Energy-Efficient Duct and Pipe Systems
Introduction
This chapter covers the fundamentals of how to design energy-efficient of duct and
piping systems in multi-zone buildings.
Duct Systems in Multi-zone Buildings
In most multi-zone buildings, duct systems deliver air from air handling units to
individual zones. The two principle types are single-duct systems and dual-duct
systems.
A single-duct system is shown below. In single-duct systems, the supply air is cooled
while passing through a cooling coil before being directed to each zone. In a single-duct
constant-air-volume system, SD-CAV, the flow of cool air to the zone remains constant;
heat is added in a reheat box at each zone to meet the cooling/heating load of the zone.
In a single-duct variable-air-volume system, SD-VAV, zone temperature is maintained by
varying the flow of cool air using an actuated damper in the VAV box. VAV system are
much more energy efficient that constant flow systems in part because fan power is
reduced at low loads and low air flows.
Return Air Fan
Qsen 1
Zone 1
Qsen 2
Zone 2
Qlat 1
Qlat 2
Tz1
Reheat or VAV Box 1
TRA
Filter
Supply Air Fan
Tz2
Reheat or VAV Box 2
Cooling Coil
TSA
TMA
TOA
3-Way Valve
3-Way Valve
CW Supply
CW Return
HW Supply
HW Return
3-Way Valve
HW supply
HW Return
Single-duct system with two zones.
A VAV box for an internal zone that never requires heating and a VAV box for an
external zone that sometime requires heating are shown in the pictures below.
Energy-Efficient Fluid Flow Systems
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VAV box without reheat
VAV box with hot-water reheat
A dual-duct heating and cooling system is shown below. In a dual-duct constant-airvolume system, DD-CAV, warm and cool air streams are mixed in a mixing box at each
zone to meet the zone cooling/heating load; although the ratio of cool to warm air
varies, the total air flow to the zone remains constant. In a dual-duct variable-air-volume
system, DD-VAV, zone temperature is maintained by varying the amount of cold or
warm air introduced into the zone. When the zone calls for heating, all cooling air is
shut off. When the zone calls for cooling, all heating air is shut off. The warm and cold
air streams are mixed only during very low-load conditions, to maintain a minimum air
flow into the zone. The result is that the total flow of air into the zone varies, rather
than remaining constant as in CAV systems.
Return Air Fan
Qsen 1
Zone 1
CW Supply
RA
Qsen 2
Zone 2
Qlat 1
CW Return
Tz1
Qlat 2
Tz2
3-Way Valve
Mixing or VAV Box 1
Mixing or VAV Box 2
CSA
Cooling Coil
MA
OA
Heating Coil
HSA
Filter
Supply Air Fan
3-Way Valve
HW Supply
HW Return
Dual-duct system with two zones.
Piping Systems in Multi-zone Buildings
In many multi-zone buildings use water-cooled chillers to cool the building. In these
systems, chilled water is pumped from the chillers to air handling units. The piping
system that distributes the chilled water to the air handlers is typically the biggest piping
system in the building. In addition, water-cooled chillers reject heat to the environment
through cooling towers. In these systems, a water is pumped through a separate piping
Energy-Efficient Fluid Flow Systems
2
system through the chiller condenser and to the cooling tower. Thus, most multi-zone
buildings have a chilled-water piping loop and a cooling-tower piping loop.
A typical constant-flow primary/secondary chilled water system is shown below. In this
system, primary pumps provide constant flows of water through each chiller whenever
the chiller is operational. Secondary pumps push the chilled water from the chiller plant
to each of the air handing units and back to the chilled water plant. 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. 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
Constant-flow chilled water system with primary/secondary chilled water pumps.
Close-ups of a typical piping configuration at the air handler cooling coils in a constantflow 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.
Energy-Efficient Fluid Flow Systems
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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 untis.
Modern chillers are designed to accommodate variable flow through the evaporators
and condensers. This enables full variable-flow chilled water plants such as shown
below. In this system, the 3-way valves at the air handlers have been replaced with 2way valves, which eliminates by-pass at the air handlers. A single set of chilled water
pumps configured in parallel supply water to the chillers and then to the air handlers.
