Module 8: Energy Payback Time - Mississippi State University

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CACHE Modules on Energy in the Curriculum
Energy Topic: Solar Energy
Module Title: Energy Payback Time
Module Author: Niraj Palsule and Jason Keith
Author Affiliation: Mississippi State University
References:
L., Lu, and Yang H.X. "Environmental Payback Time Analysis Of A Roof-Mounted BuildingIntegrated Photovoltaic (BIPV) System In Hong Kong." Applied Energy 87.(n.d.): 3625-3631.
ScienceDirect. Web.
Knapp, K., and T. Jester. "Empirical Investigation Of The Energy Payback Time For
Photovoltaic Modules." Solar Energy -Phoenix Arizona Then New York- 71.3 (2001): 165-172.
British Library Document Supply Centre Inside Serials & Conference Proceedings. Web.
H. C. Kim, et al. "Energy Payback And Life-Cycle CO2 Emissions Of The BOS In An
Optimized 3·5 MW PV Installation." Progress In Photovoltaics 14.2 (2006): 179-190.
Environment Complete. Web.
V.M., Fthenakis, and Kim H.C. "Photovoltaics: Life-Cycle Analyses." Solar Energy 85.Progress
in Solar Energy 1 (n.d.): 1609-1628. ScienceDirect. Web.
Key concepts: Energy of production, insolation, balance of system (BOS).
Introduction: Energy Payback Time (EPBT) is the period required for a system to generate an
amount of energy which is equal to energy used to produce the system. EPBT is one of the
common metrics used in life cycle analysis of a system. It is very difficult to get an exact
generalized EPBT since it is dependent on multiple variable factors like insolation, source of
electricity for production, method of manufacturing, design of system etc. A decade ago, the
EPBT for crystalline silicon solar PV systems was about 6-8 years. Due to rapid advancement in
manufacturing and efficiency of these systems, the production energy required has lowered down
and the energy output has increased. Some life cycle assessments now depict an EPBT as low as
0.8-1yr. The EPBT for residential roof-top installations is much higher than that of commercial
PV systems because of their lower size. In this module we would look at only silicon PV
system, since other PV systems using CIGS/CIS (CaInGaSe/CaInSe) or CdTe solar panels have
not been thoroughly studied yet.
EPBT =
Formulae:
EPBT =
Total primary energy requirement of the system throughout its life cycle
Annual primary energy generation by the system
Production energy requirement
Power rating × Insolation
EPBT =
𝐸𝑆,𝐸 + EBOS,E
Eoutput
𝐸𝑆,𝐸 = 𝐸𝑝 + 𝐸𝑠 + 𝐸𝑓 + 𝐸𝑑 + 𝐸𝑑
ES, E: Embodied energy of PV module
EBOS, E: Embodied energy of balance of system
BOS: All other components of a PV system except the PV modules.
Ep: Embodied energy in processing and purification of silicon.
Es: Embodied energy in slicing the silicon ingot to form wafers.
Ef : Embodied energy in formation of PV modules.
Et: Embodied energy in transport of all components.
Ed: Embodied energy in disposing the PV modules.
Example Problems:
1) What is the EPBT of a PV system having a total production energy requirement of 5500
kWh/kWp through its lifecycle and annual energy output of 2150 kWh/kWp?
Solution:
EPBT =
Total primary energy requirement of system throughout its life cycle
Annual primary energy generation by the system
5500
= 2150 = 2.6 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
2) The nationally accepted U.S. average conditions for assessment of PV installations are 1800
kWh/m2/yr insolation, rated module efficiency of 12.2% and a system efficiency of 80%. If a
system has a production energy requirement of 600MJ/m2, find its EPBT using these conditions.
Adapted from: H. C. Kim, et al. "Energy Payback And Life-Cycle CO2 Emissions Of The BOS
In An Optimized 3·5 MW PV Installation." Progress In Photovoltaics 14.2 (2006): 179-190.
Environment Complete. Web.
