Fluid Flow Systems - University of Dayton · Web viewEnergy-Efficient Fluid FlowIntroduction Pumps and fans move fluids in a wide variety of applications. Careful analysis can improve

Energy-Efficient Fluid Flow

Introduction Pumps and fans move fluids in a wide variety of applications. Careful analysis can improve the energy efficiency of most pump and fan systems. This chapter discusses fundamentals of fluid flow systems, with an emphasis on energy-efficient design, retrofit and operation.

This chapter includes: Derivation of the fluid work equation Methods to calculate pressure loss through pipes and fittings Introduction to piping and duct system design Key principles for designing and operating low-energy fluid flow systems

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

Energy-Efficient Fluid Flow Systems 1

Opportunities for Improving The Energy-Efficiency of Fluid Flow SystemsThese principles can be organized using the inside-out approach, which sequentially reduces end-use energy, distribution energy, and primary conversion energy. Using these principles, common opportunities to cost-effectively improve the energy efficiency of fluid flow systems include:

End Useo Decrease elevation heado Run pumps/fans slower and longer

Distributiono Increase pipe/duct diametero Use smoother pipe/ducto Use low friction fittingso Reduce pipe/duct length and turns

Conversiono Reduce entrance/exit head losso For fixed flow, control flow by trimming pump impellor or slowing fano For variable flow, control flow with VFDo In VFD applications, reduce static pressure requirements during off-peak

periodso In VFD applications, install control pressure sensor near the end of the

duct/piping systemo Select high efficiency pumps/fans, drives, motors, and size pumps/fans to

operate at near peak efficiency

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. 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 = 0.743h = 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 (ft-H20) = [V1 (ft/min) / 4,005]2 = (1,273/4,005)2 = 0.1010 in-H20hvelocity2 (ft-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.

Pipe roughness. (Source: Mott, 2000).

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-07f = 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 some 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 6 at 100 gpm and 3-in diameter steel pipe, the friction head loss through the pipe is:hp = 2.5 ft-H20 per 100-ft pipe x 200 ft-pipe = 5.0 ft-H20

The velocity is:

V (ft/s) = 0.4085 V (gpm) / [d (in)]2 = 0.4085 100 (gpm) / [3 (in)]2 = 4.54 ft/s


From the ASHRAE table, kf = 0.34 for a 3-inch flanged welded 90-degree standard elbow. The friction head loss through four elbow fittings is:

hf (ft-H20) = nf x kf x 0.0155[V (ft/s)]2 hf (ft-H20) = 4 x 0.34 x 0.0155[4.54 (ft/s)]2 = 0.43 ft-H20

The total head loss through ducts and fittings is:

htotal = hp + hf = 5 ft-H20 + 0.43 ft-H20 = 5.43 ft-H20

The work added to the fluid is:

Wf (hp) = V (gal/min) htotal (ft-H20) SGf / 3,960 (gal-ft-H20/min-hp)Wf (hp) = 100 gpm x 5.43 (ft-H20) 1.0 / 3,960 (gal-ft-H20/min-hp) = 0.137 hp

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.


Source: United McGill, 1990.

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



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.


Source: McQuiston and Parker, 1994.

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: McQuiston and Parker, 1994. 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). Neglecting pressure losses through elbows and fittings, determine pipe diameters.

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.

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.


A LA = 40 ft VA = 700 cfm

B LB = 200 ft VB = 500 cfm

C LC = 20 ft VC = 200 cfm

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.

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.

Pump TypesPumping applications can generally be divided into two categories: “low flow at high pressure” and “high flow at low pressure”. “Low flow at high pressure” applications include hydraulic power systems and typically employ positive-displacement pumps. The majority of fluid-flow applications are “high flow at low pressure” and use centrifugal pumps.

In centrifugal pumps, the fluid enters along the centerline of the pump, is pushed outward by the rotation of the impeller blades, and exits along the outside of the pump. A schematic of a centrifugal pump is shown below.


Centrifugal pump

Fan TypesThe two most common types of fans are axial and centrifugal fans.

Axial fans, also called propeller fans, generate high flow rates but cannot generate the high pressures required to push air through ducts. Thus, axial fans are generally used in applications such as exhausting air through ceilings or walls, or destratifying air in rooms with high ceilings.

Centrifugal fans are capable of generating the pressure required to force air through ducts and are used almost exclusively in these applications. In centrifugal fans, the shape of the blade (forward-curved, backward-curved or radial) changes the relationship between pressure and flow. In general, backward-curved blades provide the highest efficiency and are used in most HVAC applications.

Axial Fan Centrifugal Fan

Pump CurvesPumps can generate high volume flow rates when pumping against low pressure or low volume flow rates when pumping against high pressure. The possible combinations of


total pressure and volume flow rate for a specific pump can be plotted to create a pump curve. The curve defines the range of possible operating conditions for the pump.

If a pump is offered with multiple impellers with different diameters, manufacturers typically plot a separate pump curve for each size of impellor on the same pump performance chart. Smaller impellors produce less pressure at lower flow rates. Typical pump performance charts with multiple pump curves are shown below.

Typical pump performance charts. (McQuiston and Parker, 1994)

The power required to push the fluid through the pipe, Wfluid, is the product of the volume flow rate and system pressure drop.


Wfluid = V Ptotal

Graphically, fluid work is represented by the area under the rectangle defined by the operating point on a pump performance chart.

Typically, the efficiency of the pump at converting the power supplied to the pump into kinetic energy of the fluid is also plotted on the pump performance chart. Pump efficiencies typically range from about 50% to 80%. Power that is not converted into kinetic energy is lost as heat. The power required by the pump, which is called the “shaft work” or “brake horsepower”. Pump efficiency is the ratio of fluid work to shaft work. , can be calculated from the flow rate, total pressure, and efficiency values from the pump curve, using the following equation.

Wpump = Wfluid / Effpump = V Ptotal / Effpump

A useful dimensional version of this equation for pumping water at standard conditions is:

Wpump (hp) = V (gal/min) htotal (ft-H20) / [3,960 (gal-ft/min-hp) x Effpump]

Many pump performance graphs, including those shown above, also plot curves showing the work required by the pump to produce a specific flow and pressure. Note that these curves show work required by the pump including the efficiency of the pump. Calculating the work supplied to the pump using the preceding equation and comparing it to the value indicated on a pump performance graph is a useful exercise.

