Ashrae journal understanding expansion tanks - taylor

24 A S H R A E J o u r n a l a s h r a e . o r g March 2003

About the AuthorSteven T. Taylor, P.E., is a principal at TaylorEngineering in Alameda, Calif. He is a member ofTCs 4.3 and 1.4 and former chair of SSPC 62.1.

isleading and sometimes incorrect statements regarding how toselect and pipe expansion tanks can be found in design manuals

and manufacturers’ installation guides. Some of the more commonclaims follow. Claim: Expansion tanks must be connected to the sys-tem near the suction of the pump. Fact: Closed expansion tanks (thosenot vented to the atmosphere) may be connected anywhere in thesystem provided their precharge (initial) pressure and size are cor-rectly selected. However, maximum pressure ratings of some systemcomponents may limit expansion tank connection location.

est point in the system on the return pip-ing to the pump. But the “best” locationmay not be the one that results in thesmallest tank size. It may be where thetank may be most conveniently located,such as in a mechanical room where spaceis available and the tank is readily ac-cessible for service. The low-pressurepoint may, for instance, be located over ahotel guest room that is not a convenientlocation for an expansion tank.

Claim: The connection point of theexpansion tank is the “point of no pres-sure change,” i.e., the pressure at the ex-

pansion tank remains constant.Fact: This claim has resulted in unnec-

essary confusion and concern amongoperating engineers when they find sys-tem operating pressures varying widely.The pressure at the expansion tank willnot change when pumps are started andstopped (other than a brief pulse), butthe pressure will change when the tem-perature of the fluid in the systemchanges, causing the water volume toexpand or contract. The amount of pres-sure change is a function of the size ofthe tank and the change in fluid tempera-ture. As shown in the examples below,the designer can select a tank to operateover a wide range of pressures, e.g., 10 to1 or more.

To better understand these facts and toproperly select expansion tanks, we muststart with the fundamentals.

PurposeExpansion tanks are provided in

closed hydronic systems to:

By Steven T. Taylor, P.E., Member ASHRAE

Claim: The best connection point forexpansion tanks is near the suction ofthe pump.

Fact: If “best” means the point that willresult in the smallest (lowest cost) expan-sion tank, the best location is at the pointin the system that has the lowest gaugepressure when the pump is on. For a sys-tem that is distributed horizontally (e.g.,a one-story building), the low-pressurepoint will be at the pump suction. How-ever for a system that is distributed verti-cally (e.g., a multi-story building), thelow-pressure point will be near the high-

Practical Guide

M

The following article was published in ASHRAE Journal, March 2003. © Copyright 2003 American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. It is presented for educational purposes only. This article may not be copied and/or distributed electronically or inpaper form without permission of ASHRAE.

March 2003 ASHRAE Journa l 25

Hydronic

• Accept changes in system water volume as water densitychanges with temperature to keep system pressures below equip-ment and piping system component pressure rating limits.

• Maintain a positive gauge pressure in all parts of the sys-tem in order to prevent air from leaking into the system.

• Maintain sufficient pressures in all parts of the system inorder to prevent boiling, including cavitation at control valvesand similar constrictions.

• Maintain required net positive suction head (NPSHr) atthe suction of pumps.

The latter two points generally apply only to high tempera-ture hot water systems. For most HVAC applications, only thefirst two points need be considered.

TypesThere are basically two types of expansion tanks:• Tank type or “plain steel” tanks (water in contact with air),

which may be atmospheric (vented to atmosphere) or pressur-ized, and

• Diaphragm or bladder-type tanks (air and water separated bya flexible diaphragm or bladder, typically made of heavy-dutybutyl rubber). The bladders in bladder-type tanks generally arefield replaceable should they fail while failure of a diaphragmtank would require complete replacement of the tank.

Both types of expansion tanks work by allowing water tocompress a chamber of air as the water expands with increas-ing temperature. When the system is cold and the water in thetank is at the minimum level (which may be no water at all),the tank pressure is at its initial or precharge pressure Pi. As thewater in the system expands upon a rise in temperature, waterflows into the tank and the air pocket is compressed, increas-ing both the air and water system pressure.

When the system is at its highest temperature and the tankwater volume is at its design capacity, the resulting air and watersystem pressure will be equal to or less than the design maxi-mum pressure Pmax. Both Pi and Pmax are predetermined by thedesigner as part of the tank selection process outlined below.

