CORK INSTITUTE OF TECHNOLOGY
INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ
Autumn Examinations 2014
Module Title:
Sustainable Electrical Systems
Module Code:
ELEC7016
School:
Electrical & Electronic Engineering
Programme Title:
Bachelor of Engineering in Electrical Engineering
Programme Code:
EELEC_7_Y3
EELPS_8_Y3
External Examiner(s): Ms Mary Desmond, Mr. Colm Murray
Internal Examiner(s): Mr. Noel Mulcahy
Instructions:
Answer ALL questions.
Total of 50 marks.
Duration:
2 Hours
Sitting:
Autumn 2014
Requirements for this examination:
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you
have received the correct examination.
If in doubt please contact an Invigilator.
Q1.
The following graph show the Irish Total Primary Energy Requirement from 1990 – 2011
Renewables
Gas
Oil
Peat
Coal
Comment upon the general graph shape above and briefly detail the how the fuel types (Coal,
Gas, Renewables, Peat & Oil) have changed over the period detailed in the graph.
(10 Marks)
Q2.
The following graphs show the Efficiency of the Irish Efficiency of Electricity Supply and
the CO2 emissions per kWh of Electricity Supply from 1990 – 2011.
A. Briefly comment upon the Efficiency of Irish Electricity Supply and the emissions per kWh
of Electricity Supplied during this timeframe.
(4 Marks)
B. A 30 kW 4-pole, 60Hz machine rated at 400 V, drawing a full-load current of 53.5 A and
the full-load power factor is 0.75. The reactive component of this current can be taken as
the magnetizing current. Calculate from the manufacturer's data above the values of
excitation capacitance required for the generator chosen
i. Star
(3 Marks)
ii. Delta.
(3 Marks)
Q3.
A 3phase, 8 pole DFIG has a rating of 1300kW. The Stator is connected to a 50Hz supply
and the rotor is connected to a variable-frequency convertor that produces a frequency of
22Hz. The machine is operating in sub-synchronous speed and the machine develops
533kW. The wind turbine is connected to the rotor shaft by a speed raising gear box.
The losses:
Gear box and other misc losses:
2
Rotor I R losses:
2
Stator I R losses:
Stator iron losses:
Convertor losses:
Pv = 12 kW
Pjr = 5 kW
Pjs = 14 kW
Pf = 9 kW
Pc = 4 kW
Pc
P1
P2
VS
VR
50Hz
22Hz
Frequency
Convertor
Pe
Pjs
Pf
Pjr
Per
rotor
stator
Pr
f2 /f Pr
ns = 60 * f/p
Pm
n = ns (1 - f2 /f)
Pv
T m = 9.55 Pr / ns
12 kW
G
Pm = (1-f2 /f) Pr
Per = f2 /f Pr + Pjr
PL 533 kW
P2 = Per - Pc
Calculate:
1. The sub-synchronous speed of the rotor.
2. The mechanical power Pm delivered to the rotor.
3. The electromagnetic power Pr transferred from the rotor to the stator.
4. The mechanical torque Tm at the input to the rotor.
5. The electrical power Per delivered to the rotor windings.
6. The electrical power P2 absorbed by the convertor from the 50Hz supply.
7. The electrical power Pe delivered by the stator to the 50Hz supply.
8. The efficiency P1/PL of the wind turbine assembly.
(2 Marks)
(3 Marks)
(3 Marks)
(3 Marks)
(2 Marks)
(3 Marks)
(2 Marks)
(2 Marks)
Q4.
A wind speed anemometer is placed on site for a year at 50m. The site data produces an
-1
average wind speed of 6.8ms .
1. Using the Log Law equation manipulate the equation so that the von Karmen constant is
removed from the equation and that the wind speed at a known height can be calculated at a new
height.
(3 Marks)
2. Using the answer from part 1 above determine the effective wind speed by raising the wind
turbine nacelle to 85m You may assume that zo is 75cm.
(2 Marks)
3. Ireland wishes to achieve 40% of electricity generation by 2020. What challenges / issues may
impede this goal with respect to wind resource, grid infrastructure and electrical loads.
(5 Marks)
Some Useful Formula
 v*
v z  = 
 k

Log Law
  z 
 ln  
  zo 
Pm = Pin – Pmechloss
Pm = (1-f2/f1) Pr
2
Pm = IR R’R (1-s)/s
2
2
Pg = (IR R’R(1-s)/s)) + IR R’R
Pg = Pm / (1-s)
2
Ploss = IS RS
Pout = Pg - Ploss
η = - Pout/Pin
PF = -Pout / VI
R+jX = jXM(jX’LR + R’R/s)/[R’R/s + j(XM + X’LR)]
2
2
2
R = XM R’R/s / [(R’R/s) /[R’R/s +(XM + X’LR) ]
2
2
2
X = XM [(R’R/s) + X’LR(XM + X’LR)]/[(R’R/s) + (XM + X’LR) ]
2
Pm = (1-s)Is R
2
Pout = Is (R’R+R)
2
Q = Is (X’LS+X)