Fan CurvesSimilarly, the volume flow rate generated by a fan depends on the total system pressure drop. Fans can generate high volume flow rates at low system pressure drops or low volume flow rates at high system pressure drops. A fan curve shows the relationship between total pressure drop and volume flow rate for a specific fan.


Source: McQuiston et al., 2000.

It is common for fan manufacturers to publish fan performance data in terms of “static pressure” versus flow. The “static pressure’ in performance data is actually the difference between the static pressure at the fan outlet and the total pressure at the fan inlet.

Pstatic,performance data = Pstatic,2 – (Pstatic,1 + Pvelocity,1)

All fan performance calculations should be performed using total pressure. For example, the methods to calculate pressure loss through ducts and fittings, calculate total pressure loss. In addition, the power requirement of a fan is a function of total pressure loss, not static pressure loss. Thus, for fan calculations, it is important to add the velocity pressure of the air leaving the fan outlet to the static pressure reported in fan performance data to determine the relationship between total pressure and flow for the fan. In some cases manufactures list outlet velocity for the fan. In other cases, outlet velocity can be calculated from outlet dimensions of the fan and airflow. The total pressure of the fan is then:

Ptotal = Pstatic,performance data + Pvelocity,2 = (Pstatic,2 + Pvelocity,2 )– (Pstatic,1 + Pvelocity,1)

In U.S. dimensional units, the relation is:

Ptotal (in-H20) = Pstatic,performance data (in-H20) + (V2 / 4,005)2 where V2 is in ft/min


System CurvesThe total pressure that the pump must produce to move the fluid is determined by the piping system. This total pressure of the piping system is the sum of the pressure due to inlet and outlet conditions and the pressure loss due to friction. In a piping system, pressure loss due to friction increases with increasing fluid flow; thus, system curves have positive slopes on pump performance charts. The operating point of a pump is determined by the intersection of the pump and system curves.

To determine the form of a system curve, consider the equation for total pressure in a piping system. The total pressure caused by a piping system is the sum of the pressure due to inlet and outlet conditions and the pressure required to overcome friction through the pipes and fittings.

Ptotal = (Pstatic + Pvelocity + Pelevation )inlet-outlet + Pfricition

Inlet/Outlet PressureThe inlet/outlet pressure that the pump must overcome is the sum of the static, velocity and elevation pressures between the inlet and outlet of the piping system. For closed loop piping systems, the inlet and outlet are at the same location; hence, the static, velocity and elevation pressure differences are all zero. For open systems, the differences in static, velocity and elevation pressures must be calculated.

In many pumping applications, the velocity pressure difference between the inlet and outlet is zero or negligible, and inlet/outlet pressure is simply the sum of the static and elevation heads. In these cases, the inlet/outlet pressure is independent of flow and is represented on a pump performance chart as the pressure at zero flow.

Friction Pressure DropThe equations for pressure loss from friction through pipes and through fittings are:

Pp = (f L fluid V2) / (2 D)

Pf = kf fluid V2 / 2

These equations clearly show that for a given pipe system, the pressure drop is proportional to the square of the velocity, and hence the square of the volume flow rate.

Pfriction = C1 V2 = C2 V2

This quadratic relationship can be plotted on the pump curve to show the “system curve”.


Plotting System CurvesAs the preceding discussion showed, system curves have a flow-independent component (of inlet/outlet pressure) and a flow-dependent component that varies with the square of flow rate (of friction pressure).

A system curve for a closed-loop piping system with no inlet/outlet pressure difference is shown below. The curve is a parabola of the form hheadloss = C2 V2. The curve passes through the origin because the inlet/outlet pressure difference, sometimes called the static head, is zero. The coefficient C2 can be determined if the operating point is known by substituting the known pressure drop and flow rate into the equation and solving for C2. The fluid work required to push the fluid through the pipe is the product of the volume flow rate and system pressure drop and is represented graphically by the area under the rectangle defined by the operating point.

System curve for closed-loop piping system with no inlet/outlet pressure difference. Source of original pump curve: Kreider and Rabl, 1996.

A system curve for an open-loop piping system with a “static” or “inlet/outlet” pressure of 40 ft is shown below. This system curve is of the form hheadloss = A + C2 V2; where A is the “static” or “inlet/outlet” pressure drop. As before, the coefficient C2 can be determined if the operating point and inlet/outlet pressure are known by substituting the known values into the equation and solving for C2.


System curve for open-loop piping system with 40 ft-H20 inlet/outlet pressure differenceSource of original pump curve: Kreider and Rabl, 1996.

Multiple Pumps Operating In ParallelMany pumping systems employ multiple pumps in parallel rather a single large pump. One advantage of specifying multiple pumps in a parallel configuration is redundancy in case of failure. For example, it is common to design a pumping system with three pumps in parallel configuration, even though no more than two pumps would ever run simultaneously. The third pump provides redundancy in case of failure, and allows the system to function at full capacity even when one pump is being serviced. Another advantage of parallel pumping configuration is the ability to vary flow by turning one or more of the pumps on and off. Finally, in many applications, it is more energy efficient to operate multiple smaller pumps in parallel rather than operating a single large pump.

When two pumps are operated in parallel, they perform like a single pump with twice the flow rate at the same pressure drop. The figure below shows the pump curve of a single pump A, two pumps operating in parallel B, and the system curve C.


Pump and systems curves for secondary chilled water loop.

The system curve C describes the relationship of pressure drop and flow rate for the given piping system with no static head. Because head loss varies with the square of flow rate, the equation of the system curve can be estimated by fitting a quadratic equation through the origin and the design operating point:

h (ft H20) = (2.4 x 10-6) x V2

where V is the flow rate in gpm. The equation for curve B can be estimated by the fitting a regression equation through the data points on the curve:

h (ft H20) = 149 + 0.00106 x V + (3.65 x 10-7) x V2 The operating point of curve B for two pumps in parallel can be found from Equations 1 and 2 to be about:

h = 135 ft H20 V = 7,500 gpm Note that the total volume flow rate of two pumps operating in parallel is less than twice the flow rate of a single pump operating alone (at the intersection of C and A).

Fan Inlet-Outlet ConditionsCentrifugal fans produce the most airflow with the least pressure drop when the airflow is directed perpendicular to the fan inlet and evenly distributed across the inlet. Unfortunately, fan inlets are often coupled to elbows and turning boxes, which can cause uneven flow and/or flow rotation into the fan inlet. Uneven flow and flow rotation have the effect of increasing pressure drop and decreasing flow volume. In


these cases, turning vanes in fan inlet elbows and specially designed fan inlet boxes can reduce the pressure drop and increase flow.