To guarantee minimum flow to the chillers, the system employs a bypass loop in which
the position of the bypass valve is controlled based on flow measured by the flow
meter. Variable-flow systems are much more energy efficient that constant-flow
systems since pumping power is reduced at low loads and low flows.
Cooling Tower Fan
VFD
Chilled Water Supply
AHU 1
Chiller 1
VFD
AHU 2
AHU 3
Bypass
Valve
VFD
Chiller 2
VFD
Condenser
(Cooling Tower)
Pumps
Flow
Meter
Chilled Water Return
VFD
Primary
Chilled Water
Pumps
Variable-flow chilled water system with primary chilled water pumps.
Energy-Efficient Fluid Flow Systems
4
Principles of Energy Efficient Fluid Flow
Energy Balance Approach
The work required to move a fluid through a pipe or duct can be derived from an energy
balance on the system. The electrical power to a pump/fan, We, is:
We = V Ptotal / [Effpump/fan x Effdrive x Effmotor ]
where V is the volume flow rate, Ptotal is the total pressure rise created by the
pump/fan and Effpump/fan, Effdrive, and Effmotor are the efficiencies of the pump/fan,
drive and electrical motor. The energy balance equation shows that electrical power is
converted into fluid power in the form of a volume flow rate of fluid raised to a pressure
great enough to overcome friction and inlet/outlet requirements. In addition, some
electrical power is converted to heat rather than fluid work due to inefficiencies of the
pump/fan, drive and motor.
Thus, the energy balance equation is also a guide to improving the energy-efficiency of
fluid flow systems:



Reduce volume flow rate
Reduce required pump/fan head
o Pstatic
o Pvelocity
o Pelevation
o Pfriction
Increase pump/fan, drive and motor efficiency
In this chapter we focus on designing duct and pipe systems to reduce required
pump/fan head. Key savings opportunities are:
o
o
o
o
Increase pipe/duct diameter
Use smoother pipe/duct
Use low-friction fittings
Reduce pipe/duct length and turns
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:
Energy-Efficient Fluid Flow Systems
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Q
2
1
Wf
Figure 1. Control-volume diagram of pumping system.
Wf + m1(h + V2/2 + gz)1 - m2(h + V2/2 + gz)2 – Q = 0
Wf = m2(h + V2/2 + gz)2 – m1(h + V2/2 + 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.
Substituting m1 = m2 = m and h = u + Pv into Equation 1 gives:
Wf = m [ (u + Pv + V2/2 + gz)2 – (u + Pv + V2/2 + gz)1 + q ]
Wf = m [ (Pv + V2/2 + gz)2 – (Pv + V2/2 + gz)1 + (q + u2 – u1) ]
Wf = m [ (Pv + V2/2 + gz)2 – (Pv + V2/2 + 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/ + V2/2 + gz)2 – (P/ + V2/2 + gz)1 ] + (Q + U2 – U1)
Wf = V [ (P + V2/2 + gz)2 – (P + V2/2 + gz)1 ] + (Q + U2 – U1)
Wf = V [ (P2 – P1) + V22- V12) + 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)
Energy-Efficient Fluid Flow Systems
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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), V22- V21), 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
V22- V12) 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 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.
Energy-Efficient Fluid Flow Systems
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Gage pressure P = P – Po = g h
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, ftwg. 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)
[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.
Energy-Efficient Fluid Flow Systems
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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
Fan Systems
In US units, a useful dimensional equation to calculate the fluid work, in horsepower, to
move air at standard conditions (density = 0.075 lbm/ft3) through ducts is:
Wf = V Ptotal = V (g  z)
Wf (hp) = V (ft3/min) htotal (in-H20) / 6,356 (ft3-in/min-hp)
[7]
Equation [5] is for air at standard conditions when the density of air is 0.075 lbm/ft 3.
This corresponds to air at sea level at about 65 F and 40% relative humidity or 70 F and
0% relative humidity. The density of air decreases at higher elevations and
temperatures. The correction for non-standard conditions is:
Pactual = Pstandard (actual / standard)
hactual = hstandard (actual / standard)
For air at non-standard elevations and temperatures, the following dimensional
correction factors can be applied (Source: United McGill, 1990):
Pactual = Pstandard KE KT or hactual = hstandard KE KT
KE = [1 – (6.8754 x 10-6) Z (ft)]4.73
KT = [530 / (Tair (F) + 460)]0.825
Example
Calculate the work added to the fluid (hp) by a fan moving 10,000 cfm of air at 300 F if
the pressure rise across a fan is 3.0 in-H20.
