Pebble bed storage size correction

For the development of the /-chart of Figure 11.6, a standard storage capacity of 0.25 cubic meters of pebbles per square meter of collector area was considered, which corresponds to 350 kJ/m2-°C for typical void fractions and rock properties. Although the performance of air-based systems is not as sensitive to the storage capacity as in liquid-based systems, other storage capacities can be used by modifying the factor X by a storage size correction factor, Xc/X, as given by (Klein et al., 1977):

Mb,a

X where

Mb, a = actual pebble storage capacity per square meter of collector area (m3/m2). Mb, s = standard storage capacity per square meter of collector area = 0.25 m3/m2.

Equation (11.18) is applied in the range of 0.5 < (Mb,a/Mb,s) < 4.0 or 0.125 < Mb a < 1.0. The storage correction factor can also be determined from Figure 11.7 directly, obtained by plotting Eq. (11.18) for this range.

s

Mbs = 0.25 m3/m2

\

\

FIGURE 11.7 Storage size correction factor for air-based systems.

FIGURE 11.7 Storage size correction factor for air-based systems.

AIR FLow RATE correction

Air-based heating systems must also be corrected for the flow rate. An increased air flow rate tends to increase system performance by improving FR, but it tends to decrease performance by reducing the pebble bed thermal stratification. The standard collector flow rate is 10 L/s of air per square meter of collector area. The performance of systems having other collector flow rates can be estimated by using appropriate values of FR and Y, then modifying the value of X by a collector air flow rate correction factor, Xc/X, to account for the degree of stratification in the pebble bed (Klein et al., 1977):

x0.28

where ma = actual collector flow rate per square meter of collector area (L/s-m2). ms = standard collector flow rate per square meter of collector area = 10 L/s-m2.

Equation (11.19) is applied in the range of 0.5 < (ma/ms) < 2.0 or 5 < rna < 20. The air flow rate correction factor can also be determined from Figure 11.8 directly, obtained by plotting Eq. (11.19) for this range.

Air systems ms = 10 L/s-m2

/

z

z

/

FIGURE 11.8 Correction factor for air flow rate to account for stratification in pebble bed.

ma/ms

Example 11.7

If the air system of Example 11.6 uses a flow rate equal to 17 L/s-m2, estimate the solar fraction for the month of January. At the new flow rate, the performance parameters of the collector are FRUL = 3.03 W/m2-°C, Fr(to) = 0.54.

Solution

m a

U.Z.O

17'

.1s .

0.28

1.16

The increased air flow rate also affects the FR and the performance parameters, as shown in the problem definition. Therefore, the value of X to use is the value from Example 11.6 corrected for the new air flow rate through the collector and the pebble bed. Hence,

Xc = X

FRUL 1 new

Xc

= 0.70

3.03

, FRUL1 test ,

X

The dimensionless parameter Y is affected only by the FR. So, fr (to )l,

0.19

0.54

0.52

0.20

Finally, from the /-chart of Figure 11.6 or Eq. (11.17), f = 0.148 or 14.8%. Compared to the previous result of 14.7%, there is no significant reduction for the increased flow rate, but there will be an increase in fan power.

If, in a solar energy system, both air flow rate and storage size are different from the standard ones, two corrections must be done on dimensionless parameter X. In this case, the final X value to use is the uncorrected value multiplied by the two correction factors.

Example 11.8

If the air system of Example 11.6 uses a pebble storage tank equal to 0.35 m3/m2 and the flow rate is equal to 17 L/s-m2, estimate the solar fraction for the month of January. At the new flow rate, the performance parameters of the collector are as shown in Example 11.7.

Solution

The two correction factors need to be estimated first. The correction factors for X and Y for the increased flow rate are as shown in Example 11.7. For the increased pebble bed storage, from Eq. (11.18),

0.35

0.25

0.30

0.90

The correction for the air flow rate is given in Example 11.7 and is equal to 1.16. By considering also the correction for the flow rate on FR and the original value of X,

X„ = X

FRUL 1 new

Xc

Xc

, FRUL 1 test ,

X

flow X

3.03

2.92

The dimensionless parameter Y is affected only by the FR. So, the value of Example 11.7 is used here (=0.20). Therefore, from the f chart of Figure 11.6 or Eq. (11.17), f = 0.153 or 15.3%.