Initial PressureThe tank initial or precharge gauge pressure Pi must be

greater than or equal to the larger of the following:A. The minimum pressure required to prevent boiling and

to maintain a positive gauge pressure at any point in thesystem.

This pressure can be determined as follows:1. Find the low pressure point (LPP) in the system when the

pump is on. To do this, start at the highest point in elevation onthe return side closest to the pump suction (Point A in Figure1). Calculate the net pressure drop from that point to the pumpsuction with friction and dynamic head losses as negative andstatic head increase due to elevation as positive. The pointwith the lowest net pressure is the LPP. The static pressureincrease as you move down in elevation from the high point

increases 1 foot of head for every foot of elevation drop (3 kPafor every 0.3 m). In normal practice, the frictional pressuredrop rate will be almost two orders of magnitude smaller, so theincrease in pressure due to a reduction in elevation is always amuch larger factor than the pressure decrease due to friction.Hence, the LPP will almost always be the highest point of thereturn line just after it drops down to the pump.

2. Determine Pmin, the minimum pressurization required atthe LPP to maintain a positive gauge pressure (to prevent airfrom leaking into the system when a vent is opened) and toprevent boiling. A commonly recommended minimum pres-surization is 4 psig (28 kPa) plus 25% of the saturation vaporpressure when this pressure exceeds atmospheric pressure (Fig-ure 2). For chilled water, condenser water, and typical hot wa-ter (<200°F [93°C]) systems, the recommended minimumpressure is 4 psi (28 kPa).

For high temperature hot water, the minimum pressure willneed to be higher as shown in Figure 2 . For hot water systemsthat have high pressure-drop control valves located near theLPP, the minimum pressure may need to be even higher toprevent cavitation downstream of the valve.1

3. Locate the tank position. The tank will be the smallestand least expensive if located near the LPP. However, space orother considerations (e.g., convenient makeup water source,which should be connected near the same point in the system)may make another location more desirable.

4. Calculate the static pressure rise tankLPPsP →∆ , from theLPP to the point of connection. This is simply the elevationdifference between the two (for pressure in units of feet of water).

5. If the tank is upstream of the LPP, calculate the frictionalpressure drop LPPtankfP →∆ , from the connection point to theLPP when the pump is on. LPPtankfP →∆ , will be negative ifthe tank is downstream of the LPP, but if included in the calcu-lation of Pi, minimum pressures would only be maintainedwhen the pump is running. Once it stops, the pressure at theLPP will drop by LPPtankfP →∆ , below the desired minimumpressure. Thus if the tank is downstream of the LPP,

LPPtankfP →∆ , should be ignored (assumed to be zero).

‘But the “best” location may notbe the one that results in thesmallest tank size. It may bewhere the tank may be mostconveniently located, such as ina mechanical room where spaceis available and the tank is readilyaccessible for service.’

2 6 A S H R A E J o u r n a l a s h r a e . o r g M a r c h 2 0 0 3

50 100 150 200 250 300 350

RecommendedMinimum

Pressurization, Pmin

SaturationVapor

Pressure

Temperature, °F

160

140

120

100

80

60

40

20

0

–20

Gau

ge

Pre

ssu

re,

psi

g

6. Calculate the minimum tank initial or precharge gaugepressure Pi as:

LPPtankftankLPPsmini PPPP →→ ∆+∆+= ,, (1)

The tank precharge pressure Pi is often selected at a mini-mum of 12 psig (83 kPa), which is the industry standardprecharge for diaphragm and bladder tanks when no othervalue is specified.

Tanks typically are precharged to the specified pressure inthe factory. However, the precharge pressure should be fieldverified and adjusted if necessary. First, turn off any heat-pro-ducing equipment in the system, then close the tank isolationvalve so it is isolated from the system. Fully drain the tank,then check and adjust the tank air pressure to the desired Pisetpoint using an air compressor. Before opening the tank iso-lation valve, the system must be near its lowest temperature (asdetermined under Selection later). Connecting the tank to asystem that is warmer than the minimum temperature will re-sult in pressures below Pi when the system temperature dropsand water volume decreases.