KT = [530 / (Tair (F) + 460)]0.825 = [530 / (300 (F) + 460)]0.825 = 0.743
Energy-Efficient Fluid Flow Systems
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h = hstandard KT = 3.0 in-H20 x 0.743 = 2.23 in-H20
Wf (hp) = V (ft3/min) htotal (in-H20) / 6,356 (ft3-in/min-hp)
Wf (hp) = 10,000 cfm x 2.23 (in-H20) / 6,356 (ft3-in/min-hp) = 3.51 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.
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.
Energy-Efficient Fluid Flow Systems
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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
Example
If air 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 27.7 in-H20/psi = 277 in-H20
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 is zero.
Pvelocity = V22- V12) = 0
Energy-Efficient Fluid Flow Systems
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When the inlet and outlet velocities are different, the change in velocity pressure must
be calculated. Useful dimensional relationships to calculate velocity V from volume flow
rate, V, and pipe diameter, d, are:
V (ft/s) = 0.4085 V (gpm) / [d (in)]2
V (ft/min) = 183.35 V (cfm) / [d (in)]2
Useful dimensional relations to calculate the velocity head associated with a velocity, V,
for water at standard conditions are:
hvelocity = Pvelocity / H20 g = fluid V2 H20 g]
h velocity (ft-H20) = 0.0155[V (ft/s)]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[V1 (ft/s)]2 = 0.0155[2.55 (ft/s)]2 = 0.10 ft-H20
h velocity2 (ft-H20) = 0.0155[V2 (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
Example
If 1,000 cfm of air is pumped through a duct with an inlet diameter of 12 inches and
discharged from a duct with an outlet diameter of 8 inches, the required velocity head
is:
V1 = 183.35 V (cfm) / [d (in)]2 = 183.35 x 1,000 cfm / (12 in)2 = 1,273 ft/min
V2 = 183.35 V (cfm) / [d (in)]2 = 183.35 x 1,000 cfm / (8 in)2 = 2,865 ft/min
hvelocity1 (in-H20) = [V1 (ft/min) / 4,005]2 = (1,273/4,005)2 = 0.1010 in-H20
hvelocity2 (in-H20) = [V2 (ft/min) / 4,005]2 = (2,865/4,005)2 = 0.5117 in-H20
h velocity = hvelocity2 - hvelocity1 = 0.5117 in-H20 - 0.1010 in-H20 = 0.4107 in-H20
Energy-Efficient Fluid Flow Systems
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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 )

For water at standard conditions, a useful dimensional relationship is:
helevation (ft-H20) = (z2 – z1) ft
For air at standard conditions, a useful dimensional relationship is:
helevation (in-H20) = 0.01445 [z2 (ft) – 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
Example
If air at standard conditions is lifted through an elevation gain of 100 ft, the required
elevation head is:

helevation = 0.01445 x 100 (ft) = 1.445 in-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
Energy-Efficient Fluid Flow Systems
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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.
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
Energy-Efficient Fluid Flow Systems
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Typical values for pipe roughness factors, e, are shown in the figure below.
Material
Glass, plastic
Copper, brass, lead (tubing)
Cast iron – uncoated
Cast iron – asphalt coated
Commercial steel or welded steel
Wrought iron
Riveted steel
Concrete
e
r
D
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
Typical duct roughness factors, e, for common air ducts, including material, joint type,
and joint spacing are shown in the table below.
Type
Smooth (PVC plastic pipe)
Medium (Galvanized steel with spiral or longitudinal seams every 4 ft)
Average (Galvanized steel with longitudinal seams every 2.5 ft)
Medium Rough (Fiberous glass duct)
Rough (Flexible duct) = 0.01 ft
e
0.0001 ft
0.0003 ft
0.0005 ft
0.003 ft
0.01 ft
Source: ASHRAE Fundamentals 2005, Pg. 35.7
Energy-Efficient Fluid Flow Systems
15
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
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)
Example
Calculate the friction head loss (in-H20) to convey 1,000 cfm of air through 100 ft of 12in diameter medium smooth duct using the friction-factor method and Churchill
relation.
