv + V 2/2 + gz)2 – (Pv + V 2/2 + gz)1 ] + (Q + U2 – U1)
where u is the specific internal energy, U is the internal energy, P is the pressure and v is the
specific volume. Assuming the density  of the fluid does not change,
Substituting , m = V  and v = 1/ gives:
Wf = V  [ (P/ + V 2/2 + gz)2 – (P/ + V 2/2 + gz)1 ] + (Q + U2 – U1)
Wf = V [ (P + V 2/2 + gz)2 – (P + V 2/2 + gz)1 ] + (Q + U2 – U1)
Wf = V [ (P2 – P1) + V22- V 12) + g(z2 – z1) ] + (Q + U2 – U1)
where V is the volume flow rate. The term (Q + U2 – U1) represents the net energy added to
the fluid from friction with the pipe/duct walls. To be consistent with the other terms, it is
useful to write (Q + U2 – U1) in terms of pressure drop. Thus:
(Q + U2 – U1) = m (q + u2 – u1) = V  (q + u2 – u1) = V  (hl)
where hl is the “headloss” in units of specific energy (Btu/lb or J/kG) due to friction between
the fluid and pipes, ducts and fittings.
Substituting (Q + U2 – U1) = V  (hl) gives:
Wf = V [ (P2 – P1) + V22- V12) + g(z2 – z1) + (hl) ]
[2]
A number of interesting observations can be made about Equation 2. First, each of the
terms (P2 – P1), V 22- V 21), g(z2 – z1) and (hl) have units of pressure. Thus, the fluid
work necessary to propel the fluid can be written in terms of W = V P. The term (P2 – P1)
represents the “static pressure difference” between the inlet and outlet. The term V22V12) represents the “velocity pressure difference” between the inlet and outlet. The term
g(z2 – z1) represents the “elevation pressure difference” between the inlet and outlet. The
term (hl) represents the “friction pressure drop” as the fluid flows through the pipes or
ducts. Thus, the equation for the energy required to move an incompressible fluid through
pipes or ducts, Wf, can be written as:
Wf = V [ Pstatic + Pvelocity + Pelevation + Pfriction ] = V Ptotal
[3]
The first three components of the total pressure loss (Pstatic , Pvelocity , Pelevation) refer to
differences between the inlet and outlet of the system. The forth component of the total
pressure loss, Pfriction, refers to irreversible friction losses in the pipes and ducts and is
2
always present (non-zero) in all real pump/fan applications. Thus, the total pressure drop
can also be written as:
Ptotal = (Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition
And the equation for Wf can be written as:
Wf = V [(Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition]
Pressure and Head
Historically, pressure was often measured using a manometer, and the pressure difference
between a fluid and the atmosphere was expressed in terms of the difference in height
between levels of liquid in the manometer. Using a manometer, pressure difference is:

P = g h
[4]
where g is the acceleration of gravity,  is the density of the fluid in the manometer, and h
is the height of the fluid column. When a pressure difference is characterized in terms of
h, it is frequently called head. Thus, when pressure loss due to friction in pipes or ducts
measured in terms of h, it is often called friction head or head loss. Similarly, when the
pressure required to lift a fluid against the force of gravity is measured in terms of h, it is
often called elevation head. When the fluid in the manometer is water, the relationship
between pressure and head is:
h = P / (g H20
In pump systems, head is often expressed as the difference in height, h, between levels of
a water-filled manometer in units of feet of water, ft-H20 or, equivalently, ft-wg. In fan
systems, h is typically measured in inches of water, in-H20, or, equivalently, in-wg.
Common conversions between pressure and manometer height are:
1 lb/in2 = 27.7 in-H20 = 2.31 ft-H20
Dimensional Equations for Fluid Work
Pump Systems
In U.S. units, a useful dimensional equation to calculate the fluid work, in horsepower, to
move water at standard conditions (P = 1 atm, T = 60 F) through pipes is:
Wf = V Ptotal = V (g  h)
Wf (hp) = V (gal/min) htotal (ft-H20) / 3,960 (gal-ft-H20/min-hp)
3
[5]
The volume flow rate in this equation is the product of the mass flow rate and density of the
fluid. Thus, the pumping equation is easily modified for any fluid with a density different
than water at standard conditions by including term for the specific gravity of the fluid, SGf.
Specific gravity is the ratio of the density of the fluid to the density of water at standard
conditions.
SGf = fH20
Wf (hp) = V (gal/min) htotal (ft-H20) SGf / 3,960 (gal-ft-H20/min-hp)
[6]
Example
Calculate the work added to the fluid (hp) by a pump pumping 100 gpm of water at
standard conditions if the pressure rise across a pump was 30 psi.
h = P / (g 30 psi x 2.31 ft-H20 / psi = 69.3 ft-H20
Wf (hp) = V (gal/min) htotal (ft-H20) SGf / 3,960 (gal-ft-H20/min-hp)
Wf (hp) = 100 gpm x 69.3 (ft-H20) 1.0 / 3,960 (gpm-ft-H20/hp) = 1.75 hp
Inlet/Outlet Pressure Changes
The total pressure rise that a pump/fan must generate to move a fluid through a pipe/duct
system is the sum of the pressure rise required to meet inlet and outlet conditions and the
pressure rise to overcome friction in the pipe system. The pump/fan must generate a
pressure rise to meet inlet and outlet conditions whenever the pressures, fluid velocities or
elevations are different between the inlet and outlet of the pipe system. The total pressure
rise required to compensate for different inlet and outlet conditions is the sum of Pstatic ,
Pvelocity and Pelevation,. If the inlet and outlet pressures, velocities and/or elevations are the
same, the corresponding term will evaluate to zero. If the inlet and outlet fluid pressures,
velocities and/or elevations are different, the corresponding terms must be evaluated.
Closed Loop Systems
In closed-loop systems, such as the one shown below, fluid is pumped through a continuous
loop. Thus, the inlet and outlet of the system are at the same location. Hence the pressure,
velocity and elevation of the inlet and outlet are identical, and the changes in static, velocity
and elevation pressures are zero.
4
Figure 2. Closed loop piping system
Open Systems
In open systems, such as the one shown below, fluid is pumped from one location to a
different location. In open systems the change between static, elevation and velocity
pressures between the inlet and outlet to the system must be considered; however, careful
definition of the inlet and outlet locations can minimize the complexity of the calculations.
Open Tank 2
P-26
Open Tank 1
Figure 3. Open piping system.
Static Pressure and Head
In an open system, it is frequently possible to define the inlet and outlet locations so that
the inlet, 1, and outlet, 2, of the system are surfaces of open tanks. If so, both the inlet and
outlet pressures, P1 and P2, are equal to atmospheric pressure, and the change in static
pressure is zero.
Pstatic = P2 – P1 = Patm – Patm = 0
In some cases, however, the inlet and outlet pressures are different. In these cases, the
required static pressure or static head must be calculated.
Example
If water is pumped from an open tank to a pressurized tank at 10 psig, then the required
static head is:
Pstatic = P2 – P1 = Ppres tank – Patm = (10 + 14.7) psia – 14.7 psia = 10 psi
hstatic = 10 psi x 2.31 ft-H20/psi = 23.1 ft-H20
5
Velocity Pressure and Head
For internal incompressible flow, such as the flow of water through a pipe, fluid velocity is
inversely proportional to the square of the pipe diameter. Thus, if the pipe diameter
remains constant, the inlet and outlet velocities are equal, and the change in velocity
pressure, Pvelocity is zero.
Pvelocity = V22- V 12) = 0
When the inlet and outlet velocities are different, the change in velocity pressure must be
calculated. A useful dimensional relationship to calculate velocity V from volume flow rate,
V, and pipe diameter, d, is:
V (ft/s) = 0.4085 V(gpm) / [d (in)]2
Useful dimensional relations to calculate the velocity head associated with a velocity, V, for
water at standard conditions is:
hvelocity = Pvelocity / [ H20 g] = (V2 fluid ) / (H20 g)
h velocity (ft-H20) = 0.0155[V (ft/s)]2 = [V (ft/s) / 8.032]2
hvelocity (in-H20) = [V (ft/min) / 4,005]2
Example
If 100 gpm of water is pumped through a pipe with an inlet diameter of 4 inches and
discharged from a pipe with an outlet diameter of 2 inches, the required velocity head is:
V1 (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 100 (gpm) / [4 (in)]2 = 2.55 ft/s
V2 (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 100 (gpm) / [2 (in)]2 = 10.2 ft/s
h velocity1 (ft-H20) = 0.0155[V 1 (ft/s)]2 = 0.0155[2.55 (ft/s)]2 = 0.10 ft-H20
h velocity2 (ft-H20) = 0.0155[V 2 (ft/s)]2 = 0.0155[10.2 (ft/s)]2 = 1.61 ft-H20
h velocity = h velocity2 - h velocity1 = 1.61 ft-H20 - 0.10 ft-H20 = 1.51 ft-H20
Elevation Pressure and Head: The change in elevation pressure is:
Pelevation = fluidg (z2 – z1)
The change in elevation head, in terms of water filled manometer height, is:
helevation (ft-H20) = Pelevation / (H20g ) = fluid g (z2 – z1) / (H20g )
6

