Solar Panel expansion Tank Sizing: Getting It right – Part I
The most common type of solar thermal system uses a closed piping loop between the collector array and storage tank heat exchanger. BY JOhN SIeGeNThAler
his loop is filled with an antifreeze solution such (SRCC) (solar-rating.org) must undergo 30-day stagnation as a 40 per cent solution of propylene glycol. As in tests. During this time collectors are subjected to 30 con- any closed hydronic circuit, the fluid expands when secutive days where total daily solar radiation is at least heated. To avoid excessively high pressure, an expan- 1,500 Btu/ft2/day in the plane of the collector and at least sionT tank must accommodate the increase in volume. one four-hour period with a minimum solar intensity of 300 The sizing of this tank is similar, yet different, from siz- Btuh/ft2. The average ambient temperature during this four- ing an expansion tank in a typical hydronic heating system. hour severe stagnation period must be at least 80F. Determining the size of this tank requires knowledge of After this exposure, collectors are visually inspected how hot the solar collector circuit can become. That subject for any indication of insulation outgassing, pealing or flak- is analyzed here. Part II of this discussion will appear in an ing of the absorber coating, or other signs of degradation. upcoming Solar Panel column and uses this information to Evidence of such degradation would disqualify the collector size the expansion tank. for the OG-100 certification. The temperature the collector’s absorber plate reaches STAGNATION during stagnation is determined based on: the collector’s Under normal operating conditions the collector fluid sel- dom reaches temperatures in excess of 200F. However, dur- FIGure 1 Y-intercept =0.76 ing its service life, almost every solar thermal system will 1 experience stagnation conditions, where bright sunshine 0.9 0.8 strikes the collectors without flow through the absorber 0.7 plates. This can be caused by several factors including: 0.6 » loss of electrical power during the day; 0.5 slope = (0.76/0.93) = 0.817 » failure of a controller or sensor; 0.4 (decimal %) » the thermal storage tank reaching its maximum allowed 0.3 0.2
temperature; and collector thermal efficiency 0.1 » during installation before the system is put into 0 operation. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 2 T − T º F ⋅ ft ⋅ hr Solar collectors rated to the OG-100 standard estab- Inlet fluid parameter i a I Btu lished by the Solar Rating and Certification Corporation
60 HPAC | APRIL 2009 hPAcmAg.com “This temperature probably surprises some of you. It is high enough to melt 50/50 tin/lead solder.” efficiency graph, assumptions about the solar intensity and which converts to 317 Btuh/ft2. The outdoor air temper- ambient air temperature at stagnation conditions. ature is 85F. Under these conditions the absorber plate An example of a typical efficiency graph for a flat plate in the (cYo l–le icnttoerr creepptr)esented by Figure 1 (e.g. Y-intercept Tstag= I + T collector is shown in Figure 1. = 0.76 and(S sloloppee) = 0.817) awir ould reach a stagnation tem- Two numbers describe the straight line on this graph: the perature of: Y-intercept and the slope. For the collector represented in Figure 1, the Y-intercept is 0.76. The slope is the change (Y – intercept) 0.76 T = I + T = 317 + 85 = 380F in vertical distance divided by the change in horizontal stag (Slope) air 0.817 distance. In Figure 1, this is 0.76 divided by 0.93, which equals 0.817. This temperature probably surprises some of you. It is The Y-intercept and slope