4. Consider a small concentrating solar power (CSP) system consisting of a parab

Question

4. Consider a small concentrating solar power (CSP) system consisting of a parabolic dish solar collector coupled with a Stirling heat engine and electric generator. The dish-Stirling system supplies electricity for a single household. The parabolic collector concentrates solar energy onto a single receiver, heating the hot side of the heat engine generator. The average daily household electricity usage is 15 kWh. Assume the system is mounted on a two-axis heliostat enabling it to track the sun throughout the day. The average solar insolation is 220 W/m2. The capacity factor is 19% based on rated power. The solar collector has a diameter of 2.5 meters and can be modeled as a circular plane. The collector has a reflectivity of 92% , a cleanliness of 90%, and a field efficiency of 96%. The receiver efficiency is 70% and the heat transfer efficiency is 99%. The concentration ratio is 1000 which results in upper and lower cycle temperatures of 550°C and 35°C, respectively. Cycle efficiency is 70% of Carnot efficiency Assume the system has an initial investment cost of S10,000 and is purchased at an interest rate of 5% over the 15 year expected life of the system. The system is grid connected and any electricity generated and not used may be sold to the grid and purchased back at a later time at the same cost fi.e., net metering). Operations and maintenance costs for the dish-Stirling system are $0.05/kWh. Saring ongine with Recaiver and Generatof Tracking sysiem a. What is the overall efficiency of the system (%)? b· The dish-Stirling system is rated for a solar insolation of 1000 W/m2. What is the rated power of the system (kw? c. What is the average daily electrical generation by the system (kWh/day)? d. What is the levelized cost of electricity generated by the dish-Stirling solar thermal power system ($kWh)?

Explanation / Answer

This is a very simple question You just have to keep in mind few things

PART A

Here , for overall efficiency of a system always remmember that overall efficiency is the products of individual efficiency .

Hence , from given data:

Average daily household consumption =15kWh

Average solar insolation =220W/m2

Capacity factor = 19% based on rated power

diameter of collector=2.5m

Reflectivity =92%

Cleanliness=90%

Field Efficiency=96%

Reciever efficiency =70%

Heat Transfer efficiency=99%

Concentration ratio =1000

Upper temperature =550 oC

Lower Temperature=35oC

Cylinder Efficiency=70% of Carnot

carnot Efficiency =(1-Tc/Ts)x100%)

(1-35/550)x100%=93.6%

Investment=1000$

Rate =5%

time =15years

Operation and maintenance =0.05$/kWh

Overall Efficiency =Product of individual efficiency factors.

=(0.96x0.7x0.99x0.7x0.936)

=43.5%

Part B

Rated power =1000x(Area)+220x(Area)

Area=pi(2.5/2)2m2

=5.98 m2

Rated Power=1220x(5.98)kW

Rated Power=7295.6kW

Part C

Average daily electrical generation =7295.6/24=303.9kWh per day.

Part D

Total initial investment =(10000+0.05*15*10000)$

=17500$

=17500$/303.9kWh

=57.6$/kWh

Operation cost=0.05$/kWh

Levelized cost=(57.6+0.05)$/kWh

=58.1$/kWh.

Please rate my answer if you like it . But this is the best approach to this question as far as my knowledge concerned.If you need any further help or if you were expecting something else in this question . I will help you in comment section.

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