Non-uniform flow from rectangular inlet duct

Improved flow with specially-designed inlet box.

Inlet duct causing inlet spin.

Turning vane correction for inlet spin.Source: Tutterow, 199x.

Similarly, two to six diameters of duct are generally required for the airflow leaving a fan to become fully developed. Branch lines, elbows and other ductwork installed before the flow is fully developed increase pressure drop and reduce airflow.



Improper air inlet and outlet conditions can dramatically increase the total pressure drop across the fan. The increased pressure drop due to fan inlet and outlet conditions is called the “system effect”. The system effect frequently causes fans to operate at a higher pressure and lower flow rate, A, than would be anticipated based on the system curve from the duct system B. To maximize energy-efficiency, it is important to minimize these losses by proper design of fan inlet and outlet conditions.



Pump/Fan Motor WorkPumps and fans are typically driven by electrical motors. The power required by the motor is greater than the fluid work because the pump/fan, power transmission and motor all incur losses. Thus, fluid work must be divided by the product of the efficiencies of all components of the pump energy-delivery system to determine the electricity required by the motor.

Welec = Wf / ( Efficiencypump.fan x Efficiencydrive x Efficiencymotor)

Pump/fan system efficiency.

For example, if the efficiency of the motor at converting electrical energy to motor shaft work is 90%, the efficiency of belt drives at transferring motor shaft work to pump is 92%, and the efficiency of a pump at converting pump shaft work to fluid work is 75%, the electrical energy use required by the motor would be 73% greater than the required fluid work.

Welec = Wf / ( 90% x 92% x 70%) = 1.73 Wf

Pump/Fan Affinity LawsThe fundamental fluid mechanic relationships developed thus far can be modified to generate other useful relations between fan parameters. These relationships are known as fan affinity laws. The two most important relationships are derived below.

As shown in the section of system curves, friction head loss is proportional to the square of the volume flow rate.

Pfriction = C1 V2 = C2 V2

By substitution, fluid work is proportional to the cube of volume flow rate

Wf = V Pfriction = V C2 V2 = C2 V3


Since Wf / V3 is constant, it follows that:

(Wf / V3)1 = C = (Wf / V3)2

Wf2 = Wf1 (V2 / V1)3

This relation shows that a small reduction in the volume flow rate results in a large reduction in the fluid work. For example, reducing the volume flow rate by one half reduces fluid work by 88%!

Wf2 = Wf1 (1/2)3 == Wf1 (1/8)(Wf1 – Wf2) / Wf1 = [Wf1 - Wf1 (1 /8)] / Wf1 = 1 – (1/8) = 88%

Another useful relation can be derived from the relationship between volume flow rate V and the rotational speed of the pump fan. In centrifugal pumps and fans, the volume flow rate is proportional to the rotational speed of the pump fan.

V = C RPM

Since V/RPM is constant, it follows that:

(V / RPM)1 = C = (V / RPM)2

V2 = V1 (RPM2 / RPM1)

Thus, volume flow rate varies in proportion to pump/fan speed.

Inside-Out Approach to Energy-Efficient Fluid Flow 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 Elevation DifferenceMany pumping applications involve lifting fluids from lower to higher elevations. The total pressure difference through the piping system must include this elevation pressure difference. In some applications, it may be possible to reduce the elevation pressure difference by increasing the height of the fluid in the supply tank or reducing the height of the fluid in the outlet tank. Doing so reduces the total system pressure difference and pump energy use.

Example

A pump lifts 100 gpm of water 120 feet to fill an elevated reservoir. Determine the electrical power savings if the level of the reservoir could be reduced by 20 feet. The pump is 70% efficient and the pump motor is 90% efficient.

PE1 = 100 gpm x 120 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70 x 0.90] x 0.75 kW/hp = 3.61 kWPE2 = 100 gpm x 100 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70 x 0.90] x 0.75 kW/hp = 3.01 kW

Savings = PE1 – PE2 = 3.61 kW – 3.01 kW = 0.60 kW

Energy Savings from Reducing 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 (V / D2 )2 / (2 D) = f L V2 / (2 2 D5 )

Thus, friction pressure loss through pipes/ducts is inversely proportional to the fifth power of the diameter


Pp ~ C / D5

This means that doubling the pipe/duct diameter reduces friction pressure loss by about 97%!

Example

Calculate the percentage reduction in friction head loss if pumping 4 gpm of water through 0.5-inch and 1-inch diameter schedule 40 steel pipes.

From the monogram:

h 0.5-inch = 17 ft-H20/100 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%

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.


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%

Modify Pump/Fan to Realize Savings from Reducing Friction It may seem that reducing friction losses in a pipe/duct system would automatically reduce pump/fan energy use. However, reducing system pressure drop without modifying the pump/fan causes the pump/fan to move more fluid. This increased volume flow rate actually increases pump energy consumption.

Thus, it is important to reduce the diameter of the pump impellor, or slow the pump/fan so that the volume flow rate remains the same as it was in the high friction in order to realize energy savings. The following examples demonstrate this importance of modifying the pump to realize savings from reducing piping system pressure drop.

Example: Modify Pump to Realize Friction Reduction Savings


The figure below shows a set of pump curves with two system curves. The pump originally operates at point A, and the system curve for the original piping systems extends from the origin to A (which means that this is a closed loop piping system with no inlet/outlet pressure difference to overcome). The friction pressure drop through the piping system is then reduced by 40 psi by increasing pipe diameter, using low-flow fixtures or using smoother pipe. Reducing the system pressure drop from 180 ft-H20 at point A to 140 ft-H20, without altering the pump impellor or speed, would cause the pump to operate at point B. The power required to pump a fluid is the product of the volume flow rate and pressure drop; hence, the areas enclosed by the rectangles defined by each operating point represent the fluid power requirements, WfA and WfB, at the different system pressure drops.

Source of pump curve: (McQuiston and Parker, 1994)

WfA = 235 gpm x 180 ft-H20 / 3,960 gpm-ft-H20/hp = 10.7 hpWfB = 330 gpm x 140 ft-H20 / 3,960 gpm-ft-H20/hp = 11.7 hp

The power required by the pump, Wp, is the fluid power requirements divided by the pump efficiency.