numbers are determined by high enough to melt 50/50 tin/lead solder. It is also well testing and are listed for many specific collectors at SRCC’s above the temperature at which the fluid in the collector will web site. remain a liquid. During stagnation, there is no fluid flow to extract heat As vapourization of the collector fluid begins, the remain- from the collector’s absorber plate. The absorber plate’s ing liquid in the collector gets pushed out into the rest temperature will climb until the heat loss from the collec- of the piping circuit. A properly-sized expansion tank will tor housing matches the rate of incoming solar energy. This accommodate this volume, as well as the expanded vol- temperature can be estimated using Formula 1. ume of the fluid that does not vapourize elsewhere in the loop, so that the pressure relief valve on the circuit does Formula 1: not open. The details of calculating the proper size expan- sion tank will be discussed in Part II. (Y – intercept) Tstag= I + T (Slope) air John Siegenthaler, P.E. is the author of Modern Hydronic Heating. Visit hydronicpros.com for Where: reference information and software to assist in
Tstag = s(Yta –g ninateiorcne tpetm) perature of 0th.7e6 absorber plate (F) hydronic system design. He can be reached at Tstag= I + T = 317 + 85 = 380F Y – intercep(tS =lo vpael)ue where thaier co0ll.e8c1t7or’s efficiency line [email protected]. touches the vertical axis Slope = numerical slope of the efficiency line (Btu/ºF/ft2/hr) See John Siegenthaler at the I = intensity of solar radiation striking the collector Foothills Conference and Trade Show. (Btu/hr/ft2)
Tair = ambient air temperature around the collector (F) Be sure to stop by and Here is an example: Assume it is a nice bright summer day, and a power outage occurs in the early afternoon. The visit HPaC at Booth 32 solar radiation intensity is 1,000 watts per square metre,
hPAcmAg.com APRIL 2009 | H PA C 61 Solar Panel
Expansion Tank Sizing for Solar Collector Circuits – Part II
In Part I (April 2009) we calculated the temperature that the absorber plate in a solar collector could reach under stagnation conditions. BY JOHN SIEGENTHaLER
f stagnation occurs on a hot/bright summer afternoon, may even violate some mechanical codes that require the the temperature that the absorber plate reaches could pressure in the collector circuit to be no greater than the exceed 350F. We need to deal with the ramifications pressure of the domestic water. of such a temperature. Specifically, we need to select We cannot count on pressurization to suppress boiling Ia diaphragm-type expansion tank that prevents the sys- under maximum stagnation conditions, therefore there will tem’s pressure relief valve from opening under stagnation be times when the fluid in the collector will vapourize. Under conditions. these conditions the liquid in the piping leading to and from the collector array could also be very hot. WHEN BOILING OCCURS To prevent the relief valve from opening under these con- Figure 1 shows the relationship between absolute pressure ditions, the expansion tank must absorb both the fluid dis- and boiling point for a 40 per cent solution of propylene gly- placed by vapourization in the collectors, as well as the col. This is the fluid used in the collector circuit of many expanding volume of the remaining liquid in the circuit. closed-loop solar thermal systems. To maintain this solution as a liquid at a stagnation tem- SELECTING THE PRV perature of 350F would require an absolute pressure of Before calculating the volume of the expansion tank it is about 118 psi (corresponding to a