This document presents the design of a new concentrated solar power plant in Morocco. Six potential locations were considered and Ouarzazate was selected as the most suitable. A central tower configuration was chosen over a parabolic trough design. A MATLAB model was developed to size the solar field, power cycle components, and optimize the storage size and turbine capacity. The final optimized plant produces 150 MW, has 7 hours of molten salt storage, and annually produces 731 GWh at 19.3% overall efficiency. The payback period is 10 years with a levelized cost of electricity of 16.7 cents/kWh and annual CO2 emissions reductions of 403,500 tons.

1. Advanced Renewable Energy Technology
MJ2412
Concentrated Solar Power - Morocco Project
Authors:
Mattia Beretta
Camilo Diaz
Hammad Farrukh
Sunay Gupta
Rachit Kansal
Timothy Mulé
Candace Shaw
Supervisors:
Professor Andrew Martin
Lukas Aichmayer
Rafael Guédez
Monika Topel
Jorge Garrido
Monica Arnaudo
[1]
March 7, 2016

2. Abstract
In order to increase the share of renewables in the energy mix of Morocco, a new CSP power plant design proposal
was presented in this report, based on a program developed in MATLAB.
In total, six locations were considered for siting and after evaluating them using certain viability factors, Ouarza-
zate was chosen as the most suitable location. Using the irradiation data from Solar Radiation Data (SODA) for
Ouarzazate in 2005, a central tower configuration was chosen after a performance evaluation against a parabolic
trough configuration, using The United State’s National Renewable Energy Lab (NREL) System Advisor Model
(SAM) software.
MATLAB was then used to dimension the solar field, thermal power at the receiver, the power cycle and the total
electricity output, taking into account several factors and efficiencies.
Then, an optimization tool was developed in order to determine the ideal combination of storage size and turbine
capacity, considering both technical and economic performance. Transient operations were also modeled and
evaluated and associated losses were included.
The final optimized plant produces 150 MW of power through a steam turbine, with seven hours of two-tank,
molten salt storage. The plant annually produces 731 GWh of electricity, at an overall efficiency of 19.3%. The
payback period period of the plant is 10 years, with a levelized cost of electricity of 16.7
¢
kWh . The power plant
also ends up reducing annual carbon dioxide emissions by 403,500 tons.
Key Words: concentrated solar power, storage, solar tower, Morocco, Ouarzazate
1

3. Contents
List of Figures 3
List of Tables 3
1 Introduction 4
2 Selection Process 4
2.1 Location Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Technology Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 Technical Analysis 7
3.1 Gathering Meteorological Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.1.1 Solar Field Design and Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.1.2 Solar Multiple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Calculating Power Output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2.1 Choosing Thermal Storage or Hybridization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2.2 Optimization - Choosing the Right Combination of Turbines, Storage, and Hybridization . . . . . . . . 11
3.2.3 Justification of Hybridization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.3 Power Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.3.1 Heat Exchanger (Steam Generator) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3.2 Turbines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3.3 Condenser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3.4 Preheating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3.5 Feed Water Tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.3.6 Efficiency of the Power Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3.7 Operative Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3.8 Transients Considered in the CSP Plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3.9 Daily Operation on a Typical Day . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3.9.1 Phase 1: Startup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.3.9.2 Phase 2: Full Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3.9.3 Phase 3: Operation After Sunset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3.9.4 Shutdown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.3.9.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.3.10 Limitations on Receiver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4 Transients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.4.1 Impacts of Transients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.4.2 Tower Receiver Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.4.3 Operation of Plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.4.4 Energy Needed for Molten Salt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.4.5 Transients Energy Calculation for Turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4 The Algorithm Logic 24
4.1 Algorithm Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
5 Overall Performance of the Plant 29
6 Financial Analysis 30
6.1 Capital Expenditures (CAPEX) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
6.2 Operational Expenditures (OPEX) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
6.3 Tariffs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
7 Environmental Impacts 33
8 Conclusion 34
9 References 35
Appendices 39
2

4. List of Figures
1 Morocco GIS City Selection: Railways, Rivers, Electric Lines, & Lakes . . . . . . . . . . . . . . . . . 5
2 Morocco GIS City Selection: Roads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Parabolic Trough . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4 Solar Tower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
5 Block Diagram of Solar Field Design and Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
6 Solar Multiples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
7 Power Block Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
8 Receiver Temperature vs. Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
9 Turbine Start-Up Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
10 Turbine Ramp Down Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
11 Algorithm Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
12 Annual Revenue vs. Storage Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
13 Annual Operating Costs vs. Storage Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
14 Hybridization Amount vs. Storage Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
15 Capital Costs vs. Storage Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
16 Payback Period vs. Storage Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
17 Annual Heat Wasted vs. Storage Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
18 CAPEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
19 OPEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
20 Avoided Emissions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
List of Tables
1 Location Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Technology Selection from SAM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Variable Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4 Solar Field Loss Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
5 Variable Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
6 Tariff Structure for CSP Plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
7 Tariff Scheme for Various Moroccan Cities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
8 Power Cycle Efficiencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
9 Representative Day of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
10 Variable Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
11 Receiver Temperature vs Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
12 Total CSP Plant Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
13 Cost Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
14 CAPEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
15 OPEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
16 Equipment List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
17 Variable Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
18 Final Tariff Structure for CSP Plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3

5. 1 Introduction
With growing global concern for the environment, as well as a strong desire to shift away from foreign energy
dependance, Morocco has recently been increasing its availability of concentrated solar power (CSP) electricity
production. Morocco currently imports fossil fuels to satisfy 97% of its energy needs and the introduction of
new CSP plants in the country has the potential to greatly help the western kingdom on a course for energy
independence, and to set off a movement of harnessing the vast solar resource throughout other countries in
Northern Africa [2].
Long term plans have taken into account the idea of transmitting the excess power generated from CSP in Morocco
throughout Northern Africa and eventually into the European Union. With this potential source of renewable
power for Europe and figures from the International Energy Agency estimating that 11% of the world’s electricity
could come from CSP by 2050, organizations such as The World Bank, The African Development Bank, The Euro-
pean Investment Bank, as well as private investors are all getting involved with the Western Kingdom’s ambitious
CSP plans [3].
Different possible locations around Morocco and various forms of CSP technologies can be suitable for the devel-
opment of future plants. This report details a selection process that was undertaken to establish the best site and
mechanism for a new CSP plant to be installed. A full financial and power generation investigation was conducted
as well and can be seen in detail in the following sections.
2 Selection Process
In order to move forward with a project for a new CSP plant in Morocco it was necessary to determine both the
ideal location in the country, as well as the preferred CSP technology. Various methods were used to eventually
determine that Ouarzazate, Morocco was the best location for a new plant and that a tower configuration would be
the best technology to implement. The details of the selection processes can be seen in the following subsections.
2.1 Location Selection
The location selection for the CSP plant was done considering the parameters shown in Table 1. Each parameter
was given a certain weightage on a 1 to 5 scale by the analyzing team, and that parameter was assessed for the
location. Geographic Information Systems (GIS) as well as satellite imaging was used to determine the rankings
for each of the criteria. ArcGIS software was utilized for analyzing areas of high solar irradiation, road, railway
and grid infrastructure locations, water resource locations, and distance to city centers. Figure 1 is a GIS map
combining layers of datasets listed in Table 1, except for road infrastructure which had to be on a separate map
for simplicity of reading which is shown in Figure 2.
4

6. Figure 1: Morocco GIS City Selection: Railways,
Rivers, Electric Lines, & Lakes Figure 2: Morocco GIS City Selection: Roads
Solar irradiation was given the highest weightage considering the fact that it is the most important parameter
for a solar power plant because it is the primary resource for the plant. Access to water and grid accessibility
were given a weightage of 4 as the second most important criteria because they involve large infrastructure costs.
Road and railroad accessibility and distance to the consumption centers were weighted 3 and 1 respectively, the
reason being the importance of the road and rail access to facilitate the construction and operation of the plant.
The topography and the proximity to other CSP plants were considered to be least important because they have
minimal impact of the plant. The topography only affect the design process, but that impact is also small. The
table below details the location selection process:
Criteria
W
eightage
SebkhatTah
O
uarzazate
A
in
BeniM
athar
Foum
A
lO
uad
Boujdour
M
idelt
Solar Irradiation 5 3 4 2 3 3 3
Access to Water (Rivers, Lakes) 4 2 5 4 1 1 4
Grid Accessibility (from the map) 4 2 5 5 3 5 5
Road Accessibility (from the map) 3 2 4 5 3 5 5
Proximity to Existing CSP Plants 1 0 5 5 2 0 0
Distance to Consumption Centers 1 4 4 3 4 5 3
Topography (Flat or Mountainous) 1 4 4 3 4 5 3
Total - 43 87 77 48 53 77
Table 1: Location Selection
Based on the selection process in Table 1, Ouarzazate was decided to be the optimal location for a new CSP plant.
5

7. 2.2 Technology Selection
Two major technologies were compared for use in the new CSP plant: central tower receiver and parabolic trough.
Examples of each are shown in Figure 3 and 4 respectively.
Figure 3: Parabolic Trough [4] Figure 4: Solar Tower [5]
Both technologies have their share of advantages and disadvantages. In order to determine which technology
would result in the best plant layout a simulation software for the design of CSP plants was used. The United
State’s National Renewable Energy Lab (NREL) has a System Advisor Model (SAM) which was used as starting
point for the project. General reference data was used as the input into SAM which included cost, material, effi-
ciencies and other useful parameters for each technology. The simulation on SAM was run aiming at a generation
target of at least 110 megawatts of electricty (MWe) while in production. The results were evaluated on an eco-
nomic basis, so the cheaper technology able to provide the desired output was chosen. The results are seen in
Table 2, and it was apparent that tower technology was the optimal choice.
6

8. Tower Trough
Metric Value Metric Value
Annual Energy 334,093,120 kWh Annual Energy 225,247,153 kWh
Capacity Factor 38.10% Capacity Factor 25.70%
PPA Price (Year 1) 15.93 ¢/kWh PPA price (Year 1) 35.70 ¢/kWh
Levelized PPA Price (nominal) 20.82 ¢/kWh Levelized PPA Price (nominal) 48.66 ¢/kWh
Levelized COE (nominal) 19.45 ¢/kWh Levelized COE (nominal) 45.19 ¢/kWh
Net Present Value $48,216.984 Net Present Value $82,499,200
Internal Rate of Return (IRR) 10.00% Internal Rate of Return (IRR) 10.00%
Year IRR is Achieved 20 Year IRR is Achieved 20
IRR at End of Analysis Period 11.75% IRR at End of Analysis Period 11.75%
Net Capital Cost $847,133,536 Net Capital Cost $1,441,399,424
Equity $431,258,528 Equity $734,937,664
Size of Debt $415,855,008 Size of Debt $706,461,760
Table 2: Technology Selection from SAM
Compared to parabolic trough, tower technology is able to achieve a 148% higher annual energy and able to
provide a higher capacity factor of 38.10%. Purchase Price Allocation (PPA) and net capital cost were the other two
main parameters influencing the selection of the technology. Tower again showed better performances, granting
a 66% lower PPA price and almost half of the net capital cost.
3 Technical Analysis
Once it was determined that solar tower technology was the ideal solution, it was necessary to gather meteoro-
logical data for Ouarzazate to be used in the technical analysis. This data was then fed into a Matlab script to
calculate the generating output of the plant. Detailed explanations of these processes can be seen in the following
subsections.
3.1 Gathering Meteorological Data
In a Typical Meteorological Year (TMY), data is provided in terms of P-measurements, usually P50, P70, or P90 in
order to provide a criteria to judge the reliability of the solar resource. TMY values of P50 mean that there was
at least a 50% probability that the solar resource actually exceeded the value that was reported. Similarly, values
P70 and P90 TMY data mean that the probability of exceeding the reported value are 70% and 90% respectively.
Most financial backing for CSP plants require either P70 or P90 irradiation values, to ensure that there is at least
a 70% or 90% probability that the true values will exceed the given base. Incorporating these P-measurements is
very beneficial when attempting to forecast future supplies of the solar resource for a given CSP plant [7].
For the purposes of this report, the initial meteorological data used for Ouarzazate was historic data that was
gathered from SoDa-IS, therefore the P-measurements did not come into play. The data acquired was plane nor-
mal to the sun with direct beam solar irradiation. The period that was examined was from January 1, 2005 to
December 31, 2005, and this data that was based on an hourly time series. [6]. Using historic data was beneficial
as it showed what the area had already been exposed to in the past, and could therefore give the high level of
detail needed to help project the financial and technical requirements of the plant.
7