WPA = 10.7 hp / .74 = 14.5 hpWPB = 11.7 hp / .70 = 16.7 hp

Thus, decreasing system pressure drop without altering the pump impellor or speed would cause the pump to consume more energy, not less.

Savings = WPA – WPB = 14.5 hp – 16.7 hp = -2.2 hp


A

B

C

To realize energy savings from reducing pressure drop, it is necessary to slow the pump speed or decrease the size of the impellor. To determine the pump impellor size required to deliver the initial flow of 235 gpm with the new low-pressure drop pipe system, it is necessary to develop a system curve for the new pipe system. Pressure drop through piping systems varies with the square of flow rate. Thus, the equation for a system curve that passes through the origin can be written as:

h = C V2

The coefficient, C, for the new system curve can be found by substituting the values of pressure drop and volume flow rate for point B.

C = h / V2 = 140 / 3302 = 0.001286

Thus, the pressure drop through the new duct system at 235 gpm would be about:

h = C V2 = 0.001286 2352 = 71 ft-H20

The flow rate of 235 gpm and 71 ft-H20 defines point C, which would be the operating point of the pump with a 5.5-in impeller. At this operating point, the pump would be about 65% efficient, and the pump power draw would be about:

WPC = 235 gpm x 71 ft-H20 / (3,960 gpm-ft-H20/hp x 0.65) = 6.5 hp

Thus, the savings from reducing the pressure drop in the pipe system, if the pump impeller diameter were reduced, would be about:

Savings = WPA – WPC = 14.5 hp – 6.5 hp = 8.0 hp

This example demonstrates the importance of modifying the pump to realize savings from reducing system friction loss.

Example: Modify Fan to Realize Friction Reduction Savings

The figure below shows fan performance at various speeds and a system curve. Assume the fan is operated at 1,550 rpm on the system curve at point A. The volume flow rate would be about 10,500 cfm and the system pressure drop would be about 8.4 in-H20. From the chart, the required power to the fan would be about 20 hp.


Source: Nadel et al., 1991

If the pressure drop were reduced to 5.8 in-H20 at point B by increasing duct diameter, opening dampers or using low-pressure drop fittings, the operating point would shift down the fan curve. The volume flow rate would increase to about 18,000 cfm and the required power to the fan would increase to about 25 hp. Thus, reducing pressure drop without slowing the fan would increase energy use!

To realize energy savings from reducing pressure drop, it is necessary to slow the fan speed. To determine the fan speed required to deliver the initial air flow of 10,500 cfm with the new lower pressure drop duct system, it is necessary to develop a system curve for the new duct system. Pressure drop always varies with the square of flow rate. Thus, the equation for a system curve can be written as:

h = C V2

The coefficient, C, for the new system curve can be found by substituting the values of pressure drop and volume flow rate for point B.

C = h / V2 = 5.8 / 182 = 0.0179

Thus the pressure drop through the new duct system at 10,500 cfm would be about:

h = C V2 = 0.0179 10.52 = 2.0 in-H20

According to the fan curves, the fan would deliver 10,500 cfm at 2.0 in-H20 if the fan speed were slowed to about 900 rpm at point C. At this operating point, the fan would


require about 5 hp of power. Thus, the savings from reducing the pressure drop in the duct system, if the fan speed were modified, would be about:

20 hp – 5 hp = 15 hp

This example demonstrates the importance of slowing the fan to realize savings from reducing system friction loss.

Energy Savings from Energy-Efficient Flow ControlMost pump/fan systems are designed to handle peak conditions. Since peak conditions typically occur infrequently, substantial energy savings are possible by controlling fluid flow rate to match actual demand. The inside-out approach to low-energy pump/fan systems starts by reviewing all end-use applications to determine the required flow, before proceeding “upstream” with the analyses of the piping and pumping systems. Once the required flow is determined, it is necessary to determine how the flow is currently controlled and consider more energy-efficient options. Inefficient methods of flow control, in order of worst to better, are:

Bypass Throttling / Outlet dampers Fan variable inlet vanes Intermittent pump operation

More efficient methods of flow control are:

Continuous pump operation with smaller impeller or at lower speed. Reducing pump impellor diameter Slowing fan speed Continuously varying pump/fan speed with a variable speed drive.

These concepts are demonstrated in the following figures.


Typical relationships between power and flow for outlet damper control, variable-inlet vane control and speed control via variable frequency drives are shown in the figure below.


Old Inefficient Flow Control

By-pass loop(No savings)

By-pass damper (No savings)

Outlet valve/damper(Small savings)

Inlet vanes(Moderate savings)

Fan w/ Inlet Vanes

New Efficient Flow Control

Trim impellor for constant-volume

pumps

Slow fan for constant-volume

fans

VFD for variable-volume pumps or fans

Close Bypass Valve

dP

VFD

We recommend switching from inefficient methods of flow control to efficient methods whenever possible. The following examples discuss each type of flow control.

Bypass: Bypass control is sometimes employed to couple constant speed pumps/fans with variable process loads. In by-pass control, valves are opened or closed to direct excess water/air through a bypass loop or to simply reject it to the environment. Thus, the flow of water/air through the pumps/fans remains constant even as the flow of water/air through the process varies. Bypass is the least efficient method of flow control, since pump/fan power remains nearly constant even as the load varies.

Throttling/Outlet Dampers: Controlling flow by closing a flow-control valve/damper downstream of the pump/fan increases pressure drop and causes the operating point to move up and left on the pump/fan curve.


Source: Gould Pumps, GPM 7-CD, Technical Information.

This results in relatively small energy savings, since

Wf2 = V2 P2 where V2 < V1 but P2 > P1

Thus, throttling is an energy inefficient method of flow control.

Flow Control with Intermittent Pump OperationIn some pumping applications, pumps may operate at a relative high flow rate for part of the time and then be turned off until needed again. Because friction losses are proportional to the square of flow, it is more energy-efficient to pump a lower volume flow rate for a longer period of time. We call this the “Pump Long, Pump Slow” principle. “Pump long, pump slow” opportunities may exist whenever pumps run intermittently. If a single pump operates intermittently, then application of the “pump long, pump slow” principal would require installing a smaller pump, trimming the impeller or slowing the pump rotational speed. If multiple pumps operate in parallel, it may be possible to simply run fewer pumps more continuously.