gauge pressure of about necessary to know the rated opening pressure of the system’s 103 psi) in the solar collectors. This is not practical and pressure relief valve. Start by checking local codes to see if they mandate a max- FIGURE 1 aBSOLUTE PRESSURE aND BOILING imum pressure relief valve setting in solar thermal systems. POINT–THE RELaTIONSHIP If this is the case, the code mandated rating is obviously the benchmark. If there are no code restrictions on the rating of the collec- tor circuit pressure relief valve, I suggest it be determined by assuming a cold fill static pressure of 25 psi at the top of the collector array, and calculating the corresponding pressure at the relief valve location. Then, pick a relief valve setting 15 to 20 psi above this pressure. Use Formula 1 to determine the static pressure at the
relief valve location. See Figure 2 for the corresponding
terms and dimensions. FORmULa 1: (
64.9
P = P + H
@PRV @top ( 144
( Where: ( P@PRV = static cold fill pressure a6t 4p.r9essure relief v6a4lv.9e P@PRV = P@top + H = 25 + 20 = 34 psi location (psi) ( 144 ( 144
continued on page 24
64.9 (
P@top = PPRVrated – H 22 HPAC | MAY/JUNE 2009 ( 144 hpAcMAg.coM
( 64.9 ( 64.9 P = P – H = 50 – 20 = 41 psi @top PRVrated ( 144 ( 144
Va = 1.1[(Vc + Vp) 0.08 + Vc] 64.9 (
P = P + H
static @top ( 144 PRV +14.7 ( Vt = Va ( PRV – Pstatic
Va = 1.1[(Vc + Vp) 0.08 + Vc] = 1.1 [(6 + 7.95) 0.08 + 6] = 7.83 gallons
(
64.9 ( 64.9
P = P + H = 25 + 20 = 34 psi
static @top ( 144 ( 144
( * ( PRV +14.7 47 + 14.7 Vt = Va = 7.83 = 37.7 gallons ( PRV – Pstatic ( 47 – 34
64.9 (
Solar Panel continued from page 22 P = P + H @PRV @top ( 144
( 64.9 ( 64.9 P = P + H = 25 + 20 = 34 psi
@PRV @top ( 144 ( 144
P@top = static cold fill pressure at top of collector array FORmULa 2:
(psi) 64.9 ( H = vertical distance from pressure relief valve to top of P = P – H
@top PRVrated ( 144 collectors (feet)
( 64.9 = density of 40 per cent propylene glycol solution at 50F Where: ( ( 64.9 64.9 144 = units conversion factor P@top = pressureP at t=op P of colle–ctor array as relief valve 20 = 41 psi
@6to4p .9 PRVrated H = 50 –
For example, assume 25 psi static cold fill pressurization rePaches= ra Pted o+pening preHssure (ps(i)144 ( 144
@PRV @top ( 144 at the top of the array, and 2( 0 feet of vertical distance from PPRVrated = rating of pressure relief valve (psi) (
64.9 V = 1.1[( (V + V ) 0.08 + V ] theP top o=f t Phe arr+ay down to the relief valve location. The H = vertical distaance from pcressupre relief valcve to top of
@PRV @top H 64.9 64.9 static pressure at th(e1 r4e4lief valve location would be: cPollecto=rs P(feet)+
@PRV @top H = 25 + ( 20 = 34 psi
64.9 = density of 4( 01 4p4er cent propy(le1n4e4 glycol solution at ( ( 64.9
P = P + H 64.9 64.9 50F static @top 144
P = P + H = 25 + 20 = 34 psi ( ( @PRV @top 144 144 144 = units conversion factor 64.9 (
( ( P = P –
@top PRVrated HP +14.7
V =( V144 RV
Based on this, select a reli( ef valve rated at 50 psi. In our example tthis waouldP be:– P 64.9 ( RV static ( UPse F=or mPula 2 t–o determHine the pressure at the top (
@top PRVrated 144 64.9 64.9
( P = P – 20 = 41 psi
of the collector array just as the selected pressure relief @top PRVrateVd = 1.1[(VH =+ V50) –0.08 + V ] = 1.1 [(6 + 7.95) 0.08 + 6] = 7.83 gallons
a ( 144 c p ( 144 c
( valve is about to open—a c( ondition we want to avoid at ( stagnation. 64.9 64.9 20 = 41 psi The gauge pressure of 41 psi correspo( nds to an absolute P@top = PPRVrated – H = 50 – 64.9 64.9
144 144 Va = 1.1[(Vc +P Vp) 0=.0 P8 + V+c] H = 25 + 20 = 34 psi
( ( pressure of 41+14s.t7ati c= 55.@7t oppsi a(t1 t4he4 top of the co(lle1c4t4ors.
