fluid flow systems - university of...

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 BuildingsIn 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.

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 1

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-air-volume 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.

Dual-duct system with two zones.

Piping Systems in Multi-zone BuildingsIn 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


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.

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 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.


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 2-way 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.

Variable-flow chilled water system with primary chilled water pumps.


Principles of Energy Efficient Fluid FlowEnergy Balance ApproachThe 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 Increase pipe/duct diametero Use smoother pipe/ducto Use low-friction fittingso 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:


Figure 1. Control-volume diagram of pumping system.

Wf + m1(h + V2/2 + gz)1 - m2(h + V2/2 + gz)2 – Q = 0Wf = 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 = mh = 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) + V2

2- 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)


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- V1

2) + g(z2 – z1) + (hl) ] [2]

A number of interesting observations can be made about Equation 2. First, each of the terms (P2 – P1), V2

2- 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 V2

2- 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.


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, 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 WorkPump SystemsIn 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.


SGf = fH20

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-H20Wf (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 SystemsIn 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/ft3. 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


h = hstandard KT = 3.0 in-H20 x 0.743 = 2.23 in-H20Wf (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 ChangesThe 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 SystemsIn 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 SystemsIn 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.


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- V1

2) = 0


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 V 2 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/sV2 (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-H20h 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/minV2 = 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-H20hvelocity2 (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


Elevation Pressure and Head: The change in elevation pressure is:

Pelevation = fluidg (z2 – z1)

The change in elevation head, in terms of water filled manometer height, is:

helevation (ft-H20) = Pelevation / (H20g ) = fluid g (z2 – z1) / (H20g )

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 FrictionTotal 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 DuctsFriction Factor MethodThe 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 andwater 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.

Material Roughness (m) Roughness (ft)Glass, plastic Smooth SmoothCopper, brass, lead (tubing) 1.5 x 10-6 5 x 10-6

Cast iron – uncoated 2.4 x 10-4 8 x 10-4

Cast iron – asphalt coated 1.2 x 10-4 4 x 10-4

Commercial steel or welded steel 4.6 x 10-5 1.5 x 10-4

Wrought iron 4.6 x 10-5 1.5 x 10-4

Riveted steel 1.8 x 10-3 6 x 10-3

Concrete 1.2 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 eSmooth (PVC plastic pipe) 0.0001 ftMedium (Galvanized steel with spiral or longitudinal seams every 4 ft) 0.0003 ftAverage (Galvanized steel with longitudinal seams every 2.5 ft) 0.0005 ftMedium Rough (Fiberous glass duct) 0.003 ftRough (Flexible duct) = 0.01 ft 0.01 ft

Source: ASHRAE Fundamentals 2005, Pg. 35.7

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/minRe = 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 = .000600A = [2.457 ln(((7/Re)^0.9 + (0.27e/D))^-1)]^16 = 4.3862E+20B = [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 12-in 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/minRe = 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 ftA = [2.457 ln(((7/Re)^0.9 + (0.27e/D))^-1)]^16 = 1.11595 E21B = [37,530/Re]^16 = 1.51151E-09f = 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 MethodAlternately, head loss due to friction for water flow through pipes, hp, can be determined from monographs such as shown below.


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.


Source: ASHRAE Handbook: Fundamentals, 2005, ASHRAE.


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 FittingsThe 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:


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.


ExampleFind 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.

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

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.

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

Loss Coefficients for 90o elbows

Diameter (in)

Loss

Coe

fficie

nt, K

Mitered without turning vanes

Mitered with turning vanes

5 gore elbow

7 gore elbow


Mitered elbow without turning vanes

Mitered elbow with turning vanes

5-gore elbowr/D = 1.5

7-gore elbowr/D = 2.5

Source data from ASHRAE Handbook: Fundamentals 2002

Pressure Loss through 90-Degree Branch FittingsValues 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 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 ReductionsLoss 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


Source Data: United McGill, 1990.

Pressure Loss through DampersDampers 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.


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.

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/minhf (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:

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 DiagramsDiagrams 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.

System Schematic and System Pressure Grade Line

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) + V2

2- V12)

htotal = hstatic in-H20 + [(V2 / 4,005)2 - (V1 / 4,005)2] in-H20 (where V is in ft/min)

Pipe/Duct System DesignWhen 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 DiameterThe 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 PipingMany 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.

