aq1.1.1 costs of pv technology generations · first generation module was changed from 17% to 18%....

AQ1.1.1 COSTS OF PV TECHNOLOGY GENERATIONS

(1 point possible) (Note: due to requests from many students, to clarify the exercise and enhance the differences between the two options, the efficiency of the first generation module was changed from 17% to 18%. The correct answer to AQ1.1.3 remains the same as before).

We are going to build a PV system on a roof of 10m2, and we have two possible PV modules: (1) A first generation module with efficiency η=18% and cost of 0.80€/Wp. (2) A second generation PV module with efficiency η=10% and cost of 0.40€/Wp. The non-modular costs are 100€/m2. Note that the irradiance under standard test conditions is 1000W/m2.

What is the cost in € of implementing a PV system for the entire roof using first generation technologies?

If you’re populating the entire area, the cost is

C1 = [ (0.18)(1000 W/m2)(0.80€/Wp) + (100 €/m2) ] (10 m2) = €2,440

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(1 point possible) What is the cost in € of implementing a PV system for the entire roof using second generation technologies?

C2 = [ (1000 W/m2)(0.40€/Wp) + (100 €/m2) ] (10 m2) = €1,400

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(1 point possible) If the power demand is 700Wp, which will be the most cost-effective (or cheapest) option? (Note that for meeting the power demand, you may not need the whole roof area.) For gen 1, the cost to install a 700 W system is

C

1= (0.80 €/W) + (100 €/m2)

0.18(1000 W/m2)⎡

⎣⎢

⎤

⎦⎥(700 W) = €949

For gen 2, the cost to install a 700 W system is

C

1= (0.40 €/W) + (100 €/m2)

0.10(1000 W/m2)⎡

⎣⎢

⎤

⎦⎥(700 W) = €980

So, the first generation is slightly more cost effective.

First generation technologies Second generation technologies Both options have

exactly the same cost FINAL CHECKYOUR ANSWER SAVEYOUR ANSWER You have used 0 of 1 submissions

AQ1.2.1 RENEWABLES IN GLOBAL ELECTRICITY SUPPLY MIX

(1 point possible) The worldwide installed wind power in 2012 was 280GW. Assume that the total electricity generation worldwide is 20200TWh per year, and a capacity factor for wind power of 30%.

What percentage of the total electricity generation worldwide was covered by wind energy in 2012?

(0.280 TW)(365 × 24 hours)(0.30)20,200 TW

= 3.64%


AQ1.2.2 RENEWABLES IN GLOBAL ELECTRICITY SUPPLY MIX

(1 point possible) The same question for solar power. The worldwide installed solar power in 2012 was 102GW. Assume the same total electricity generation worldwide that was given in the previous question. Assume a capacity factor for solar power of 15%.

What percentage of the total electricity generation worldwide was covered by solar energy in 2012?

(0.102 TW)(365 × 24 hours)(0.15)20,200 TW

= 0.663%

AQ1.3.1 POWER SPECTRAL DENSITY AND PHOTON FLUX

(1 point possible) Solar simulators are used to study the performance of solar cells and modules. A solar simulator is a lamp which has to simulate the solar spectrum under standard test conditions (STC). The figure below shows the spectral power density of a solar simulator.

The spectral power density is divided in two spectral ranges:

P(λ)=4∗1015λ−1.2∗109 Wm−2m−1 for 300nm<λ<800nm P(λ)=5.2∗109−4∗1015λ Wm−2m−1 for 800nm<λ<1300nm Where the wavelength λ is expressed in meters.