9. 3.1.1 Solar Field Design and Simulation
The total land available for the project was set to 579 hectares, or 5,790,000m2. Of this total 904,355m2 was the
suggested mirror area. Part of the total available land had to set aside to account for the land requirements of the
tower receiver, storage system, and power block; which was calculated to be 4,531.7m2. The first ring of mirrors
was then placed 38 meters from the center point of the field. The solar field design was made using a Matlab script
which was able to divide the available land into different cells. The script could then evaluate the performance of
each cell in the field.
The field was set to have a circular shape and was divided in 12 equidistant radial and azimuthal regions. This
division of the field resulted in a total of 144 cells in the field. This was considered adequate as it helped reduced
the computational time required to describe the solar field, but was still able to provide accurate results of how
the field operated.
Each cell was assigned with a characteristic heliostat density, which decreased as the distance from the solar re-
ceiver grew larger. This was because mirrors located in the outer rings have lower performance compared to those
located in the inner rings, so it was not beneficial to have a very high mirror density in the outskirts of the field.
A known expression for calculating the heliostat density in a solar tower configuration can be seen below [16]:
ρF = 0.721 ∗ exp(−0.29
rH
hT
) + 0.03
Where:
Variable Description Units
ρF Heliostat Density
m2
helio
m2
land
rH Radial Position of the Heliostat m
hT Height of the Tower m
Table 3: Variable Definitions
However, when using this formula the total mirror area of the field was considerably lower than the reference
value that was mentioned in the project description for this CSP plant design. In order to determine a heliostat
density that was comparable with the reference plant, a Matlab code was developed to iterate through different
densities for each circular ring in the field. A line of best fit was developed for the initial part of the exponential
function of the given formula, and the iterations in Matlab were executed by applying the slope of this line. The
results were adjusted to resemble the heliostat densities and field area of the given reference plant.
After dividing the solar field in cells, it was necessary to define the performance of a representative heliostat for
each cell. In order to accomplish this, a series of loss parameters such as surface & cosine effectiveness, shadowing,
blocking, attenuation, and spillage factor were all taken into account. These factors are reported in Table 4
8

10. Variable Value
Surface Effectiveness 0.92 ∗ 0.966 = 0.88872 [17]
Shadowing Factor 0.05 [16]
Blockage Factor 0.05 [16]
Table 4: Solar Field Loss Parameters
The calculation of the attenuation factor was dependent on the distance of the heliostats from the central receiver.
Due to this varying parameter the attenuation losses could not be given as an overall factor as the previous losses.
Therefore, the estimation of the attenuation losses was done using the following formulas [19]:
AttenuationFactor = 0.99326 − 0.1046 ∗ S + 0.017 ∗ S2
− 0.002845 ∗ S3
S = r2
cell + (htower − hheliostat)2
Where:
Variable Description Units
r2
cell Radial Distance of the Cell m
htower Height of the Solar Tower m
hheliostat Height of the Heliostat Pedestal m
S Minimum Distance of the Representing Heliostat
from the Receiver Tower
m
Table 5: Variable Definitions
The heliostat height was assumed to be 6 meters, whereas the tower height was given in the project description
and it was equal to 163.2 meters [18, 17].
The next step in the design process was the importation of an excel file containing hourly incoming solar radiation
for each day of the year 2005 and all of the angles required to calculate the solar azimuth and zenith angles. The
solar azimuth and zenith angles were then further used to define the vectors vs and vt. The vector vs describes
the solar position and vt is the vector between the mirror pivot and the center of the receiver. Using vs and vt, it
was possible to define nh, the normal vector to the pivot of the mirrors. nh multiplied by vt resulted in the cosine
effectiveness of the cell.
After calculating all of the data about the solar radiation and the relevant angles, it was possible to run the sim-
ulation for the entire solar field during a year of operation. The net electrical power output of the plant for each
hour of the year was then calculated using the efficiencies of receiver, thermal cycle and the generator. In order
to optimize the storage system, a different type of output was created which was not the electrical output power,
but the thermal power. This was done in order to work more easily with the thermal energy storage systems in
the plant. A block diagram showing the solar field design and simulation process can be seen in Figure 5.
9

11. Figure 5: Block Diagram of Solar Field Design and Simulation
3.1.2 Solar Multiple
The solar multiple of the plant was calculated to be 1.83 on summer solstice (21st June). In order to have a com-
parative view of the plant around the year, solar multiple was also calculated for other days of the year as shown
in Figure 6. The meteorological data revealed that there was a lack of solar irradiation on the winter solstice, most
likely due to cloud cover. Because of this the solar multiple was calculated for 24th of December for comparison.
Figure 6: Solar Multiples
3.2 Calculating Power Output
Since the output of the CSP plant is dependent on the Sun, it means that it varies with the rise and fall of the
Sun through the day and the year. However, the minimum requirement from the grid is always 110 MW for all
the hours the plant is operating during. If the plant were to not employ any storage or hybridization and solely
10

12. rely on the field’s output, it would not be able to meet the requirement for 2,775 hours in a year, or the equivalent
of 115 days. Hence, there is a clear need for either storage or hybridization to meet the grid requirement for the
duration of plant operation.
3.2.1 Choosing Thermal Storage or Hybridization
There is now a choice between storage or hybridization. Storage in this case would entail a two-tank, molten ni-
trate salt system with a cold tank at 290oC and a hot tank at 565oC. The storage would be daily, meaning that it
would retain heat for a maximum of 10-12 hours and would be sized accordingly. At times when the production
from the solar field far exceeds the requirements from the grid, the extra heat would be used to heat the salt. Then,
at times when the production falls short, the same heat would be used to cover the supply shortfall and meet the
demand requirement.
The hybridization option involves burning natural gas for all the times the plant is not able to produce enough
electricity to meet the minimum requirement. It does not involve utilizing the excess energy produced by the
field, in any way.
Given the above description, it should not be surprising that with just a hybridization option, 761.4 GWh of elec-
tricity would be wasted in a year, an average of 2.09 GWh per day. With 6 hours of storage and the same turbine
specifications, 19.7 GWh of electricity could be saved, averaging 54 MWh per day.
Moreover, environmental stewardship was one of the main goals of the power plant design, which is why associ-
ated carbon dioxide emissions were an important consideration in making the decision. Given how much energy
is saved via storage, and how it is carbon-neutral in its operation, storage was chosen as the main driver for the
plant. Hybridization was included as a back-up option, for very cloudy days or weeks, where neither the plant
nor the storage would be enough.
3.2.2 Optimization - Choosing the Right Combination of Turbines, Storage, and Hybridization
The main requirement for this plant was to provide a minimum of 110 MW of electric power output during all
hours of operation. The hours of operation for the plant were not defined or constrained, but there were certain
hours where it was more favorable to produce electricity than others. Table 6 below shows the rate structure for
this plant.
Clock Time CSP Tariff ( USD
MWh )
00:00 - 05:00 0
05:00 - 17:00 X (Base Price)
17:00 - 21:00 αX (Where 1<α<3)
21:00 - 22:00 X
22:00 - 24:00 0
Table 6: Tariff Structure for CSP Plant
As can be seen from Table 6, it was favorable to produce electricity anytime between 5:00 and 22:00. It was particu-
larly profitable to produce between 17:00 and 19:00. The base price for concentrated solar power plants was higher
than the grid electricity price, based on the country’s renewable energy policies. It was found to be $140/MWh
11

13. of electricity [23].
To find α, the multiplier for the time period of 17:00 to 21:00, current Moroccan time-of-use grid policies were
examined. While no policies were found for the city of Ouarzazate, policies were found for various Moroccan
cities, including its capital Rabat as seen in Table 7.
Region Ratio of Mid-Peak/Off-Peak Ratio of Peak/ Off-Peak
Casablanca 1.532 1.521
Rabat 1.526 1.517
Tanger 1.437 1.452
Table 7: Tariff Scheme for Various Moroccan Cities [24]
Here, off-peak hours range from 22:00 to 7:00, mid-peak hours range from 7:00 to 17:00 and peak hours range
from 17:00 to 22:00. From Table 6, it can be seen that the multiplier α ranges from 17:00 to 21:00 and is a multiple
of the base price, which ranges from 5:00 to 17:00. From the above information, it was clear that the ratio needed
to calculate α was the ratio of peak power to mid-peak power. The average of the ratios from the three cities was
taken and was found to be 1.497.
Since the only constraint was to produce a minimum of 110 MW and the main objective was to maximize revenue
and minimize payback period, the turbine capacity and the size of the storage were both taken as variables. To
carry out the optimization, a range of possible values for both variables needed to be created, which would then
be iterated to give the best possible result.
For determining the possible turbine capacities, different turbine manufacturers were considered. Siemens, a
worldwide leading turbine manufacturer, was finally chosen to ensure the highest quality for the price paid. The
major Siemens turbines that were examined were the SST-600 series of 150 MW capacity, the SST-700 series of 175
MW capacity, and SST-900 series of 250 MW capacity [25, 26, 27]. Hence, 150, 175 and 250 MW formed the range
of possible turbine sizes.
From the tariff scheme in Table 6, it was determined that the size of the thermal storage could range between 1
to 10 hours. This was aimed at capturing the maximum and minimum possible storage sizes that would be eco-
nomically feasible. A case with no storage whatsoever was also included for comparison purposes.
From the earlier evaluation of the power produced by the solar field, a table was created which detailed the amount
of thermal power generated in each of the 144 cells of the field, along with the exact hour in which this production
were to happen. This table was then compiled to provide the hourly data of the total thermal power production
of the entire plant, for the whole year.
Before the start of the optimization, certain variables and costs were defined, to allow for the financial parameters
of each combination to be determined. First, the thermal power requirement corresponding to a minimum elec-
trical power output of 110 MW, was defined. Based on the power cycle efficiency defined earlier in the report, this
thermal power was found to be 427 MW. Also, the base price of electricity was found to be $140/MWh and the
multiplier, α, was 1.4997 as previously mentioned. The tariff scheme followed is the one in Table 6.
12

14. 3.2.3 Justification of Hybridization
For the choice made using the optimization, hybridization was used for 260 hours in the whole year, which is
equivalent to approximately 11 days. From and economic standpoint, hybridization is not justified as it costs 6.95
million USD (including capital and operation costs) and brings in revenue of 2.61 million USD.
However, the true justification depends on the power plant’s and grid’s policy. If the grid and/or the plant decides
to guarantee a power output every day of the year no matter what, hybridization is definitely justified, as it allows
for that to happen at a reasonable cost. If, for example, the grid allows the plant to produce electricity whenever
it likes and vice versa, it would work in favor of the plant to not employ hybridization.
3.3 Power Block
The power cycle used for the CSP plant was a steam Rankine cycle. After an analysis (details attached in the ap-
pendix) of the requirements to get a minimal power output of 110 MWe, the Siemens SST600 turbine was proposed
for the plant and the following layout was developed.
Figure 7: Power Block Design
As shown in the Figure 7, the power cycle will run via the heat given by the molten salts. This heat will evaporate
and superheat the water, and then this superheated vapor will run through two turbine stages to allow for a
reheating process and improved cycle efficiency. The cycle is capable of producing 150MW of electric power. The
thermal power input required for the cycle is 428MW in order to produce 110MW of electricity.
The overall efficiency of the power cycle was calculated to be 27% after assessing the efficiencies of the different
stages in the process. Each step of the power cycle is explained below.
13

15. 3.3.1 Heat Exchanger (Steam Generator)
At the steam generator, the heat from the molten salt will be transferred to the water in the power cycle. The
efficiency of the heat exchanger is 75%, with reference to a cross flow heat exchanger used in CSP plants. This
heat transfer process usually occurs in different stages. For simplifying the design, it was assumed that there
would only be a single heat exchanger, as shown in the layout. There will also be hybridization in the form of
an auxiliary natural gas boiler that can provide heat to the heat exchanger in the event that there is not enough
molten salt to keep up electricity production. After the molten salt passes through the steam generator it will
return at 260oC to the low temperature thermal storage tank. The now vaporized steam will then continue on
through the turbines in the system.
3.3.2 Turbines
The turbine consists of high pressure (HP) and low pressure (LP) stages connected with a common generator
shaft. The layout was designed so that the steam input to the HP stage of the turbine is at 160 bar and 560oC and
the expansion ends at 30 bar. The isentropic efficiency of high pressure turbine stage was assumed to be 85% [11].
The steam then goes to reheating after the HP stage, and heated up to 560oC. This steam then flows into the LP
stage of the turbine and the second expansion will occur, with an assumed isentropic efficiency of 80% [11]. The
exit of this LP turbine will be at 0.15 bar, and then the condensation process will start. At this LP turbine, there are
four steam extractions, the first one will go the feed water tank, and the next three will be used in three preheaters
for the working fluid after the condenser.
3.3.3 Condenser
After the low pressure turbine, the working fluid will be at 60oC and 0.15 bar. These conditions were selected to
allow the condensation with dry cooling. Dry cooling is being used because of the need to reduce the total water
consumption of the plant. Although Ouarzazate is not as dry as other areas in Morocco, there is a competition for
water with near population centers, so it is ideal to reduce the water consumption of the plant . At the condenser,
the mass flow from all the three extractions for the preheaters is added back together.
3.3.4 Preheating
After the condenser, the working fluid is saturated liquid water at 54oC. After this point the water will pass
through the 3 preheaters and heat will be transferred from the exit flows of the LP turbine to the working fluid.
The reason for choosing three preheaters for the water was to reduce the boiler load, or the amount of heat needed
for the steam to reach the requirements needed for the first pass through the HP turbine. Extracting some steam
from the turbines will mean less power output from the turbine but the gains in reducing the boiler load (in this
case less heat needed from the salts for unit of electrical power output) will increase the overall efficiency of the
cycle. As a tradeoff, for any additional preheater, the increase in efficiency becomes lower, so after analyzing the
power output needed, the decision for an optimal performance was to install 3 preheaters.
3.3.5 Feed Water Tank
The feed water tank is essential to the power cycle in order to keep operating the cycle at controlled conditions.
Also, it is needed as deaerator (in order to get rid of the gases that may be present in the fluid before going to the
heat exchanger i.e. oxygen from air in the condensation process. The feed water tank operates at 6 bar and at this
stage the inlet water from the preheaters will be mixed with the first steam extraction at the low pressure turbine.
At the feed water tank exit, all the water will be saturated liquid at this pressure.
14