B. The minimum pressure required to maintain the availablenet positive suction head (NPSHa ) at the pump suction above thepump’s minimum required net positive suction head (NPSHr ):

This criterion is only a factor in determining Pi for high tem-perature (>200°F [93°C]) hot water systems where the pump is along distance (hydraulically) from the expansion tank or lo-cated well above the tank. In almost all other applications, theminimum pressurization resulting from Step A will result in pumpsuction pressures well above the NPSHr . Hence, for the largemajority of common HVAC applications, the calculations inStep B may be skipped. This step is listed here for completeness.

The precharge pressure required to maintain NPSHr is deter-mined as follows:

1. Find the required pump net positive suction head (NPSHr)from the manufacturer’s pump curve or selection software.

2. Calculate the frictional pressure drop

suctiontankfP →∆ , from the tank to the pump suction in the di-rection of flow.

3. Determine the gauge saturation vapor pressure of the fluidPv at its maximum expected temperature. See discussion underSelection later for determining the maximum fluid tempera-ture. See Figure 2 for a curve of Pv in psig versus temperature.(Pv is more commonly listed in absolute pressure rather thangauge pressure, but gauge pressure is used here since we arecalculating Pi in gauge pressure.)

4. Calculate the static pressure difference suctiontanksP →∆ ,

from the tank to the pump suction. This is simply the elevationdifference between the two (for pressure in units of feet of water).

5. Calculate the difference in velocity pressure

suctiontankVP →∆ , in the piping at the point where the tank isconnected to that at the pump suction. The velocity pressure isproportional to the square of the velocity in the pipe (equal to

3642 .V in units of psi with velocity V expressed in ft/s).Typically this term is negligible and may be ignored, particu-larly when the pipe size at the expansion tank connectionpoint is close to the pump suction pipe size, resulting in nearlyequal velocity pressures.


suctions,tankvsuctiontankfri PPPNPSHP →→ ∆−+∆+= ,suctiontankVP →∆− , (2)

Maximum PressureTank maximum pressure Pmax is determined as follows:1. Determine the maximum allowable system pressure

Pma and critical pressure point, CPP. The CPP is the “weakest

Figure 1 (left): Typical system in a multistory building. Figure 2 (right): Recommended minimum pressurization andsaturation vapor pressure of water.

Point A

100 ft(30.5 m)

Point CPoint B

Pump

Chiller orBoiler

Hydronic


link” in the system. It is a function of the pressure ratings atthe maximum expected operating temperature of componentsand equipment (obtained from the manufacturer) and theirlocation in the system in elevation and relative to the pump.To find the CPP, make a list of the components and equip-ment with the lowest pressure ratings (often boilers and otherpressure vessels) and calculate the difference between theirrating and their vertical elevation, in consistent units(e.g., convert the psig pressure ratings into units of head (feetof water) and subtract the elevation of the component).The component with the smallest difference is the “weakestlink.” Its location on discharge side of the pump is the CPPand the maximum pressure Pma is the rating pressure of theequipment.

2. Locate the pressure relief valve. Typically, the best loca-tion is near the CPP and the component the relief valve isintended to protect, but a common location is near the expan-sion tank connection to the system, on the system side of thetank isolation valve.

3. Calculate the static pressure difference

PRVCPPsP →∆ , from the CPP to the connection point of thepressure relief valve. This is simply the elevation differencebetween the two (for pressure in units of feet of water)and may be positive (CPP above PRV) or negative (CPPbelow PRV).

4. If the relief valve is downstream of the CPP, calculate thefrictional pressure drop PRVCPPfP →∆ , from the CPP to therelief valve when the pump is on. (This term is ignored if therelief valve is upstream of the CPP because maximum pressurewill need to be maintained even when the pump is off.)

5. Calculate the pressure relief valve setpoint Prv as:

PRVCPPfPRVCPPsmarv PPPP →→ ∆−∆+= ,, (3)

For most single-story systems, 30 psig (207 kPa) is com-monly used as the relief valve setpoint even when the abovecalculation results in a larger value. This is the standard ratingof many low-pressure boilers, although most boilers are avail-able at higher pressure ratings (e.g., 60 psig [414 kPa]) at lowor no cost.

6. Calculate the static pressure difference

tankPRVsP →∆ , from the connection point of the pressure reliefvalve to that of the expansion tank. This is simply the eleva-tion difference between the two (for pressure in units of feet ofwater) and may be positive (PRV above tank) or negative (PRVbelow tank).