A =  D2 / 4 = 3.14159 x (12/12)2 / 4 = 0.7854 ft2
V = V / A = (1,000 ft3/min / 0.7854 ft2 = 1,273 ft/min
Re = V D / = 1,273 ft/min x 60 min/hr x (12/12) ft / 0.572 ft2/hr = 133,557
e/D = (0.0003) ft / (12/12) ft = .0003 ft
A = [2.457 ln(((7/Re)^0.9 + (0.27e/D))^-1)]^16 = 1.11595 E21
B = [37,530/Re]^16 = 1.51151E-09
f = 8[(8/Re)^12 + (A+B)^-1.5]^(1/12) = 0.0187 (from Churchill)
P = (f L  V2) / (2 D) = 0.0187 x 100 ft x 0.075 lbm/ft3 x (1,273 ft/min)2 / (2 x 12/12 ft x
(60 s/min)2) = 31.5997 lbm/ft-s2
h = P / (g w lbm/ft-s2 / (32.2 ft/s2 x 62.27 lbm/ft3) x 12 in/ft
h = 0.1891 in-H20 (per 100-ft duct)
Monograph Method
Alternately, head loss due to friction for water flow through pipes, hp, can be
determined from monographs such as shown below.
Energy-Efficient Fluid Flow Systems
16
Pipe friction loss (ASHRAE, 2005)
Similarly, head loss due to friction for air flow through ducts, hp, can be determined
from monographs such as shown below.
Energy-Efficient Fluid Flow Systems
17
Source: ASHRAE Handbook: Fundamentals, 2005, ASHRAE.
Energy-Efficient Fluid Flow Systems
18
Example
Calculate the friction head loss in ft-H20 for pumping 100 gpm of water through 200 ft of
3-in 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:
h = 2.5 ft-H20 per 100-ft pipe x 200 ft-pipe = 5.0 ft-H20
Example
Calculate the friction head loss (in-H20) from moving 20,000 cfm of air through 200 ft of
34-inch diameter duct using the ASHRAE monograph.
From monograph, h/L at 20,000 cfm and 34-inch duct is 0.3 in-H20 per 100-ft duct.
h = h/L x L = [3 in-H20 per 100-ft duct] x 200 ft-duct = 0.6 in-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 V2 / 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 V2 / (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 V2 / (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:
Energy-Efficient Fluid Flow Systems
19
V (ft/s) = 0.4085 V (gpm) / [d (in)]2
In US units, this height is commonly measured in in-H20 for fan systems. For air flow
through duct fittings, a useful dimensional relationship is:
hf (in-H20) = fluid kf V2 / (2  g) = kf [V (ft/min) / 4,005]2
A useful dimensional relationship to calculate velocity, V, from volume flow rate, V, and
duct diameter, d, is:
V (ft/min) = 183.35 V (cfm) / [d (in)]2
Loss coefficient data, kf, for common pipe fittings are shown in the Tables below.
Source: ASHRAE Fundamentals 2005, Pg 36.2.
Energy-Efficient Fluid Flow Systems
20
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
From monograph in Figure 4 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
Energy-Efficient Fluid Flow Systems
21
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
Values of loss coefficient, kf, for elbows in ducts are shown in the figure below, where
Vu is upstream velocity in ft/min. Note that kf, and hence pressure drop, varies by a
factor of more than 10 for the different types of elbows shown. This demonstrates the
importance of selecting low pressure-drop fittings when designing energy efficient
duct/fan systems.
Energy-Efficient Fluid Flow Systems
22
Loss Coefficients for 90o elbows
1.5
1.4
1.3
1.2
1.1
Loss Coefficient, K
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
5 gore elbow
0.1
7 gore elbow
0
0
3
6
9
12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60
Diameter (in)
Mitered elbow without
turning vanes
Mitered elbow with
turning vanes
5-gore elbow
r/D = 1.5
7-gore elbow
r/D = 2.5
Source data from ASHRAE Handbook: Fundamentals 2002
Pressure Loss through 90-Degree Branch Fittings
Values of loss coefficient, Kb, for 90-degree branch fittings are shown in the figure
below. To enter this table, calculate the ratio of the velocity in the branch, Vb, and the
velocity upstream of the branch, Vu. The branch loss coefficient, Kb, can then be
Energy-Efficient Fluid Flow Systems
23
determined from the table. Note that the loss coefficients vary significantly as a
function of the design of the branch. For example, at Vb/Vu = 1.5, the branch loss
coefficient for a straight T is about four times greater than for a LoLoss-T. This
emphasizes the importance of fitting selection for energy–efficient duct design.
