Table of Contents

### DOWNLOAD SOLAR PV SYSTEM DESIGN REPORT PDF

Get **Mechanical Engineering Project Ideas** for Engineering and Diploma. we * Mechanical Farm* provide the widest list of

**Mechanical projects/Automobile /Production engineering projects topics**to help students, researchers & engineers in their R&D. Also, we have a great variety of pre-made project kits using hydraulics, gears, energy generation systems for you to use in your projects.

### A solar PV system design can be done in four steps:

- Load estimation
- Estimation of the number of PV panels
- Estimation of the battery bank
- Cost estimation of the system.
- Base condition:2 CFLs(18 watts each),2 fans (60 watts each) for 6hrs a day.

The total energy requirement of the system (total load) i.e Total connected the load to the PV panel system = No. of units × rating of equipment = 2 × 18 + 2 × 60 = 156 watts

- Total watt-hours rating of the system= Total connected load (watts) × Operating hours = 156 × 6 = 936 watt-hours

- Actual power output of a PV panel = Peak power rating × operating factor = 40 × 0.75 = 30 watt

** The power used at the end-use is less (due to lower combined efficiency of the system**

= Actual power output of a panel × combined efficiency

= 30 × 0.81 = 24.3 watts (VA)

= 24.3 watts

** The energy produced by one 40 Wp panel in a day**

= Actual power output × 8 hours/day (peak equivalent)

= 24.3 × 8 = 194.4 watts-hour

** The number of solar panels required to satisfy the given estimated daily load :**

= (Total watt-hour rating (daily load)/(Daily energy produced by a panel)

=936/194.4 = 4.81 = 5 (round figure)

**Inverter size is to be calculated as :**

- Total connected load to PV panel system = 156 watts
- The inverter is available with a rating of 100, 200, 500 VA, etc.
- Therefore, the choice of inverter should be 200 VA.

### COST ESTIMATION OF A PV SYSTEM

(a) Cost of arrays = No. of PV modules × Cost/Module

= 5 × 8000 (for a 40 Wp panel @ Rs.200/Wp) = Rs.40000

(b) Cost of batteries = No. of Batteries × Cost/Module

=1 × 7500= Rs.700

(c) Cost of Inverter = No. of inverters × Cost/Inverter = 1 × 5000

= 1 × 5000 = Rs.5000** Total cost of system = A + B + C = 40000 + 7500 + 5000 = Rs.52500**

[Additional cost of wiring may be taken as 5% of total system cost]

#### ASSUMPTIONS TAKEN FOR DESIGN:

- Inverter converts DC into AC power with efficiency of about 90%.
- Battery voltage used for operation = 12 volts
- The combined efficiency of inverter and battery will be calculated as : combined efficiency = inverter efficiency × battery efficiency = 0.9 × 0.9 = 0.81 = 81%
- Sunlight available in a day = 8 hours/day (equivalent of peak radiation.
- Operation of lights and fan = 6 hours/day of PV panels.
- PV panel power rating = 40 Wp (Wp, meaning, watt (peak), gives only peak power output of a PV panel)
- A factor called „ operating factor‟ is used to estimate the actual output from a PV module.
- [The operating factor between 0.60 and 0.90 (implying the output power is 60 to 80% lower than rated output power) in normal operating conditions, depending on temperature, dust on module, etc.]

#### DESIGN OF BIOGAS PLANT

Biogas system design for cooking for a family of six members is considered here. The system design includes the estimation of total gas required, amount of feedstock (or dung), required and the number of animals required to have feedstock of a given amount.

#### Following assumptions are made for the design:

About 350-450 litres of biogas required per day per person for cooking (in engine about 450 liters/hp/hour of biogas is required)

Average production of dung per animal per day:

Cow– 10 kg/day

Bullock – 14 kg/day

Buffalo – 15 kg/day

Other biomass night soil, sugar cane bagasse, maize straw, etc. can also be used)

1000 litres of gas is equivalent to 1 m3 of gas.

Average gas production from dung is about 40 litres/kg of fresh dung.

Retention period of dung slurry in digester is 50 days.

AMOUNT OF GAS REQUIRED PER DAY

I. Number of family members =6 (adult). Considering 400 litre/day/person for cooking, total gas required = 6 × 400 = 2400 litres/day or about 2.4 m3 gas/day.

II. Number of animals required to fulfil daily gas requirement = Amount of gas produced from a kg of fresh dung = 40 litre/kg.

III. Total amount of dung required = Total gas required/Gas per kg of dung

= 2400/40 = 60 kg.

IV. Thus, in order to have 60 kg of dung no. of cows required = in order to have 60 kg of dung no. of cows required = 60/10 = 6 cows.

#### DESIGN OF DIGESTER AND GAS HOLDER

In order to make slurry, water should be added to equal amount of dung.

Total mass of slurry – Dung+ Water = 60+60 = 120 kg.

Volume of slurry per day (specific gravity of slurry is about 1090 kg/m3)

= Total mass of slurry/Specific gravity

= 120/1090 = 0.11 m3

Retention period of slurry = 50 days. Total volume of the digester = per day volume of slurry × retention period = 0.11 × 50 = 5.5 m3 ~ 6 m3.

Dimensions of the digester : Depth to diameter ratio should be between 1 and 1.3

Gas holder tank dimension : The volume of the gas holder tank should be about 60% of the per day gas volume.

Various possible costs incurred in manufacturing of a biogas plant is given below:

Get **Mechanical Engineering Project Ideas** for study and research. we * Mechanical Farm* provide the widest list of

**Mechanical projects/Automobile /Production engineering projects topics**to help students, researchers & engineers in their R&D. Also, we have a great variety of pre-made project kits using hydraulics, gears, energy generation systems for you to use in your projects.