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Combien coûte 1 kWh d'électricité avec un système PV? (document en anglais)

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Par   •  3 Mai 2013  •  851 Mots (4 Pages)  •  1 449 Vues

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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 ;

...

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