GLASGOW CALEDONIAN UNIVERSITY
SCHOOL OF ENGINEERING & BUILT ENVIRONMENT
Example Diet
2 HOURS
Attempt FOUR questions
All questions carry equal marks
SOLUTIONS
To be supplied:
Question Paper (Supplied)
Formula sheet (Supplied)
Lined Examination Script (Supplied)
Calculator (Allowed)
Module: Renewable Power Integration
Module No: MHH623513
Question No.: 1
Year: 4
DIET: 2nd
SESSION: 18/19
Author: D. Chen
(a)
(i)
The student should clearly indicate ⑤ and ⑥ are DG.
[2]
[Each irrelevant nominate will result in 1 mark deduction.]
(ii)
The student should clearly indicate that:
Unplanned islanding operation may energize a part of the network meant to be deenergized for maintenance.
[2]
Un expected energizing may lead to hazards of electric shock to maintenance
personnel.
[2]
(b)
(i)
The equivalent circuit
[4]
For any missing essential element, 0.5 mark will be deducted.
(ii)
The equivalent impedance in total:
(iii)
=0.22+0.83/2=0.635 p.u.
Considering
Therefore
sinδ =
=
𝑃𝑒 𝑋0
[1]
𝐸𝑉
1×0.635
1×1
= 0.635
[1]
o
the power angleδ=39.4
[1]
(iv)
The student should clearly indicate that If a three phase fault happens at point A, the
voltage at Point A will be 0.
[1]
No power can be delivered from the generator to the grid.
[2]
(v)
The new impedance:
[2]
=0.22+0.83=1.05 p.u. [1]
(vi)
The maximum electrical power output is
𝐸𝑉
𝑋1
=
1×1
1.05
= 0.95
[1]
The student should clearly indicate that
The maximum electrical power can be delivered is less than the input mechanical power. [1]
This means that the electrical torque will not be able to decelerate the rotor after the
clearance so the frequency or the power angle will not be maintained. The system will lose
synchronization and become unstable.
[1]
Module: Renewable Power Integration
Module No: MHH623513
Question No.: 2
Year: 4
DIET: 2nd
SESSION: 18/19
Author: D. Chen
(a)
(i)
The student should clearly indicate that
Roof-top PV is more close to the loads and can be consumed locally.
[1]
The distribution distance between roof-top generation and loads is much shorter
than off-short windfarm; hence the distribution/transmission losses are lower when
delivering the same amount of power.
[1]
(ii)
The student should clearly indicate that
in the event of a fault within the prospected network, the hydro plant will produce
significant fault current;
[1]
such fault current will make the fault current distribution deviating from the preset values in
the relays causing undesirable operation of protective devices.
[2]
Such extra fault current may exceed the rating of switch gears.
[1]
(b)
(i)
(ii)
(iii)
(iv)
for feeder section CD the first hosting capacity is:
P gen.max= P cons,max+ P cons,min
[1]
=2 MW+600 kW=2.6 MW
[1]
For feeder section BC, the whole downstream load should be considered, that is, load D plus load C:
P gen,max= (2 MW+2.5 MW)+ (600 kW+700 kW) [1]
= 5.8 MW
[1]
For feeder section AB:
P gen,max= (2 MW+2.5 MW +3 MW)+ (600 kW+700 kW+800 kW)
[1]
= 9.6 MW
[1]
The second hosting capacity for feeder section CD is :
P gen,max= P max,limit +P cons,min
[1]
= 4 MW+500kW= 4.5 MW
[1]
The second hosting capacity for feeder section BC is :
P gen,max= 7 MW+(700 kW+600 kW)
[1]
= 8.3 MW
[1]
The second hosting capacity for feeder section AB is :
P gen,max= 10 MW+(600 kW+700 kW+800 kW)
[1]
= 12.1 MW
[1]
The student should clearly indicate that when the second hosting capacity for the first feeder section
is exceeded
[2]
(above 12.1 MW), the circuit breaker at location A will trip. [the correct value for 1 mark]
The 2nd capacity at Location D is 4.5 MW
[1]
The 2nd capacity at Location C is 8.3 -4.5 = 3.8 MW [1]
The 2nd capacity at Location B is 12.1 -8.3 = 3.8 MW [1]
Module: Renewable Power Integration
Module No: MHH623513
Question No.: 3
Year: 4
DIET: 2nd
SESSION: 18/19
Author: D. Chen
(a)
(i)
The student should clearly indicate that
PE interfaced DG can contribute to a prolonged fault current hence can impact
the protection system.