Example

Two pumps operate in parallel to lift water 40 vertical feet and generate 5,800 gpm at 140 ft H20 as shown in the figure below. One pump operating alone would generate 5,200 gpm at 123 ft H20. The pumps are 75% efficient and the pump motors are 90% efficient. When operating in parallel, the pumps operate 10 hours per day. Determine the energy savings from “pumping slow, pumping long” and running one pump longer to generate the same daily flow rate.


The daily flow rate is:

5,700 gpm x 60 minutes/hour x 10 hours/day = 3,420,000 gal/day

The electrical power and energy required to pump this quantity of water using two pumps in parallel is:

P2 Pumps = 5,700 gpm x 140 ft-H20 / (3,960 gpm-ft-H20/hp x 0.75 x 0.90) x 0.75 kW/hpP2 Pumps = 223.9 kWE2 Pumps = 223.9 kW x 10 hours/day = 2,239 kWh/day

The time required to generate the same daily flow rate while operating only one pump is:

3,420,000 gal/day / [5,700 gpm x 60 minutes/hour] = 10.96 hours/day

The energy required to pump this quantity of water using one pumps is:

P1 Pump = 5,200 gpm x 123 ft-H20 / (3,960 gpm-ft-H20/hp x 0.75 x 0.90) x 0.75 kW/hpP1 Pump = 179.5 kWE1 Pump = 179.5 kW x 10.96 hours/day = 1,967 kWh/day

The fraction savings would be about:

(2,239 kWh/day – 1,967 kWh/day) / 2,239 kWh/day = 14%


0

40

80

120

160

200

0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000

V (gpm)

dh (f

t H20

)

A (one pump) B (tw o pumps in parallel) C (system curve)

A

B

C

Flow Control with Variable Inlet VanesThe fraction of full-load blower power as a function of flow rate for variable-inlet vane control is given in ASHRAE Systems and Equipment Handbook, 2000, page 18.9. This relation is summarized as:

Wn = -0.042487 + 3.6573 x Vn -6.4957 x Vn2 + 3.8760 x Vn3 (Vn >= 50%) (1)Wn = 0.589 + 0.116 x Vn (Vn < 50%) (2)

Where Vn is the fraction of full-load flow rate and Wn is the fraction of full-load power. The power draw of the blower motor with VIV control is:

W = Wfl x Wn (3)

Example

At full flow, a blower powered by an 800-hp motor draws 466 kW of power when delivering 25,000 cfm at full flow. Flow is controlled using variable inlet vanes. The flow rate is 18,000 cfm for 80% of the time, 20,000 cfm for 10% of the time and 16,000 cfm for 10% of the time. Based on these data and Equations 2 and 3, the average power draw of the blowers during blast mode is about:

Flow Control by Trimming Pump ImpellorThe most energy-efficient methods of varying flow are controlling pump/fan speed or decreasing the size of the impellor. If the required flow is constant and less than the design flow, more cost effectively alternatives are to slow a fan or trim a pump impellor. The following example illustrates energy savings from reducing flow by trimming a pump impellor.

Example: Trim Pump Impellor in Closed-Loop System

The figure below shows a set of pump curves for different impellor diameters. The pump is equipped with a 7-inch impellor. At design conditions, the pump would operate at point A. The system curve extends from the origin to A, which means this is a closed-loop piping system with no inlet/outlet pressure difference to overcome.


%Time CFM Vn (%) Wn (%) W (kw)

10% 20000 0.8 71% 331.180% 18000 0.72 67% 312.310% 16000 0.64 65% 304.6

Weighted Average 316.0

The actual flow required was less than the design flow. Thus, flow was reduced by a throttling valve to operating point B. A more energy efficient method of reducing flow is to open the throttling valve, and trim the impellor to a smaller diameter. This would cause the pump to operate at point C.

The power required to pump a fluid is the product of the volume flow rate and pressure drop; hence, the areas enclosed by the rectangles defined by the vertical and horizontal axes and each operating point represent the fluid power requirements, WfA , WfB, , and WfC,. Comparison of these rectangles clearly indicates that reducing flow by trimming the pump impellor uses much less energy than reducing flow by throttling.

Source of pump curve: McQuiston and Parker, 1994

The power required by the pump at A, WPA, is the fluid power requirement divided by the pump efficiency.

WPA = 330 gpm x 140 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70] = 16.7 hp

The power required by the pump for throttled flow at B, WPB, is the fluid power requirement divided by the pump efficiency.

WPB = 235 gpm x 180 ft-H20 / [3,960 gpm-ft-H20/hp x 0.74] = 14.4 hp

To determine the pump impellor size required to deliver a flow of 235 gpm with a trimmed impellor, it is necessary to develop the design system curve with the throttling valve wide open and the pump operating at A. Pressure drop through piping systems


A

B

C

varies with the square of flow rate. Thus, the equation for a system curve that passes through the origin can be written as:

h = C V2

The coefficient, C, for the new system curve can be found by substituting the values of pressure drop and volume flow rate for point A.

C = h / V2 = 140 / 3302 = 0.001286

Thus, the required head at 235 gpm would be about:

h = C V2 = 0.001286 2352 = 71 ft-H20

The flow rate of 235 gpm and 71 ft-H20 defines point C. This indicates that the pump impellor should be trimmed to a diameter of 5.5-inches. At this operating point, the pump would be about 65% efficient, and the pump power draw would be about:


Thus, the savings from controlling flow by trimming the pump impellor rather than throttling flow would be about:

Trim Impellor Savings = WPB – WPC = 14.4 hp – 6.5 hp = 8.0 hp

Alternately, power savings from reducing the volume flow rate can be estimated from the pump affinity laws. Theoretically, pump work varies with the cube of volume flow rate. Use of the cubic relationship would predict:

WPC = WPA (VC/VA)3 = 16.7 hp x (235 gpm / 330 gpm) 3 = 6.0 hp

The 6.0 hp predicted by the pump-affinity law is less than the 6.5 hp predicted by the pump curve. This example demonstrates how use of the cubic relationship typically exaggerates savings. In practice, the efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of flow. Using this relationship, if PA is 16.7 hp at 330 gpm, PC at 235 gpm would be about:

WPC = WPA (VC/VA)2.5 = 16.7 hp x (235 gpm / 330 gpm) 2.5 = 7.1 hp

Using the 2.5 exponent, savings would be about:

Savings = WPB – WPC = 14.4 hp – 7.1 hp = 7.3 hp


This slightly lower estimate of savings incorporates the reduction in motor efficiency, and power loss by the VSD.

