This document discusses various types of losses that can occur in solar photovoltaic power systems, including module-level losses from shading, dust, reflection; thermal losses from heat generated by the modules; incidence angle losses from the angle of sunlight hitting the modules; and cable losses from resistance in the wiring. It provides percentages for typical losses from each category, noting that overall system losses can range from 10% to 70% of total energy produced, with about 24-25% lost on average due to these loss factors.

1. LOSSES IN SOLAR POWER
PLANT
PHOTOVOLTAIC POWER PLANT
ENGINEERING
M.TECH(SOLAR),PDPU
BY VANSH PANDYA
17MSE010

2. MODULE LEVEL LOSSES
DC CABLE LOSSES
INVERTER LOSSES
AC CABLE LOSSES
TRANSMISSION LOSSES

3. SHADING LOSSES
DUST LOSSES
REFLECTION
SPECTRAL
LOSSES
IRRADIANCE LOSSES
THERMAL LOSSES
ARRAY MISSMATCH
LOSSES
DC CABLE
LOSSES
INVERTER
LOSSES
AC CABLE
LOSSES
• It is necessary to minimize the losses by eliminating the
factors that cause the losses occurred in PV systems to
get high energy yield.
• Factors that may cause losses in PV systems are
environmental factors such as shade, dust, snow, rain,
temperature and losses due to system components
such as cables, inverters and batteries.
• PV systems should be installed taking into account the
losses and the produced energy should be consumed in
local areas where it was produced as much as possible.
• The power loss can vary between 10% to 70%.
About 24%-25% of the produced energy by the PV
system is lost due to some system losses as shown in
the figure.

4. While the PV
module absorbs
some of the solar
radiation, a certain
amount of solar
radiation is
reflected back
from the module
surface.
To reduce
reflection, the
module surfaces
are coated with an
anti-reflecting
film.
REFLECTION
LOSSES
Nh
These losses are
caused by pv module
surface due to any
reason and
decreasing the
radiation
One study says that
this losses reach
15% in
extraordinary case
For big plant to
improve efficiency
clean module
regularly.
DUST LOSSES
•SHADOW ON MODULE
•TREE SHADOW
•NEAREST BUILDING
SHADOW
•WATER TANK SHADOW
ETC
•So we have to reduce
this shadow effect on
module
•Therefore for selection
of location for installing
carefully
SHADING LOSSES

5. Thermal Losses
• Producing electricity starts with receiving solar radiation on the PV module surface. While some of the
incident solar radiation is transformed into electrical energy, a portion of solar radiation is converted into
heat energy. PV performance is decreasing with increasing temperatures that occur in the panel.
• The conversion rate of PV panels is about in the range of 5-25%. Therefore, more energy that the solar
modules can’t convert it to the electrical energy causes heating of the modules and so thermal losses.
• Nominal operating cell temperature:-Temperature reached by open circuit cell in module under this
condition:-
Irradiance 800 W/m^2
Air temp 20 C
Wind velocity 1 m/s
Mounting Open back side
𝑻𝒄𝒆𝒍𝒍 = 𝑻 𝑨𝑰𝑹 +
𝑵𝑶𝑪𝑻 − 𝟐𝟎
𝟖𝟎
𝒙 𝑺
Best Module at NOCT of 33 C
Worst module at NOCT of 58 C
Typically module at NOCT of 48 C

6. Losses due to module temperature:-
𝐼 𝑇 = 𝐸𝐴𝑇 − 𝐸𝐴
𝐸𝐴𝑇 =
𝐸 𝐴
1 + 𝛼 𝑇 𝑚 − 25
Where, = Temperature Co-Efficient
EA= Output energy of PV array
EAT = Output energy of module
Tm = Module temperature
Temperature
Voltage
Current

7. Incidence angle Losses:-
• 𝐹𝐼𝐴𝑀 = 1 −
𝑏0
cos i−1
• Where, i = incidence angle
Real crystalline b0= 0.05
Single glazed thermal solar b0= 0.1
Example ,
1 Nov ,Mumbai incidence loss on module
by calculation i =0.824
FIAM = 0.98
If 1000 W/m2 irradiation is there we get only 980 W/m2 after incidence
losses.