According to Figure 1, this ( pressure would maintain the 40 per (
FIGURE 2 64.9 ( Va = 1.1[(Vc + Vp) 0.08 + Vc] cent propylene glycol solution in the collectors, as a l*iquid,
Pstatic = P@top + H PRV +14.7 47 + 14.7
to a temperatureV( to1=f4 aV4baout 297F. This is =de7fi.n8i3tely higher = 37.7 gallons ( ( PRV – Pstatic ( 47 – 34 64.9 than the fluid would see i( n normal operation, but not high
P = P + H P +14.7
static @top ( 144 eVno=ug hV to prevReV nt vapour formation under summer stagna-
t a P – P tion conditi(onsR.V static ( P +14.7 (
V = V RV t a P – P SVIZI=NG1 .T1H[(EV T a+N VK ) 0.08 + V6]4 =.9 1.1 [(6 + 7.95) 0.08 + 6] = 7.83 gallons
( RV static a Pc p= P + c H
@PRV @top 144
( Since the collector fluid( will l(ikely vapourize during stagna-
tion, we need to si6z4e .9the expansion 6ta4n.k9 to accommodate (
Va = 1.1[(Vc + Vp) 0.08 + Vc] = 1.1 [(6 + 7.95) 0.08 + 6] P=s ta7ti.c8=3 Pga@ltolpo+ns H = 25 + ( 20 = 34 psi
this expansion. ( 144 64.9( 144 64.9 (
( P = P + H = 25 + 20 = 34 psi Step 1: Calc@uPlRaVt e the@ tvoop lume1 4t4he expansion ta1n4k4 must
( 64.9 64.9 ( ( ( P = P +
static @top H = 25 + 20 = 34 psi P +14.7 47* + 14.7
( 144 ( 144 aVbs=or bV at stagRnVa tion using Formula 3. This assumes vapour
t a = 7.83 ( = 37.7 gallons will form in (thPeR Vc o–l lePcsttoatric during stag(na4tio7n – a 3nd4 that the liquid ( ( 64.9 P = P –
P +14.7 47* + 14.7 temperature in@ tthoep remPaRinVrdateedr of the sysHtem will reach 200F
V = V RV ( 144 t a = 7.83 = 37.7 galloanbsove the temperature at which the system was filled. The
( PRV – Pstatic ( 47 – 34 ( latter is a conservatively safe assumpti( on. 64.9 64.9 20 = 41 psi P@top = PPRVrated – H = 50 – FORmULa 3: ( 144 ( 144
Va = 1.1[(Vc + Vp) 0.08 + Vc]
Where: Va = expansion volume to be a6c4c.o9mm( odated (gallons).
Pstatic = P@top + H Vc = total volume of collector( a1r4ra4y (gallons)
V = total volume of collector piping other than p (
collectors (gallons) PRV +14.7 Vt = Va ( PRV – Pstatic continued on page 26
24 HPAC | MAY/JUNE 2009 Va = 1.1[(Vc + Vp) 0.08 + Vc] h=p A1c.1M A[(g6.c +o M7.95) 0.08 + 6] = 7.83 gallons
(
64.9 ( 64.9
P = P + H = 25 + 20 = 34 psi
static @top ( 144 ( 144
( * ( PRV +14.7 47 + 14.7 Vt = Va = 7.83 = 37.7 gallons ( PRV – Pstatic ( 47 – 34
64.9 (
P = P + H @PRV @top ( 144
( 64.9 ( 64.9
P = P + H = 25 + 20 = 34 psi
@PRV @top ( 144 ( 144
64.9 (
P = P – H @top PRVrated ( 144
( Solar Panel continued from page 24 64.9 ( 64.9 P = P – H = 50 – 20 = 41 psi @top PRVrated ( 144 ( 144
V = 1.1[(V + V ) 0.08 + V ] 0.08 = expansion factor for 40 per cent propylene glycol Step 3: Calculatae the minimc um prequired expcansion tank solution for 200F temperature rise volume using Formula 5, which is derived ( from Boyle’s law.