Solution:
Production energy requirement
EPBT =
=
Power rating × Insolation
600MJ
𝐽
π‘˜π‘Šβ„Ž
× 106 𝑀𝐽 × 2.78 × 10−7 𝐽
m2
kWh
2
0.122 × 0.8 × 1800 m
yr
= 0.95 years
3) 1m2 of mono-crystalline silicon solar cell contains 1.5 kg of mono-crystalline silicon. The
conversion of silica into mono-crystalline silicon requires the following steps: Conversion of
silica into MG-Si (metallurgical grade silicon), conversion of MG-Si to EG-Si (electronic grade
silicon) and conversion of EG-Si to mono-crystalline silicon. Consider that the energy required
to produce MG-Si is 21kWh/kg. The energy required to produce EG-Si is 105kWh/kg with a
89% loss of mass in the process. The energy required to make mono-crystalline silicon from this
EG-Si is 280kWh/kg with 70% loss of mass.
a) Find the embodied energy for silicon purification and processing (Ep) in units of kWh/m2.
b) After purification and processing, the silicon ingot needs to be sliced to form wafers. This
process requires 115kWh/m2 of energy (Es). Interconnecting these wafers to form PV modules
requires around 185kWh/m2 of energy (Ef). Neglecting the energy required in transportation (Et)
and end-of-life disposal(Ed), calculate the total embodied energy of PV modules (ES, E)
c) If the energy required to make its support structure for rooftop installation is 180kWh/m2, in
production of inverters would be 30kWh/m2 and for operation and maintenance is 120kWh/m2,
then find the total embodied energy of BOS. Also, find the EPBT of the entire PV system under
average U.S. conditions.
Adapted from: L., Lu, and Yang H.X. "Environmental Payback Time Analysis Of A RoofMounted Building-Integrated Photovoltaic (BIPV) System In Hong Kong." Applied Energy
87.(n.d.): 3625-3631. ScienceDirect. Web.
Solution:
1.5
a)π‘€π‘Žπ‘ π‘  π‘œπ‘“ 𝐸𝐺 − 𝑆𝑖 π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ π‘‘π‘œ π‘šπ‘Žπ‘˜π‘’ π‘šπ‘œπ‘›π‘œ − π‘π‘Ÿπ‘¦π‘ π‘‘π‘Žπ‘™π‘™π‘–π‘›π‘’ 𝑆𝑖 π‘–π‘›π‘”π‘œπ‘‘ = 0.70 π‘˜π‘” = 2.14 π‘˜π‘”
π‘€π‘Žπ‘ π‘  π‘œπ‘“ 𝑀𝐺 − 𝑆𝑖 π‘Ÿπ‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ π‘‘π‘œ π‘šπ‘Žπ‘˜π‘’ 𝐸𝐺 − 𝑆𝑖 =
2.14
π‘˜π‘” = 2.40 π‘˜π‘”
0.89
𝐸𝑝 (π‘π‘’π‘Ÿ π‘š2 ) = 21
π‘˜π‘Šβ„Ž
π‘˜π‘”
× 2.40 π‘˜π‘” + 105
π‘˜π‘Šβ„Ž
π‘˜π‘”
× 2.14 π‘˜π‘” + 280
b) 𝐸𝑆,𝐸 = 𝐸𝑝 + 𝐸𝑠 + 𝐸𝑓 + 𝐸𝑑 + 𝐸𝑑 = (695.1 + 115 + 185)
c) 𝐸𝐡𝑂𝑆,𝐸 = (180 + 30 + 120)
π‘˜π‘Šβ„Ž
π‘š2
= 330
π‘˜π‘Šβ„Ž
π‘˜π‘Šβ„Ž
π‘š2
π‘˜π‘”
× 1.5 π‘˜π‘” = 695.1
= 995.1
π‘˜π‘Šβ„Ž
π‘š2
π‘˜π‘Šβ„Ž
π‘š2
π‘˜π‘Šβ„Ž
π‘˜π‘Šβ„Ž
𝐸𝑆,𝐸 + EBOS,E (995.1 π‘š2 + 330 π‘š2 )
𝐸𝑃𝐡𝑇 =
=
kWh
Eoutput
2
0.122 × 0.8 × 1800 m
yr
= 7.5 π‘¦π‘’π‘Žπ‘Ÿπ‘ 
π‘˜π‘Šβ„Ž
π‘š2
Homework Problem Statement:
1) Find the amount of insolation for a system in kWh/m2/year if its EPBT is 1.5years,
production energy requirement is 1000MJ/m2, module efficiency is 85% and cell
efficiency is 16%.
2) Consider example problem 3. Find the percentage of mass lost in the two conversion
processes if the percentage of mass lost in converting MG-Si to EG-Si is 20% more than the
percentage of mass lost in conversion of EG-Si to mono-crystalline silicon ingot. Take Ep as
900 kWh/m2.
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