16. 3.3.6 Efficiency of the Power Cycle
The overall efficiencies are summarized in the next chart:
Component Efficiency
Heat Exchanger (Steam Generator) Efficiency 75% [12]
Isentropic Efficiency of High Pressure Turbine 85% [11]
Isentropic Efficiency of Low Pressure Turbine 80% [11]
Mechanical Efficiency of each Turbine 95% [11]
Electrical Efficiency of the Generator 97% [11]
Overall Efficiency of the Cycle 27%
Table 8: Power Cycle Efficiencies
These efficiencies were used during the calculations of the electrical power output from the Rankine cycle used
in the CSP plant.
3.3.7 Operative Strategies
The proposed CSP plant, in order to be as profitable as possible, will produce electricity during the times where
the price is higher (higher demand) whenever is possible. In order to do that, storage is a key part of the plant.
The storage will allow to keep producing after the sunset and reduce the impacts of transients.
All the considerations and guidelines for optimal storage in the CSP plant are presented in the technical analysis.
3.3.8 Transients Considered in the CSP Plant
For the proposed CSP plant, there will be basically transients related with:
• Everyday a startup and shut down process due to the fact the plant is producing only when the energy is
paid.
• Transients due to clouds or any other meteorological event.
There is a clear difference between the first and second case. The first one can be analyzed and optimized, as well
that operational issues can be solved, in summary it is a controlled process. Also, these processes can be divided
into transients at the solar field and transients in the power cycle.
For the second case, however, it is quite more complicated to solve the issues related with them. Moreover, there
is uncertainty about when exactly they will occur, and to what extent (i. e. a cloud event that will affect for an
hour half of the field and for half an hour the other side). The transients associated with these events will have a
significant impact, so for example, the power output will not be constant. Considering all these potential impacts,
storage will allow to minimize their impact.
In the daily operation of the CSP plant, we focuse in the daily startup and shut down process. The steps of the
process will not change: although the lenght of the day is different each day, both storage and alternative boiler
will allow to produce full power when the energy is paid.
The startup of the plant consists of:
15

17. • Feeding the power cycle. With Molten salts from the hot salt tank (this process will not be always available,
and will differ on time, since the total amount of heat stored will vary according with the day) or with the
heat from the alternative boiler. The losses associated with the turbine are presented in the transient impacts
part.
• Tracking the heliostats to the initial position, so the heat can be transfer to the receiver as soon as posible
after the sunrise time.
• Heating up the receiver and pipes. Once the heliostats receive solar irradiation and heat up the receiver,
temperature of both pipes and receiver will increase in order to allow the salt being heated up and feed the
hot tank (and consequently, to run the power cycle).
The shutdown process consist of:
• Shut down the turbine. Once it is decided to shut down the turbine, the power output will be reduced
rapidly. The shut down process of the turbine is also 40 minutes. After the turbine is shut down, its tem-
perature will decrease, so a cool down process will take place. The turbine needs to be kept at a minimum
temperature, therefore there is a minimum steam flow requirement.
• Cooling down of solar field. The heliostats, receiver and pipes will have a surface temperature drop after
the sunset.
As it has been decided that the startup will happen everyday, the shut down process becomes important in terms
of the lowest temperature reached by the receiver.
If the shutdown times of the plant are too long, the amount of heat losses in the system will be very high. However,
for the plant in question, the shutdown time is only for single nights. Therefore all of the startups are assumed to
be hot startups because there is no excessive heat loss during one night.
It should be considered that the transients of startup and shut down processes have impacts both in time and
energy. In terms of time, there will be 90 minutes where the turbine will be working and not producing at full
capacity. In terms of energy, the startup and shut down will require energy that will come from the storage and
therefore is an energy loss. The impacts in terms of energy loss are explained in the transients impact section.
3.3.9 Daily Operation on a Typical Day
Since electricity sold throughout the day will only generate revenue for certain times, it will be necessary to plan
for a startup and a shut down process in the plant everyday. In order to see the details of how this will happen, a
specific day was chosen to show how the plant will operate. The details of this day can be seen in Table 9.
Representative Day June 22nd
Length of the Day 14.007 hours
Sunrise Time 4:53
Sunset Time 18:53
Table 9: Representative Day of Operation
3.3.9.1 Phase 1: Startup
The proposed configuration has storage capabilities and an additional gas fired boiler that will allow the cycle
to run in the event that the thermal storage is not enough to reach the minimum. With these available options
16

18. other than the power directly from the solar field, the initial time for full production is set at 5:00. For the day
in question, there is thermal storage in the hot salt tank which can be used for the startup of the turbine at this time.
The total startup time for the plant is set at 45 minutes. The most significant portion of this time is devoted to the
startup process of the turbine. The turbine startup time was determined to be 40 minutes 1. The rest of the time
is left for the process of heating the pipes from the hot tank to the heat exchanger (steam generator) to run the
power cycle. It is important to note that if there was no remaining thermal storage at the beginning of the day, the
startup time would change.
Therefore, the the warm-up operations will start at 4:15 to be on schedule to produce full power at 5:00. There
is no day of the year where there is enough solar irradiance at 4:15 to run the cycle without taking heat from
the hot salt tank, or from the gas fired boiler. Regarding the solar field, the receiver and pipes along the central
tower require a certain amount of time to reach the minimum temperature and to allow the salt to be heated up.
However, this heating process of the receiver will run after the turbine startup process and it will not reduce the
power output.
Due to the tracking systems of the heliostats in the solar field, it can be assumed that the warm-up process of
the solar receiver will not take a substantial amount of time because all available heat will be concentrated to this
effort. The initial temperature can be either the ambient temperature or higher 2. Once the receiver has reached a
higher temperature than the cold tank (which is at 290oC), the salt can be pumped to the receiver for the heating
process. However, the salt needs to reach 560oC before it can run the power cycle. For this representative day, the
minimum required temperature of the receiver to run the cycle, should be achieved by 8:00.
3.3.9.2 Phase 2: Full Operation
Once the receiver temperature has reached 600oC with the incoming solar irradiance, the molten salt flow from
the cold to the hot tank will first pass through the tower and receiver. The flow will then travel from the hot
thermal tank to the power cycle. By keeping the mass flows equivalent, this can be done without incurring any
losses in the amount of heat stored. In the event that more thermal energy is coming in than is necessary to meet
the electricity demand, then the excess can build up in the thermal storage tank. When the electricity is produced
with the energy collected at the solar field, the plant is in its optimal performance. If there is more energy coming
from the solar field and the storage is full, the excess energy will be wasted. In the Section 3.4 it can be seen that
the thermal power at the receiver was enough to store a significant amount of energy in the hot tank.
As a last step into this process, it should be seen that since the maximum irradiance to the receiver is reached in
the solar noon, and after that the irradiance will be reduced, there will be a certain point where the irradiance
may not be enough to keep the receiver at the necessary temperature.
3.3.9.3 Phase 3: Operation After Sunset
The time at which the sun sets will define the time when the energy produced will come from the storage instead
of the solar field. If the storage is exhausted, then the auxillary boiler will produce the necessary steam to keep
power cycle in operation.
For June 22nd, the sunset happens at 18:53. From this time onward, another transient phase for the receiver will
start. Once the incoming solar irradiance on the field is no longer enough to heat up the salt to the necessary
temperature, it becomes pointless to keep the heliostats focused on the receiver. Therefore, the heliostats will be
1
See Section 3.4 for more information
2
See Section 3.4 for more details
17

19. defocused in order to be ready for the next day. The receiver then needs to be drained (it still has some salt inside)
and this salt will be directed to the cold tank. At this time, stored heat from the hot tank will feed the cycle. For
this representative day, there was no need to operate the auxillary boiler.
3.3.9.4 Shutdown
The shut down time for the plant is defined by an automatic shut down process at 22:00 everyday. At 22:00, there
will be no more solar irradiance at any point in the year. For this particular day, the energy needed to produce the
steam at the power cycle comes from the storage reserves. The steam flow to the turbine eventually stops and the
power produced by the turbine drops. A cool down process of the turbine will subsequently start, and the turbine
will have a minimum steam flow requirement throughout the night that will allow to start operations next day.
3.3.9.5 Summary
In summary, the operational strategy follows these guidelines:
• The storage was set according with an optimization method, that accounts for both the energy needs of
the cycle, and the cost and profitability of the storage. Storage will minimize the impacts on electricity
generation due to meteorological events, and will allow production when the solar resource is not enough
(and therefore will do a much better job of controlling the daily startup process).
• Electricity will be produced only when it is profitable to do so.
• Preventive maintenance can be done during the time intervals where there is no energy production.
• Since the higher selling price of electricity is from 17:00 to 21:00, the optimal performance will be reached if
the plant is able to deliver full power at these times.
• Hybridization is considered as an option in order to produce when there is not enough solar resource and
there is also a storage deficiency. According to the optimization method developed and explained in Sec-
tion 3.2.2, there will be 18 days during the year where this alternative boiler is used to reach the full power
output.
• Startup and shut down processes will impact the energy losses throughout the year. Also, due to the daily
startups, the storage optimization becomes a crucial process. And since there is a fixed time for these tran-
sients, operation can be planned in a much better way.
3.3.10 Limitations on Receiver
The receiver had some physical constraints that reduced the maximum amount of solar radiation that could be
harnessed from the solar field. The maximum heat flux was equal to 1.5 MWth
m2 [22], so considering the area of the
receiver as 805 m2, the heat flux resulted in 1207.5 MWth. Once this limit is reached, defocusing of the mirrors
is done to reduce the incoming heat flux. The nature of the limit is related to the fatigue of the material and the
thermal stresses that can be produced by very high heat fluxes.
Therefore, in order to account for this technical limit, a constraint was added to the Matlab code to dump the
additional power and at the same time optimize the storage and hybridization of the plant. It was because of this
constraint, the turbine of 150MWe capacity was chosen since the turbine installed with higher capacity (250MWe)
would be operating at a considerably lower rated power leading to loss of efficiency.
18

20. 3.4 Transients
3.4.1 Impacts of Transients
In terms of the impacts of transients, the following constraints were identified:
• Within the solar field, the temperature of both the heliostats and the solar receiver at the tower will be a
function of the solar irradiance at that time of the day. Therefore, in a situation with no storage, the power
output of the cycle will not meet the minimum requirement at the start of the day, because the incoming solar
radiation in the morning will not be enough. However, if thermal storage were implemented that necessary
heat during startup would be available.
• Throughout the year, there are some cloudy or troublesome weather events that may cause reduction in
power generated in a situation where storage is not implemented. It was identified that for the chosen
location, there were some drops in solar irradiance that may be related with these types of events. For these
reasons, storage implementation was analyzed in depth.
3.4.2 Tower Receiver Temperature
When analyzing the tower receiver temperature it was deduced that the amount of time required to heat up the
receiver would not impact the power output as long as storage or the hybridization of a gas fired boiler were used.
Therefore, there would be no need to wait until enough solar irradiation is available for the receiver to reach the
operational temperature needed to run the power cycle. Despite this fact, it is still important to know the temper-
ature at the receiver in order to have an understanding of how long it will take after the sunrise to run the cycle
exclusively with incoming concentrated solar radiation.
The receiver can increase by 50oC in 300 seconds after sunrise, and takes 60 seconds to drop 45oC when the solar
energy is no longer concentrated on it [20]. Therefore, the temperature will be related with the input power to the
receiver. It has been determined that the molten salts will operate between the temperatures of 290oC to 565oC.
Based on these parameters the receiver temperature had a desired set point of 600oC in order to run the power
cycle with molten salts. The receiver temperature has an upper limit, because if the temperature of the salt reaches
levels higher than 600oC, then salt degradation may occur [21].
The salts will perform optimally at 565oC which coincides with a receiver temperature of 600oC. If the receiver is
heated up further than this, the heat transfer efficiency will start to decrease [21]. Ideally, the receiver temperature
will reach 600oC as soon as possible. The critical parameter to take into account is then the initial temperature of
the receiver, which can be assumed to be the average ambient temperature of the area. However, in order to keep
consistency with the hot start up conditions, this temperature was set according to the efficiency curve for a solar
tower plant [21]. This value is 77oC and is the temperature where there is a receiver efficiency of zero, therefore,
there is no solar irradiance.
The solar receiver will increase its temperature until 600oC. The heat transfer at that temperature is the maximum
heat transfer rate at the receiver. After this point the receiver does not need more heat. In order to avoid very high
temperatures that may damage the receiver or cause degradation of the salt, some heliostats will be need to be
defocused. The receiver has a peak flux limit, or a limit to the amount of thermal power that is transferred from
the receiver to the salt.
The receiver gives the salt a thermal power of 483 MW with an efficiency of 91%, which means that the required
heat rate that makes the receiver able to run the cycle is 530.77 MW [17]. If the receiver gets this heat, the salt can
be heated from its base of 290o to it’s ideal performance temperature of 565oC. The area of the receiver was set to
19