1.1 = 10 per cent added safety factor to allow for system 64.9
Pstatic = P@top + H volume estimates FORmULa 5: ( 144 Collector fluid volume is usually listed in manufacturer’s ( P +14.7 specifications, as is the volume of the tank’s internal heat V = V RV t a P – P exchanger. The volume of copper collector piping can be ( RV static estimated using data from Figure 3. If other types of tubing Where:
are used for the collector circuit, obtain volume data from VT = minimum reVqau=ire1d. 1ex[p(Vanc s+io Vn pt)a 0nk.0 v8o l+u mVec] ( g=a l1lo.1n s[)(6 + 7.95) 0.08 + 6] = 7.83 gallons
( the manufacturer. Va = expansion volume to be accommod( ated (from Step 1) 64.9 64.9
(gallons) P = P + H = 25 + 20 = 34 psi
static @top ( 144 ( 144
Pstatic = cold fill static pressure at the relief valve location
FIGURE 3 COPPER COLLECTOR PIPING ( (from Step 2) (psi) ( P +14.7 * Tube type/size Gallons/foot PRV = maximum allowed presRsVu re at the relief va4lv7e l+oc 1a4- .7 Vt = Va = 7.83 = 37.7 gallons 3/8" type M copper: 0.008272 tion (psig). Recommended (vPalRuVe – i sP pstraeticssure relief (va4lv7e –ra 3t-4 1/2" type M copper: 0.0132 ing minus three psi. This allows for a slight safety factor against relief valve “dribbling" as the pressure approaches
3/4" type M copper: 0.0269
the valve’s rating. 1" type M copper: 0.0454 1.25" type M copper: 0( .068 64.9 Here is a final example that pulls this all together. Assume
P = P + H
1.5" type M co@ppPReVr : @top ( 144 0.095 a residential solar water heating system has the following
2" type M copper: 0.165 components: ( 2.5" type M copper: 64.90( .2543 64.9 » Four collectors, each having a volume of 1.5 gallons
P = P + H = 25 + 20 = 34 psi
3" type M copp@ePrR:V @top ( 144 0.3630 ( 144 » Total of 120 feet of one-inch copper tubing between heat exchanger and collector array
( » Heat exchanger volume = 2.5 gallons 64.9 64.9 (
Step 2: CalcuPlate =th eP cold fi–ll static pHressure at the » HPeigh=t oPf to+p of colHlector array above relief valve = 20 feet
@top PRVrated @PRV @top 144
location of the pressure relief valv(e.1 4Th4is is the pressure » Pressure relie( f valve rating = 50 psi
( ( caused by the weight of fluid in the collector circuit above » Collector circuit flu( id = 40 per cent solution of ( ( 64.9 64.9 P = P + 64.9 P@PRV = P@top + H = 25 + 20 = 34 psi
the pressure relief valve location, pl6u4s .t9he static pres6s4ur.e9 prop@yPleRVn e g@ltyopco(l144 H ( 144
P = P – H = 50 – 20 = 41 psi ( 144
@top PRVrated 144 144 maintained at the top of the system. It can be calculated Determine the minim( um size of a diaphragm-type expansion ( ( ( ( 6644.9.9 64.9 using Formula 4. tanPPk fo=r = tPh Pe sy+s–tem sHuHc =h 2t5h a+t the re2l0ie =f v3a4 lpvsei does not open
@@toPpRV PR@Vrtoapted ( (11444 ( 144
under stagnation conditions. The cold fill pressure at the
Va = 1.1[(Vc + Vp) 0.08 + Vc] ( (
FORmULa 4: top of the system is 25( psi. 6644..99 64.9 20 = 41 psi ( – PP@top == PPPRVrated – HH = 50 –
64.9 @top PRVrated ((114444 ( 144
P = P + H ( static @top Solution: (
( 144 V = 1.1[(V + V ) 0.08 + V ]
a c p 64.9 c 64.9 20 = 41 psi TPh@eto pt=o tPaPlR Vcroateldl–ector arHra =y 5v0o –lume is 4 x 1.5 = 6 gallons ( ( 14( 4 ( 144 Where: P +14.7 The total pipi6n4g.9 plus heat exchanger volume is 120 ft x
RV Pstatic = P@top + H
Vt = Va ( 144
Pstatic*= static pressure at the relief valve location (psi) (0.0Va4=541. 1g[a(Vllco +n V/pf)t )0 .+0 82 +.5 V c=] 7.95 gallons ( PRV – Pstatic ( H = height of collector circuit above location of pressure ( PRV +14.7