Pressure Loss through Reductions
Loss coefficient data, Kd, for straight-through reductions, such as those on the
downstream side of a 90-degree branch fitting, are shown in the figure below. To enter
this table, calculate the ratio of the velocity downstream, Vd, and upstream, Vu, of the
reduction. The branch loss coefficient, Kd, can then be determined from the table. Use
the upstream velocity Vu to calculate the velocity pressure when calculating head loss
through the reducer:
hf (in-H20) = kd [Vu (ft/min) / 4,005]2
Energy-Efficient Fluid Flow Systems
24
Source Data: United McGill, 1990.
Pressure Loss through Dampers
Dampers are used to balance airflow in ducts and regulate the quantity of outside air
introduced into air-distribution systems. A schematic of parallel and opposed blade
damper are shown below. Even when fully open, dampers result in significant pressure
loss. A typical loss coefficient, k, for a fully open damper is 0.52.
Frame
Frame
Stop
Stop
Parallel
Operation
Energy-Efficient Fluid Flow Systems
Opposed
Operation
25
Example
Find the fluid power, Wf, required to push 1,000 cfm of air through 200 ft of 12-inch
duct with four 5-gore 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
From the monograph, for 1,000 cfm through 12-inch duct, Pduct ~ 0.2 in-H20/100 ft.
Hence, the total head loss through the duct is:
hp = 0.2 in-H20/100 ft x 200 ft = 0.4 in-H20
The velocity is:
V = V / A = 1,000 ft3/min / (3.143 x 12 ft2 / 4) = 1,273 ft/min
From the fittings chart, for a 12-inch 5-gore elbow, kf = 0.18. For four elbows, the total
head loss is:
hf (in-H20) = Nf x kf (V / 4,005)2 where V is in ft/min
hf (in-H20) = 4 x 0.18 x (1,273/4,005)2 = 0.073 in-H20
The total head loss through ducts and fittings is:
htotal = hp + hf = 0.4 in-H20 + 0.073 in-H20 = 0.473 in-H20
The work added to the air is:
Energy-Efficient Fluid Flow Systems
26
Wf (hp) = V (ft3/min) htotal (in-H20) / 6,356 (ft3-in/min-hp)
Wf (hp) = 1,000 (ft3/min) 0.473 (in-H20) / 6,356 (ft3-in/min-hp) = 0.074 hp
Pressure-Gradient Diagrams
Diagrams of total pressure throughout the fan/duct system are often instructive and
useful for avoiding design mistakes. The total pressure, which is a measure of energy,
always decreases in the direction of flow due to friction and other losses. A total
pressure diagram for a typical system is shown below.
Heat
Element Heat and
Cool Coils
Fan
Return
grille
Return
Duct
Supply Ducts
Diffusers
Filter
1
4
3
Fan
5
6
8
7
2
0.42
Total Pressure, inches of water
5
0.18
6
2
Atmospheric Pressure
7
0.04
8
Distance
1
— 0.05
3
— 0.12
4
—0.22
System Schematic and System Pressure Grade Line
Energy-Efficient Fluid Flow Systems
27
The total pressure drop is the sum of the static and velocity pressure drops:

Ptotal = Pstatic + Pvelocity
Although total pressure always decreases, the static and velocity pressures will vary if
duct diameter changes. The following pressure diagrams illustrate how static pressure
can be converted into velocity pressure, and vise versa, when duct diameter changes.
Source: ASHRAE Handbook: Fundamentals 2005.
.
When a fitting’s inlet and outlet areas are different, the inlet and outlet velocity will be
different. Thus, the total pressure drop through the fitting must include the difference
in velocity pressures. If a fitting manufacturer only lists the static pressure drop across
the fitting, be sure to add the velocity pressure drop when computing total pressure
drop.