21. be 805.36 m2. In order to estimate the amount of time required to reach the desired temperatures the following
equations were used:
Q = hA∆T
h =
Q
A∆T
=
530.7692x106[W]
(805.36466[m2])(600 − 77)[oC]
h = 1260.1186[
W
m2C
]
Where:
Variable Description Units
Q Heat Going Through the Receiver W
A Area of the Receiver m2
h Heat Transfer Coefficient W
m2C
∆T Temperature Difference oC
Table 10: Variable Definitions
This is the heat transfer coefficient of the receiver for the proposed plant. With this value, the receiver temperature
curve for the plant can be generated. At 600oC, the salt can be pumped to the receiver to acquire the thermal energy
needed for the Rankine cycle. It is possible to look into the power at the receiver after each hour as well as the
temperature at its surface. When this value is reached, there is no need to use the storage or the alternative boiler.
The data for the energy harvested by the receiver and the associated temperatures can be seen in Table 11.
Time Power at the Receiver (MW) Receiver Temperature (oC)
5:00 0 77
6:00 12.11 92.75
7:00 208.11 363.30
8:00 479.52 600 3
9:00 675.27 600
Table 11: Receiver Temperature vs Time
After 8:00 the power required to run the cycle is provided entirely from the field, and from this time forward
the plant is able to store any amount of excess energy that is collected. Even though more power is being given
by the field, the heat rate needed is only to keep the temperature of the receiver at 600oC. Also at this time the
temperature will allow the heating of the pipes and the maintenance of the salt flow along the receiver.
Figure 8 shows that two hours after the sunrise, the receiver is able to work at full operation and run the cycle.
During that interim time, the electricity can be produced with the stored heat or gas boiler. It is important to take
into account the fact that the heating process of the field can be done simultaneously with the startup process of
the turbine. However, after the first hour, at 7:00 the temperature is high enough to heat up the salt partially and
together with the storage reach the needed temperature for the cycle. In a no storage situation the transient losses
would be reflected in a partial load operation. Another interesting observation in Figure 8 is that from 7:00 to 8:00,
approximately every 10 minutes the receiver temperature increases by roughly 40oC.
3
At 8:00 the temperature reached 623.38o
C, but it is kept at 600o
C to ensure ideal performance
20

22. Figure 8: Receiver Temperature vs. Time
The major losses due to transients in the solar field are significant in a no storage situation. If this is the case, the
startup time is longer, and the time of startup and shut down will change everyday. The production will also not
be the maximum when the demand is higher, and it will definitely not be possible to produce from 5:00 to 22:00
everyday. The estimation of the losses due to transients becomes much more complex in a no storage situation.
Therefore, it is possible to conclude that for the proposed CSP plant, storage will reduce the impacts of transients
related with solar field.
3.4.3 Operation of Plant
Due to the fact that any electricity sold between the hours of 22:00 and 5:00 will not generate any revenue for the
plant, there will be at least one startup and one cool down everyday. However, a benefit of this is that some of the
maintenance can be scheduled during these times throughout the year. This maintenance can be either preventive
or corrective. During these shutdowns, when the plant is not running in full operation, energy still needs to be
available for the following reasons.
Operational Requirements During Shutdown Periods:
• To keep a minimum steam flow rate at the turbine in order to maintain a minimum temperature.
• To start up the turbine and allow it to reach full operation levels.
• To preheat the receiver, pipes from the hot tank to the receiver, from the receiver to the cold tank, as well the
pipes that go to the steam generator. It is also necessary to keep the hot salt tank at a minimum temperature
level.
• To track the heliostats from the last position at sunset to the following position at sunrise, so the receiver can
be warmed in an efficient way.
All of these actions require energy and therefore can be viewed as a cost, since that energy would have to be pro-
vided for through some sort of hybridization with fossil fuels or an external backup. However, if the plant were
to have storage capabilities, the additional energy that is reserved can be used for these operational needs. This
represents a considerable advantage, since the excess energy would no longer be lost but would be able to aid in
these operational requirements.
Another major advantage of storage is that it helps reduce the shutdowns and subsequent startups of the turbine
due to cloudiness throughout the day. In the event of passing cloud cover, the electricity production can remain
constant by drawing from the thermal storage of the molten salts. Also, having the ability to store some energy
21

23. when the irradiance is high and then produce and sell electricity later in the day when the price is optimized, can
help increase the profitability of the plant.
The startup time of the turbine was chosen as a governing parameter during the analysis of the plant. The startup
process will take longer if the starting temperature of the receiver and other components are at the ambient tem-
perature of nighttime in the desert. Therefore, based on using energy for the above operational requirements
during the shutdown periods, it was assumed that all startups would be "hot startups". Therefore, the key aspects
were considered to estimate the impact due to transients were the startup, cool down, and the energy needed to
keep the salt and plant components at their minimum temperatures.
3.4.4 Energy Needed for Molten Salt
The plant was designed so that during operation, the molten salt will be heated up to 565oC, and after the steam
generation process the temperature will drop to 290oC. This will then be the temperature of the salts in the cold
reserve tank. Heat losses of the storage tanks have been reported to be 1oF (0.55oC) each day Then, it can be
assumed that the heat losses at the tanks are not significant and no additional energy needs to be added, since the
salt will not reach a temperature lower than 260oC during the night period [8].
3.4.5 Transients Energy Calculation for Turbine
Depending on the style and manufacturer of the turbine used in the power cycle, the startup time can vary, as
seen in Figure 9.
Figure 9: Turbine Start-Up Curve [9]
It can be seen that within a 40 minute window, the model turbines shown were up to full load. Therefore it was
assumed that the plant would also have a turbine running at full load within 40 minutes of startup. However,
22

24. before the startup process of the turbine, additional time would be needed for the receiver and the heliostats to
get into full load conditions as well. Therefore it has been assumed that startup time for the entire plant would be
45 minutes [10].
As an example, the startup time can be chosen to be 4:15 and the cool down time at 22:00. Choosing the Siemens
turbine from Figure 9, the process can be divided into three steps. For 12 minutes, there is no power output at the
turbine. From 12 minutes to 28 minutes, the output power goes from 0% to 97%. Finally from 28 minutes to 40
minutes after startup, the power output reaches its full load of 150 MW
The energy produced can be calculated by determining the area under the curve in the power vs. time chart.
Therefore, at minute 28 the power output would be 145.5 MW and the function describing the linear relationship
from this point would be:
y = 6.6687x
Integrating from minute 12 to minute 28 the total amount of energy produced can be calculated as 19.4 MWh.
Calculated in a similar fashion from minute 28 to minute 40, the energy output would be 0.45 MWh.Therefore the
total energy for each turbine startup would be 19.85 MWh.
If storage is implemented, then it would be possible to keep the turbine running after sunset and sell the electricity
when the price is increased. However like previously mentioned, there is a point where no profit will be generated
so the turbine will be forced to stop for economic reasons. For this turbine stop, or cool down process, there will
be a similar shape, but with different phases. It can assumed that the over the first 16 minutes of powering down
the power output will be reduced by 97%. Between 16 minutes and 28 minutes the last 3 percent of power output
will be scaled down [13]. This cool down process can be see in Figure 10.
Figure 10: Turbine Ramp Down Curve
Calculating the power output that is generated but not paid for can be done via:
y = −6.6687x
23

25. The energy output will be the same as during the start up process. Integrating along the 16 minute process, yields
a total of 19.4 MWh. In the next 12 minutes the energy output will be 0.45 MWh. So, for the stopping process (and
consequently, a partial cool down process) the energy loss is 19.85 MWh.
In total, each day the energy produced by the turbine and not sold would be 39.7 MWh. Over the course of a year
there will be a total of 14,490.5 MWh (14.49 GWh). This energy lost is equivalent to % of our total output.
In order to know how accurate this answer is, is was necessary to research the startup losses for existing CSP plants.
According with the National Renewable Energy Laboratory of the US Department of Energy, for a 330MW plant
with 3 hours of storage, the losses due to startup processes are around 23 GWh throughout the year. Considering
the fact that the proposed plant is 45% of that capacity, the losses should be roughly this percentage as well, which
would result in 10.45 GWh of losses, compared to the predicted losses of 14.49 GWh).
4 The Algorithm Logic
Figure 11 below highlights the logic of the optimization algorithm used to determine the optimal combination of
storage size and turbine capacity, to give the lowest payback period. The outermost loop of the algorithm iterates
between 150, 175 and 250 MW turbine capacities. For each turbine capacity, another loop iterates through storage
sizes ranging from 1 to 10 hours, along with an option of zero storage. As Figure 11 shows, within both of these
loops, the algorithm then moves through each day of the year and subsequently, each hour of the day.
As it goes through each hour, it first checks to see if the hour falls within the range of 5:00 to 22:00. If not, according
to the tariff scheme, there is no opportunity for revenue generation. Outside of that range there is no chance for
power production and is therefore of no interest to the algorithm. If the hour is within the required range, the
algorithm then compares the thermal power output of the plant to the minimum thermal requirement of 427 MW.
This is done to see how the power produced is to be managed.
24

26. Figure 11: Algorithm Logic
As Figure 11 shows, if the power production is less than the power requirement, the algorithm checks the state
of the storage. In the algorithm, the storage has been designed with the ability to carry over energy from one day
to another, with negligible heat loss. Thus, if the algorithm finds that the storage has enough energy to cover the
deficit between power production and the minimum requirement, it does so and generates revenue in the process.
The amount of storage used is then subtracted from the real-time storage state, to ensure it stays updated. If how-
ever, there is not enough energy in the storage, hybridization via natural gas is used to meet minimum demand.
The operating costs associated with hybridization are also included every time it used.
Now, if the power production is more than the minimum requirement instead, the algorithm checks the real-time
storage state. If the storage has not reached its limit, the excess energy is added to it and the current storage state is
updated. If, on the other hand, the storage has reached its limit, the algorithm checks how much turbine capacity
is still available for production, after having been used to meet the minimum requirement. If there is still enough
capacity left, the turbine runs and generates electricity and revenue. If there is not enough capacity left, the excess
heat is wasted.
The above logic is performed for every hour of every day in the whole year. The total revenues associated with
storage and the solar field are recorded into tables for later analysis.
25

27. 4.1 Algorithm Assumptions
One of the assumptions made for this algorithm is that as long as production is between 5:00 and 22:00, the mar-
ket would be willing to buy as much electricity as the plant produces, as long as it is more than the minimum
requirement. Another assumption is that the revenue from hybridization is not a part of the optimization algo-
rithm. The reason behind this is that hybridization is available as a last-resort, but not as an active part of the
plant. Optimizing while including hybridization revenue would skew the results in favor of maximizing natural
gas output and minimizing solar production, which is counter-productive to the overall goal.
4.2 Results
To complete the optimization, the revenues, operating costs and capital costs were all calculated for each combi-
nation of storage size and turbine capacity. Figure 12 presents the variation of revenue with each of these combi-
nations.
Figure 12: Annual Revenue vs. Storage Size
As was expected, Figure 12 shows that revenues for all three turbine capacities increase with increasing storage
size. The increase is initially linear, with it eventually slowing down significantly after 6 hours or so. The reason
behind the initial increase is that the availability of storage allows excess energy produced during the day, to be
harnessed in the evening during peak hours, drastically increasing the opportunity for revenue generation. How-
ever, as the storage size is increased, the increase in revenue reaches a limit as there is only so much excess energy
to capture and sell. Therefore, after a point, increasing the storage size would be economically unfavorable as the
return on the extra kWh that storage offers would be small. Another trend to be observed in Figure 12 is that the
larger the turbine capacity, the more the revenue generated. This was also expected, as a larger capacity allows
for more of the excess energy to be harnessed without reaching the storage limit. If the capacity were to keep
increasing, it would eventually also become economically unfavorable.
Figure 13 illustrates the annual operating costs for different turbine capacities and storage sizes. The operating
cost for the plant includes the operating costs of the turbine, the storage, the hybridization and the rest of the
plant. As can be seen, there is a significant initial decrease in costs, which eventually even out to become fairly
constant. The primary reason behind this is that the cost of hybridization decreases significantly as the storage
size is increased and natural gas is required less and less. At large storage sizes of 7 hours or more, the use of
natural gas is low and fairly steady, which is why the costs level out.
26

28. Figure 13: Annual Operating Costs vs. Storage Size
Figure 14, shows how much hybridization was used in the whole year, and highlights the same trend. The reason
the amount of hybridization is exactly the same for all three turbines is that hybridization is only used when the
solar field or the storage are not able to reach the minimum requirement. It does not in any way depend on the
size of the turbine itself.
Figure 14: Hybridization Amount vs. Storage Size
The difference in costs for different turbine capacities in Figure 13, are due to different the turbine operating costs
which scale linearly with size. This is why the difference in the curves aligns well with the difference in turbine
capacities.
Figure 15 shows the variation of total annual capital costs with storage sizes and turbine capacities. The capital
cost includes the turbine equipment cost, the storage equipment cost, the hybridization equipment cost and the
equipment costs for the rest of the plant. It also includes the cost of project development and site installation. As
all these costs are linearly dependent on the size of the system, they increase linearly with both the storage size
and the turbine capacity.
27