[1]
The impact to protection system is limited as the fault current is normally
controlled within nominal value and can be further suppressed by it control
strategy if necessary.
(ii)
[2]
The student should clearly indicate the follows or equivalent
the DG is not “fit-and-forget” anymore, the operator directly involved in DG
operation;
[2]
Extra control, monitoring and real time communication channels become
essential, which adds up to system complexity and can lower overall reliability.
[2]
(b)
(i)
(ii)
(iii)
(iv)
Calculate the reactive power at no load
The reactive power at no load is 10/5 = 2 MVar
[3]
Calculate the reactive power at full load.
The reactive power at full load is 2 + (0.15+0.1+0.05)X10 = 5 MVar [3]
Calculate the desirable range of capacitance connected to MV network and
corresponding resonance frequencies.
To compensate the full reactive power, the size of the reactive power
compensation is between 2 [1.5] and 5 Mvar [1.5].
Calculate the maximum and minimum resonance frequencies.
For 2 MVA
𝑆
1
𝑓𝑟𝑒𝑠,𝑚𝑎𝑥 = 𝑓0 × √𝑋 𝑡𝑟𝑄 = 50 × √0.1×0.2=354Hz
𝑡𝑟
[1.5]
For 5 MVA
𝑆
1
𝑓𝑟𝑒𝑠,𝑚𝑖𝑛 = 𝑓0 × √𝑋 𝑡𝑟𝑄 = 50 × √0.1×0.5=223.6Hz
𝑡𝑟
[1.5]
(v)
Find out the order of harmonics that can be potentially induced by the resonance.
5th harmonics (at 250 Hz)
[1.5]
and 7th harmonics (at 350 Hz) [1.5]
are between 223.6 Hz and 354 Hz.
(vi)
𝑆
𝑓0 × √𝑋 𝑡𝑟𝑄 = 𝑓𝑟𝑒𝑠
𝑡𝑟
1
√0.1×𝑄 = 5  𝑄 = 0.4 𝑝. 𝑢.
[1]
[1]
1
√0.1×𝑄 = 7  𝑄 = 0.2 𝑝. 𝑢.
[1]
The risky reactive compensation values for 5th and 7th order hramonics are 10X0.4 =4 MVar
and 10X0.2 =2.0 MVar
Module: Renewable Power Integration
Module No: MHH623513
Question No.: 4
Year: 4
DIET: 2nd
SESSION: 18/19
Author: A. Gowaid
(a) 12 points, 6 for (i) and 6 for (ii)
(i)
- Feed-in Tariffs (FiT) are a financial incentive to support distributed and smallscale renewable energy generation (up to 5 MW in the UK). [1]
FiT is available for the following generation technologies: Wind power - Solar
PV power
- Hydro power - Anaerobic digestion - Micro-CHP (≤ 2kW) [2]
There are three sources of financial benefit from a generation project
receiving FiTs:
Generation tariff:
A fixed price for each unit of electricity generated by the renewable
system.
[1]
Export tariff:
A guaranteed price for each unit of electricity exported to the grid. [1]
Import reduction:
Reducing your import from the grid by using your own electricity. [1]
(ii)
Renewable Energy Certificate (REC) is a concept brought forward by
regulators to incentivise the installation of more renewable energy plants by
requesting licensed electricity suppliers to buy a certain percentage of their
supply from eligible renewable generation plants.