Ptotal = Pstatic + Pvelocity
Ptotal = (P2 – P1) + V22- V12)
htotal = hstatic in-H20 + [(V2 / 4,005)2 - (V1 / 4,005)2] in-H20 (where V is in ft/min)
Energy-Efficient Fluid Flow Systems
28
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
To generate a starting place for sizing ducts carrying 40,000 cfm and less, select the duct
diameter such that:
0.08 in-H20/100 ft < hp < 0.15 in-H20/100 ft
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 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.
Energy-Efficient Fluid Flow Systems
29
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 equal-friction 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 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).
A
E (20 ft H20)
Energy-Efficient Fluid Flow Systems
B
C
CC
CC
CC (10 ft H20)
30
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

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]
Energy-Efficient Fluid Flow Systems
31
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
Parallel Flow Duct System Design
In commercial buildings, duct systems almost always include multiple branches, which
result in parallel flow. Several methods exist for designing parallel-flow 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 duct system shown below. The ducts can be sized to deliver the specified
volume flow rates using the Equal Friction or Equal Pressure method. For both
methods, assume a design friction head loss of 0.10 in-H20/100 ft. Neglect pressure
losses through elbows, fittings and inlet dampers and outlet dampers.
A
LA = 40 ft
VA = 700 cfm
B
LB = 200 ft
VB = 500 cfm
C
LC = 20 ft
VC = 200 cfm
Equal Friction Method
Enter the monograph with the design friction loss of and volume flow rate to determine
duct diameters.
DA (700 cfm at 0.10 in-H20/100 ft) = 12 inches
DB (500 cfm at 0.10 in-H20/100 ft) = 10.5 inches
DC (200 cfm at 0.10 in-H20/100 ft) = 7.5 inches
Calculate the head loss from the fan to the outlet of each duct using these duct
diameters.
Energy-Efficient Fluid Flow Systems
32
hAB = hA + hB = (0.10 in-H20/100 ft x 40 ft) + (0.10 in-H20/100 ft x 200 ft)
hAB = hA + hB = 0.04 in-H20 + 0.20 in-H20 = 0.24 in-H20
hAC = hA + hC = (0.10 in-H20/100 ft x 40 ft) + (0.10 in-H20/100 ft x 20 ft)
hAC = hA + hC = 0.04 in-H20 + 0.02 in-H20 = 0.06 in-H20
Note that to deliver the required flow through duct A, the pressure at exit of the fan
must be equal to the maximum of these pressure drops. Thus, the pressure at exit of
the fan must be hAB = 0.24 in-H20. If so, then the pressure from the exit of the fan to
the exit of duct C would also be 0.24 in-H20. This pressure would cause a much higher
flow rate than the desired 200 cfm. Thus, a balancing valve must be added to the end of
duct C to generate the required flow.
Equal Pressure Method
The Equal Pressure method also begins by entering the monograph with the design
friction loss of and volume flow rate to determine duct diameters.
DA (700 cfm at 0.10 in-H20/100 ft) = 12 inches
DB (500 cfm at 0.10 in-H20/100 ft) = 10.5 inches
DC (200 cfm at 0.10 in-H20/100 ft) = 7.5 inches
The method then calculates the pressure drop through each branch duct using these
duct diameters.
hB = (0.10 in-H20/100 ft x 200 ft) = 0.20 in-H20
hC = (0.10 in-H20/100 ft x 20 ft) = 0.02 in-H20
As before, to deliver the required flows the pressure at the entrance to branch ducts
must be equal to the maximum of these pressure drops. Thus, the pressure at entrance
of the branch ducts must be h = hB = 0.20 in-H20. If so, then the pressure drop from
this point to the exit of duct C will also need to be 0.20 in-H20. Using this relation, the
required pressure drop per 100 feet of duct length can be found.
hB = hC
0.20 in-H20 = (X in-H20/100 ft x 20 ft)
X = 1.0 in-H20/100 ft
Finally, enter the monograph with this pressure drop and flow rate to determine the
required duct diameter.
DC (200 cfm at 1.0 in-H20/100 ft) = 4.8 inches
Energy-Efficient Fluid Flow Systems
33
Note that no balancing valve is required. The duct was sized to generate the required
flow.