For water at standard conditions, a useful dimensional relationship is:
helevation (ft-H20) = (z2 – z1) ft
Example
If water at standard conditions is pumped from one open tank to another open tank with a
surface 10 feet higher than the first open tank, then the required elevation head is:
helevation = 10 ft – 0 ft = 10 ft-H20
Pressure Loss Due to Friction
Total pressure loss due to friction, Pfriction, is the sum of the total pressure loss from friction
with the pipes, Pp, and the total pressure loss from friction through the fittings, Pf.
Pfriction = Pp + Pf
Similarly, the total friction loss, hfriction, as fluid flows through pipes is the sum of the head
loss from friction with the pipes, hp, and the head loss from friction through the fittings,
hf.
hfriction = hp + hf
The next two sections describe how to calculate pressure loss due to friction through pipes
and fittings.
Pressure Loss Due to Friction through Pipes and Ducts
Friction Factor Method
The total pressure loss from friction with the pipes and ducts, Pp, can be calculated from
Pp = (f L fluid V2) / (2 D)
where f is the friction factor, L is the pipe/duct length, is the fluid density and D is the
pipe/duct diameter. An explicit algebraic expression for friction factor, f, was developed by
Churchill. This relationship is valid for all ranges of Reynolds numbers.
7
 8
f  8 
 ReDh

12


1
 

1.5



A

B





 2.457 ln
 7 / ReD

h



0.9
1 / 12

1

 0.27 / Dh  
16
16
 37,530 

B
 ReD 
h


Source: ASHRAE, 2005, pg 2.7.
The Reynolds number is:
Re = V D / = fluid V D / 
where  is the kinematic viscosity (air at 60 F = 0.572 ft2/h and water at 60 F = 0.044 ft2/h), and 
is the dynamic viscosity (air at 60 F = 0.043 lbm/h-ft andwater at 60 F = 2.71 lbm/h-ft).
The equivalent diameter, De, for a rectangular duct with dimensions L and W is:
De = 1.3 (L W)5/8 / (L + W)1/4
Typical values for pipe roughness factors, e, are shown in the figure below.
e
r
D
Material
Glass, plastic
Copper, brass, lead (tubing)
Cast iron – uncoated
Cast iron – asphalt coated
Commercial steel or welded steel
Wrought iron
Riveted steel
Concrete
Roughness (m)
Smooth
1.5 x 10-6
2.4 x 10-4
1.2 x 10-4
4.6 x 10-5
4.6 x 10-5
1.8 x 10-3
1.2 x 10-3
Roughness (ft)
Smooth
5 x 10-6
8 x 10-4
4 x 10-4
1.5 x 10-4
1.5 x 10-4
6 x 10-3
4 x 10-3
Example
Calculate the friction head loss (ft-H20) to pump 100 gpm of water through 100 ft of 3-in
diameter steel pipe using the friction-factor method and Churchill relation.
A =  D2 / 4 = 3.14159 x (3/12)2 / 4 = 0.049087 ft2
V = V / A = (100 gal/min / 7.481 gal/ft3) / 0.049087 ft2 = 272.316 ft/min
8
Re = V D / = 272.316 ft/min x 60 min/hr x (3/12) ft / 0.044 ft2/hr = 92,835
e/D(steel pipe) = (1.5 x 10-4) ft / (3/12) ft = .000600
A = [2.457 ln(((7/Re)^0.9 + (0.27e/D))^-1)]^16 = 4.3862E+20
B = [37,530/Re]^16 = 5.08943E-07
f = 8[(8/Re)^12 + (A+B)^-1.5]^(1/12) = 0.0210 (from Churchill)
P = (f L  V2) / (2 D) = 0.021 x 100 ft x 62.27 lbm/ft3 x (272.316 ft/min)2 / (2 x 3/12 ft x (60
s/min)2) = 5,387 lbm/ft-s2
h = P / (g  lbm/ft-s2 / (32.2 ft/s2 x 62.27 lbm/ft3)
h = 2.69 ft-H20 (per 100-ft pipe)
Monograph Method
Alternately, head loss due to friction for water flow through pipes, hp, can be determined
from monographs such as shown below.
9
Pipe friction loss (ASHRAE, 2005)
Example
Calculate the friction head loss in ft-H20 for pumping 100 gpm of water through 200 ft of 3in diameter steel pipe using the ASHRAE monographs.
From the monograph in Figure 6, the head loss for a flow rate of 100 gpm through a 3-in
diameter steel pipe is 2.5 ft-H20 per 100-ft pipe. Thus, the head loss through 200 ft of pipe
is:
10
h = 2.5 ft-H20 per 100-ft pipe x 200 ft-pipe = 5.0 ft-H20
Pressure Loss Due to Friction through Fittings
The total pressure loss from friction through the fittings, Pf, is proportional to the velocity
pressure. The constant of proportionality depends on the fitting. Thus, total pressure loss
from friction through a fitting is calculated as:
Pf = kf fluid V 2 / 2
where V is velocity and kf is measured empirically and reported by fitting manufacturers.
The head loss from friction through the fittings, hf, can be calculated from:
hf = Pf / ( g ) = kf fluid V 2 / (2  g)
In US units, this height is commonly measured in ft-H20 for pumping systems. For water
flow through pipe fittings, a useful dimensional relationship is:
hf (ft-H20) = kf V 2 / (2 g) = kf [V (ft/s)]2 / 64.4 ft/s2 = kf 0.0155[ V (ft/s)]2
A useful dimensional relationship to calculate velocity, V, from volume flow rate, V, and
pipe diameter, d, is:
V (ft/s) = 0.4085 V (gpm) / [d (in)]2
Loss coefficient data, kf, for some pipe fittings are shown in the Tables below.
11
Source: ASHRAE Handbook: Fundamentals 2005
Example
Find the fluid work, Wf, required to move 100 gpm of water through 200 ft of 3-in diameter
steel pipe with for four flanged welded 90-degree standard elbows assuming that 1 and 2
are open to the atmosphere and at the same elevation.
2
1
Wf
P2 = P1 because 1 and 2 are open the atmosphere.
V2= V1 because the area of duct at 1 and 2 are the same.
z2 = z1 because 1 and 2 are at the same elevation.
Thus:
hpres = hvel = helev = 0
htotal = hpres + hvel + helev + hp + hf = hp + hf
12
From monograph in Figure 6 at 100 gpm and 3-in diameter steel pipe, the friction head loss
through the pipe is:

hp = 2.5 ft-H20 per 100-ft pipe x 200 ft-pipe = 5.0 ft-H20
The velocity is:
V (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 100 (gpm) / [3 (in)]2 = 4.54 ft/s
From the ASHRAE table, kf = 0.34 for a 3-inch flanged welded 90-degree standard elbow.
The friction head loss through four elbow fittings is:
hf (ft-H20) = nf x kf x 0.0155[V (ft/s)]2
hf (ft-H20) = 4 x 0.34 x 0.0155[4.54 (ft/s)]2 = 0.43 ft-H20
The total head loss through ducts and fittings is:
htotal = hp + hf = 5 ft-H20 + 0.43 ft-H20 = 5.43 ft-H20
The work added to the fluid is:
Wf (hp) = V (gal/min) htotal (ft-H20) SGf / 3,960 (gal-ft-H20/min-hp)
Wf (hp) = 100 gpm x 5.43 (ft-H20) 1.0 / 3,960 (gal-ft-H20/min-hp) = 0.137 hp
Pipe/Duct System Design
When designing a piping/duct system, flow requirements and piping/duct distances are
typically known. Based on this information, the engineer must then must select pipe/duct
diameter, select fittings, determine a piping/duct configuration that results in sufficient flow
to the end uses, and determine the total pressure drop caused by the piping/duct system.
Initial Selection of Pipe/Duct Diameter
The selection of pipe/duct diameter generally involves a tradeoff between the first cost of
the pipe/duct and pump/fan energy costs of the lifetime of the system, both of which are
highly dependent on pipe/duct diameter. Large diameter pipes/ducts have a higher initial
cost, but result in reduced friction losses and pump/fan costs. A rule of thumb that is often
used as a starting place for selecting pipe diameters is to select the pipe diameter such that:
hfriction ~ 4.0 to 2.5 ft-H20 / 100 ft-pipe
These design guidelines insures that the fluid velocity is low enough to avoid pipe erosion
and excess noise, and provide a reasonable balance between the cost of the pipes/ducts
13
and pump/fan energy costs. Using this as a starting place, subsequent design iterations can
identify economically optimum pipe/duct diameters. In many cases, the economically
optimum pipe/duct diameter will be larger than that suggested by the design guideline.
Parallel Flow Piping
Many piping designs employ parallel flow. In parallel flow designs, the total pressure drop
for sizing the pump and calculating pump energy costs is the total pressure drop for the
path with the highest pressure drop. The figure below shows two common piping
configurations that employ parallel flow.
A
B
C
D
A
B
C
D
Parallel flow piping systems. a) Direct return. b) Indirect return.
The configuration on the left is called direct return. In this configuration, the total pressure
drop for flow through leg A is less than the total pressure drop for flow through leg D. Thus,
if no balancing valves were installed, more fluid would flow through leg A than D, and the
total pressure drop across the pump would be set by the pressure drop through leg D.
The configuration on the right is called indirect return. In this configuration, the pressure
drop and flow through all legs are equal. Thus, indirect return guarantees equal flow
through all legs in the absence of balancing or flow control valves.
Parallel Flow Pipe System Design
In commercial buildings, duct systems almost always include multiple branches, which
result in parallel flow. Several methods exist for designing parallel-flow pipe/duct systems.
Most are variants of the equal friction and equal pressure methods. Use of the equalfriction and equal-pressure methods is demonstrated in the following example.
Example
Consider the piping system shown below. The head loss though each cooling coil, CC, is 10
ft-H20 and head loss though the evaporator, E, is 20 ft-H20. The flow though each cooling
coil, CC, is 100 gpm. The length of pipe run A is 50 ft, pipe run B is 25 ft and pipe run C is 25
ft. The distance between the supply and return headers is negligible. Piping connections
are flanged welded and elbows are long radius. The pump is 75% efficient and the pump
motor is 90% efficient. Using a design friction head loss of 3.0 ft-H20/100 ft, a) determine
pipe diameters to the nearest nominal diameter, b) determine the total head and flow
14
required by the pump, c) determine the required size of the pump motor to the nearest hp
(round up!), and d) determine annual pumping electricity use (kWh/year). Neglecting
pressure losses through elbows and fittings, determine pipe diameters.
A
E (20 ft H20)
B
C
CC
CC
CC (10 ft H20)
a) Enter the monograph with the design friction loss of and volume flow rate to determine
pipe diameters, D.
DA (300 gpm at 3.0 ft-H20/100 ft) = 5 inches with actual dh = 1.7 ft-H20/100 ft
DB (200 gpm at 3.0 ft-H20/100 ft) = 4 inches with actual dh = 3.5 ft-H20/100 ft
DC (100 gpm at 3.0 ft-H20/100 ft) = 3 inches with actual dh = 2.5 ft-H20/100 ft
b) Calculate the head loss through the pipes.
hA,pipe = 1.7 ft-H20/100 ft x 100 ft = 1.70 ft-H20
hB,pipe = 3.5 ft-H20/100 ft x 50 ft = 1.75 ft-H20
hC,pipe = 2.5 ft-H20/100 ft x 50 ft = 1.25 ft-H20
Calculate the head loss through the tees (line) and elbows from the ASHRAE table.
The velocity is:
VA (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 x 300 (gpm) / [5 (in)]2 = 4.90 ft/s
VB (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 x 200 (gpm) / [4 (in)]2 = 5.11 ft/s
VC (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 x 100 (gpm) / [3 (in)]2 = 4.54 ft/s
hA,line (ft-H20)= nf x kf x 0.0155[V (ft/s)]2 = 1 x 0.13 x 0.0155[4.90 (ft/s)]2 = 0.048 ft-H20
hA,elbow = nf x kf x 0.0155[V (ft/s)]2 = 2 x 0.20 x 0.0155[4.90 (ft/s)]2 = 0.149 ft-H20
hB,line = 2 x 0.15 x 0.0155[5.10 (ft/s)]2 = 0.121 ft-H20