7. If the tank is downstream of the relief valve, calculate thefrictional pressure drop tankPRVfP →∆ , from the relief valve to

the tank when the pump is on. (This term is ignored if the tankis upstream of the relief valve because maximum pressure willneed to be maintained even when the pump is off.)

8. Calculate the tank maximum gauge pressure Pmax as:

tankPRVftankPRVsrvmax PPPP →→ ∆−∆+= ,, (4)

Tank SelectionTypically, expansion tanks are selected using manufactur-

ers’ software or selection charts based on the following data:• The minimum temperature the system will see, Tc . For

heating systems, this would generally be the initial fill tem-perature, e.g., 50°F (10°C). For cooling systems, this wouldbe the design chilled water temperature, e.g., 40°F (4°C).

• The maximum temperature the system will see, Th. For heat-ing systems, this would be the design hot water temperature,e.g., 180°F (82°C). For cooling systems, this would generallybe the temperature the system may rise to when it is off, e.g.,80°F (27°C) or so depending on the location of piping (in-doors or outdoors).

• The total volume of water in the system, Vs , including allpiping and vessels.

• The precharge pressure Pi and the maximum pressure Pmaxdetermined previously.

Tank volume also may be selected from fundamental equa-tions such as Equations 5 and 6,2 which apply tobladder/diaphragm type tanks:

ea VV ≥

−≥ 1

c

hs v

vV (5)

( ) ( )maxaia

et PPPP

VV

++−≥

1 (6)

whereVt = the tank volumeVa = the tank “acceptance” volume. This is the capacity of

the bladder (for bladder tanks) or the volume of the watersideof the tank when the diaphragm is fully extended (for dia-phragm tanks). With so-called “full acceptance” tanks, the blad-der can open to the full shape of the tank, so the tank’sacceptance volume and total volume, Vt , are equal.

Ve = the increase in volume of water as it expands from itsminimum temperature to its maximum temperature.

vc = the specific volume of water at the minimum tempera-ture, Tc.

vh = the specific volume of water at the minimum tempera-ture, Th.

°F 40 60 80 100 120 140 160 180 200 220 240 260 280 300ft3/ lbm 0.01602 0.01604 0.01608 0.01613 0.01620 0.01629 0.01639 0.01651 0.01663 0.01677 0.01692 0.01709 0.01726 0.01745

°C 4 16 27 38 49 60 71 82 93 104 116 127 138 149cm3/g 1.000 1.001 1.004 1.007 1.011 1.017 1.023 1.031 1.038 1.047 1.056 1.067 1.078 1.089

Table 1: Specific volume of saturated water at various temperatures.


Point A

15 ft(4.6 m)Point C

Point B

PumpBoiler

Equation 5 ensures that the acceptance volume Va exceedsthe expanded water volume Ve to avoid damage to the blad-der or diaphragm when the system is at its highest tempera-ture and pressure. Equation 6 ensures that the tank volume Vtis sufficient for both the expanded water Ve and the air cush-ion necessary to maintain pressures in the tank between Piand Pmax .

Equation 6 conservatively ignores the expansion of thesystem piping since the resulting increase in volume is rela-tively small and the calculation is complicated in systemswith various piping materials, each with different coefficientsof expansion.

Specific volume at various temperatures is shown in Table 1.

ExamplesExample 1: Chilled Water System

Assume the system is a chilled water system with a designchilled water temperature of 40°F and system volume of 1,000gallons (3785 L). The layout is as shown in Figure 1 with thepump at the bottom of a multistory building. Pump head is 80ft (240 kPa).

First calculate the initial precharge pressure Pi . Since thesystem is chilled water, there is no concern about net positivesuction head, so only Step A needs to be followed:

1. The LPP in the system is the highest point on the returnline just as it drops down to the pump (Point A in Figure 1).

2. Pmin is 4 psig (28 kPa) as shown in Figure 2 .3. The tank will be the smallest and least expensive if lo-

cated near the LPP. However, for this example, assume the tankis located near Point B at the pump suction, which is often themost convenient location since space is usually available.

4. The static pressure rise tankLPPsP →∆ , from the LPP to thetank is 100 ft (or 100/2.31= 43 psi [296 kPa]).

5. The frictional pressure drop LPPtankfP →∆ , from tank tothe LPP is taken as zero since the tank is downstream of theLPP.