Duct System Design Software
In practice, these methods are frequently built into duct design software. DuctDesigner
(Kissock, 2004) is one such program. The use of the DuctDesigner software to
determine duct diameters and the required pressure at the exit of the fan is
demonstrated in the examples that follow.
Example
Calculate duct diameters for the following system using the DuctDesigner Equal
Pressure Method
1500 ft3/min
1000 ft3/min
5
3
50 ft
50 ft
50 ft
1
50 ft
4
50 ft
2
50 ft
Fan
3200 ft3/min
700 ft3/min
50 ft
6
DDInput.XLS
Instructions for Equal Pressure Method
Use this template to create an input file for "DuctDesigner.exe".
Draw duct layout. Specify duct lengths (ft) and flow rates (cfm)
Number all locations where ducts split or discharge air as nodes.
Node 1 should be exit of supply air fan.
Based on layout, fill in the fields below for all ducts.
Duct order is not important.
ENP is "End Node Pressure" for duct. Enter 0 if node opens to atmosphere, otherwise enter -1.
'Select' and 'Copy' numeric values below (NOT HEADINGS) and 'Paste' into Notepad text editor.
Save Notepad file as "DDInputEP.txt"
Run "DuctDesigner.exe"
Select Equal Pressure Method" and specify "DDInputEP.txt" as input file.
"DuctDesigner.exe" will calculate duct diameters and node pressures.
Part Num Str Node End Node Len (ft)
1
1
2
50
2
2
3
50
3
2
4
50
4
4
5
50
5
4
6
50
Energy-Efficient Fluid Flow Systems
V (cfm)
ENP (inH20)
3200
-1
1500
0
1700
-1
1000
0
700
0
34
Example
Calculate duct diameters for the following system using the DuctDesigner Equal Friction
method. Each discharge damper has an equivalent length of 20 feet.
6
20
1500 ft3/min
1000 ft3/min
5
4
19
18
50 ft
3
2 7
50 ft
50 ft
1
50 ft
8 15
9
17
50 ft
16
50 ft
Fan
3200 ft3/min
14
Energy-Efficient Fluid Flow Systems
700 ft3/min
13 12
10
50 ft
11
35
DDInput.XLS
Instructions for Equal Friction Method
Use this template to create an input file for "DuctDesigner.exe", "Equal Friction Method".
Draw duct layout. Specify duct lengths (ft) and flow rates (cfm)
Number all locations where ducts branch, turn, or discharge air as nodes.
Node 1 should be exit of supply air fan.
Based on layout, fill in the fields below for all ducts.
Duct order is not important.
Specify "Type" for each part (i.e. Discharge, Circ Duct, Square Elbow, 7-Gore Elbow, etc.)
'Select' and 'Copy' numeric values below starting in row 29 (NOT HEADINGS)
Paste' into Notepad text editor.
Save Notepad file as "DDInputEF.txt"
Start "DuctDesigner.exe"
Select "Equal Friction Method", and specify "DDInputEF.txt" as input file.
"DuctDesigner.exe" will calculate duct diameters and total pressure for fan.
Type
0
1
2
3
4
5
6
7
Discharge To Atm
Circular Duct
Square Elbow
7-Gore Elbow
90 - Straight T Branch
90 - LoLoss Branch
Straight Reduction
Equiv Length Fitting
Part Num Str Node End Node
1
1
2
2
2
3
3
3
4
4
4
5
5
5
6
6
2
7
7
7
8
8
8
9
9
9
10
10
10
11
11
11
12
12
12
13
13
13
14
14
8
15
15
15
16
16
16
17
17
17
18
18
18
19
19
19
20
Energy-Efficient Fluid Flow Systems
Type
1
5
1
7
0
6
1
5
1
3
1
7
0
6
1
3
1
7
0
Len (ft)
V (cfm)
50
3200
0
1500
50
1500
20
1500
0
1500
0
1700
50
1700
0
700
50
700
0
700
50
700
20
700
0
700
0
1000
50
1000
0
1000
50
1000
20
1000
0
1000
36
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
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 pump is placed near the inlet to the system, as shown below, then the minimum
pressure of the fluid is 0 psig.
Energy-Efficient Fluid Flow Systems
37
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.
Energy-Efficient Fluid Flow Systems
38
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.