[1]
When a renewable energy generator — a wind farm or solar power plant, for
example — generates a megawatt-hour (MWh) of power, it receives one REC,
a certificate saying that it generated one MWh of electricity from clean
sources, which it can also sell in the open market. [2]
This means every renewable generator has two revenue streams, electricity
sold at prevailing market rates, and REC sold at rates set by the regulator
(normally updated annually).
[1]
Suppliers prove compliance to the regulator by presenting a sufficient number
of RECs to cover their obligation. [1]
Suppliers can purchase RECs directly from renewable power plants, or from
other suppliers. So, RECs is in reality a commodity. [1]
(b)
Total of 13 points
(i)
P = S cosφ = 10000*0.8 = 8kW [2]
Q = S sinφ = 10000* sin (cos-1φ) = 6kVAr [2]
(ii)
(iii)
P = 3VI cosφ  I = P/(3Vcosφ) = 8000/3/220/0.8 = 15.15A [2]
Po = 0.5 A ρ Vw3 Cp1*η [2]
Po1/Po2 = Vw13 Cp1 η1 / Vw23 Cp2 η2
[1]
But Cp2 = 0.8 Cp1
[1]
and η1 = η2
[1]
3
Then Po2 = Po1 *Vw2 /vw13 *0.8 = 8000*(8/11)3 *0.8 = 2.462 kW [2]
Module: Renewable Power Integration
Module No: MHH623513
Question No.: 5
Year: 4
DIET: 2nd
SESSION: 18/19
Author: A. Gowaid
(a) 6 points in total
(b) 6 points in total 3 for advantages and 3 for disadvantages (1 each)
Advantages of DFIG-WT
1- Variable speed wind turbine, so more output power and less mechanical stresses on
WT drive train. [1]
2- Decoupled control of active and reactive power allows grid connection at unity
power factor. [1]
3- Utilization of part-rated power electronics converter which reduces cost. [1]
Disadvantages of DFIG-WT
1- Use of brushes and gearbox reduce reliability and increase maintenance frequency.
[1]
2- Prone to damage under sever voltage dips and does not inherently contribute to
system inertia. [1]
3- Cannot scale easily to higher power or voltage levels due to several induction
machine design constraints. [1]
(c) Total 7 points = 1 point for first equation + 6 spread equally on iterations.
(ii)
The current (I) in the circuit may be calculated from:
I =S1*/V1* = (0.5 + j0.101)/(1.0117 - j0.0551) = 0.4873 + j0.1264 per unit or 0.503pu.
[2]
With a connection voltage of 33 kV the base current is given by
Ibase = VAbase/sqrt(3) Vbase = 50 ∙ 106/1.732 ∙ 33 ∙ 103
= 875 A [2]
Therefore, I = 0.503x875 = 440A
Power loss: I2R = 0.5032x0.05=0.0127 per unit
[2]
Module: Renewable Power Integration
Module No: MHH623513
Question No.: 6
Year: 4
DIET: 2nd
SESSION: 18/19
Author: A. Gowaid
(a) Total 8 points 1 point for each correct axis, 1 mark for each key point shown on graph.
3 points for correct curves.
(b) Power system inertia is the combined inertia of all rotating generators and motors
connected to the grid. [1]
Conventional generation is controllable and easy to plan output power ahead of
operation to meet forecasted demand by loads.
Most renewable generation is stochastic in nature due to the intermittency of
renewable resource (wind, solar, etc) [2]
Renewable power sources interface to the grid via power converters, reducing
overall system inertia. [1]
With lower inertia, and sudden supply/demand imbalance leads to wider frequency
swings due to increased ROCOF. [1]
Grid codes requirement:
1- Provide primary frequency control power to support grid frequency [1]
2- Provide synthetic inertia to reduce ROCOF in the occasion of supply/demand
imbalance. [1]
(C)
Nightly requirement= 500X 12= 6kwh [2]
Solar daily peak production in least sunny day= 140 X 2.4= 336 W/m2
Power output = 336 X 0.8 X 0.95= 255.36 W/m2
[2]
[2]
Solar module area required to sufficiently charge batteries = 6000/255.36= 26.6 m2
[2]
PV system power = 336*26.6 = 8.94 kW [2]