Example: Trim Pump Impellor in Open System

The figure below shows a set of pump curves for different impellor diameters. The pump is equipped with a 7-inch impellor. At design conditions, the pump operates at point A. The system curve extends from an inlet/outlet pressure difference of 40 ft H20 from lifting water over an elevation change of 40 ft in an open system.

The actual flow required was less than the design flow. Thus, flow was reduced by a throttling valve to operating point B. A more energy efficient method of reducing flow is to open the throttling valve, and trim the impellor to a smaller diameter. This would cause the pump to operate at point C.

The power required to pump a fluid is the product of the volume flow rate and pressure drop; hence, the areas enclosed by the rectangles defined by the vertical and horizontal axes and each operating point represent the fluid power requirements, WfA , WfB, , and WfC,. Comparison of these rectangles clearly indicates that reducing flow by trimming the pump impellor uses much less energy than reducing flow by throttling.


The power required by the pump at A, WPA, is the fluid power requirement divided by the pump efficiency.


A

B

C

WPA = 330 gpm x 140 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70] = 16.7 hp

The power required by the pump for throttled flow at B, WPB, is the fluid power requirement divided by the pump efficiency.

WPB = 235 gpm x 180 ft-H20 / [3,960 gpm-ft-H20/hp x 0.74] = 14.4 hp

To determine the pump impellor size required to deliver a flow of 235 gpm with a trimmed impellor, it is necessary to develop the design system curve with the throttling valve wide open and the pump operating at A. Pressure drop through piping systems varies with the square of flow rate. Thus, the equation for a system curve that passes through the origin can be written as:

h = h0 + C V2

Where h0 is the pump head at zero flow. The coefficient, C, for the new system curve can be found by substituting the values of pressure drop and volume flow rate for point A.

C = (h – h0) / V2 = (140 – 40) / 3302 = 0.00091827


h = 40 + C V2 = 40 + 0.00091827 2352 = 91 ft-H20

The flow rate of 235 gpm and 91 ft-H20 defines point C. This indicates that the pump impellor should be trimmed to a diameter of 5.75-inches. At this operating point, the pump would be about 69% efficient, and the pump power draw would be about:


Thus, the savings from controlling flow by trimming the pump impellor rather than throttling flow would be about:

Savings = WPB – WPC = 14.5 hp – 7.8 hp = 6.6 hp

Alternately, power savings from reducing the volume flow rate can be estimated from the pump affinity laws. However the cubic relationship between pump work and volume flow rate applies only to the friction head, not the inlet/outlet head. Thus, the component of total power due to friction must be calculated. The component of total power due to friction is the difference between the total power and the inlet/outlet power.


The inlet-outlet power at point A, WPAO, is:

WPAO = 330 gpm x 40 ft-H20 / [3,960 gpm-ft-H20/hp x 0.70] = 4.8 hp

Thus, the component of total head due to friction is:

WPF = WPA - WPAO = 16.7 hp - 4.8 hp = 11.9 hp

Theoretically, pump work varies with the cube of volume flow rate. In practice, the efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of flow. Using this relationship, the component of power at C due to reduced friction at 235 gpm would be about:

WPCF = WPAF (VC/VA)2.5 = 11.9 hp x (235 gpm / 330 gpm) 2.5 = 5.1 hp

Using the 2.5 exponent, power savings would be about:

Savings = WPF – WPCF = 11.9 hp – 5.1 hp = 6.8 hp

This estimate of savings incorporates the reduction in motor efficiency, and power loss by the VSD.

Flow Control by Varying Pump/Fan Rotational SpeedThe most energy efficient method of varying flow is by controlling pump/fan speed. If the required flow varies over time, speed control is best facilitated by an electronic variable speed drive (VSD) which can continuously and smoothly adjust pump/fan speed as needed. One time reductions in flow are more cost effectively accommodated by replacing the pump/fan pulley with a larger diameter pulley to slow pump/fan rotation or by reducing pump impellor diameter. The following example illustrates energy savings from reducing flow by slowing pump/fan speed.

Example: Slow Pump Speed in Closed-Loop System

The figure below shows pump performance at various speeds and a system curve. At design conditions, the pump operates at 1,200 RPM at point A. The system curve extends from the origin to A, which means this is a closed-loop piping system with no inlet/outlet pressure difference to overcome.


Source of pump curve: Nadel et al., 1991

At A, the volume flow rate is 1,200 gpm, the total head is 55 ft-H20 and the pump efficiency is 74%. According to the chart, the required power to the pump at this operating point is about 23 hp. Alternately, the required pump input power could be calculated as:

WPA = 1,200 gpm x 55 ft-H20 / (3,960 gpm-ft-H20/hp x 0.74) = 22.5 hp

If the flow were reduced to 900 gpm with a flow-control valve, the operating point would move along the pump curve to point B at 900 gpm, 62 ft-H20 with pump efficiency of 70%. According to the chart, the required power to the pump at this operating point is about 20 hp. Alternately, the required pump input power could be calculated as:

WPB = 900 gpm x 62 ft-H20 / (3,960 gpm-ft-H20/hp x 0.70) = 20.1 hp

Thus, the pump power savings from reducing flow from 1,200 gpm to 900 gpm with a flow-control valve would be about:

Flow-control Valve Savings = WPA – WPB = 22.5 hp – 20.1 hp = 2.4 hp

To determine the pump speed required to deliver a flow of 900 gpm, it is necessary to develop the design system curve with the flow-control valve wide open and the pump


AB

C

operating at A. Pressure drop through piping systems varies with the square of flow rate. Thus, the equation for a system curve that passes through the origin can be written as:

h = C V2

The coefficient, C, for the new system curve can be found by substituting the values of pressure drop and volume flow rate for point A.

C = h / V2 = 55 / 1,2002 = 0.000038194

Thus, the pressure drop at 900 gpm would be about:

h = C V2 = 0.000038194 9002 = 31 ft-H20

The flow rate of 900 gpm and 31 ft-H20 defines point C. This indicates that the pump speed should be reduced to 900. At this operating point, the pump would be about 67% efficient, and the required pump input power would be:


Alternately, the power required by the pump at point C, PC, can be read from the chart to be about 10 hp.

Pump power savings are the difference between PA and PB.