11.1.3 Performance and Design of Solar Service Water Systems

The f-chart of Figure 11.2 or Eq. (11.13) can also be used to estimate the performance of solar service water heating systems with a configuration like that shown in Figure 11.9. Although a liquid-based system is shown in Figure 11.9, air or water collectors can be used in the system with the appropriate heat exchanger to transfer heat to the pre-heat storage tank. Hot water from the pre-heat storage tank is then fed to a water heater where its temperature can be increased, if required. A tempering valve may also be used to maintain the supply temperature below a maximum temperature, but this mixing can also be done at the point of use by the user.

The performance of solar water heating systems is affected by the public mains water temperature, Tm, and the minimum acceptable hot water temperature, Tw; both affect the average system operating temperature and thus the collector energy losses. Therefore, the parameter X, which accounts for the collector energy losses, needs to be corrected. The additional correction factor for the parameter X is given by (Beckman et al., 1977):

Relief valves

Collector array

Collector pump

Tempering valve

Hot water supply

Relief valves

Collector array

Tempering valve

Collector pump

Hot water supply

Water supply

FIGURE 11.9 Schematic diagram of the standard of water heating system configuration.

Water supply

FIGURE 11.9 Schematic diagram of the standard of water heating system configuration.

where

Tw = minimum acceptable hot water temperature (°C).

Ta = monthly average ambient temperature (°C).

The correction factor Xc/X is based on the assumption that the solar pre-heat tank is well insulated. Tank losses from the auxiliary tank are not included in the /-chart correlations. Therefore, for systems supplying hot water only, the load should also include the losses from the auxiliary tank. Tank losses can be estimated from the heat loss coefficient and tank area (UA) based on the assumption that the entire tank is at the minimum acceptable hot water temperature, Tw.

The solar water heater performance is based on storage capacity of 75 L/m2 of collector aperture area and a typical hot water load profile, with little effect by other distributions on the system performance. For different storage capacities, the correction given by Eq. (11.14) applies.

Example 11.9

A solar water heating system is installed in an area where, for the 31-day month under investigation, the average daily total radiation on the tilted collector surface is 19.3 MJ/m2, average ambient temperature is 18.1°C, and it uses a 5 m2 aperture area collector that has FR(ja)n = 0.79 and FRUL = 6.56 W/ m2-°C, both determined from the standard collector tests. The water heating load is 200 L/d, the public mains water temperature, Tm, is 12.5°C, and the minimum acceptable hot water temperature, Tw, is 60°C. The storage capacity of the pre-heat tank is 75 L/m2 and auxiliary tank has a capacity of 150 L, a loss coefficient of 0.59 W/m2-°C, diameter 0.4 m, and height of 1.1 m; it is located indoors, where the environment temperature is 20°C. The flow rate in the collector is the same as the flow rate used in testing the collector, the FR /Fr = 0.98, and the (Ta)/(Ta)n = 0.94. Estimate the solar fraction.

Solution

The monthly water heating load is the energy required to heat the water from Tm to Tw plus the auxiliary tank losses. For the month investigated, the water heating load is

The auxiliary tank loss rate is given by UA(Tw - Ta). The area of the tank is nd2/2 + ndl = n(0.4)2/2 + n X 0.4 X 1.1 = 1.63m2

Thus, auxiliary tank loss = 0.59 X 1.63(60 - 20) = 38.5 W. The energy required to cover this loss in a month is

Therefore,

= 6.56 X 0.98(100 - 18.1) X 31 X 24 X 3600 X = 5.27

1.337 X 109

1.63

1.337 X 109

From Eq. (11.20), the correction for X is

1.08

Therefore, the corrected value of X is

From Fig. 11.2 or Eq. (11.13), for Xc and Y, we get f = 0.808 or 80.8%.

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  • adaldrida
    What is pebble bed storage?
    7 years ago

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