15
hC,line = 1 x 0.17 x 0.0155[4.54 (ft/s)]2 = 0.054 ft-H20
hC,elbow = 2 x 0.25 x 0.0155[4.54 (ft/s)]2 = 0.160 ft-H20
Calculate the friction head loss across the pump as the head loss through the path with the
greatest resistance, which in this case is through path C .
htotal = [1.70 + 1.75 + 1.25 + 0.048 + 0.149 + 0.121 + 0.054 + 0.160 + 10 + 20] ft-H20
= 35.23 ft-H20
Specify pump:
V = 300 gpm, hTotal = 35.23 ft-H20
c) Calculate pump output power.
Wpump (hp) = V (gal/min) htotal (ft-H20) / [3,960 (gal-ft/min-hp) x Effpump]
Wpump (hp) = 300 gpm 35.23 (ft-H20) / [3,960 (gal-ft/min-hp) x 0.75] = 3.56 hp
Specify 4 hp motor
d) Calculate annual pump electricity use
E = Wpump / Effmotor x dt
E = 3.56 hp / 0.90 x 0.75 kW/hp x 8,760 hours/year = 25,980 kWh/yr
Pump and Expansion Tank Placement
If the pressure of a fluid becomes too low, cavitation (boiling) will result. Cavitation causes
excess turbulence in the pump, which increases noise, decreases efficiency and may
damage a pump. Cavitation can be avoided by proper pump placement.
Open Systems
Consider a pump operating between two open reservoirs with a total head loss of 23.1 ftH20. The pressure rise to across the pump to compensate for this head loss is:
P = 23.1 ft-H20 / 2.31 ft-H20/psi = 10 psi
If the pump is placed near the inlet to the system, as shown below, then the minimum
pressure of the fluid is 0 psig.
16
P
P=0 psig
P=0 psig
10
P(psig)
0
If the pump is placed near the outlet to the system, as shown below, then the minimum
fluid pressure is -10 psig. This low pressure may cause the fluid to cavitate.
P=0 psig
P
P=0 psig
0
P(psig)
-10
Therefore, to avoid cavitation in open pumping systems, position the pump so it pushes
rather than pulls the fluid.
Closed-loop Systems
Most closed loop systems in which the fluid may undergo a large temperature change
employ an expansion tank to handle increased fluid volume with increased temperature.
The simplest expansion tank has a diaphragm with compressed air above. The tank
maintains the fluid pressure at the air pressure wherever the tank is located.
Consider a closed-loop system with an expansion tank set to 5 psig and a total friction head
loss of 23.1 ft-H20. The pressure rise to across the pump to compensate for this head loss
is:
P = 23.1 ft-H20 / 2.31 ft-H20/psi = 10 psi
If the expansion tank is placed before the pump, as shown below, then the minimum
pressure of the fluid is 5 psig.
17
P=5 psig
P=15 psig
If the expansion tank is placed before the pump, as shown below, then the minimum fluid
pressure is -5 psig. This low pressure may cause the fluid to cavitate.
P=-5 psig
P=5 psig
Therefore, to avoid cavitation in closed-loop pumping systems, position the expansion tank
in front of the pump to fix the pressure at that point. Never put more than one expansion
tank in a system.
Pump Types
Pumping applications can generally be divided into two categories: “low flow at high
pressure” and “high flow at low pressure”. “Low flow at high pressure” applications include
hydraulic power systems and typically employ positive-displacement pumps. The majority
of fluid-flow applications are “high flow at low pressure” and use centrifugal pumps.
In centrifugal pumps, the fluid enters along the centerline of the pump, is pushed outward
by the rotation of the impeller blades, and exits along the outside of the pump. A schematic
of a centrifugal pump is shown below.
18
Centrifugal pump
Source: ASHRAE Handbook: HVAC Systems and Equipment 2008
Pump Curves
Pumps can generate high volume flow rates when pumping against low pressure or low
volume flow rates when pumping against high pressure. The possible combinations of total
pressure and volume flow rate for a specific pump can be plotted to create a pump curve.
The curve defines the range of possible operating conditions for the pump.
If a pump is offered with multiple impellers with different diameters, manufacturers
typically plot a separate pump curve for each size of impellor on the same pump
performance chart. Smaller impellors produce less pressure at lower flow rates. Typical
pump performance charts with multiple pump curves are shown below.
19
Typical pump performance curves.
Source: ASHRAE Handbook: HVAC Systems and Equipment 2008
The power required to push the fluid through the pipe, Wfluid, is the product of the volume
flow rate and system pressure drop.
Wfluid = V Ptotal
Graphically, fluid work is represented by the area under the rectangle defined by the
operating point on a pump performance chart.
Typically, the efficiency of the pump at converting the power supplied to the pump into
kinetic energy of the fluid is also plotted on the pump performance chart. Pump efficiencies
typically range from about 50% to 80%. Power that is not converted into kinetic energy is
lost as heat. The power required by the pump, which is called the “shaft work” or “brake
horsepower”. Pump efficiency is the ratio of fluid work to shaft work. , can be calculated
from the flow rate, total pressure, and efficiency values from the pump curve, using the
following equation.
Wpump = Wfluid / Effpump = V Ptotal / Effpump
A useful dimensional version of this equation for pumping water at standard conditions is:
Wpump (hp) = V (gal/min) htotal (ft-H20) / [3,960 (gal-ft/min-hp) x Effpump]
20
Many pump performance graphs, including those shown above, also plot curves showing
the work required by the pump to produce a specific flow and pressure. Note that these
curves show work required by the pump including the efficiency of the pump. Calculating
the work supplied to the pump using the preceding equation and comparing it to the value
indicated on a pump performance graph is a useful exercise.
System Curves
The total pressure that the pump must produce to move the fluid is determined by the
piping system. This total pressure of the piping system is the sum of the pressure due to
inlet and outlet conditions and the pressure loss due to friction. In a piping system,
pressure loss due to friction increases with increasing fluid flow; thus, system curves have
positive slopes on pump performance charts. The operating point of a pump is determined
by the intersection of the pump and system curves.
To determine the form of a system curve, consider the equation for total pressure in a
piping system. The total pressure caused by a piping system is the sum of the pressure due
to inlet and outlet conditions and the pressure required to overcome friction through the
pipes and fittings.
Ptotal = (Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition
Inlet/Outlet Pressure
The inlet/outlet pressure that the pump must overcome is the sum of the static, velocity
and elevation pressures between the inlet and outlet of the piping system. For closed loop
piping systems, the inlet and outlet are at the same location; hence, the static, velocity and
elevation pressure differences are all zero. For open systems, the differences in static,
velocity and elevation pressures must be calculated.
In many pumping applications, the velocity pressure difference between the inlet and outlet
is zero or negligible, and inlet/outlet pressure is simply the sum of the static and elevation
heads. In these cases, the inlet/outlet pressure is independent of flow and is represented
on a pump performance chart as the pressure at zero flow.
Friction Pressure Drop
The equations for pressure loss from friction through pipes and through fittings are:
Pp = (f L fluid V 2) / (2 D)
Pf = kf fluid V 2 / 2
These equations clearly show that for a given pipe system, the pressure drop is proportional
to the square of the velocity, and hence the square of the volume flow rate.
21
Pfriction = C1 V 2 = C2 V2
This quadratic relationship can be plotted on the pump curve to show the “system curve”.
Plotting System Curves
As the preceding discussion showed, system curves have a flow-independent component (of
inlet/outlet pressure) and a flow-dependent component that varies with the square of flow
rate (of friction pressure).
A system curve for a closed-loop piping system with no inlet/outlet pressure difference is
shown below. The curve is a parabola of the form hheadloss = C2 V2. The curve passes
through the origin because the inlet/outlet pressure difference, sometimes called the static
head, is zero. The coefficient C2 can be determined if the operating point is known by
substituting the known pressure drop and flow rate into the equation and solving for C 2.
The fluid work required to push the fluid through the pipe is the product of the volume flow
rate and system pressure drop and is represented graphically by the area under the
rectangle defined by the operating point.
System curve for closed-loop piping system with no inlet/outlet pressure difference.
22
A system curve for an open-loop piping system with a “static” or “inlet/outlet” pressure of
20 ft H20 is shown below. This system curve is of the form hheadloss = A + C2 V2; where A is
the “static” or “inlet/outlet” pressure drop. As before, the coefficient C2 can be determined
if the operating point and inlet/outlet pressure are known by substituting the known values
into the equation and solving for C2.
System curve for open-loop piping system with 20 ft-H20 inlet/outlet pressure difference
Multiple Pumps Operating In Parallel
Many pumping systems employ multiple pumps in parallel rather a single large pump. One
advantage of specifying multiple pumps in a parallel configuration is redundancy in case of
failure. For example, it is common to design a pumping system with three pumps in parallel
configuration, even though no more than two pumps would ever run simultaneously. The
third pump provides redundancy in case of failure, and allows the system to function at full
capacity even when one pump is being serviced. Another advantage of parallel pumping
configuration is the ability to vary flow by turning one or more of the pumps on and off.
Finally, in many applications, it is more energy efficient to operate multiple smaller pumps
in parallel rather than operating a single large pump.
23
Head ft-H20
When two pumps are operated in parallel, they perform like a single pump with twice the
flow rate at the same pressure drop. The figure below shows the pump curve of a single
pump A, two pumps operating in parallel B, and the system curve C.
180
160
140
120
100
80
60
40
20
0
A
B
C
0
2,000
4,000
6,000
8,000
10,000 12,000 14,000 16,000
V gpm
A (One pump)
B (Two pumps in parallel)
C (System)
Pump and systems curves for secondary chilled water loop.
The system curve C describes the relationship of pressure drop and flow rate for the given
piping system with no static head. Because head loss varies with the square of flow rate,
the equation of the system curve can be estimated by fitting a quadratic equation through
the origin and the design operating point:
h (ft H20) = (2.4 x 10-6) x V2
1)
where V is the flow rate in gpm. The equation for curve B can be estimated by the fitting a
regression equation through the data points on the curve:
h (ft H20) = 149 + 0.00106 x V + (3.65 x 10-7) x V2
(2)
The operating point of curve B for two pumps in parallel can be found from Equations 1 and
2 to be about:
h = 137 ft H20
V = 7,500 gpm
Note that the total volume flow rate of two pumps operating in parallel is less than twice
the flow rate of a single pump operating alone (at the intersection of C and A).
24
Pump Motor Work
Pumps and fans are typically driven by electrical motors. The power required by the motor
is greater than the fluid work because the pump/fan, power transmission and motor all
incur losses. Thus, fluid work must be divided by the product of the efficiencies of all
components of the pump energy-delivery system to determine the electricity required by
the motor.
Welec = Wf / ( Efficiencypump.fan x Efficiencydrive x Efficiencymotor)
Qloss = 8
Qloss =10
Wmotor
100
Wdrive
90
Qloss = 25
Wpump
83
Wfluid
58
motor = 90%
drive = 92%
pump = 70%
Pump/fan system efficiency.
For example, if the efficiency of the motor at converting electrical energy to motor shaft
work is 90%, the efficiency of belt drives at transferring motor shaft work to pump is 92%,
and the efficiency of a pump at converting pump shaft work to fluid work is 75%, the
electrical energy use required by the motor would be 73% greater than the required fluid
work.
Welec = Wf / ( 90% x 92% x 70%) = 1.73 Wf
Pump/Fan Affinity Laws
The fundamental fluid mechanic relationships developed thus far can be modified to
generate other useful relations between fan parameters. These relationships are known as
pump/fan affinity laws. The two most important relationships are derived below.
As shown in the section of system curves, friction head loss is proportional to the square of
the volume flow rate.
Pfriction = C1 V2 = C2 V2
By substitution, fluid work is proportional to the cube of volume flow rate
Wf = V Pfriction = V C2 V2 = C2 V3
25
Since Wf / V3 is constant, it follows that:
(Wf / V3)1 = C = (Wf / V3)2
Wf2 = Wf1 (V2 / V1)3
This relation shows that in systems where all fluid work added by the pump/fan is to
overcome pipe/duct friction, a small reduction in the volume flow rate results in a large
reduction in the fluid work. For example, reducing the volume flow rate by one half reduces
fluid work by 88%!
Wf2 = Wf1 (1/2)3 == Wf1 (1/8)
(Wf1 – Wf2) / Wf1 = [Wf1 - Wf1 (1 /8)] / Wf1 = 1 – (1/8) = 88%
If systems where part of the fluid work added by the pump/fan is to overcome pipe/duct
friction and part is overcome inlet/outlet pressure differences, the cubic relation between
fluid work and volume flow rate applies to the fluid work for overcoming friction.
Another useful relation can be derived from the relationship between volume flow rate V
and the rotational speed of the pump fan. In centrifugal pumps and fans, the volume flow
rate is proportional to the rotational speed of the pump fan.
V = C RPM
Since V/RPM is constant, it follows that:
(V / RPM)1 = C = (V / RPM)2
V2 = V1 (RPM2 / RPM1)
Thus, volume flow rate varies in proportion to pump/fan speed.
Inside-Out Approach to Energy-Efficient Fluid Flow Systems
The most effective approach for designing energy efficient pump/fan systems and for
identifying energy savings opportunities in existing fluid flow systems is the “whole-system,
inside-out approach”. The “whole-system” part of this approach emphasizes the
importance of considering the entire conversion, delivery and end-use system. The “insideout” part of the approach describes the preferred sequence of analysis, which begins at the
point of the energy’s final use “inside” of the process, followed by sequential investigations
the energy distribution and primary energy conversion systems. This approach can tends to
multiply savings and result in smaller, more efficient and less costly systems.
26
The fluid work equation shows that energy required by fan/pump systems is a function of
the volume flow rate, inlet/outlet conditions, and system friction.
Wf = V [(Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition]
This provides a useful guide for characterizing energy efficiency opportunities.
Energy Savings from Reducing Elevation Difference
Many pumping applications involve lifting fluids from lower to higher elevations. The total
pressure difference through the piping system must include this elevation pressure
difference. In some applications, it may be possible to reduce the elevation pressure
difference by increasing the height of the fluid in the supply tank or reducing the height of
the fluid in the outlet tank. Doing so reduces the total system pressure difference and
pump energy use.
Example
A pump lifts 100 gpm of water 120 feet to fill an elevated reservoir. Determine the electrical
power savings if the level of the reservoir could be reduced by 20 feet. The pump is 70%
efficient and the pump motor is 90% efficient.
PE1 = 100 gpm x 120 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70 x 0.90] x 0.75 kW/hp = 3.61 kW
PE2 = 100 gpm x 100 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70 x 0.90] x 0.75 kW/hp = 3.01 kW
Savings = PE1 – PE2 = 3.61 kW – 3.01 kW = 0.60 kW
Energy Savings from Reducing Friction
The primary methods to reduce friction in a pipe/duct system are:



Increase pipe diameter
Use smooth pipes
Use fewer low pressure-drop fittings
Reduce System Pressure Drop: Increase Pipe/Duct Diameter
Friction head loss in internal flow is strongly related to the diameter of the pipe/duct. Small
pipes/ducts dramatically increase the velocity of the fluid and friction pressure loss. The
friction pressure loss through pipes and ducts is:
Pp =  f L V 2 / (2 D)
27
The velocity V is the quotient of volume flow rate V and area A, thus
Pp =  f L (V / A)2 / (2 D) =  f L (V /  D2 )2 / (2 D) =  f L V2 / (2  2 D5 )
Thus, friction pressure loss through pipes/ducts is inversely proportional to the fifth power
of the diameter
Pp ~ C / D5
This means that doubling the pipe/duct diameter reduces friction pressure loss by about
97%!
Example
Calculate the percentage reduction in friction head loss if pumping 4 gpm of water through
0.5-inch and 1-inch diameter schedule 40 steel pipes.
From the monogram:
h 0.5-inch = 17 ft-H20/100 ft
h1-inch = 1.3 ft-H20/100 ft
The percent reduction in friction head loss from doubling the diameter of the pipe would be
about:
(17 – 1.3) / 17 = 92%
600
600
500
500
400
300
Total Cost
200
Piping Cost
100
Life cycle cost (x1000 USD)
Life cycle cost (x1000 USD)
Optimum pipe diameter is often calculated based on the net present value of the cost of the
pipe plus pumping energy costs. Using this method in the figure below (Larson and Nilsson,
1991) optimum pipe diameter was found to be 200 mm. When the cost of the pump was
also included in the analysis, the optimum diameter was found to be 250 mm and energy
use was reduced by 50%. This illustrates the importance of considering the whole system.
Total Cost
400
300
200
Motor/Pump &
Piping Cost
100
Electricity Cost
Electricity Cost
0
0
125
150
200
250
300
Pipe Diameter (mm)
28
350
400
500
125
150
200
250
300
Pipe Diameter (mm)
350
400
500
Reduce System Pressure Drop: Use Smooth Pipes/Ducts
Similarly, use of the smoothest pipe/duct possible for a given application reduces pipe
friction losses. The progression from smoothest to roughest pipe is: plastic, copper, steel,
concrete.
Reduce System Pressure Drop: Use Fewer Low Pressure-Drop Fittings
Minimizing fittings, including turns, using welded connections instead of threaded
connections, and the use of low-pressure drop fittings can significantly reduce friction head
loss. Consider for example, these examples using data from ASHRAE fittings head loss table.
Piping Examples:
Using welded connections instead of threaded connections reduces friction head through a
2-inch 90-degree standard elbow by:
(h,thread –h,weld) / h,thead = (kf,thread –kf,weld) / kf,thead = ( 1.00 – 0.38 ) / 1.00
= 62%
Using long radius elbows instead of standard elbows reduces friction head through a 2-inch
welded elbow by:
(h,std –h,long) / h,std = (kf,,std –kf,,long) / kf,,std = ( 0.38 – 0.30 ) / 0.38 = 21%
Using gate valves instead of globe valves reduces friction head through a 2-inch welded
valve by:
(dh,globe – dh,gate) / dh,globe = (kf,,globe –kf,,gate) / kf,,globe = ( 9.00 – 0.34 ) / 9.00
= 96%
Modify Pump/Fan to Realize Savings from Reducing Friction
It may seem that reducing friction losses in a pipe/duct system would automatically reduce
pump/fan energy use. However, reducing system pressure drop without modifying the
pump/fan causes the pump/fan to move more fluid. This increased volume flow rate
actually increases pump energy consumption.
Thus, it is important to reduce the diameter of the pump impellor, or slow the pump/fan so
that the volume flow rate remains the same as it was in the high friction in order to realize
energy savings. The following examples demonstrate this importance of modifying the
pump to realize savings from reducing piping system pressure drop.
Example: Modify Pump to Realize Friction Reduction Savings
29
The figure below shows a set of pump curves with two system curves. The pump originally
operates at point A, and the system curve for the original piping systems extends from the
origin to A (which means that this is a closed loop piping system with no inlet/outlet
pressure difference to overcome). The friction pressure drop through the piping system is
then reduced by 25 ft-H20 by increasing pipe diameter, using low-flow fixtures or using
smoother pipe. Reducing the system pressure drop from 70 ft-H20 at point A to 45 ft-H20,
without altering the pump impellor or speed, would cause the pump to operate at point B.
The power required to pump a fluid is the product of the volume flow rate and pressure
drop; hence, the areas enclosed by the rectangles defined by each operating point
represent the fluid power requirements, WfA and WfB, at the different system pressure
drops.
Source of pump curve: ASHRAE Handbook, HVAC systems and Equipment, 2008
WfA = 175 gpm x 70 ft-H20 / 3,960 gpm-ft-H20/hp = 3.09 hp
WfB = 350 gpm x 45 ft-H20 / 3,960 gpm-ft-H20/hp = 3.98 hp
The power required by the pump, Wp, is the fluid power requirements divided by the pump
efficiency.
WPA = 3.09 hp / .73 = 4.24 hp
WPB = 3.98 hp / .57 = 6.98 hp
30
Thus, decreasing system pressure drop without altering the pump impellor or speed would
cause the pump to consume more energy, not less.
Savings = WPA – WPB = 4.24 hp –6.98 hp = -2.74 hp
To realize energy savings from reducing pressure drop, it is necessary to slow the pump
speed or decrease the size of the impellor. To determine the pump impellor size required
to deliver the initial flow of 175 gpm with the new low-pressure drop pipe system, it is
necessary to develop a system curve for the new pipe system. Pressure drop through piping
systems varies with the square of flow rate. Thus, the equation for a system curve that
passes through the origin can be written as:
h = C V2
The coefficient, C, for the new system curve can be found by substituting the values of
pressure drop and volume flow rate for point B.
C = h / V2 = 45 / 3502 = 0.000367
Thus, the pressure drop through the new duct system at 175 gpm would be about:
h = C V2 = 0.000367 1752 = 11 ft-H20
The flow rate of 175 gpm and 11 ft-H20 defines point C, which would be the operating point
of the pump with a 5-in impeller. At this operating point, the pump would be about 52%
efficient, and the pump power draw would be about:
WPC = 175 gpm x 11 ft-H20 / (3,960 gpm-ft-H20/hp x 0.52) = 0.96 hp
Thus, the savings from reducing the pressure drop in the pipe system, if the pump impeller
diameter were reduced, would be about:
Savings = WPA – WPC = 4.24 hp – 0.96 hp = 3.28 hp
This example demonstrates the importance of modifying the pump to realize savings from
reducing system friction loss.
Energy Savings from Energy-Efficient Flow Control
Most pump/fan systems are designed to handle peak conditions. Since peak conditions
typically occur infrequently, substantial energy savings are possible by controlling fluid flow
rate to match actual demand. The inside-out approach to low-energy pump/fan systems
31
starts by reviewing all end-use applications to determine the required flow, before
proceeding “upstream” with the analyses of the piping and pumping systems.
Once the required flow is determined, it is necessary to determine how the flow is currently
controlled and consider more energy-efficient options. Inefficient methods of flow control,
in order of worst to better, are:




Bypass
Throttling / Outlet dampers
Fan variable inlet vanes
Intermittent pump operation
More efficient methods of flow control are:




Continuous pump operation with smaller impeller or at lower speed.
Reducing pump impellor diameter
Slowing fan speed
Continuously varying pump/fan speed with a variable speed drive.
These concepts are demonstrated in the following figures.
Old Inefficient Flow Control
Fan w/ Inlet Vanes
By-pass loop By-pass damper Outlet valve/damper
Inlet vanes
(No savings)
(No savings)
(Small savings) (Moderate savings)
32
New Efficient Flow Control
Close Bypass Valve
dP
VFD
Trim impellor for
constant-volume
pumps
Slow fan for
constant-volume
fans
VFD for
variable-volume
pumps or fans
Bypass: Bypass control is sometimes employed to couple constant speed pumps/fans with
variable process loads. In by-pass control, valves are opened or closed to direct excess
water/air through a bypass loop or to simply reject it to the environment. Thus, the flow of
water/air through the pumps/fans remains constant even as the flow of water/air through
the process varies. Bypass is the least efficient method of flow control, since pump/fan
power remains nearly constant even as the load varies.
Throttling/Outlet Dampers: Controlling flow by closing a flow-control valve/damper
downstream of the pump/fan increases pressure drop and causes the operating point to
move up and left on the pump/fan curve.
Pump Curve
Head
Throttled
System Curve
Flow Rate
This results in relatively small energy savings, since
Wf2 = V2 P2
where
33
V2 < V1 but
P2 > P1
Thus, throttling is an energy inefficient method of flow control.
Flow Control with Intermittent Pump/Fan Operation
In some pumping applications, pumps may operate at a relative high flow rate for part of
the time and then be turned off until needed again. Because friction losses are proportional
to the square of flow, it is more energy-efficient to pump a lower volume flow rate for a
longer period of time. We call this the “Pump Long, Pump Slow” principle. “Pump long,
pump slow” opportunities may exist whenever pumps run intermittently. If a single pump
operates intermittently, then application of the “pump long, pump slow” principle would
require installing a smaller pump, trimming the pump impeller, or slowing the pump
rotational speed. If multiple pumps operate in parallel, it may be possible to simply run
fewer pumps more continuously.
Example
Two pumps operate in parallel to lift water 40 vertical feet and generate 5,800 gpm at 140
ft H20 as shown in the figure below. One pump operating alone would generate 5,200 gpm
at 123 ft H20. The pumps are 75% efficient and the pump motors are 90% efficient. When
operating in parallel, the pumps operate 10 hours per day. Determine the energy savings
from “pumping slow, pumping long” and running one pump longer to generate the same
daily flow rate.
200
160
dh (ft H20)
B
120
A
80
C
40
0
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
V (gpm )
A (one pump)
B (tw o pumps in parallel)
C (system curve)
The daily flow rate is:
5,700 gpm x 60 minutes/hour x 10 hours/day = 3,420,000 gal/day
34
9,000
The electrical power and energy required to pump this quantity of water using two pumps
in parallel is:
P2 Pumps = 5,700 gpm x 140 ft-H20 / (3,960 gpm-ft-H20/hp x 0.75 x 0.90) x 0.75 kW/hp
P2 Pumps = 223.9 kW
E2 Pumps = 223.9 kW x 10 hours/day = 2,239 kWh/day
The time required to generate the same daily flow rate while operating only one pump is:
3,420,000 gal/day / [5,700 gpm x 60 minutes/hour] = 10.96 hours/day
The energy required to pump this quantity of water using one pumps is:
P1 Pump = 5,200 gpm x 123 ft-H20 / (3,960 gpm-ft-H20/hp x 0.75 x 0.90) x 0.75 kW/hp
P1 Pump = 179.5 kW
E1 Pump = 179.5 kW x 10.96 hours/day = 1,967 kWh/day
The fraction savings would be about:
(2,239 kWh/day – 1,967 kWh/day) / 2,239 kWh/day = 14%
Flow Control by Trimming Pump Impellor
The most energy-efficient methods of varying flow are controlling pump/fan speed or
decreasing the size of the impellor. If the required flow is constant and less than the design
flow, more cost effectively alternatives are to slow a fan or trim a pump impellor. The
following example illustrates energy savings from reducing flow by trimming a pump
impellor.
Example: Trim Pump Impellor in Closed-Loop System
The figure below shows a set of pump curves for different impellor diameters. The pump is
equipped with an 8 1/16-inch impellor. At design conditions, the pump would operate at
point A. The system curve extends from the origin to A, which means this is a closed-loop
piping system with no inlet/outlet pressure difference to overcome.
The actual flow required was less than the design flow. Thus, flow was reduced by a
throttling valve to operating point B. A more energy efficient method of reducing flow is to
open the throttling valve, and trim the impellor to a smaller diameter. This would cause the
pump to operate at point C.
35
The power required to pump a fluid is the product of the volume flow rate and pressure
drop; hence, the areas enclosed by the rectangles defined by the vertical and horizontal
axes and each operating point represent the fluid power requirements, WfA , WfB, , and WfC,.
Comparison of these rectangles clearly indicates that reducing flow by trimming the pump
impellor uses much less energy than reducing flow by throttling.
Source of original pump curve: ASHRAE Handbook, HVAC systems and Equipment 2008
The power required by the pump at A, WPA, is the fluid power requirement divided by the
pump efficiency.
WPA = 350 gpm x 45 ft-H20 / [3,960 gpm-ft-H20/hp x 0.57] = 6.98 hp
The power required by the pump for throttled flow at B, WPB, is the fluid power
requirement divided by the pump efficiency.
WPB = 175 gpm x 70 ft-H20 / [3,960 gpm-ft-H20/hp x 0.73] = 4.24 hp
To determine the pump impellor size required to deliver a flow of 175 gpm with a trimmed
impellor, it is necessary to develop the design system curve with the throttling valve wide
open and the pump operating at A. Pressure drop through piping systems varies with the
square of flow rate. Thus, the equation for a system curve that passes through the origin
can be written as:
h = C V2
36
The coefficient, C, for the new system curve can be found by substituting the values of
pressure drop and volume flow rate for point A.
C = h / V2 = 45 / 3502 = 0.000367
Thus, the required head at 175 gpm would be about:
h = C V2 = 0.00036 1752 = 11 ft-H20
The flow rate of 175 gpm and 11 ft-H20 defines point C. This indicates that the pump
impellor should be trimmed to a diameter of 5-inches. At this operating point, the pump
would be about 52% efficient, and the pump power draw would be about:
WPC = 175 gpm x 11 ft-H20 / (3,960 gpm-ft-H20/hp x 0.52) = 0.96 hp
Thus, the savings from controlling flow by trimming the pump impellor rather than
throttling flow would be about:
Trim Impellor Savings = WPB – WPC = 4.24 hp – 0.96 hp = 3.28 hp
Alternately, power savings from reducing the volume flow rate can be estimated from the
pump affinity laws. Theoretically, pump work varies with the cube of volume flow rate. Use
of the cubic relationship would predict:
WPC = WPA (VC/VA)3 = 6.98 hp x (175 gpm / 350 gpm) 3 = 0.87 hp
The 0.87 hp predicted by the pump-affinity law is less than the 0.96 hp predicted by the
pump curve. This example demonstrates how use of the cubic relationship typically
exaggerates savings. In practice, the efficiencies of the VSD, pump and motor typically
decline as flow rate decreases, resulting in slightly less savings than would be predicted
using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5
power of flow rather than the cube of flow. Using this relationship, if P A is 6.98 hp at 350
gpm, PC at 175 gpm would be about:
WPC = WPA (VC/VA)2.5 = 6.98 hp x (175 gpm / 350 gpm) 2.5 = 1.23 hp
Using the 2.5 exponent, savings would be about:
Savings = WPB – WPC = 4.24 hp – 1.23 hp = 3.01 hp
This slightly lower estimate of savings incorporates the reduction in motor efficiency, and
power loss by the VSD.
37
Example: Trim Pump Impellor in Open System
The figure below shows a set of pump curves for different impellor diameters. The pump is
equipped with an 8 1/16-inch impellor. At design conditions, the pump operates at point A.
The system curve extends from an inlet/outlet pressure difference of 25 ft-H20 from lifting
water over an elevation change of 25 ft in an open system.
The actual flow required was less than the design flow. Thus, flow was reduced by a
throttling valve to operating point B. A more energy efficient method of reducing flow is to
open the throttling valve, and trim the impellor to a smaller diameter. This would cause the
pump to operate at point C.
The power required to pump a fluid is the product of the volume flow rate and pressure
drop; hence, the areas enclosed by the rectangles defined by the vertical and horizontal
axes and each operating point represent the fluid power requirements, WfA , WfB, , and WfC,.
Comparison of these rectangles clearly indicates that reducing flow by trimming the pump
impellor uses much less energy than reducing flow by throttling.
Source of pump curve: ASHRAE Handbook, HVAC systems and Equipment, 2008
The power required by the pump at A, WPA, is the fluid power requirement divided by the
pump efficiency.
38
WPA = 350 gpm x 45 ft-H20 / [3,960 gpm-ft-H20/hp x 0.57] = 6.98 hp
The power required by the pump for throttled flow at B, WPB, is the fluid power
requirement divided by the pump efficiency.
WPB = 175 gpm x 70 ft-H20 / [3,960 gpm-ft-H20/hp x 0.73] = 4.24 hp
To determine the pump impellor size required to deliver a flow of 175 gpm with a trimmed
impellor, it is necessary to develop the design system curve with the throttling valve wide
open and the pump operating at A. Pressure drop through piping systems varies with the
square of flow rate. Thus, the equation for a system curve that passes through the origin
can be written as:
h = h0 + C V2
Where h0 is the pump head at zero flow. The coefficient, C, for the new system curve can
be found by substituting the values of pressure drop and volume flow rate for point A.
C = (h – h0) / V2 = (45 – 20) / 3502 = 0.00020408
Thus, the pressure drop through the new duct system at 175 gpm would be about:
h = 40 + C V2 = 20 + 0.00020408 1752 = 26.3 ft-H20
The flow rate of 175 gpm and 26.3 ft-H20 defines point C. This indicates that the pump
impellor should be trimmed to a diameter of 5.75-inches. At this operating point, the pump
would be about 66.5% efficient, and the pump power draw would be about:
WPC = 175 gpm x 26.3 ft-H20 / (3,960 gpm-ft-H20/hp x 0.665) = 1.74 hp
Thus, the savings from controlling flow by trimming the pump impellor rather than
throttling flow would be about:
Savings = WPB – WPC = 4.24 hp – 1.74 hp = 2.49 hp
Alternately, power savings from reducing the volume flow rate can be estimated from the
pump affinity laws. However the cubic relationship between pump work and volume flow
rate applies only to the friction head, not the inlet/outlet head. Thus, the component of
total power due to friction must be calculated. The component of total power due to
friction is the difference between the total power and the inlet/outlet power.
The inlet-outlet power at point A, WPAO, is:
39
WPAO = 350 gpm x 20 ft-H20 / [3,960 gpm-ft-H20/hp x 0.57] = 3.1 hp
Thus, the component of total head due to friction is:
WPF = WPA - WPAO = 6.98 hp – 3.1 hp = 3.88 hp
Theoretically, pump work varies with the cube of volume flow rate. In practice, the
efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in
slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we
estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of
flow. Using this relationship, the component of power at C due to reduced friction at 175
gpm would be about:
WPCF = WPAF (VC/VA)2.5 = 3.88 hp x (175 gpm / 350 gpm) 2.5 = 0.69 hp
Using the 2.5 exponent, power savings would be about:
Savings = WPF – WPCF = 3.88 hp – 0.69 hp = 3.19 hp
This estimate of savings incorporates the reduction in motor efficiency, and power loss by
the VSD.
Flow Control by Varying Pump/Fan Rotational Speed
If the required flow varies over time, speed control is best facilitated by an electronic
variable speed drive (VSD) which can continuously and smoothly adjust pump/fan speed as
needed. One time reductions in flow are more cost effectively accommodated by replacing
the pump/fan pulley with a larger diameter pulley to slow pump/fan rotation or by reducing
pump impellor diameter. The following examples illustrate how to calculate energy savings
from reducing flow by slowing pump/fan speed.
Example: Slow Pump Speed in Closed-Loop System
The figure below shows pump performance at various speeds and a system curve. At design
conditions, the pump operates at 1,200 RPM at point A. The system curve extends from the
origin to A, which means this is a closed-loop piping system with no inlet/outlet pressure