Variable Speed Drive Savings = WPA – WPC = 22.5 hp – 10.5 hp = 12.0 hp

Alternately, power savings from reducing the volume flow rate can be estimated from the pump affinity laws. Theoretically, pump work varies with the cube of volume flow rate. Use of the cubic relationship would predict:

WPC = WPA (VC/VA)3 = 22.5 hp x (900 gpm / 1200 gpm) 3 = 9.5 hp

The 9.5 hp predicted by the pump-affinity law is less than the 10.5 hp predicted by the pump curve. This example demonstrates how use of the cubic relationship typically exaggerates savings. In practice, the efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of flow. Using this relationship, the required pump input power at 900 gpm would be:

WPC = WPA (VC/VA)2.5 = 22.5 hp x (900 gpm / 1200 gpm) 2.5 = 11.0 hp


Thus, savings would be:



Example: Slow Pump Speed with Inlet/Outlet Head

The figure below shows pump performance at various speeds and a system curve. At design conditions, the pump operates at 1,200 RPM at point A. The system curve extends from an inlet/outlet pressure difference of 20 ft H20. This inlet/outlet head can result from lifting water over an elevation change of 20 ft in an open system, or as the required head between the supply and return pipes for controlling a variable speed drive.

Source of pump curve: Nadel et al., 1991

At A, the volume flow rate is 1,200 gpm, the total head is 55 ft-H20 and the pump efficiency is 74%. According to the chart, the required power to the pump at this operating point is about 23 hp. Alternately, the required pump input power could be calculated as:


AB

C

WPA = 1,200 gpm x 55 ft-H20 / (3,960 gpm-ft-H20/hp x 0.74) = 22.5 hp

If the flow were reduced to 900 gpm with a flow-control valve, the operating point would move along the pump curve to point B at 900 gpm, 62 ft-H20 with pump efficiency of 70%. According to the chart, the required power to the pump at this operating point is about 20 hp. Alternately, the required pump input power could be calculated as:

WPB = 900 gpm x 62 ft-H20 / (3,960 gpm-ft-H20/hp x 0.70) = 20.1 hp

Thus, the pump power savings from reducing flow from 1,200 gpm to 900 gpm with a flow-control valve would be about:

Flow-control Valve Savings = WPA – WPB = 22.5 hp – 20.1 hp = 2.4 hp

To determine the pump speed required to deliver a flow of 900 gpm, it is necessary to develop the design system curve with the flow-control valve wide open and the pump operating at A. Pressure drop through piping systems varies with the square of flow rate. Thus, the equation for a system curve that passes through the origin can be written as:

h = h0 + C V2

Where h0 is the pump head at zero flow. The coefficient, C, for the new system curve can be found by substituting the values of pressure drop and volume flow rate for point A.

C = (h – h0) / V2 = (55 – 20) / 1,2002 = 0.00002430556


h = 40 + C V2 = 20 + 0.00002430556 9002 = 40 ft-H20

The flow rate of 900 gpm and 40 ft-H20 defines point C. This indicates that the pump speed should be reduced to 900. At this operating point, the pump would be about 70% efficient, and the required pump input power would be about:


Alternately, the power required by the pump at point C, PC, can be read from the chart to be about 13 hp. Pump power savings are the difference between WPA and WPC.



Alternately, power savings from reducing the volume flow rate can be estimated from the pump affinity laws. However the cubic relationship between pump work and volume flow rate applies only to the friction head, not the inlet/outlet head. Thus, the component of total power due to friction must be calculated. The component of total power due to friction is the difference between the total power and the inlet/outlet power.

The inlet-outlet power at point A, WPA0, is:

WPA0 = 1,200 gpm x 20 ft-H20 / [3,960 gpm-ft-H20/hp x 0.74] = 8.2 hp

Thus, the component of total head due to friction is:

WPAF = WPA - WPA0 = 22.5 hp – 8.2 hp = 14.3 hp

Theoretically, pump work varies with the cube of volume flow rate. In practice, the efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of flow. Using this relationship, the component of power at C due to reduced friction PCF at 235 gpm would be about:

WPCF = WPAF (VC/VA)2.5 = 14.3 hp x (900 gpm / 1,200 gpm) 2.5 = 7.0 hp

Using the 2.5 exponent, power savings would be about:

Savings = WPAF – WPCF = 14.3 hp – 7.0 hp = 7.3 hp


Example: Slow Fan Speed with No Inlet/Outlet Head

The figure below shows fan performance at various speeds and a system curve.


Source: Nadel et al., 1991

Reducing the fan speed from 1,550 rpm to 1,050 rpm on the system curve would reduce the volume flow rate from 10,500 cfm to 7,000 cfm. The reduced volume flow rate would also generate less friction, and the system pressure drop would be reduced from 8.4 to 3.9 in-H20. The power required to push the air is the product of the volume flow rate and pressure drop; hence, the areas enclosed by the rectangles defined by each operating point represent the fluid power requirements, WA and WB, at the different flow rates.

WA = 10,500 cfm x 8.4 in-H20 / 6,356 (ft3-in/min-hp) = 13.9 hpWB = 7,000 cfm x 3.9 ft-H20 / 6,356 (ft3-in/min-hp) = 4.3 hp

The fluid work calculated above does not take into account fan efficiency. Taking into account fan efficiency, the shaft power required by the fan, which is plotted on the chart, accounts for fan efficiency. From the chart, shaft power is:

PA = 20 hp PB = 6.3 hp

Thus, the fan efficiency at each operating point is:

EffA = 13.9 hp / 20 hp = 0.70EffB = 4.3 hp / 6.3 hp = 0.68

The shaft power savings are:


PA – PB = 20 hp – 6.3 hp = 13.7 hp

Alternately, power savings from reducing the volume flow rate can be estimated using the fan affinity laws. Theoretically, fan work varies with the cube of volume flow rate. Use of the cubic relationship would predict:

PB = PA (VB/VA)3 = 20 hp x (7,000 cfm / 10,500 cfm) 3 = 5.9 hp

Using this method, the shaft power savings would be:

PA – PB = 20 hp – 5.9 hp = 14.1 hp

The small discrepancy is due to the fact that the fan affinity method does not take into account the slight decrease in fan efficiency at the lower speed. In practice, the efficiencies of the VSD, pump and motor typically decline as flow rate decreases, resulting in slightly less savings than would be predicted using this ‘cubic’ relationship. Thus, we estimate that pump/fan work varies with the 2.5 power of flow rather than the cube of flow. Using this relationship, if we measured PA to be 20.0 hp at 10,500 cfm, we would estimate PB at 7,000 cfm to be about:

PB = PA (VB/VA)2.5 = 20 hp x (7,000 cfm / 10,500 cfm) 2.5 = 7.3 hp

Thus, savings would be about:

Savings = PA – PB = 20.0 hp – 7.3 hp = 12.7 hp

This slightly lower estimate of savings incorporates the reduction in motor efficiency, and power use by the VSD.

