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We made the approximate calculation for a vacuum tube solar collector. Because of the features of the collectors, You can't choose the case where the 100% of the DHW demand is served by solar collectors in every month. We recommend an average of 60% per a year. You can find the explanation below the calculation.

 

Number of occupants
What percentage of DHW demand would you like to served by vacuum tube solar collectors?
%   (We recommend 60%.)
Heat amunt needed for DHW making
 
Orientation
Tilt angle
Absorbing surface of the vacuum tubes
 
Month Monthly heat amount produced by solar collectors (kWh) What percentage this is of the whole DHW demand of the month?
January
 
 
February
 
 
March
 
 
April
 
 
May
 
 
June
 
 
July
 
 
August
 
 
September
 
 
October
 
 
November
 
 
December
 
 
TOTAL
 
 

 

 

 

 

The vacuum tube collectors are more beneficial than the flat plate collectors for the DHW of detached houses. Observing the efficiency diagram, we can see that the vacuum tube collector's performance for a surface unit (y-axis) is the biggest in the range of the temperature (x-axis), required for hot water making.

source: László Fülöp PTE PMMF

Because of the cylindrical shape of the vacuum tubes the sun gets to the glass surface vertically almost for the whole day and the performance values are increasing constantly from the vertical incidence, so the thermal efficiency is pretty good from morning till late afternoon. The flat plate collector only gives its full performance at noon, when the sun is peripendicular to it's surface. On the other hand, the vacuum tube prevents the losing heate from the inside, thanks to this, the wind and cold weather have a smaller affect on the vacuum tube collector.

It is worth to measure the solar collector to serve the 60% of the yearly DHW demand. If we choose a bigger surface than this, more hot water is produced in the summer than what is really needed.

Formulas and quantities used in the calculation:

V= n x V1=number of occupants x 50 liter/day = 0,05 m3/day/person

QDHW = 1,2 x c x m x ?T = 1,2 x c x ?  x V (tm - th) =1,2 x 4180 J/kg K x 1000 kg/m3 x 0,05 m3/day (45K - 10K) =8 778 000 J/day = 2,438 kWh/day x 365 day= 890 kWh/year/person

where:

Specific heat of water [J/kg/K]
V Daily consumption [l]
? desity of water [kg/m3]
th cold tap water temperature °C
tm DHW temperature °C
1,2 Estimated loss coefficient

Qcollector yearly = k x Ncoll x Aabs x ?Q collector monthly

where:

k performance coefficient
Ncoll number of collectors
Aabs absorbing surface
?Q collector monthly the heat amount produced by the collectors in a month

source: http://www.energiakozpont.hu/index.php?p=212

The calculation is an estimation, the performance generated by the unit of the absorbing surface is different depending on the type of the collector. For any further question or exact calculation, call us!

Related articles

     
How to select the appropriate solar collector system?  Szeminárium sorozat 2012 Passzívház, alacsony energiaigényû ház, A kategóriás ház  95 m2 alacsony energiaigényû családi ház 13 millió forintból 

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Comments:

  • - 2015. ápr. 15 22:10:59

  • Igor - 2014. jún. 20 13:30:38

    chief@babichdesign.com

  • - 2014. ápr. 25 19:26:51

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