As we saw in block 1.6, the irradiance can be calculated via the spectral power density as:

I=∫λ0P(λ)dλ Calculate the irradiance of the solar simulator in Wm−2 The area under the curve (i.e. the integral of P with respect to lambda) gives the irradiance:

(1.0 ×109 m W m−2 m−1) (1300 − 300) ×10−9 m⎡

⎣⎤⎦ = 1,000 W m−2

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AQ1.3.2 POWER SPECTRAL DENSITY AND PHOTON FLUX

(1 point possible) The photon flux can be calculated by integrating the spectral photon flux over a certain wavelength range:

ϕ=∫λ0Φ(λ)dλ=∫λ0P(λ)λhcdλ What is the photon flux of the solar simulator in 1021m−2s−1? (explicit integration gives a quadratic in lambda and is probably beyond the scope of this class) Taking midpoint wavelength for conversion gives

φ =λ

mid

hcP

= (800 ×10−9 m)(1000 W/m2)(6.626 ×10−34 W s)(2.998 ×109 m/s)

= 4.027 ×1021 photons s−1 m−2

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AQ1.4.1 PV POTENTIAL AROUND THE WORLD

(1 point possible) The table below shows the total area, sun hours per day and energy consumption of five different countries: the United States, India, Brazil, Spain and the United Kingdom.

Country Area (km2) Sun hours per day

Energy Consumption (MWh/year)

United States 9,826,675 4 3,886,400,000

India 3,287,263 5 959,070,000

Brazil 8,515,767 4.5 455,700,000

Spain 505,992 4.5 267,500,000

UK 243,610 2.5 344,700,000

If a PV system of 270Wp is installed in each of these locations, in which country would be best to place the panels to obtain the maximum output? Of these locations, a PV system in India would produce the most energy since India has the most sunlight hours.

United States India Brazil Spain United Kingdom - FINAL CHECKYOUR ANSWER SAVEYOUR ANSWER You have used 0 of 1 submissions


(1 point possible) Now let's take the case of the United States. How much area, as a percentage of the total area of the US, will be needed to cover the total annual energy demand of the country with solar panels having an efficiency of 15%? The available energy per square km per year for the US is

Eavail

= 1 GWkm2

⎛

⎝⎜⎞

⎠⎟356 days

year⎛

⎝⎜⎞

⎠⎟4 hours

day⎛

⎝⎜⎞

⎠⎟(9.827 ×106 km2)

= 14.35 GWh km−2 yr−1

To achieve the energy production of the US with a 15% efficient PV would require

A = 3886.4 TWh yr−1

(0.15)(0.01435 TWh km−2 yr−1)= 1.806 ×106 km2

As a fraction of the area of the US, this is

f = A

Atotal

= 1.8069.827

= 18.38%



(1 point possible) Which country needs to cover the highest percentage of their area with solar panels to supply all their electricity demand with solar energy?

We can follow the same calculation for all countries. The United Kingdom is by far the worst case requiring 4244% of the available area!



(1 point possible) Which country needs to cover the lowest percentage of their area with solar panels to supply all their electricity demand with solar energy?

Brazil, which would require only 2.55% of its area


AQ1.5.1 PV SYSTEM FOR THE SMITH FAMILY

(1 point possible) Family Smith is interested in buying a solar energy system. After some research on the Internet they have found two different systems that they are considering. System A is a PV system based on multicrystalline silicon solar cells and system B is a PV system based on amorphous silicon solar cells.

System A: The efficiency of the multicrystalline silicon module amounts to 15%. The dimensions of the solar module are 0.5m by 1.0m. Each module has an output of 75 Wp. The modules cost €60 each.

System B: The amorphous silicon solar modules have an efficiency of 6%. The dimensions of the solar modules amount to 0.5m by 1.0m. The output of each module is 30 Wp. The modules cost €20 each. The advantage of the amorphous silicon solar modules is that they perform better on cloudy days in which there is no direct sunlight. Installed in the Netherlands, this system gives, on a yearly basis, 10% more output per installed Wp than the multicrystalline silicon modules.

Both systems are grid-connected using a 900 Wp inverter. The total price of the inverter, the cables, the installation and other costs amounts to €1000. The solar modules are to be installed on a shed. The roof of the shed can only support 10 m² of solar modules.