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 DesignIn 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) 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 ftDB (200 gpm at 3.0 ft-H20/100 ft) = 4 inches with actual dh = 3.5 ft-H20/100 ftDC (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-H20hB,pipe = 3.5 ft-H20/100 ft x 50 ft = 1.75 ft-H20hC,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/sVB (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 x 200 (gpm) / [4 (in)]2 = 5.11 ft/sVC (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]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 dtE = 3.56 hp / 0.90 x 0.75 kW/hp x 8,760 hours/year = 25,980 kWh/yr

Parallel Flow Duct System DesignIn 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 equal-friction 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 MethodEnter 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 inchesDB (500 cfm at 0.10 in-H20/100 ft) = 10.5 inchesDC (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.


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 MethodThe 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 inchesDB (500 cfm at 0.10 in-H20/100 ft) = 10.5 inchesDC (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-H20hC = (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

Note that no balancing valve is required. The duct was sized to generate the required flow.


Duct System Design SoftwareIn 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.

ExampleCalculate duct diameters for the following system using the DuctDesigner Equal Pressure Method

DDInput.XLS

Instructions for Equal Pressure MethodUse 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) V (cfm) ENP (inH20)1 1 2 50 3200 -12 2 3 50 1500 03 2 4 50 1700 -14 4 5 50 1000 05 4 6 50 700 0


ExampleCalculate duct diameters for the following system using the DuctDesigner Equal Friction method. Each discharge damper has an equivalent length of 20 feet.


DDInput.XLS

Instructions for Equal Friction MethodUse 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.

Type0 Discharge To Atm1 Circular Duct2 Square Elbow3 7-Gore Elbow4 90 - Straight T Branch5 90 - LoLoss Branch6 Straight Reduction7 Equiv Length Fitting

Part Num Str Node End Node Type Len (ft) V (cfm)1 1 2 1 50 32002 2 3 5 0 15003 3 4 1 50 15004 4 5 7 20 15005 5 6 0 0 15006 2 7 6 0 17007 7 8 1 50 17008 8 9 5 0 7009 9 10 1 50 700

10 10 11 3 0 70011 11 12 1 50 70012 12 13 7 20 70013 13 14 0 0 70014 8 15 6 0 100015 15 16 1 50 100016 16 17 3 0 100017 17 18 1 50 100018 18 19 7 20 100019 19 20 0 0 1000


Pump and Expansion Tank PlacementIf 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 SystemsConsider 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.


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.

Therefore, to avoid cavitation in open pumping systems, position the pump so it pushes rather than pulls the fluid.

Closed-loop SystemsMost 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.


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.

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 SystemsThe 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 “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.


Wf = V [(Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition] This provides a useful guide for characterizing energy efficiency opportunities.

Energy Savings from Reducing FrictionThe 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 DiameterFriction 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%!


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 fth1-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 fth12-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.


Reduce System Pressure Drop: Use Smooth Pipes/DuctsSimilarly, 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 FittingsMinimizing 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 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%


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%.

Typical TCES – UC Davis

Air-Handing Unit: Clean Filters Including System Effect

2.2 in. w.g. (548 Pa) 0.68 in. w.g. (169 Pa)

Dirty Filter Allowance 1.3 in. w.g. (324 Pa) 1.45 in. w.g. (361 Pa)Heat Recovery 0.5 in. w.g. (125 Pa) 0.56 in. w.g. (139 Pa)Silencer 1.0 in. w.g. (249 Pa) 0Supply Duct Work, Diffusers 2.5 in. w.g. (623 Pa) 0.65 in. w.g. (162 Pa)VAV Device 0.5 in. w.g. (125 Pa) 0.30 in. w.g. (75 Pa)Zone Coils 0.4 in. w.g. (100 Pa) 0.20 in. w.g. (50 Pa)Safety Factor 0.6 in. w.g. (149 Pa) 0.60 in. w.g. (149 Pa)Total Supply 9.0 in. w.g. (2241 Pa) 2.2 in. w.g. (1096 Pa)Hood 0.50 in. w.g. (125 Pa) 0.50 in. w.g. (125 Pa)Flow Device 0.45 in. w.g. (112 Pa) 0.30 in. w.g. (75 Pa)Exhaust Ductwork 2.00 in. w.g. (498 Pa) 0.55 in. w.g. (137 Pa)Heat Recovery With Filter 0.50 in. w.g. (125 Pa) 0.50 in. w.g. (125 Pa)Exhaust Outlet (Incl. Velocity Pressure)

0.70 in. w.g. (174 Pa) 0.70 in. w.g. (174 Pa)

Total Exhaust 4.15 in. w.g. (1033 Pa) 2.55 in. w.g. (635 Pa)Total Static Supply Plus Exhaust 13.15 in. w.g. (3275 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.

ReferencesASHRAE 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.

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”,


fluid flow systems - university of...

Documents