29. Figure 15: Capital Costs vs. Storage Size
Given the costs provided before, a non-discounted payback period was then computed for the various storage
sizes and turbine capacities. It was maintained as non-discounted, as it was only a means to choose the optimum
combination. From Figure 16, it can be seen that the combination that yields the smallest payback period is a 150
MW steam turbine with 7 hours of thermal storage.
Figure 16: Payback Period vs. Storage Size
The choice of 150 MW and 7 hours of storage reflects the most economically feasible choice given the technical
specifications and constraints of the solar field. This, however, does not mean that it is the most efficient choice.
This can be seen in Figure 17, which shows the annual amount of heat wasted for different combinations. As can
be seen, storage does reduce wastage for all turbine capacities, but the losses for a 150 MW turbine are around 740
GWh while those for a 250 MW turbine are drastically lower, at 45 GWh.
28

30. Figure 17: Annual Heat Wasted vs. Storage Size
5 Overall Performance of the Plant
From the irradiation over each hour and day of the year, and together with the code developed in Matlab for
dimensioning the field, it was possible to calculate the total thermal power output each hour at each cell in the
field. This was done for the total 365 days. Once the power cycle was designed (and therefore the efficiencies
were known), the operative strategies were defined, and the transient losses were considered, the total energy
produced was calculated. The optimization algorithm explained before was also applied.
At this stage, it was also possible to estimate the amount of time where the plant will operate at different times.
This will be the base to calculate the revenue and evaluate the profitability of the plant in the next chapter. The
total operational hours at the peak time (where the electricity is better paid) and the operational hours at normal
tariff were calculated. As explained before, there is no production at the times electricity is not being paid.
The total performance in the proposed CSP plant is summarized in Table 12.
29

31. Overall Efficiency of the Plant 19.27%
Design Day for Solar Multiple 21st June at Solar Noon
Solar Multiple 1.83
Optimal Storage 7 hours
Total Annual Hybridization Hours 209
Turbine Electrical Power Output 150 MW
Capacity Factor 55.7%
Time of Operation During Peak 1460 hours
Time of Regular Operation 3,321.93 hours
Total Operational Hours 4,781.93 hours
Total Transient Operational Losses 14.5 GWh
Total Annual Electricity Produced 731.78 GWh
Table 12: Total CSP Plant Performance
6 Financial Analysis
6.1 Capital Expenditures (CAPEX)
The capital expenditure and capital investment for a 150MWe solar tower power plant designed in Ouarzazate
was determined to be $652M. The estimated breakdown for the capital expenditure includes the cost of equip-
ment, engineering, land, installment, construction, and the plant itself. The cost of engineering, land installment,
construction and the plant are based on the total cost of equipment and corresponding factors for each. The factors
listed in Table 13, were established based on information provided by engineers working in the power industry.
Factors
Installation 15%
Engineering 5%
O&M 10%
Construction 13%
Decommissioning 5%
CAPEX 52%
Table 13: Cost Factors
The CAPEX breakdown can be seen in more detail in Figure 18 and Table 14
30

32. Figure 18: CAPEX
Cost of Equipment $461,745,150.95
Cost of Installment $26,550,346.18
Cost of Engineering $66,375,865.45
Cost of Construction $66,375,865.45
Cost of Land $28,261,000.00
Total $652,194,135.22
Table 14: CAPEX
6.2 Operational Expenditures (OPEX)
The OPEX cost of the plant was estimated to be $2,663,887 per annum. The OPEX was determined using the
operation strategy mentioned earlier in the report. The major cost in OPEX were operation & maintenance, labor
and the fuel for hybridization.
Figure 19: OPEX
31

33. Cost of O & M $1,846,980
Cost of Labor $215,930
Cost of Fuel $600,976
Total $2,663,886
Table 15: OPEX
The OPEX cost of for hybridization was estimated using the fuel cost per annum. The Matlab code iterations allow
us to calculate the extent of hybridization required for optimum operation.The thermal power requirement for the
150MW SST600 allows us to find out the amount of gas required after taking into account the system efficiencies.
Hence we know the fuel flow of the gas per kW of electric power which comes out to be 0.025 L
sec−kW . The annual
cost of hybridization is then calculated using the number of hours required for hybridization and flow of fuel
(natural gas) required per kW of electric power.
The equipment cost is composed of solar field, power block, instrument and controls, the plant balance systems,
hybridization, solar receiver, and storage equipment. Table 16 has the breakdown of equipment used for the
conceptual design of the concentrated solar power plant.
Solar Field Power Block Balance of Plant
Mirrors Heat Exchanger Auxiliary Fuel Treatment Plant
Tracking System, Drives, and VFDs Structures and Foundations Piping, Insulation, Valves & Fittings
Heliostat Structure Steam Turbine and Generator Water Treatment System
Field Control and Wiring Piping, Insulation, Valves & Fittings Waste Water System
Instrumentation & Controls Feed Water System Chemical Closage
Total Plant Area Condensate System Sampling System
Instrumentation & Controls Blowdown System HVAC
Distributed Control System (DCS) Cooling System Compressed Air
Storage LP/HP Preheaters Fire Protection System
Pumps Gland Steam Nitrogen Blanketing System
Piping, Insulation, Valves & Fittings Deareator Water Collection
Tanks Instrumentation & Controls Water Disposal
Tanks Insulation Medium-Voltage Cells Auxiliary Cooling System
Tank Foundation Auxiliary Diesel Generator Instrumentation & Controls
Molten Salts Switchyard Solar Receiver
HEXs Structure Electric Heaters High-Voltage Line Receiver
Melting Stations for Commissioning Motor Control Center Tower
Instrumentation & Controls VFDs Riser/Downcomer Piping
Main Transformer Spares Instrumentation & Controls
- Contingency -
Table 16: Equipment List
32

34. The levelized cost of electricity based on the capital expenditure, operational expenditure, and annual perfor-
mance was 167.45 $
MWh . The following equation was used to calculate the levelized cost of electricity:
LCOE =
α ∗ Cinv + Cf + CO&M + β ∗ Cdec
Enet
Where:
Variable Description Units
LCOE Levelized Cost of Electricity $/kWh
Cinv Initial Cost of Investment $
CO&M Costs for Operation and Maintenance $
Cdec Cost for Decommissioning $
n Lifetime of the Power Plant Years
α Annualization Factor -
β Discount Factor -
Table 17: Variable Definitions
The macro-environmental economic conditions used in the financial analysis of the plant were inflation, corporate
tax, and interest on investment for CSP plants. Inflation in Morocco is 0.3% and the corporate tax is set at 8% for
foreign contractors completing technical work in engineering, construction or assembly projects [14, 15]. The
levelized cost of electricity is affected by these macro-environmental economic conditions because it incorporates
a discount and annualization factor that includes the interest rate of an investment and inflation, therefore the
LCOE value is decreased. These fiscal conditions of Morocco are important to consider in order to fully assess the
viability of the CSP project.
6.3 Tariffs
Proposed schemes for tariffs based on an internal rate of return of 10% can then be seen in Table 18.
Clock Time CSP Tariff ( USD
MWh )
00:00 - 05:00 0
05:00 - 17:00 140
17:00 - 21:00 140 x1.4996 = 209.94
21:00 - 22:00 140
22:00 - 24:00 0
Table 18: Final Tariff Structure for CSP Plant
7 Environmental Impacts
In order to evaluate the environmental benefits by producing clean electricity from solar power instead of fossil
fuels, the analysis focused in two major indicator; water consumption of the plant and its CO2 emissions.
33

35. In terms of water consumption, the dry cooling process reduces the water needed of the power cycle and make up
water requirement also reduces as a result. Since the steam cycle is a closed system, the major water requirement
is for cleaning the mirrors. The estimated water consumption each year is only 55,402 m3
year [28]. Another important
thing is that there is no competition with the nearest population centers. It is also to be noted that once the plant is
under operation, the water requirement is very low as compared with the water required at the construction stage.
Regarding the CO2 emissions, the proposed power plant has CO2 emission of 1.88 kg CO2eq
MWh . In total, the CO2
emissions are 1,375.07 tons. If the same amount of electricity was to be generated using natural gas, the emissions
would be 404,956 tons. And if the electricity was to be produced with a coal power plant, the total emissions
would be 697,098 tons. It is possible to conclude that the proposed CSP plant has a significant environmental
benefits. In the graph below is clearly seen that a CSP plant is a very good alternative in terms of climate change
mitigation. The detailed calculations are attached in the appendix.
Figure 20: Avoided Emissions
8 Conclusion
The proposed CSP plant had an overall sun-to-electricity efficiency of 19.27% and an annual electricity produc-
tion of 731.78 GWh. The final payback period was determined to be 10 years, making this project economically
feasible. The Levelized Cost of Electricity was 167.45 USD
MWh , allowing for a competitive tariff scheme for electricity
distribution.
The optimization algorithm determined that a combination of 150 MW and 7 hours of molten salt storage, would
give the most viable power plant. The presence of storage proved to be hugely beneficial to the power plant, as
it allowed the plant to store excess energy during the day and use it during peak periods, generating sizable rev-
enues as a result.
The presence of storage also had major environmental benefits, as it significantly reduced the amount of hybridiza-
tion employed the plant. Overall, the plant ends up reducing carbon emissions by 403,500 tons. Also, the power
plant consumes relatively less water, with annual consumption at 55,402 m3.
Hence, the proposed power tower CSP plant is a great solution as it manages to exceed the minimum technical
requirement while being economically feasible to an investor, and environmentally friendly.
34

36. 9 References
[1] Specimen News
World’s Largest Concentrated Solar Power Plant Installed in Morocco
http://specimennews.com/2016/02/05/worlds-largest-concentrated-solar-power-plant-installed-
in-morocco/
Date Published: February 5, 2016
Date Accessed: February 25, 2016
[2] CNN
World’s Largest Concentrated Solar Plant Switches on in the Sahara
http://edition.cnn.com/2016/02/08/africa/ouarzazate-morocco-solar-plant/
Date Published: February 8, 2016
Date Accessed: February 27, 2016
[3] AIChE - ChEnected
King of Morocco Plans to Export Solar Power to Europe
http://www.aiche.org/chenected/2016/02/king-morocco-plans-export-solar-power-europe
Date Published: February 19, 2016
Date Accessed: February 27, 2016
[4] NREL
Parabolic Trough Solar Field Technology
http://www.nrel.gov/csp/troughnet/solar_field.html
Date Published: January 28, 2010
Date Accessed: February 27, 2016
[5] Solar Tower
Meteorological Reactors
http://www.solar-tower.org.uk/quick-start.php
Date Accessed: February 27, 2016
[6] SoDa
Time Series of Solar Radiation Data
http://www.soda-is.com/eng/services/services_radiation_free_eng.php
Date Published: October 2, 2015
Date Accessed: February 27, 2016
[7] CSP Today
TMY - Typical Meteorological Year - What Should it Represent and What Means P50, P70, and P90
http://social.csptoday.com/technology/tmy-%E2%80%93-typical-meteorological-year-what-
should-it-represent-and-what-means-p50-p70-and-p90
Date Published: March 7, 2011
Date Accessed: March 1, 2016
[8] Solar Reserve
Molten Salt Energy Storage
http://www.solarreserve.com/en/technology/molten-salt-energy-storage
35

37. Date Accessed: March 2, 2016
[9] Wärtsilä
Combustion Engine vs Gas Turbine: Startup Time
http://www.wartsila.com/energy/learning-center/technical-comparisons/combustion-engine-vs-
gas-turbine-startup-time
Date Accessed: March 2, 2016
[10] ACWA Power
Noor III Tower CSP Plant, Ouarzazate, Morocco Specific Environmental and Social Impact Assessment
Volume 1 pg. 47 - 52
http://www.masen.org.ma/upload/environnement/Masen_NOORoIII_SESIA_Volume_1Main_Text.pdf
Date Published: March 2015
Date Accessed: March 2, 2016
[11] KTH Bilda - SPG Exercises
Steam Cycle
https://bilda.kth.se/courseId/12692/content.do?id=24482514
Date Accessed: March 2, 2016
[12] Chalmers University
Methods of Increasing Thermal Efficiency of a Counter Flow Air to Air Heat Exchanger
Authors: Karl Larsson and Fredrik Pihlquist http://publications.lib.chalmers.se/records/fulltext/
145340.pdf
Date Published: 2011
Date Accessed: March 2, 2016
[13] Transactions of the Institute of Fluid-Flow Machinery
Methods of Increasing Thermal Efficiency of a Counter Flow Air to Air Heat Exchanger pg. 175
Author: Mariusz Banaszkiewicz http://yadda.icm.edu.pl/yadda/element/bwmeta1.element.baztech-
484a0f98-081c-46a9-8716-cb06f56ffc03/c/Trans169-198.pdf
Date Published: October 15, 2014
Date Accessed: March 4, 2016
[14] Trading Economics
Morocco Inflation Rate
http://www.tradingeconomics.com/morocco/inflation-cpi
Date Accessed: March 5, 2016
[15] Santander Trade Portal
Morocco: Tax System
https://en.santandertrade.com/establish-overseas/morocco/tax-system
Date Published: 2016
Date Accessed: March 5, 2016
[16] KTH
Design of Solar Tower CSP Plants
Author: Rafael E. Guédez
Renewable Energy Technology - Advanced Course Lecture Slides
36