Variable Speed Drive Applications Electronic variable speed drives (VSDs) control the speed of AC motors by converting the frequency and voltage of the AC line supply from fixed to variable values. VSDs are used in both constant and variable torque applications. In variable torque applications, such as pumping and fan systems, slowing the motor speed reduces the torque on the motor and can result in significant energy savings. These savings can be estimated using the Pump/Fan Affinity Laws.

VSDs subject motors to voltage spikes and fast voltage rise and fall times. These voltage spikes can “punch” through traditional winding insulation. Because of this, VSDs should only be coupled to motors that the manufacturer specifies as suitable for PWM VSDs. If


a motor is to be rewound, be sure to specify rewinding characteristics for PWM VSD motors. Most energy-efficient motors are suitable PWM VSDs.

In some applications, it may be possible to simply reduce the flow to a fixed rate rather than vary it continuously. In these cases, slowing the pump by increasing the diameter of the pump pulley or decreasing the diameter or the motor pulley would generate the same savings.

Because VSDs work best with premium efficiency motors, the motor may need to be upgraded if it is not a premium efficiency motor. Typical installed costs of VSDs are shown below.

Horsepower w/bypass w/o bypass Wiring Installation Controls w/bypassw/o

bypass5 $1,653 $500 $400 $2,000 $4,553

10 $2,008 $1,515 $579 $463 $2,000 $5,050 $4,55715 $2,233 $1,771 $658 $526 $2,000 $5,417 $4,95520 $2,458 $2,028 $737 $589 $2,000 $5,784 $5,35425 $2,772 $2,332 $816 $653 $2,000 $6,240 $5,80050 $4,344 $3,854 $1,211 $968 $2,000 $8,523 $8,033

100 $7,021 $6,049 $2,000 $1,600 $2,000 $12,621 $11,649Typical 2006 costs with 25% contractor markup on drives assuming existing motor is suitable for inverter use.

VSD Pump Retrofits VSD pump retrofits typically require making three changes to the existing pumping system:

1) Install a VSD on power supply to the pump motor. In parallel pumping configurations, one VSD is generally needed for each operational pump, but not for the backup pump.

2) Close valves on all by-pass pipes.3) Install a differential-pressure sensor between the supply and return headers at

the process load located the farthest distance from the pump. Determine the pressure drop needed to guarantee sufficient flow through the farthest process load at this point. Control the speed of the VSD to maintain this differential pressure.

If these procedures are followed, the inlet-outlet head across the pump approaches the differential-pressure between the supply and return headers as flow decreases. For example, if the differential-pressure between the supply and return headers were set to 20 ft-H20, the system curve would run from 20 ft-H20 at zero flow through C to A in the figure below. Thus, when modeling energy savings in VSD applications, use the procedure demonstrated in “Example: Slow Pump Speed with Inlet/Outlet Head”. In


addition, significant energy savings can result from minimizing the differential-pressure set point.


VSD Pumping: Industrial ExampleFor example, the Figure 18 A shows a typical industrial cooling configuration using a constant speed pump. Figure 18 B shows the system after a VSD retrofit. With the retrofit, the by-pass pressure relief valve would be closed, flow through each process load would be controlled at the load, and the VSD would modulate pump speed based on the differential pressure between the supply and return headers. The device marked dP is a differential-pressure sensor which would control the speed of the VSD.

A) constant volume and B) variable volume pumping system


AB

C

warm water

cool water

cooling tower

city water make-up

7.5 hp pump

25 hp pump

reservoir

process water return

bypass / pressure

relief valve

cooling water to process loads

dP

VSD

VSD Pumping: Commercial Building ExampleA typical constant-flow commercial-building chilled water system is shown below. This system provides constant flows through the condenser and evaporator of each chiller whenever a chiller is operational. The primary chilled water pumps are typically much smaller than the secondary pumps, since a primary pumps only have to move water through the chiller evaporators and not through the entire building. The secondary chilled water pumps are much larger than the primary pumps and provide a constant flow of chilled water to the AHUs. Each AHU varies the quantity of chilled water through the coil and bypasses unneeded chilled water.

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.

The greatest pump energy savings come from changing the secondary chilled-water loop from constant to variable flow. This is done by:

Removing or blocking the bypass piping on each AHU Replacing 3-way valves with 2-way valves on each AHU Adding VFDs to the secondary chilled water pumps Controlling the VFDs based on the differential pressure between the supply and

return headers

A typical variable flow secondary chilled water loop system is shown below.


Modern chillers are designed to accommodate variable flow through the evaporators and condensers. This enables full variable flow chilled water plants. Two variants of full variable-flow chilled water plants are shown below. The first system uses a primary-only design with a flow control and bypass valve to guarantee minimum flow to the chillers.

Source: Taylor, S., 2002, “Primary-Only vs. Primary Secondary Variable Flow Systems”, ASHRAE Journal, February, 2002, pgs 25-29.

The figure below shows an all variable-speed cooling plant (Erpelding, Ben, 2008, “Monitoring Chiller Plant Performance”, ASHRAE Journal, April, pp. 48-52.) The plant was converted from a primary-secondary constant speed plant to the variable-speed plant because of low dT (significant mixing), inefficient constant speed operation and low-load instability


Source: “Monitoring Chiller Plant Performance”, ASHRAE Journal, April, pp. 48-52.

The following figures show measured performance of the system. The two sets of values show differences between two independent measurement systems. During the monitoring period, the 300-ton chiller never operated. The data show total cooling plant power varying from about 0.5 kW/ton at high loads to about 0.3 kW/ton at low loads. In comparison, constant flow cooling plants have higher specific power requirements at full load, with constant or increasing specific power requirements as the low decreases.




VSD Fan Retrofits: Commercial Building ExampleThe figure below shows a schematic of VSD retrofit on a constant-volume supply air system in a building. Each box represents a different zone within the building. Each X represents a VAV box, which varies the supply air to maintain zone temperature. The pressure sensor would control the speed of the VSD.


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”,


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