In the Netherlands a PV system having multicrystalline silicon module generates on average 850 Wh per Wp in one year. The performance of both types of modules is guaranteed for 20 years. The price of electricity from the grid is €0.23/kWh. Assume that all the power produced by the PV system is completely consumed by the Smith family.

How much (peak) power (in Wp) can be installed for system A given the peak power of the inverter and the area available on the shed?

(1000 W/m2)(10 m2) = 10,000 W ß incident on PV (0.15)(10,000 W = 1,500 W ß possible output from unlimited models (75 W/module)(20 modules) = 1,500 W ß limit from actual modules

Although 1,500 W are possible, the power will be limited by the inverter to 900 W.



(1 point possible) How much (peak) power (in Wp) can be installed for system B given the peak power of the inverter and the area available on the shed?

(1000 W/m2)(10 m2) = 10,000 W ß incident on PV (0.06)(10,000 W = 600 W ß possible output from unlimited models (30 W/module)(20 modules) = 600 W ß limit from actual modules So 600 W are possible and will not be limited by the 900 W inverter.



(1 point possible) What is the price of a kWh of electricity (in €/kWh) generated by system A?

Cost of system: 1000 + (60 €/module)(20 modules) = €2200

Amortized cost per year: €2200 / 20 years = €110/yr

Energy per year: (850 Wh per year per Wp)(900 Wp) = 765 kWh/yr

Cost per kWh: [€110/yr] / [765 kWh/yr] = 0.14379 €/kWh



(1 point possible) What is the price of a kWh of electricity (in €/kWh) generated by system B?

Cost of system: 1000 + (20 €/module)(20 modules) = €1400

Amortized cost per year: €1400 / 20 years = €20/yr

Energy per year: (850 Wh per year per Wp)(600 Wp) = 510 kWh/yr

Cost per kWh: [€20/yr] / [510 kWh/yr] = 0.03922 €/kWh



(1 point possible) Family Smith is eligible for a municipal subsidy for sustainable energy that amounts to 15% of the initial costs of the PV system. Using this subsidy, how many years does system A need to be operational to earn the own investment back (assume that the electricity price of €0.23/kWh does not change)?

Subsidized price: (0.85)(€0.14379 / kWh) = €0.1222 / kWh

Savings per kWh: €0.23/kWh - €0.1222/kWh = €0.10778 / kWh

Savings per year: (€0.10778 / kWh)(765 kWh/yr) = €82.45 / yr

Years to recoup investment: €2200 / (€82.45 / yr) = 26.7 yr



(1 point possible) Using the same subsidy of 15% as the previous question, how many years does system B need to be operational to earn the own investment back (assume that the electricity price of €0.23/kWh does not change)?

Subsidized price: (0.85)(€0.03922 / kWh) = €0.03333 / kWh

Savings per kWh: €0.23/kWh - €0.03333/kWh= €0.19667 / kWh

Savings per year: (€0.19667 / kWh)(510 kWh/yr) = €100.30 / yr

Years to recoup investment: €1400 / (€100.30 / yr) = 14.0 yr


USEFUL CONSTANTS

Constant Symbol Value Units Speed of light in vacuum c 2.998 108 m/s

Elementary charge q 1.6 10-‐19 C Planck’s constant h 6.626 10-‐34 J s

Boltzmann’s constant k 1.38 10-‐23 J/K

USEFUL FORMULAS

Formula Speed of light in vacuum = 𝜆ν

Energy of a photon ℎ 𝐸 = ℎ𝜈 =

𝜆 Irradiance 𝜆

𝐼 = ∫ 𝑃(𝜆)𝜆 0

Air mass 1 𝐴𝑀 =

𝑜 𝑠𝜃 Spectral photon flux 𝜆

Φ(𝜆) = 𝑃(𝜆) ℎ

Photon flux 𝜆 𝜙 = ∫ Φ(𝜆)𝜆

0

aq1.1.1 costs of pv technology generations · first generation module was changed from 17% to 18%....

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