8. Cable losses
• There are two types of cable used in PV plant :i) DC cable losses and ii) AC
cable losses
• DC cable is used between module & inverter and AC cable is used between
inverter & grid.
• A way to limit this losses is to minimize voltage drop in cables. A drop of 1%
is suitable but should not be more than 3%.
• Voltage drop changes with cross-section area of cable, if area increases
voltage drop occurs less.
• As cross-section area increases, cost of cable also increases and hence
optimum cable cross-section should be chosen.
• http://photovoltaic-
software.com/DC_AC_drop_voltage_energy_losses_calculator.php

9. • Where :
ΔU : voltage drop in Volt (V)
b : lenght cable factor, b=2 for single phase wiring, b=1 for three-phased wiring.
ρ1 : resistivity in ohm.mm2/m of the material conductor for a given temperature. At 20 celcius degree °C the
resistivity value is 0.017 for copper and 0.0265 for aluminium.
Note that resistivity increases with temperature. Resistivity of copper reaches around 0.023 ohm.mm2/m at
100 °C and resistivity of copper reaches around 0.037 ohm.mm2/m at 100 °C.
Usually for voltage drop calculation according to electrical standards it is the resistivity at 100°C that is used
ρ1 = ρ0*(1+alpha(T1-T0)), here ρ0 = resistivity at 20°C (T0) and alpha = Temperature coefficient per degree C
and T1 = temperature of the cable.
T1 : Temperature of the cable (default value = 100°C).
Note that from experience, a wire with a correct sizing should not have an external temperature over 50°C,
but it can correspond to an internal temperature of the material around 100°C.
L : simple lenght of the cable (distance between the source and the appliance), in meters (m).
S : cross section of the cable in mm2
Cos φ : power factor, Cos φ = 1 for pure resistive load, Cos φ< 1 for inductive charge, (usually 0.8).
λ : reactance per lenght unit (default value 0.00008 ohm/m)
Sin φ : sinus (acos(cos φ)).
Ib : current in Ampere (A)
NB : For DC circuit, cos φ=1, so sin φ=0.
Voltage drop calculation

10. •
Voltage drop in percent :
ΔU(%) = 100 x ΔU/U0
Where :
ΔU : voltage drop in V
U0 : voltage between phase and neutral (example : 230 V in 3-phase 400 V system)
• ENERGY LOSSES
•
Energy losses in a cable is mainly due to resistive heating of the cable.
It is given by the following formula :
• E = a x R x Ib²
Where :
E : energy losses in wires, Watt (W)
a : number of line coefficient, a=1 for single line, a = 3 for 3-phase circuit.
R : resistance of one active line
Ib : current in Ampere (A)
R is given by the next formula :
R = b x ρ1 x L / S
b : lenght cable factor, b=2 for single phase wiring, b=1 for three-phased wiring.
ρ1 : resistivity of the material conductor, 0.017 for copper and 0.0265 for aluminium (temperature of the wire of 20°C) in
ohm.mm2/m.
L : simple lenght of the cable (distance between the source and the appliance), in meters (m).
S : cross section of the cable in mm2
• NB : for direct current the energy losses in percent is equal to the voltage drop in percent.

11. Inverter losses
• 𝐼𝐿 = 𝐸 𝑜 𝐼
− 𝐸𝐴
• By this equation we can calculate this losses
• Where,
EA = output of array
EOI = output of inverter
IL= losses of inverter
Generally efficiency of inverter is 90%-98% so there is only 1%-2%
losses in conversion of DC to AC.

12. SHADING LOSSES 7%
DUST LOSSES 2%
REFLECTION 2.5%
SPECTRAL
LOSSES 1%
IRRADIANCE LOSSES
1.5%
THERMAL LOSSES
4.6%
ARRAY MISMATCH
LOSSES 0.7%
DC CABLE
LOSSES 1%
INVERTER
LOSSES 3%
AC CABLE
LOSSES 0.5%
Example :-
1621 kwh/m2
1236 kwh/m2
+14.7%
horizontal
and global
irradiation
-33%
optical
losses(iam
,shading)
14.90% PV
conversion
efficiency
185.4
kwh/m2
After this losses calculation
7+2+2.5+1+1+.7+4.6+
1.5+3+.5=23.8%
185.4*0.238=44.1
185.4-44.1=141.3kwh
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