Combien coûte 1 kWh d'électricité avec un système PV? (document en anglais)
Mémoire : Combien coûte 1 kWh d'électricité avec un système PV? (document en anglais). Recherche parmi 300 000+ dissertationsPar tonio69002 • 3 Mai 2013 • 851 Mots (4 Pages) • 1 449 Vues
Profitability and operating efficiency of PV Systems :
How much is 1kWh electricity using a PV system ?
1) Non-commercial perspective, no interest rates are considered.
Consider the total costs.
C(tot)=A0 + A1 + A2 +…+An
Where A0 ist the investment cost of the system and Ai are operating expenses like maintenance, repair, insurance,….
C(tot)= A0 + ∑(n ; i=1)Ai
n is the lifetime of the system. Typical for PV system sis 25 years.
The costs per year are : Cyear = Ctot/n
Assuming an average yield of the PV system of Eyear the production costs of 1kWh are :
C(E) = Cyear/Eyear = Ctot/ n x Eyear
Components of a PV system :
- Solar modules
- Charge controller (protection of overload and deep discharge)
- Cables
- Batteries
- Inverter (transforms 12V (DC) to 220 V (AC))
Exemple :
a) Investment for a 1kWh
PV system : A0=4500€
After 10 years we assume a repair (e.g of the inverter) A10 =1000€
Ctot=4500€+1000€=5500€
Slope= orientation of the panel => it’s the tilt angel
Energy yield can be obtained from databases.
e.g. PVGIS
Eyear =800 kWh
for the northern part of Europe.
Then the costs of energy are :
C(E)= Ctot/ n.Eyear = 5500€/25 years . 800 kWh/year
= 0,28€/kWh
The cost of 1 kWh electricity are 0,28€.
b) The same system but another région
Cf : re.jrc.ec.europa.eu/pvgis/apps4/pvest.php#
Angola = 5500/25x1620
= 5500/40500=0,14€ kWh
Sudan= 5500/25 x 1720 = 0,13€/kWh
Italy = 5500/25 x 1480 = 0,15kWh
Berlin = 0,23 €/kwh
2) Commercial perspective take into account the interest rate of the investment.
Assume an investment A0 with a period of n years and an interest rate p (in percent).
Then the capital at the end of the period is
Cn = A0 x ( 1 + p/100)^n
Cn= A0 x q^n q= 1+p/100
Example : A0= 4500€ ; n= 25 years ; p =6%
C25 = 4500€ x (1 + 6/100)^25
= 19.313€
If there are further investments needed (eg. Repair, maintenance…), they have to take into account with an interest rate as well but with a shorter period.
Example : After 10 years, 1000€ are needed for repair. For this, the remaining period is 25years – 10 years = 15 years.
The value of the capital is then :
C25 = A0.q^25 + A10. q^15
= 4500€ x (1+6/100)^25 + 1000€ . (1+ 6/100)^15
= 21710€
The investment and operating costs of 5500€ are rising to 21710€ due to the interest rate.
Generalization of the result :
Cn = A0.q^n + A1 . q^(n-1) + A2. q(n-2)…
= A0.q^n + ∑(n ; i=1)Ai. q^(n-i)
In Financial economics and for investment décisions it is common to consider the net present value by discounting all values.
The value of the capital is then :
C(0) = Cn/q^n = Cn. q^(-n)
Co = A0.q^n + A1.q^(-1) + A2.q^(-2)+….
Co = A0 + ∑(n ; i=1)Ai. q^(-i)
For constant investments Ai during the period this formula simplifies to
Co = A0 + A ∑(n ;
...