38. Date Published: January 22, 2016
Date Accessed: March 6, 2016
[17] KTH
Project Description and Guidelines
Author: Concentrating Solar Power Group
Renewable Energy Technology - Advanced Course Project Description
Date Published: February 2016
Date Accessed: March 6, 2016
[18] KTH
Heliostat Design pg. 22
Master of Science Thesis Author: Nils Björkman
http://www.diva-portal.se/smash/get/diva2:769446/FULLTEXT01.pdf Date Published: 2014
Date Accessed: March 6, 2016
[19] Power from the Sun
Central Receiver Systems
http://www.powerfromthesun.net/Book/chapter10/chapter10.html Date Accessed: February 27, 2016
[20] Science Direct
Transient Analysis of a Molten Salt Cavity Receiver pg. 604
Authors: Q.Q. Zhang, X. Li, C. Chang, H. Liu, Z.F. Wang
http://ac.els-cdn.com/S1876610214005190/1-s2.0-S1876610214005190-main.pdf?_tid=184d90de-
e3c9-11e5-a927-00000aab0f02&acdnat=1457289034_d982356116d99fc80c4731759896fa3f
Date Published: 2013
Date Accessed: February 28, 2016
[21] DLR
Receivers for Solar Tower Systems
Author: Prof. Dr. Bernhard Hoffschmidt
http://sfera2.sollab.eu/uploads/images/networking/SFERA%20SUMMER%20SCHOOL%202014%20-
%20PRESENTATIONS/SolarTowerReceivers%20-%20Bernhard%20Hoffschmidt.pdf
Date Published: June 27, 2014
Date Accessed: March 6, 2016
[22] École Polytechnique Fédérale de Lausanne
Thermo-Economic Optimisation of Large Solar Tower Power Plants
Author: Germain Augsburger
http://infoscience.epfl.ch/record/183139/files/EPFL_TH5648.pdf
Date Published: January 24, 2013
Date Accessed: March 6, 2016
[23] CSP Today
Morocco’s First CSP Plant Forges Path to Tech-led Cost Cuts
http://social.csptoday.com/markets/moroccos-first-csp-plant-forges-path-tech-led-cost-cuts
Date Published: December 2, 2015
Date Accessed: March 6, 2016
37

39. [24] Invest in Morocco
Electricity Tariffs
http://www.invest.gov.ma/index.php?Id=34503&lang=en
Date Accessed: March 6, 2016
[25] Siemens
Siemens Steam Turbine SST-600
http://www.energy.siemens.com/br/en/fossil-power-generation/steam-turbines/sst-600.htm
Date Accessed: March 6, 2016
[26] Siemens
Siemens Steam Turbine SST-700
http://www.energy.siemens.com/hq/en/fossil-power-generation/steam-turbines/sst-700.htm
Date Accessed: March 6, 2016
[27] Siemens
Siemens Steam Turbine SST-900
http://www.energy.siemens.com/mx/en/fossil-power-generation/steam-turbines/sst-900.htm
Date Accessed: March 6, 2016
[28] Solar Energy Industries Association
Water Use Management
http://www.seia.org/policy/power-plant-development/utility-scale-solar-power/water-use-
management
Date Accessed: March 6, 2016
38

40. Appendices
Figure 21: Financial Metrics 1
39

41. Figure 22: Financial Metrics 2
Figure 23: Power Block
40

42. Figure 24: Power Block 2
Figure 25: Environmental Calculation
41

43. 1 %CSP Tower Calculation Script
2
3 Nrows=8760; %input
4 inputfilename = ’ Solar Values_v05_2005 . csv ’ ; %input
5
6 num_helio =8177; %number of h e l i o s t a t s in solar f i e l d
7 h_T =163.2; %height of tower (m)
8 D_reciever =15.16; %diameter of receiver on tower (m)
9 A_total =5.79∗10^6; %t o t a l land area required (m2) ? source
10 TES_D=30.07; %diameter of thermal storage tank (m)
11 Azi_Zones =12; %number of azimuthal zones
12 Rad_Zones =12; %number of radial zones
13
14 A_TES=(TES_D/2)^2∗ pi ∗2; %area of TES tanks (m2)
15 A_PowerBlock=A_TES ; %area required for Power Block (m2) ? Tower Area not included
in power block ?
16 A_Tower=( D_reciever /2)^2∗ pi ; %land area required for Tower (m2)
17
18 A_Inner =(A_TES+A_PowerBlock+A_Tower) ∗ 1 . 5 ; %Area of the inner c i r c l e (m2) ? Reason
for 1.5
19 r_Inner=sqrt ( ( A_Inner )/pi ) ; %radius of the inner c i r c l e (m)
20 A_SolarField=A_total −((A_TES+A_PowerBlock+A_Tower) ∗ 1 . 5 ) ; %Area of the solar f i e l d
(m2)
21 r_Outer=sqrt ( ( A_total )/pi ) ; %raidus of the entire solar f i e l d (m)
22
23 Stepwidth =( r_Outer−r_Inner ) /12; %Height of a single c e l l
24
25 helio_density = [ 0 . 8 3 ; 0.7875; 0.7450; 0.7025; 0.6600; 0.6175; 0.5750; 0.5325;
0.4900; 0.4475; 0.4050; 0 . 3 6 2 5 ] ; %setup a blank vector for the h e l i o s t a t f i e l d
density
26
27 % helio_density = [ ] ; %setup a blank vector for the h e l i o s t a t f i e l d density
28 % for r_H=( r_Inner+Stepwidth /2) : Stepwidth : ( r_Outer−Stepwidth /2) %radial distance (
m)
29 % helio_density =[ helio_density ; 0.721∗ exp ( −0.29∗(r_H/h_T ) +0.03) ] ; %h e l i o s t a t
f i e l d density
30 % end
31
32 Cell_Height =( r_Outer−r_Inner ) /12; %height of each c e l l in a radial direction (m)
33
34 A_cell = [ ] ; %setup a blank vector for the areas of each c e l l in a radial direction
35 r _ c e l l =[ r_Inner+Cell_Height : Cell_Height : r_Outer ] ; %s p e c i f i y the radius of a
s p e c i f i c ring of c e l l s
36
37 for i =1:11 %for loop for calculating the area of each ring of c e l l s
38 A_cell ( 1 ) =( r _ c e l l ( 1 ) ^2∗pi −(A_Inner ) ) /12; %calculating the area of the f i r s t
ring
39 A_cell ( i +1) =( r _ c e l l ( i +1)^2∗pi−r _ c e l l ( i ) ^2∗ pi ) /12; %calculating the area of
subsequent rings
40 end
41
42 A_cell ; %Area of 1 c e l l in each radial zone
43
42

44. 44 A_helio_cell=helio_density . ∗ transpose ( A_cell ) ; %area of helio s t a t s in 1 c e l l in a
radial ring
45
46 E_surface =0.92∗0.966; %calculation of surface e f f i c t i v e n e s s with reference values
47 f_shad =0.05; %shadowing f a c t or − given in lecture s l i d e s
48 f_block =0.05; %blocking f a ct o r − given in lecture s l i d e s
49
50 h_Helio =6; %height of a h e l i o s t a t from the ground to the panel surface ( calculated
via scaling assumption )
51
52 d_r=sqrt ( r _ c e l l .^2+(h_T−h_Helio ) .^2) ; %di r ec t distance from tower to h e l i o s t a t (
distance l i g h t will travel )
53
54 f _ a t t =0.99326 −0.1046∗( d_r /1000) +(0.017∗( d_r /1000) .^2) −(0.002845∗( d_r /1000) .^3) ; %
attenuation losses c o e f f i c i e n t s found from powerfromthesun . net
55
56 %−−−− Sunay −−−−−;
57
58 % Importing solar irradiance data at hourly level
59 f i l e I D = fopen ( inputfilename ) ;
60 solar_data = textscan ( fileID , ’%f %f %f %f %f %f %f %f %f %f %f %f %f %f ’ , ’
Delimiter ’ , ’ , ’ , ’ Headerlines ’ , 1) ;
61 f c l o s e ( f i l e I D ) ;
62
63 %Cosine Effectiveness Calculation
64 %1. Vs vector − 16800x3 matrix
65 solar_data { : , 1 5 } = −sin ( solar_data { : , 1 3 } ) . ∗ cos ( solar_data { : , 1 4 } ) ;
66 solar_data { : , 1 6 } = −sin ( solar_data { : , 1 3 } ) . ∗ sin ( solar_data { : , 1 4 } ) ;
67 solar_data { : , 1 7 } = cos ( solar_data { : , 1 3 } ) ;
68
69 v_s = [ solar_data { : , 1 5 } solar_data { : , 1 6 } solar_data { : , 1 7 } ] ;
70
71 %2. Vt vector − 144x5 matrix
72
73 theta_T = atan ( ( h_T−h_Helio ) ./ r _ c e l l ) ; %for each 12 radial zones
74 theta_H = [ pi /12: pi /6:2∗ pi−pi /12]; %per radial zone , for each 12 azimuthal zones
75 v_T = [ ] ;
76 position_dependent_factors = ( A_helio_cell . ∗ ( f_att ’ ) ) ’ ; %depends on radial
position
77
78 for i =1:12;
79 for j =1:12;
80 temp_T = [ r _ c e l l ( i ) theta_H ( j ) sin ( theta_H ( j ) ) . ∗ cos ( theta_T ( i ) ) −sin (
theta_H ( j ) ) . ∗ cos ( theta_T ( i ) ) sin ( theta_T ( i ) ) position_dependent_factors
( i ) ] ;
81 v_T = [ v_T ; temp_T ] ;
82 end ;
83 end ;
84
85 %3. NH vector − (16800 x144 ) x3 matrix
86
87 vs_vt = zeros (Nrows∗144 ,4) ; %CHANGE IT ACCORDING TO EXCEL
88 NH = zeros (Nrows∗144 ,3) ; %CHANGE IT ACCORDING TO EXCEL
43

45. 89
90 for i =1:Nrows ; %Change 12 to 16800/2 − Do i t only for 2004 , no need to c a l c u l a t e
separately for 2005
91 for j =1:144;
92 vs_vt ( ( i −1)∗144+ j , : ) = [ v_s ( i , 1 ) +v_T ( j , 3 ) v_s ( i , 2 ) +v_T ( j , 4 ) v_s ( i , 3 ) +v_T
( j , 5 ) sqrt ( ( v_s ( i , 1 ) + v_T ( j , 3 ) ) ^2+(v_s ( i , 2 ) +v_T ( j , 4 ) ) ^2+(v_s ( i , 3 ) +v_T ( j
, 5 ) ) ^2) ] ;
93 NH( ( i −1)∗144+ j , : ) = [ vs_vt ( ( i −1)∗144+ j , 1 ) ./ vs_vt ( ( i −1)∗144+ j , 4 ) vs_vt ( ( i
−1)∗144+ j , 2 ) ./ vs_vt ( ( i −1)∗144+ j , 4 ) vs_vt ( ( i −1)∗144+ j , 3 ) ./ vs_vt ( ( i −1)
∗144+ j , 4 ) ] ;
94 end ;
95 end ;
96
97 %4. Cosine Effectiveness − Dot Product of v_T and NH;
98
99 v_T_for_dot = [ v_T ( : , 3 ) v_T ( : , 4 ) v_T ( : , 5 ) ] ;
100 E_cosine = ones (Nrows , 1 ) ;
101
102 for i =1:Nrows ;%Change 12 to 16800/2 − Do i t only for 2004 , no need to c a l c u l a t e
separately for 2005
103 for j =1:144;
104 E_cosine ( ( i −1)∗144+ j ) = [ abs ( dot ( v_T_for_dot ( j , : ) , NH( ( i −1)∗144+ j , : ) )/
sqrt ( v_T_for_dot ( j , 1 ) ^2 + v_T_for_dot ( j , 2 ) ^2 + v_T_for_dot ( j , 3 ) ^2) ) ] ;
105 end ;
106 end ;
107
108 %5. Merging ( Horizontal Concatenation ) Cosine Effectiveness with Solar Data ;
109 temp_data = [ solar_data { : , 1 } solar_data { : , 2 } solar_data { : , 3 } solar_data { : , 4 }
solar_data { : , 5 } solar_data { : , 6 } solar_data { : , 7 } ] ;
110 master_data = zeros (Nrows∗144 ,11) ;
111 for i =1:Nrows ;%Change 12 to 16800/2 − Do i t only for 2004 , no need to c a l c u l a t e
separately for 2005
112 for j =1:144;
113 master_data ( ( i −1)∗144+ j , : ) = [ temp_data ( i , : ) v_T ( j , 1 ) v_T ( j , 2 ) v_T ( j , 6 )
E_cosine ( ( i −1)∗144+ j , 1 ) ] ;
114 end ;
115 end ;
116
117 %Field power ca l cu la t io ns
118 f _ s p i l l = 0.0284;
119 time_independent_factors = E_surface ∗(1−f_shad ) ∗(1− f_block ) ∗(1− f _ s p i l l ) ;
120 master_data ( : , 1 2 ) = time_independent_factors ( : , 1 ) . ∗ master_data ( : , 5 ) . ∗ master_data
( : , 1 0 ) . ∗ master_data ( : , 1 1 ) ; %master_data ( : , 1 0 ) i s the position dependent f a ct o r
and master_data ( : , 1 1 ) i s cosine e f f e c t i v e n e s s
121 Power_Cell_heliostat = master_data ;
122
123 Helio_Density_Area_ratio = helio_density . ∗ A_cell ’ . / A_helio_cell ; % turns out to be
unity always
124 Power_field = Power_Cell_heliostat ;
125
126 %Efficiency of CSP Plants − Taking reference from PPT and substantiating i t
127 %from https ://www. irena . org/DocumentDownloads/Publications/IRENA−ETSAP%20Tech%20
Brief%20E10%20Concentrating%20Solar%20Power . pdf
44