difference to overcome.
40
At A, the volume flow rate is 1,200 gpm, the total head is 55 ft-H20 and the pump efficiency
is 74%. According to the chart, the required power to the pump at this operating point is
about 23 hp. Alternately, the required pump input power could be calculated as:
WPA = 1,200 gpm x 55 ft-H20 / (3,960 gpm-ft-H20/hp x 0.74) = 22.5 hp
If the flow were reduced to 900 gpm with a flow-control valve, the operating point would
move along the pump curve to point B at 900 gpm, 62 ft-H20 with pump efficiency of 70%.
According to the chart, the required power to the pump at this operating point is about 20
hp. Alternately, the required pump input power could be calculated as:
WPB = 900 gpm x 62 ft-H20 / (3,960 gpm-ft-H20/hp x 0.70) = 20.1 hp
Thus, the pump power savings from reducing flow from 1,200 gpm to 900 gpm with a flowcontrol valve would be about:
Flow-control Valve Savings = WPA – WPB = 22.5 hp – 20.1 hp = 2.4 hp
To determine the pump speed required to deliver a flow of 900 gpm, it is necessary to
develop the design system curve with the flow-control valve wide open and the pump
operating at A. Pressure drop through piping systems varies with the square of flow rate.
Thus, the equation for a system curve that passes through the origin can be written as:
h = C V2
41
The coefficient, C, for the new system curve can be found by substituting the values of
pressure drop and volume flow rate for point A.
C = h / V2 = 55 / 1,2002 = 0.000038194
Thus, the pressure drop at 900 gpm would be about:
h = C V2 = 0.000038194 9002 = 31 ft-H20
The flow rate of 900 gpm and 31 ft-H20 defines point C. This indicates that the pump speed
should be reduced to 900. At this operating point, the pump would be about 67% efficient,
and the required pump input power would be:
WPC = 900 gpm x 31 ft-H20 / (3,960 gpm-ft-H20/hp x 0.67) = 10.5 hp
Alternately, the power required by the pump at point C, PC, can be read from the chart to
be about 10 hp.
Pump power savings are the difference between PA and PB.
Variable Speed Drive Savings = WPA – WPC = 22.5 hp – 10.5 hp = 12.0 hp
Alternately, power savings from reducing the volume flow rate can be estimated from the
pump affinity laws. Theoretically, pump work varies with the cube of volume flow rate. Use
of the cubic relationship would predict:
WPC = WPA (VC/VA)3 = 22.5 hp x (900 gpm / 1200 gpm) 3 = 9.5 hp
The 9.5 hp predicted by the pump-affinity law is less than the 10.5 hp predicted by the
pump curve. This example demonstrates how use of the cubic relationship typically
exaggerates savings. In practice, the efficiencies of the VSD, pump and motor typically
decline as flow rate decreases, resulting in slightly less savings than would be predicted
using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5
power of flow rather than the cube of flow. Using this relationship, the required pump
input power at 900 gpm would be:
WPC = WPA (VC/VA)2.5 = 22.5 hp x (900 gpm / 1200 gpm) 2.5 = 11.0 hp
Thus, savings would be:
Variable Speed Drive Savings = WPA – WPC = 22.5 hp – 11.0 hp = 11.5 hp
42
This estimate of savings incorporates the reduction in motor efficiency, and power loss by
the VSD.
Example: Slow Pump Speed with Inlet/Outlet Head
The figure below shows pump performance at various speeds and a system curve. At design
conditions, the pump operates at 1,200 RPM at point A. The system curve extends from an
inlet/outlet pressure difference of 20 ft H20. This inlet/outlet head can result from lifting
water over an elevation change of 20 ft in an open system, or as the required head between
the supply and return pipes for controlling a variable speed drive.
At A, the volume flow rate is 1,200 gpm, the total head is 55 ft-H20 and the pump efficiency
is 74%. According to the chart, the required power to the pump at this operating point is
about 23 hp. Alternately, the required pump input power could be calculated as:
WPA = 1,200 gpm x 55 ft-H20 / (3,960 gpm-ft-H20/hp x 0.74) = 22.5 hp
If the flow were reduced to 900 gpm with a flow-control valve, the operating point would
move along the pump curve to point B at 900 gpm, 62 ft-H20 with pump efficiency of 70%.
According to the chart, the required power to the pump at this operating point is about 20
hp. Alternately, the required pump input power could be calculated as:
WPB = 900 gpm x 62 ft-H20 / (3,960 gpm-ft-H20/hp x 0.70) = 20.1 hp
43
Thus, the pump power savings from reducing flow from 1,200 gpm to 900 gpm with a flowcontrol valve would be about:
Flow-control Valve Savings = WPA – WPB = 22.5 hp – 20.1 hp = 2.4 hp
To determine the pump speed required to deliver a flow of 900 gpm, it is necessary to
develop the design system curve with the flow-control valve wide open and the pump
operating at A. Pressure drop through piping systems varies with the square of flow rate.
Thus, the equation for a system curve that passes through the origin can be written as:
h = h0 + C V2
Where h0 is the pump head at zero flow. The coefficient, C, for the new system curve can
be found by substituting the values of pressure drop and volume flow rate for point A.
C = (h – h0) / V2 = (55 – 20) / 1,2002 = 0.00002430556
Thus, the pressure drop through the new duct system at 235 gpm would be about:
h = 40 + C V2 = 20 + 0.00002430556 9002 = 40 ft-H20
The flow rate of 900 gpm and 40 ft-H20 defines point C. This indicates that the pump speed
should be reduced to 900. At this operating point, the pump would be about 70% efficient,
and the required pump input power would be about:
WPC = 900 gpm x 40 ft-H20 / (3,960 gpm-ft-H20/hp x 0.70) = 12.9 hp
Alternately, the power required by the pump at point C, PC, can be read from the chart to
be about 13 hp. Pump power savings are the difference between WPA and WPC.
Variable Speed Drive Savings = WPA – WPC = 22.5 hp – 12.9 hp = 9.6 hp
Alternately, power savings from reducing the volume flow rate can be estimated from the
pump affinity laws. However the cubic relationship between pump work and volume flow
rate applies only to the friction head, not the inlet/outlet head. Thus, the component of
total power due to friction must be calculated. The component of total power due to
friction is the difference between the total power and the inlet/outlet power.
The inlet-outlet power at point A, WPA0, is:
WPA0 = 1,200 gpm x 20 ft-H20 / [3,960 gpm-ft-H20/hp x 0.74] = 8.2 hp
44
Thus, the component of total head due to friction is:
WPAF = WPA - WPA0 = 22.5 hp – 8.2 hp = 14.3 hp
Theoretically, pump work varies with the cube of volume flow rate. In practice, the
efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in
slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we
estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of
flow. Using this relationship, the component of power at C due to reduced friction P CF at
235 gpm would be about:
WPCF = WPAF (VC/VA)2.5 = 14.3 hp x (900 gpm / 1,200 gpm) 2.5 = 7.0 hp
Using the 2.5 exponent, power savings would be about:
Savings = WPAF – WPCF = 14.3 hp – 7.0 hp = 7.3 hp
This estimate of savings incorporates the reduction in motor efficiency, and power loss by
the VSD.
Variable Speed Drive Applications
Electronic variable speed drives (VSDs) control the speed of AC motors by converting the
frequency and voltage of the AC line supply from fixed to variable values. VSDs are used in
both constant and variable torque applications. In variable torque applications, such as
pumping and fan systems, slowing the motor speed reduces the torque on the motor and
can result in significant energy savings. These savings can be estimated using the Pump/Fan
Affinity Laws.
VSDs subject motors to voltage spikes and fast voltage rise and fall times. These voltage
spikes can “punch” through traditional winding insulation. Because of this, VSDs should only
be coupled to motors that the manufacturer specifies as suitable for PWM VSDs. If a motor
is to be rewound, be sure to specify rewinding characteristics for PWM VSD motors. Most
energy-efficient motors are suitable PWM VSDs. Because VSDs work best with premium
efficiency motors, the motor may need to be upgraded if it is not a premium efficiency
motor. Typical installed costs of VSDs are shown below.
45
Horsepower
5
10
15
20
25
50
100
w/bypass
$1,653
$2,008
$2,233
$2,458
$2,772
$4,344
$7,021
w/o bypass
$1,515
$1,771
$2,028
$2,332
$3,854
$6,049
Wiring
$500
$579
$658
$737
$816
$1,211
$2,000
Installation
$400
$463
$526
$589
$653
$968
$1,600
Controls
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
$2,000
w/bypass
$4,553
$5,050
$5,417
$5,784
$6,240
$8,523
$12,621
w/o
bypass
$4,557
$4,955
$5,354
$5,800
$8,033
$11,649
Typical 2006 costs with 25% contractor markup on drives assuming existing motor is
suitable for inverter use.
VSD pump retrofits typically require making three changes to the existing pumping system:
1) Install a VSD on power supply to the pump motor. In parallel pumping
configurations, one VSD is generally needed for each operational pump, but not for
the backup pump.
2) Close valves on all by-pass pipes.
3) Install a differential-pressure sensor between the supply and return headers at the
process load located the farthest distance from the pump or the process load with
the greatest required head to generate flow through the load. Determine the
pressure drop needed to guarantee sufficient flow through the farthest process load
at this point. Control the speed of the VSD to maintain this differential pressure.
If these procedures are followed, the inlet-outlet head across the pump approaches the
differential-pressure between the supply and return headers as flow decreases. For
example, if the differential-pressure between the supply and return headers were set to 20
ft-H20, the system curve would run from 20 ft-H20 at zero flow through C to A in the figure
below. Thus, when modeling energy savings in VSD applications, use the procedure
demonstrated in “Example: Slow Pump Speed with Inlet/Outlet Head”. In addition,
significant energy savings can result from minimizing the differential-pressure set point.
46
VSD Pumping: Industrial Example
A typical industrial cooling configuration using a constant speed pump is shown below. The
second figure shows the system after a VSD retrofit. With the retrofit, the by-pass pressure
relief valve would be closed, flow through each process load would be controlled at the
load, and the VSD would modulate pump speed based on the differential pressure between
the supply and return headers. The device marked dP is a differential-pressure sensor
which would control the speed of the VSD.
bypass /
pressure
relief
valve
cooling
tower
dP
cooling
water to
process
loads
7.5 hp
pump
city water
make-up
25 hp
pump
reservoir
warm
water
cool
water
VSD
process water return
Constant volume
47
Variable volume pumping system
VSD Pumping: Commercial Building Example
A typical constant-flow commercial-building chilled water system is shown below. This
system provides constant flows through the condenser and evaporator of each chiller
whenever a chiller is operational. The primary chilled water pumps are typically much
smaller than the secondary pumps, since a primary pumps only have to move water
through the chiller evaporators and not through the entire building. The secondary chilled
water pumps are much larger than the primary pumps and provide a constant flow of
chilled water to the AHUs. Each AHU varies the quantity of chilled water through the coil
and bypasses unneeded chilled water.
Cooling Tower Fan
Chilled Water Supply
AHU 1
Chiller 1
Chiller 2
Condenser
(Cooling Tower)
Pumps
AHU 2
AHU 3
Secondary
Chilled Water
Pumps
Chilled Water Return
Primary
Chilled Water
Pumps
Close-ups of a typical piping configuration at the air handler cooling coils in a constant-flow
chilled-water supply system are shown below. The three-way valves direct chilled water
either through the cooling coil or around the cooling coil via the bypass loop. The flow of
chilled water through the cooling coils is varied to maintain the temperature of the air
leaving the cooling coils at a constant temperature. In a VSD retrofit, the bypass valves
would be closed, and a differential-pressure sensor would be installed between the supply
and return headers at the air handler located farthest from the pump. In some cases, it may
be necessary to replace the three-way valves with two-way valves if the three-way valves
were not designed to handle larger pressure drops in a VSD situation.
48
Cooling Coil
Manual
2-Way Valve
Tsa
Automatic
3-Way Valve
Cooling Coil
Manual
2-Way Valve
Tsa
Automatic
3-Way Valve
Chilled Water Supply
Chilled Water Return
Piping configuration at air handling units.
The greatest pump energy savings come from changing the secondary chilled-water loop
from constant to variable flow. This is done by:




Removing or blocking the bypass piping on each AHU
Replacing 3-way valves with 2-way valves on each AHU
Adding VFDs to the secondary chilled water pumps
Controlling the VFDs based on the differential pressure between the supply and
return headers
A typical variable flow secondary chilled water loop system is shown below.
Cooling Tower Fan
VFD
Chilled Water Supply
VFD
AHU 1
AHU 2
AHU 3
Chiller 1
dP
Secondary
Chilled Water
Pumps
Chiller 2
Condenser
(Cooling Tower)
Pumps
49
Chilled Water Return
Primary
Chilled Water
Pumps
Modern chillers are designed to accommodate variable flow through the evaporators and
condensers. This enables full variable flow chilled water plants. A variable-flow chilled
water plant with a flow control and bypass valve to guarantee minimum flow to the chillers
is shown below.
Cooling Tower Fan
VFD
Chilled Water Supply
AHU 1
AHU 2
AHU 3
Chiller 1
dP
VFD
VFD
Bypass
Valve
Chiller 2
VFD
Condenser
(Cooling Tower)
Pumps
Flow
Meter
Chilled Water Return
VFD
Primary
Chilled Water
Pumps
References
ASHRAE Handbook: Fundamentals, 1977, 1985, 1996, 2005, ASHRAE.
Bernier and Bourret, 1999, “Pumping Energy and Variable Frequency Drives”, ASHRAE
Journal, December.
Gould Pumps, GPM 7-CD, Technical Information.
Incropera and DeWitt, 1985, Fundamentals of Heat and Mass Transfer, John Wiley and
Sons.
Kreider and Rabl, 1994, Heating and Cooling of Buildings, McGraw-Hill Inc.
Larson, E.D. and Nilsson, L.J., 1991, “Electricity Use and Efficiency in Pumping and Air
Handling Systems, ASHRAE Transactions, pgs. 363-377.
McQuiston and Parker, 1994, Heating Ventilating and Air Conditioning, John Wiley and Sons,
Inc.
McQuiston, F., Parker, J. and Spitler, J., 2000, “Heating, Ventilating and Air Conditioning:
Analysis and Design”. John Wiley and Sons, Inc.
50
Mott, R. L, 2000, Applied Fluid Mechanics, Prentice Hall, Inc.
Nadel, S., Shepard, M., Greenberg, S., Katz, G., and Almeida, A., 1991, “Energy Efficient
Motor Systems”, American Counsel for an Energy Efficient Economy, Washington D.C.
Taylor, S., 2002, “Primary-Only vs. Primary Secondary Variable Flow Systems”, ASHRAE
Journal, February, 2002, pgs 25-29.
Tutterow, 199x, Energy Efficient Fan Systems, Industrial Energy Technology Conference,
Houston, TX
United McGill, 1990, Engineering Design Reference Manual.
U.S. Department of Energy, 2002m “Pumping Systems Field Monitoring and Application of
the Pumping System Assessment Tool PSAT”,
51