Vt = Va 64.9 relief valve (feet) SPtep =1 P:( PRV+ – Pstatic H
V = 1.1[(V + V ) 0.08 + V ] = 1.1 [(6 + 7.95) s0tat.ic08 +@t op6] (=1 474.83 gallons
64.9 = density of 40a per cent pcropylep ne glycol soclution at 50F (
(
( V = 1.1[(VP + +V1)4 0.7.08 + V ] = 1.1 [(6 + 7.95) 0.08 + 6] = 7.83 gallons
144 = units conversion factor a c RV p c Vt = Va ( 64.9 64.9 ( ( PRV – Pstatic
P = P + H = 25 + 20 = 34 psi 64.9 64.9
static @top P = P +
( 144 ( 144 static @top H = 25 + 20 = 34 psi
*Note: The air chamber in the diaphragm expansion tank must be pressur- Step 2: ( 144 ( 144
Va = 1.1[(Vc + Vp) 0.08 + Vc] = 1.1 [(6 + 7.95) 0.08 + 6] = 7.83 gallons
( ( ( ( ( ized to this calculated static pressure before fluid is added to the c*ollector P +14.7 ( 47* + 14.7 P +14.7 47 + 14.7 RV 64.9 64.9 RV Vt = Va = 7.83 = 37.7 gallons circuit. This ensures tVhe =dia Vphragm is fully expanded =aga7in.s8t3 the tank shell = P3s7tati.c7= gPa@Ptloplo+–n Ps H = 25 + 47 – 3420 = 34 psi
t a ( RV ( 1s4tat4ic ( ( 144
( PRV – Pstatic ( 47 – 34 before the fluid begins to warm. ( * ( PRV +14.7 47 + 14.7 Vt = Va = 7.83 = 37.7 gallons ( PRV – Pstatic ( 47 – 34 26 HPAC | MAY/JUNE 2009 hpAcMAg.coM
64.9 (
P = P + H @PRV @top ( 144
( 64.9 ( 64.9
P = P + H = 25 + 20 = 34 psi
@PRV @top ( 144 ( 144
64.9 (
P = P – H @top PRVrated ( 144
( 64.9 ( 64.9 P = P – H = 50 – 20 = 41 psi @top PRVrated ( 144 ( 144
Va = 1.1[(Vc + Vp) 0.08 + Vc] 64.9 (
P = P + H
static @top ( 144 PRV +14.7 ( Vt = Va ( PRV – Pstatic
Va = 1.1[(Vc + Vp) 0.08 + Vc] = 1.1 [(6 + 7.95) 0.08 + 6] = 7.83 gallons
( 64.9 ( 64.9 P = P + H = 25 + 20 = 34 psi
stSatitcep 3@t:op ( 144 ( 144 or increasing the pressure rating of the relief valve. These
changes all have consequences, but may be possible in ( P +14.7 * ( RV 47 + 14.7 some applications. Vt = Va = 7.83 = 37.7 gallons ( PRV – Pstatic ( 47 – 34 In cases where the minimum required expansion tank volume exceeds the volume of available tanks, it is accept- *Note: The PRV value in Step 3 was set to 50-3=47 psi to guard against able to use multiple tanks connected in parallel. Be sure “dribbling" of the relief valve as it approaches its rated pressure. the air side pressure in each tank is set to the calculated static pressure prior to filling the collector circuit. As you can see, the expansion tank for the system is Finally, be sure to locate the check valve in the collec- significantly larger than a hydronic heating system of similar tor circuit so that fluid can be pushed out of the bottom volume. This is partially the result of very conservative and top of the collectors as vapourization occurs (see assumptions and partially the result of steam flash in the Figure 2). This is an important detail to avoid “slugging" collectors at stagnation. As in hydronic heating systems, of the liquid if it can only exit at the top of the collector a conservatively sized expansion tank is good insurance array. against the system requiring servicing following a stagna- tion condition. John Siegenthaler, P.E. is the author of Modern The volume of the expansion tank could be reduced by Hydronic Heating. Visit hydronicpros.com for reducing system volume (e.g. smaller tubing, smaller heat reference information and software to assist in exchanger and so on). It could also be reduced by lowering hydronic system design. He can be reached at the static pressure at the top of the collector array and/ [email protected].
Innovative system technology for modern •Solar stations solar thermal systems •Solar heat transfer
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