46. 128
129 Incident_Solar_Energy = 2319.9;
130 Net_Energy = 3 8 0 . 9 ;
131 %Net_Efficiency = Net_Energy/Incident_Solar_Energy ;
132 Net_Efficiency = 0 . 7 5 ;
133
134 %Net Power Output from the CSP Plant
135 P_elec_output = Power_field ;
136 P_elec_output ( : , 1 2 ) = Net_Efficiency ∗ P_elec_output ( : , 1 2 ) ;
137
138 P_elec_output_f = num2cell ( P_elec_output ) ;
139 P_elec_output_f = vertcat ( { ’ Year ’ ’Month ’ ’Day ’ ’Time ’ ’ Irradiance ’ ’
Lower bound ’ ’Upper bound ’ ’ Radial Position ’ ’ Azimuthal Position ’ ’ Position
dep f a c t o r s ’ ’ Cosine Effectiveness ’ ’Net Power Output ’ } , P_elec_output_f ) ;
140
141 fid = fopen ( ’ Solar Thermal Power Output . csv ’ , ’w’ ) ;
142 f p r i n t f ( fid , ’%s , ’ , P_elec_output_f { 1 , 1 : end } ) ;
143 f p r i n t f ( fid , ’%sn ’ , P_elec_output_f {1 , end } ) ;
144 f c l o s e ( fid ) ;
145 dlmwrite ( ’ Solar Thermal Power Output . csv ’ , P_elec_output_f ( 2 : end , : ) , ’−append ’ )
146
147 %−−−− Sunay −−−−−;
1 %Running the whole year of 2005 through the storage optimization code
2
3 %Importing the f i l e requested and putting i t into an array
4 fid = fopen ( ’ Solar Thermal Power Output . csv ’ ) ;
5 input_arr = textscan ( fid , ’%f %f %f %f %f %f %f %f %f %f %f %f ’ , ’ Delimiter ’ , ’ , ’ , ’
Headerlines ’ ,1) ;
6
7 %Only extracting the year , month , day , time and net power output from
8 %the f i e l d
9 power_output_arr = [ input_arr { 1 } , input_arr { 2 } , input_arr { 3 } , input_arr { 4 } ,
input_arr { end } ] ;
10
11 %I n i t i a l i z a t i o n
12 strt_indx = 1;
13 new_arr = ones (8760 ,5) ;
14 count = 0;
15
16 %Combining the power outputs of 144 c e l l s for each hour into one single
17 %value
18 while strt_indx < length ( power_output_arr )
19 count = count +1;
20 temp_arr = power_output_arr ( strt_indx : strt_indx +143 ,:) ; %s p l i t t i n g the rows
into hours , by i t e r a t i n g through every 144 rows ( which are the number of
c e l l s )
21 new_arr ( count , : ) = [ temp_arr ( 1 , 1 : 4 ) , sum( temp_arr ( : , 5 ) ,1) ] ; %making a new
array that imports the ( year , month , day , time ) stamp along with the sum of
power outputs from a l l c e l l s for that hour
22 strt_indx = strt_indx + 144;
23 end
24
45

47. 25 %I n i t i a l i z a t i o n
26 previous_day = 1;
27 P_out_daily = 0;
28 day_count = 1;
29 days_arr = c e l l (365 ,1) ;
30 operation_count = 0;
31
32 %Breaking up the hour−wise array into a day−wise c e l l array with each day
33 %in i t s own c e l l
34 for i = 1: length ( new_arr )
35 today = new_arr ( i , 3 ) ;
36 i f new_arr ( i , 5 ) >0 %checking how many hours the plant ( minus storage ) i s
operating for
37 operation_count = operation_count + 1;
38 end
39 i f abs ( today − previous_day ) >0 || i== length ( new_arr ) %i f we move to the next
day now, store the previous day ’ s data . Or i f we’ re at the end of the f i l e
40 i f i < length ( new_arr )
41 day_count = day_count +1;
42 end
43 days_arr { day_count , 1 } = [ days_arr { day_count , 1 } ; new_arr ( i , : ) ] ;
44 else
45 days_arr { day_count , 1 } = [ days_arr { day_count , 1 } ; new_arr ( i , : ) ] ;
46 end
47 previous_day = today ;
48 end
49
50 %I n i t i a l i z a t i o n
51 minimum_req = 427; %in Thermal MWh, minimum requirement for an hour
52 minimum_elec = 110; %in E l e c t r i c MWh, minimum requirement for an hour , for s e l l i n g
.
53 thermtoelec_ratio = minimum_elec/minimum_req ; %Thermal to e l e c t r i c
54 storage_revenue = 140; %$/MWh, earned from storing and s e l l i n g the e l e c t r i c i t y
55 plant_revenue = 140; %$/MWh, earned from producing and s e l l i n g the e l e c t r i c i t y
56 hybrid_cost = 40; %$/MWh, cost of hybridization cost
57 date_store = ones (365 ,3) ;
58 storage_waste_store = [ ] ;
59 max_heat = 0.75∗1083; %in MWH, maximum heat
60
61 turbine_vec = [150 , 175 , 2 5 0 ] ;%d i f f e r e n t turbine sizes for optimization , in MW
62
63 %We’ l l need these vectors at the end
64 master_store = ones (365 ,33) ;
65 hybrid_store = ones (365 ,33) ;
66 hybrid_cost_store = ones (365 ,33) ;
67 elec_store = ones (365 ,33) ;
68 hybrid_revenue_store = ones (365 ,33) ;
69 storage_waste_store = ones (365 ,33) ;
70 no_stg_hyb_store = ones (365 ,33) ;
71
72 column_count = 0;
73 hybrid_amount = 0;
74 hybrid_hours = 0;
46

48. 75
76 for turbine_capacity = turbine_vec %i t e r a t i n g for d i f f e r e n t turbine sizes and
c a p a c i t i e s
77 for storage_size = 0:10 %i t e r a t i n g between 1 and 10 hours of storage
78 stuff_happened = 0;
79 column_count = column_count +1;
80 storage_limit = storage_size ∗ minimum_req ; %in Thermal MWh, the maximum
amount the storage can have
81
82 %Going through each day of the year
83 for each_day = 1: length ( days_arr )
84 hour_arr = days_arr { each_day , 1 } ; %extracting the hour−by−hour
information for that day
85
86 %For the f i r s t day of the year
87 i f each_day==1
88 %I n i t i a l i z a t i o n
89 storage_accum = 0; %the amount of thermal MWh in the storage right
now, it ’ s constantly accumulating
90 total_hybrid_cost = 0; %t o t a l cost of hybridization
91 total_storage_revenue = 0; %t o t a l revenue earned from s e l l i n g
e l e c t r i c i t y from storage
92 total_plant_revenue = 0; %t o t a l revenue earned from s e l l i n g
e l e c t r i c i t y from the plant
93 storage_waste = 0; %how much excess energy can ’ t be stored as
storage i s f u l l
94 total_elec_prodn = 0;
95 total_hybrid_revenue = 0;
96 no_stg_hyb_count = 0;
97
98 %I t e r a t i n g through each hour of the day
99 for each_hour = 1: length ( hour_arr )
100 time_hour = hour_arr ( each_hour , 4 ) ; %that pa rt icu lar hour of
the day
101 power_hour = ( hour_arr ( each_hour , 5 ) ) /(10^6) ; %Thermal power , in
MWh, for that hour
102
103 i f power_hour > max_heat
104 power_hour = max_heat ;
105 end
106
107 %Only entering in that period when we’ re getting money for i t
108 i f each_hour >= 5 && each_hour <=22
109 i f each_hour >=17 && each_hour <=21
110 alpha = 1 . 4 9 7 ;
111 e l s e i f each_hour == 22
112 alpha = 1;
113 e l s e i f each_hour >= 5 && each_hour <17
114 alpha = 1;
115 else
116 alpha = 0;
117 end
118
47

49. 119 %I f there was more power produced than required , i t i s
stored
120 i f power_hour > minimum_req
121
122 storage_accum = storage_accum + ( power_hour −
minimum_req ) ; %Excess thermal power i s stored
123 total_plant_revenue = total_plant_revenue +
minimum_req∗ alpha ∗ thermtoelec_ratio ∗ plant_revenue ;
124 turbine_left_to_produce = turbine_capacity −
minimum_elec ; %how much the e l e c t r i c MWh can the
turbine s t i l l produce , given i t produces the
minimum amount f i r s t
125
126 total_elec_prodn = total_elec_prodn + minimum_elec ;
127
128 i f storage_size == 0
129 storage_limit = 10^6;
130 storage_accum = 0;
131 end
132
133 %i f the amount of storage exceeds how much the
134 %storage l i m i t i s
135 i f storage_accum > storage_limit
136 storage_excess = storage_accum − storage_limit ; %
how much heat can ’ t be stored
137 total_waste = ( storage_excess ∗ thermtoelec_ratio )−
turbine_left_to_produce ; %how much of the "
waste " heat will not be harnessed by the
turbine
138
139 i f total_waste >= 0 %in the case where the turbine
i s not able to harness a l l the waste heat/ j u s t
able to harness i t
140 total_plant_revenue = total_plant_revenue + (
turbine_left_to_produce ∗ plant_revenue ) ;
141 total_elec_prodn = total_elec_prodn +
turbine_left_to_produce ;
142 storage_waste = storage_waste + total_waste/
thermtoelec_ratio ;
143
144 else %in the case where the waste heat i s l e s s
than the capacity l e f t over in the turbine
145 total_plant_revenue = total_plant_revenue + (
storage_excess ∗ thermtoelec_ratio ∗
plant_revenue ) ;
146 total_elec_prodn = total_elec_prodn + (
storage_excess ∗ thermtoelec_ratio ) ;
147 end
148 storage_accum = storage_limit ; %s e t t i n g the amount
of storage back to the l i m i t i s supposed to be
at
149 end
150
48

50. 151 %I f there isn ’ t enough power production and i f storage has
enough to meet that d e f i c i t
152 e l s e i f power_hour < minimum_req && storage_accum > (
minimum_req−power_hour )
153 storage_accum = storage_accum − ( minimum_req−
power_hour ) ; %the d e f i c i t for that hour i s removed
from the accumulated storage
154 total_storage_revenue = total_storage_revenue + ( (
minimum_req−power_hour ) ∗ thermtoelec_ratio ∗
storage_revenue ) ; %revenue from storage for f i l l i n g
that d e f i c i t
155 total_plant_revenue = total_plant_revenue + power_hour
∗ thermtoelec_ratio ∗ plant_revenue ; %revenue from
e l e c t r i c i t y production
156
157 total_elec_prodn = total_elec_prodn + minimum_elec ;
158 no_stg_hyb_count = no_stg_hyb_count + 1;
159 else
160 hybrid_amount = hybrid_amount + ( minimum_req −
power_hour ) ∗ thermtoelec_ratio ; %amount of
hybridization
161 total_hybrid_cost = total_hybrid_cost + ( minimum_req −
power_hour ) ∗ hybrid_cost ; %cost of hybridization
162 hybrid_hours = hybrid_hours + 1;
163 total_hybrid_revenue = total_hybrid_revenue + ( (
minimum_req − power_hour ) ∗ thermtoelec_ratio ∗
plant_revenue ) ;
164
165 total_elec_prodn = total_elec_prodn + minimum_elec ;
166 no_stg_hyb_count = no_stg_hyb_count + 1 ;
167 end
168 end
169 end
170
171 %Now t h i s looks at every day besides the f i r s t one
172 else
173 power_prodn_vec = hour_arr ( : , 5 ) ; %vector with a l l the hourly power
production
174 non_zero_prodns = find ( power_prodn_vec~=0) ; %finding the f i r s t non
−zero hour of power production
175 strt_prodn_time = non_zero_prodns ( 1 ) ; %finding that hour of f i r s t
non−zero power production
176
177 strt_prodn = ( hour_arr ( strt_prodn_time , 5 ) ) /(10^6) ; %production
during f i r s t hour of sunshine
178 i f strt_prodn > max_heat
179 strt_prodn = max_heat ;
180 end
181
182 i f strt_prodn < minimum_req %i f there isn ’ t enough power
production in the f i r s t hour of sunshine
183 %going through the hours in that day , u n t i l production
184 %goes above the minimum amount required
49