6. The minimum tank initial or precharge gauge pressurePi is:

LPPtankftankLPPsmini PPPP →→ ∆+∆+= ,,

0434 ++= psig47= (324 kPa)

Now find the maximum pressure:1. The standard pressure rating of all components in the

system will be 125 psig (862 kPa) or higher. Hence Pma istaken as 125 psig (862 kPa) and the CPP is the lowest point inthe system on the discharge side of the pump, Point C.

2. Assume the pressure relief valve will be located near thechiller and the expansion tank, Point B.

3. The static pressure difference PRVCPPsP →∆ , from the CPPto the pressure relief valve is zero since they are at the sameelevation.

4. The relief valve (Point B) is downstream of the CPP (PointC). PRVCPPfP →∆ , is the roughly equal to pump head (80 ft or

35 psi [241 kPa]).5. The pressure relief valve setpoint Prv is then:

PRVCPPfPRVCPPsmarv PPPP →→ ∆−∆+= ,,

350125 −+= psig90=

This setting assures that the pressure at the discharge side ofthe pump will not exceed 125 psig (862 kPa). If the relief valvesetpoint were 125 psig (862 kPa), the pressure downstream ofthe pump could be as high as 160 psig (1100 kPa), possiblyabove the equipment rating.

6. The static pressure difference tankPRVsP →∆ , from the re-lief valve to the tank is zero since they are at the same location.

7. The frictional pressure drop tankPRVfP →∆ , from the re-lief valve to the tank is zero since they are at the same location.

8. The tank maximum gauge pressure Pmax is:

tankPRVftankPRVsrvmax PPPP →→ ∆−∆+= ,,0090 −+=

psig90=The tank minimum volume (assuming a maximum tempera-

ture of 80°F (27°C) with specific volume values taken fromTable 1) is then calculated as:

ea VV ≥

−≥ 1

c

hs v

vV

−≥ 1016020016080

1000.

.

gallons 75.3≥

( ) maxaiat PPPP

V++−

≥1

75.3

( ) ( )90714477141753

++−≥

..

.

gallons .19≥

Hence, the expansion tank must have an acceptance volumegreater than 3.75 gallons (14 L) and an overall volume greaterthan 9.1 gallons (34 L).

Example 2: Chilled Water System

Assume the system is as in Example 1, but instead of locat-ing the tank at the pump, assume it is located at the LPP (PointA). The precharge pressure would then be calculated as (start-

Figure 3: Typical system in a single story building.

Hydronic


ing at Step 4):4. The static pressure rise tankLPPsP →∆ , from the LPP to the

tank is zero since the tank is located at this point.5. The frictional pressure drop LPPtankfP →∆ , from the tank

to the LPP is zero since the tank is located at this point.6. The minimum tank initial or precharge gauge pressure Pi is:


004 ++= psig4=

Now find the maximum pressure (starting at Step 6):6. The static pressure difference tankPRVsP →∆ , from the re-

lief valve to the tank is –100 ft (or –100/2.31= –43 psi [–296kPa]). It is negative since the PRV is below the tank.

7. The frictional pressure drop tankPRVfP →∆ , from the re-lief valve to the tank is taken as zero since the tank is upstreamof the relief valve.

8. The tank maximum gauge pressure Pmax is:

tankPRVftankPRVsrvmax PPPP →→ ∆−∆+= ,,

04390 −−= psig47=

The tank volume is then calculated as:

( ) ( )maxaiat PPPP

V++−

≥1

753.

( ) ( )4771447141753

++−≥

..

.

gallons .45≥

As in Example 1, the expansion tank must have an accep-tance volume greater than 3.75 gpm (0.24 L/s) (the volumeof expanded water is unchanged), but the total volume re-quired falls to 5.4 gallons (20 L). This demonstrates that atank located at the LPP in a multi-story system is smaller andtherefore less expensive than a tank located at the pump suc-tion (Example 1).

Example 3: High Temperature Hot Water System

Assume the system is a high temperature hot water systemwith a design hot water temperature of 300°F (149°C) andsystem volume of 1,000 gallons (3785 L). The pumps’ requiredNPSHr is 5 ft (15 kPa) and its head is 50 ft (150 kPa). Thelayout is horizontal as shown in Figure 3 .

