Dear Negi,

The solar irradiation can be taken equal to 800W/m2

The 300 sunny days are not a useful value about power, but the for the yearly produced energy.

You can assume a thermal effiency of 65% (this varies with the geographical poisition for the incident angle and the operational temeprature).

You can say that 10MW=107Watt

Area*800*0.65=107

so: Area=19231m2

But, the land will be greater, may be 50% greater or double. This is something that varies from plant to plant.

As final conclusion, 30.000-40.000 m2 of land will be teh total area for a 10MW plant.

The solar power is 1000 W/m², if there are no clouds. At least, you will need a circle with a diameter of 113 m. If the efficiency is lower than 100%, the collecting area must be ever larger.

Dear Negi,

The solar irradiation can be taken equal to 800W/m2

The 300 sunny days are not a useful value about power, but the for the yearly produced energy.

You can assume a thermal effiency of 65% (this varies with the geographical poisition for the incident angle and the operational temeprature).

You can say that 10MW=107Watt

Area*800*0.65=107

so: Area=19231m2

But, the land will be greater, may be 50% greater or double. This is something that varies from plant to plant.

As final conclusion, 30.000-40.000 m2 of land will be teh total area for a 10MW plant.

Thank you Evangelos A Bellos sir

Check this out Ashutosh!

http://www.nrel.gov/docs/fy13osti/56290.pdf

Dear Ashutosh,

I would recommend you to make a simple simulation of your CSP plant using the System Advisor Model (SAM), developed by NREL. This is a free software. In there, you can calculate, for specfic climate conditions which is the required area of the plant.

From preliminar work, I have calculated that a 30MWe CSP-parabolic through power plant would require 0,84 km^2 (83,7 ha). I considered wet cooling, no thermal energy storage and a back-up system based in a biomass boiler. I used an irradiation for design of 750 W/m^2.

The relation for 10 MW is not linear, take care!. Finally, of course the required area will depend of the thermal energy storage capacity, the irradiation at design, and the strategy of operation. It is possible to oversize the solar field to store more energy and dispatch only at peak hours. In that case the solar multiple and required area will be larger.

Good luck!

Rafael

Thanks Rafael, for providing precious information. SAM NREl , i will check it .

Good responses above, required area probably best achieved for that capacity with linear parabolic systems, parabolic trough or small scale heliostats. See also Solar Tracking, Sun Tracking, Sun Tracker, Solar Tracker, Follow Sun, Sun Position 

https://www.researchgate.net/publication/263128579_Solar_Tracking_Sun_Tracking_Sun_Tracker_Solar_Tracker_Follow_Sun_Sun_Position

Gerro

Book Solar Tracking, Sun Tracking, Sun Tracker, Solar Tracker, Fo...

Dear Ashutosh,

It is better for you to do a simulation instead on making assumptions based on average solar irradiance; there a plenty of software that do similar simulations of the solar system, and you can also develop your own one; I would suggest Solar Engineering of Thermal Processes, for Duffie and Beckman (as a reference.)

Furthermore, the required megawatts are thermal or electric? are you integrating your system with a power plant to generate electricity or using it to just generate steam? 

One more thing, there are tracking systems installed on parabolic collectors that changes the orientation of the collector and maximize the energy collection; installing such systems will also reduce the required area.

Hope my answer helped; in case you require addition information, please let me know.

Regards,

Patrick Kattan, LEED AP BD+C, LEED AP O+M

Mechanical Engineer and Applied Energy Specialist

Owner at Patrick Kattan Engineering Office, Lebanon

Thanks, Patrick.

A has been mentioned previously, approximately 1000 W/m^2 of direct normal irradiance falls each hour in the middle of a clear day near noon.  To be on a more realistic and conservative side, put the number at about 900 W/m^2.  With a parabolic trough, you only collect the direct normal irradiance.  The density of the installation is determined by the minimum inter-row shading that is desired.  As a crude estimate, 40% of the area is covered by the parabolic troughs.  If the efficiency is 15%, one can then estimate the size of the plant. 

Area of power plant in meters^2 is 10MW/(900W*.4*.15).

http://www.bine.info/fileadmin/content/Publikationen/Englische_Infos/themen_0213_engl_Internetx.pdf