51. 185 iter_time = strt_prodn_time ;
186 while strt_prodn < minimum_req
187 no_stg_hyb_count = no_stg_hyb_count + 1;
188 i f storage_accum >= ( minimum_req−strt_prodn ) %i f there i s
enough storage to make up for the d e f i c i t in power
production
189 storage_accum = storage_accum − ( minimum_req−
strt_prodn ) ;
190 total_storage_revenue = total_storage_revenue + ( (
minimum_req−strt_prodn ) ∗ thermtoelec_ratio ∗
storage_revenue ) ;
191 total_plant_revenue = total_plant_revenue +
strt_prodn ∗ thermtoelec_ratio ∗ plant_revenue ;
192
193 total_elec_prodn = total_elec_prodn + minimum_elec ;
194 else %i f there isn ’ t enough storage , hybridize .
195 hybrid_hours = hybrid_hours +1;
196 hybrid_amount = hybrid_amount + ( minimum_req −
strt_prodn ) ∗ thermtoelec_ratio ;
197 total_hybrid_cost = total_hybrid_cost + ( minimum_req −
strt_prodn ) ∗ hybrid_cost ;
198 total_elec_prodn = total_elec_prodn + minimum_elec ;
199 total_hybrid_revenue = total_hybrid_revenue + ( (
minimum_req − strt_prodn ) ∗ thermtoelec_ratio ∗
plant_revenue ) ;
200 end
201
202 iter_time = iter_time + 1; %going to the next hour
203 i f iter_time >24 %t h i s i s incase of days with j u s t bad
production a l l day
204 break
205 end
206 strt_prodn = ( hour_arr ( iter_time , 5 ) ) /(10^6) ; %getting the
207 stuff_happened = 1;
208 end
209
210 %This i s the case when there i s enough energy in the f i r s t hour
211 %of sunlight . So there i s no need for l a s t night ’ s storage to
212 %f i l l that hour .
213 else
214 %Now we try to see i f there i s enough storage to s e l l
215 %e l e c t r i c i t y between 5 am and the f i r s t hour of
216 %sunshine
217 i f storage_accum >= ( ( strt_prodn_time −5)∗minimum_req ) %
Checking i f there ’ s enough for prod ’ n between 5 am and
s t a r t of day
218 storage_accum = storage_accum − ( ( strt_prodn_time −5)∗
minimum_req ) ;
219 total_storage_revenue = ( ( strt_prodn_time −5)∗minimum_req ) ∗
thermtoelec_ratio ∗ storage_revenue ;
220 total_elec_prodn = total_elec_prodn + ( ( strt_prodn_time
−5)∗minimum_elec ) ;
221
50

52. 222 e l s e i f storage_accum >= ( ( strt_prodn_time −6)∗minimum_req )%
Checking i f there ’ s enough for prod ’ n between 6 am and
s t a r t of day
223 storage_accum = storage_accum − ( ( strt_prodn_time −6)∗
minimum_req ) ;
224 total_storage_revenue = ( ( strt_prodn_time −6)∗minimum_req ) ∗
thermtoelec_ratio ∗ storage_revenue ;
225 total_elec_prodn = total_elec_prodn + ( ( strt_prodn_time
−6)∗minimum_elec ) ;
226
227 e l s e i f storage_accum >= ( ( strt_prodn_time −7)∗minimum_req )%
Checking i f there ’ s enough for prod ’ n between 7 am and
s t a r t of day
228 storage_accum = storage_accum − ( ( strt_prodn_time −7)∗
minimum_req ) ;
229 total_storage_revenue = ( ( strt_prodn_time −7)∗minimum_req ) ∗
thermtoelec_ratio ∗ storage_revenue ;
230 total_elec_prodn = total_elec_prodn + ( ( strt_prodn_time
−7)∗minimum_elec ) ;
231 else
232 %don ’ t do anything , keep that storage accumulated for
l a t e r
233 end
234 end
235
236 %Now we s t a r t i t e r a t i n g through each hour in the day , accumulating
storage or suplying i t
237 %Also keep in mind to not get excess
238
239 %The f i r s t case i s i f in that day , there wasn ’ t enough
240 %production in the morning and that was adjusted for using
241 %storage .
242 i f stuff_happened == 1
243 %i t e r a t i n g through the r e s t of the day s t a r t i n g from where
244 %we l e f t off l a s t time
245 for each_hour = ( iter_time ) : length ( hour_arr )
246 time_hour = hour_arr ( each_hour , 4 ) ;
247 power_hour = ( hour_arr ( each_hour , 5 ) ) /(10^6) ;
248
249 i f power_hour > max_heat
250 power_hour = max_heat ;
251 end
252
253 %Only entering in that period when we’ re getting money for
i t
254 i f each_hour >=5 && each_hour <= 22
255 i f each_hour >=17 && each_hour <=21
256 alpha = 1 . 4 9 7 ;
257 e l s e i f each_hour == 22
258 alpha = 1;
259 e l s e i f each_hour >= 5 && each_hour <17
260 alpha = 1;
261 else
51

53. 262 alpha = 0;
263 end
264
265 i f power_hour > minimum_req
266 storage_accum = storage_accum + ( power_hour −
minimum_req ) ; %Excess thermal power i s stored
267 total_plant_revenue = total_plant_revenue +
minimum_elec∗ alpha ∗ plant_revenue ;
268 turbine_left_to_produce = turbine_capacity −
minimum_elec ; %how much the turbine can s t i l l
produce , given i t produces the minimum amount
f i r s t
269 total_elec_prodn = total_elec_prodn + minimum_elec
;
270
271 i f storage_size == 0
272 storage_limit = 10^6;
273 storage_accum = 0;
274 end
275
276 i f storage_accum > storage_limit
277 storage_excess = storage_accum − storage_limit
; %how much " waste " heat i s available
278 total_waste = ( storage_excess ∗
thermtoelec_ratio )− turbine_left_to_produce
; %how much of the " waste " heat will not be
harnessed by the turbine
279
280
281 i f total_waste >= 0 %in the case where the
turbine i s not able to harness a l l the
waste heat/ j u s t able to harness i t
282 total_plant_revenue = total_plant_revenue
+ ( turbine_left_to_produce ∗
plant_revenue ) ;
283 total_elec_prodn = total_elec_prodn +
turbine_left_to_produce ;
284 storage_waste = storage_waste +
total_waste/thermtoelec_ratio ;
285 else %in the case where the waste heat i s l e s s
than the capacity l e f t over in the turbine
286 total_plant_revenue = total_plant_revenue
+ ( storage_excess ∗ thermtoelec_ratio ∗
plant_revenue ) ;
287 total_elec_prodn = total_elec_prodn + (
storage_excess ∗ thermtoelec_ratio ) ;
288 end
289 end
290
291 storage_accum = storage_limit ;
292
293 e l s e i f power_hour <minimum_req && storage_accum > (
minimum_req−power_hour )
52

54. 294 storage_accum = storage_accum − ( minimum_req−
power_hour ) ;
295 total_storage_revenue = total_storage_revenue + ( (
minimum_req−power_hour ) ∗ storage_revenue ∗
thermtoelec_ratio ∗ alpha ) ;
296 total_plant_revenue = total_plant_revenue +
power_hour∗ alpha ∗ thermtoelec_ratio ∗
plant_revenue ;
297 total_elec_prodn = total_elec_prodn + minimum_elec
;
298 no_stg_hyb_count = no_stg_hyb_count + 1;
299 else
300 hybrid_hours = hybrid_hours +1;
301 hybrid_amount = hybrid_amount + ( minimum_req −
power_hour ) ∗ thermtoelec_ratio ;
302 total_hybrid_cost = total_hybrid_cost + ( (
minimum_req − power_hour ) ∗ hybrid_cost ) ;
303 total_elec_prodn = total_elec_prodn + minimum_elec
;
304 total_hybrid_revenue = total_hybrid_revenue + ( (
minimum_req − power_hour ) ∗ thermtoelec_ratio ∗
alpha ∗ plant_revenue ) ;
305 no_stg_hyb_count = no_stg_hyb_count + 1;
306 end
307 end
308 end
309 %s t a r t the i t e r a t i o n from that time where i t e r a t i o n
stopped
310 else
311 for each_hour = 1: length ( hour_arr )
312 time_hour = hour_arr ( each_hour , 4 ) ;
313 power_hour = ( hour_arr ( each_hour , 5 ) ) /(10^6) ;
314
315 i f power_hour > max_heat
316 power_hour = max_heat ;
317 end
318
319 i f each_hour >=5 && each_hour <= 22
320
321 i f each_hour >=17 && each_hour <=21
322 alpha = 1 . 4 9 7 ;
323 e l s e i f each_hour == 22
324 alpha = 1;
325 e l s e i f each_hour >= 5 && each_hour <17
326 alpha = 1;
327 else
328 alpha = 0;
329 end
330
331 %I f there was more power produced than required , i t i s
stored
332 i f power_hour > minimum_req
333 storage_accum = storage_accum + ( power_hour −
53

55. minimum_req ) ; %Excess thermal power i s stored
334 total_plant_revenue = total_plant_revenue +
minimum_req∗ alpha ∗ thermtoelec_ratio ∗
plant_revenue ;
335 turbine_left_to_produce = turbine_capacity −
minimum_elec ; %how much the turbine can s t i l l
produce , given i t produces the minimum amount
f i r s t
336 total_elec_prodn = total_elec_prodn + minimum_elec
;
337
338 i f storage_size == 0
339 storage_limit = 10^6;
340 storage_accum = 0;
341 end
342
343 %i f the amount of storage exceeds how much the
344 %storage l i m i t i s
345 i f storage_accum > storage_limit
346 storage_excess = storage_accum − storage_limit
; %how much " waste " heat i s available
347 total_waste = ( storage_excess ∗
thermtoelec_ratio )− turbine_left_to_produce
; %how much of the " waste " heat will not be
harnessed by the turbine
348
349 i f total_waste >= 0 %in the case where the
turbine i s not able to harness a l l the
waste heat/ j u s t able to harness i t
350 total_plant_revenue = total_plant_revenue
+ ( turbine_left_to_produce ∗
plant_revenue ) ;
351 total_elec_prodn = total_elec_prodn +
turbine_left_to_produce ;
352 storage_waste = storage_waste +
total_waste/thermtoelec_ratio ;
353
354 else %in the case where the waste heat i s l e s s
than the capacity l e f t over in the turbine
355 total_plant_revenue = total_plant_revenue
+ ( storage_excess ∗ thermtoelec_ratio ∗
plant_revenue ) ;
356 total_elec_prodn = total_elec_prodn + (
storage_excess ∗ thermtoelec_ratio ) ;
357 end
358 storage_accum = storage_limit ;
359 end
360
361 e l s e i f power_hour <minimum_req && storage_accum > (
minimum_req−power_hour )
362 storage_accum = storage_accum − ( minimum_req−
power_hour ) ;
363 total_storage_revenue = total_storage_revenue + ( (
54

56. minimum_req−power_hour ) ∗ thermtoelec_ratio ∗
storage_revenue ∗ alpha ) ;
364 total_plant_revenue = total_plant_revenue +
power_hour∗ alpha ∗ thermtoelec_ratio ∗
plant_revenue ;
365 total_elec_prodn = total_elec_prodn + minimum_elec
;
366 no_stg_hyb_count = no_stg_hyb_count + 1;
367 else
368 hybrid_hours = hybrid_hours + 1;
369 hybrid_amount = hybrid_amount + ( minimum_req −
power_hour ) ∗ thermtoelec_ratio ;
370 total_hybrid_cost = total_hybrid_cost + ( (
minimum_req − power_hour ) ∗ hybrid_cost ) ;
371 total_elec_prodn = total_elec_prodn + minimum_elec
;
372 total_hybrid_revenue = total_hybrid_revenue + ( (
minimum_req − power_hour ) ∗ thermtoelec_ratio ∗
alpha ∗ plant_revenue ) ;
373 no_stg_hyb_count = no_stg_hyb_count + 1;
374 end
375 end
376 end
377 end
378 end
379
380 total_revenue = total_plant_revenue+total_storage_revenue ;
381
382 i f turbine_capacity == 150 && storage_size == 1
383 date_store ( each_day , : ) = hour_arr ( 1 , 1 : 3 ) ; %creating the f i r s t
three columns of year , month and day
384 end
385
386 elec_store ( each_day , column_count ) = total_elec_prodn ;
387 master_store ( each_day , column_count ) = total_revenue ;
388 hybrid_store ( each_day , column_count ) = ( hybrid_hours ) ; %Average MW of
hybridization required in the day
389 hybrid_cost_store ( each_day , column_count ) = total_hybrid_cost ;
390 hybrid_revenue_store ( each_day , column_count ) = total_hybrid_revenue ;
391 storage_waste_store ( each_day , column_count ) = storage_waste ;
392 no_stg_hyb_store ( each_day , column_count ) = no_stg_hyb_count ;
393
394 total_storage_revenue = 0;
395 total_hybrid_cost = 0;
396 total_plant_revenue = 0;
397 total_revenue = 0;
398 storage_waste = 0;
399 hybrid_amount = 0;
400 hybrid_hours = 0;
401 total_elec_prodn = 0;
402 total_hybrid_revenue = 0;
403 no_stg_hyb_count = 0;
404 end
55

57. 405 end
406 end
407 final_hybrid_store = [ date_store , hybrid_store ] ;
408 final_hybrid_cost_store = [ date_store , hybrid_cost_store ] ;
409 f i n a l _ s t o r e = [ date_store , master_store ] ;
410 f i n a l _ e l e c _ s t o r e = [ date_store , elec_store ] ;
411 final_hybrid_revenue_store = [ date_store , hybrid_revenue_store ] ;
412 final_storage_waste_store = [ date_store , storage_waste_store ] ;
413 final_stg_hyb_store = [ date_store , no_stg_hyb_store ] ;
56