First calculate the initial precharge pressure Pi , which will bethe larger of that required to prevent boiling and that required tomaintain adequate net positive suction head at the pump. Thepressure required to prevent boiling is determined as:

1. The LPP in the system is the highest point on the returnline just after it drops down to the pump (Point A in Figure 3 ).

2. Pmin is recommended to be 70 psig (482 kPa) as shown inFigure 2. If there are control valves located near the LPP, the

minimum pressure may need to be higher.1

3. For this example, assume the tank is located near Point Bat the pump suction.

4. The static pressure rise tankLPPsP →∆ , from the LPP to thetank is 15 ft (or 15/2.31= 6.5 psi [45 kPa]).

5. The frictional pressure drop LPPtankfP →∆ , from the tankto the LPP is taken as zero since the tank is downstream of theLPP.

6. The minimum tank initial or precharge gauge pressure Piis:


05670 ++= . psig.576=

The initial pressure must also be sufficient to maintainthe required net positive suction head (NPSHr) at the pumpinlet:

1. The required net positive suction head (NPSHr) is 5 ft (or5/2.31 = 2 psi [15 kPa]).

2. The frictional pressure drop suctiontankfP →∆ , from thetank to the pump suction is zero since the tank is located at thepump suction.

3. The gauge vapor pressure of the fluid Pv is 53 psig (365kPa) per Figure 2 .

4. The static pressure difference suctiontanksP →∆ , from thetank to the pump suction is zero since the tank is located atthis point.

5. Assume velocity pressure difference suctiontankVP →∆ , isnegligible.


suctiontanksvsuctiontankfri PPPNPSHP →→ ∆−+∆+= ,,

suctiontankVP →∆− ,

005302 −−++= (2) psig55=

This pressure is lower than that required to prevent boilingat the LPP, so it may be ignored. As noted above, this is typi-cally the case unless the pump is a long distance from theexpansion tank or located well above the tank.

Now find the maximum pressure:1. The standard pressure rating of all components in the

system will be 125 psig (862 kPa) or higher. Hence Pma istaken as 125 psig and the CPP is the lowest point in the systemon the discharge side of the pump, Point C.

2. The pressure relief valve will be located at the boiler.3. The static pressure difference PRVCPPsP →∆ , from the CPP

to the pressure relief valve is zero since they are roughly at thesame elevation.

4. The relief valve (at the boiler) is downstream of the CPP(Point C). PRVCPPfP →∆ , is the then approximately equal topump head (50 ft or 22 psi [152 kPa]).

5. The pressure relief valve setpoint Prv is then:


PRVCPPfPRVCPPsmarv PPPP →→ ∆−∆+= ,,

220125 −+= psig103=

6. The static pressure difference

tankPRVsP →∆ , from the relief valve to the tankis zero since they are roughly at the same el-evation.

7. The frictional pressure drop

tankPRVfP →∆ , from the relief valve to thetank may be ignored since they are near thesame location.

8. The tank maximum gauge pressure Pmaxis:

tankPRVftankPRVsrvmax PPPP →→ ∆−∆+= ,,

00103 −+= psig103=

The tank minimum acceptance volume (as-suming a minimum temperature of 60°F[16°C] with specific volume values taken fromTable 1) is then calculated as

ea VV ≥

−≥ 1

c

hs v

vV

−≥ 1

016040017450

1000.

.

gallons 88≥

( ) ( )maxaiat PPPP

V++−

≥1

88

( ) ( )103714576714188

++−≥

...tV

gallons 390≥

Hence, the tank must have an acceptancevolume 88 gallons (333 L) or larger and atotal volume 390 gallons (1476 L) or larger.

As a home exercise, redo the above calcu-lations with the tank at the pump discharge,Point C. You will find that the precharge pres-sure increases while all other variables re-main the same, resulting in a somewhat largertank. Nevertheless, the tank will still meetall of its intended functions so this is a per-fectly acceptable, if unusual, location.

References1. Carlson, B. 2001. “Avoiding cavitation

in control valves.” ASHRAE Journal 43(6).2. 2000 ASHRAE Handbook—HVAC Sys-

tems and Equipment.

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Ashrae journal understanding expansion tanks - taylor