Energy-Efficient Duct and Piping 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 “wholesystem, inside-out approach”. The “whole-system” part of this approach emphasizes
the importance of considering the entire conversion, delivery and end-use system. The
“inside-out” 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.
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.
Energy-Efficient Fluid Flow Systems
39
Wf = V [(Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition]
This provides a useful guide for characterizing energy efficiency opportunities.
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 V2 / (2 D)
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 ( 4V /  D2 )2 / (2 D) =8  f L V2 / (  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%!
Energy-Efficient Fluid Flow Systems
40
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%
Example
Calculate the percentage reduction in friction head loss if 1,000 cfm of air is forced
through 6-inch and 12-inch diameter ducts.
From the monogram:
h 6-inch = 6.0 in-H20/100 ft
h12-inch = 0.18 in-H20/100 ft
The percent reduction in friction head loss from doubling the diameter of the pipe
would be about:
(6 – 0.18) / 6 = 97%
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.
Energy-Efficient Fluid Flow Systems
41
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. Similarly, ducts with spiral seams have lower friction losses than ducts
with longitudinal seams, and rigid ducts have much lower friction losses than flexible or
fiberglass ducts.
Reduce System Pressure Drop: Use Fewer Low Pressure-Drop Fittings
Minimizing fittings, including turns, and the use of low-pressure drop fittings can
significantly reduce friction head loss. Consider for example, the table below. The use
of fully-open gate valves instead of globe valves reduces the friction head loss through
the valve by 98%. Similarly, the use of swing type check valves instead of butterfly
valves reduces the friction head loss through the valve by 33%, and long radius elbows
reduce the friction head loss by 50% compared to standard radius elbows.
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 2inch 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%
Energy-Efficient Fluid Flow Systems
42
Similarly, minimizing fittings, including turns, and the use of low-pressure drop fittings in
duct systems also reduces head loss. For example, the table below shows an example of
how attention to low-pressure drop design can reduce the overall head loss of the
system. In this example, duct system head loss was reduced by 47%.
Air-Handing Unit: Clean Filters
Including System Effect
Dirty Filter Allowance
Heat Recovery
Silencer
Supply Duct Work, Diffusers
VAV Device
Zone Coils
Safety Factor
Total Supply
Hood
Flow Device
Exhaust Ductwork
Heat Recovery With Filter
Exhaust Outlet (Incl. Velocity
Pressure)
Total Exhaust
Total Static Supply Plus Exhaust
Typical
TCES – UC Davis
2.2 in. w.g. (548 Pa)
0.68 in. w.g. (169 Pa)
1.3 in. w.g. (324 Pa)
0.5 in. w.g. (125 Pa)
1.0 in. w.g. (249 Pa)
2.5 in. w.g. (623 Pa)
0.5 in. w.g. (125 Pa)
0.4 in. w.g. (100 Pa)
0.6 in. w.g. (149 Pa)
9.0 in. w.g. (2241 Pa)
0.50 in. w.g. (125 Pa)
0.45 in. w.g. (112 Pa)
2.00 in. w.g. (498 Pa)
0.50 in. w.g. (125 Pa)
0.70 in. w.g. (174 Pa)
1.45 in. w.g. (361 Pa)
0.56 in. w.g. (139 Pa)
0
0.65 in. w.g. (162 Pa)
0.30 in. w.g. (75 Pa)
0.20 in. w.g. (50 Pa)
0.60 in. w.g. (149 Pa)
2.2 in. w.g. (1096 Pa)
0.50 in. w.g. (125 Pa)
0.30 in. w.g. (75 Pa)
0.55 in. w.g. (137 Pa)
0.50 in. w.g. (125 Pa)
0.70 in. w.g. (174 Pa)
4.15 in. w.g. (1033 Pa)
13.15 in. w.g. (3275 Pa)
2.55 in. w.g. (635 Pa)
6.95 in. w.g. (1731 Pa)
Source Data: Mathew, P., Greenberg, S., Sartor, D., P Rumsey, P., and Weale, J., 2008,
“Metrics for Energy Performance: Laboratory Performance”, ASHRAE Journal, April, pp.
40-47.
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.
Energy-Efficient Fluid Flow Systems
43
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.
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”,
Energy-Efficient Fluid Flow Systems
44