THE COPPERBELT UNIVERSITY
SCHOOL OF ENGINEERING
ELECTRICAL DEPARTMENT
COURSE: ENERGY UTILIZATION AND CONSERVATION (EE 570)
DATE: 5 May 2016
DURATION: 2 hours
TERM TEST 2
INSTRUCTIONS:
 Attempt all questions.
 Illustrate your answers with suitable diagrams where necessary and show all essential working
equations. Failure to do so will be penalized.
 Include units with your solutions, if applicable.
 Underline your answers.
REFERENCE EQUATIONS

  d  
Gon  d   1367 1  0.033 cos  2 
 
  365   

 360  284  d  
  23.5 sin 


365


R 
f

4
0.02
(a person's foot)


2  r 
ln 
 (for ground rods)
2
 r 

Rg 
(for a hemisphere)
2 r
Rg 
Angle of Incidence:
cos()=sin()sin()cos()-sin()cos()sin()+cos()cos()cos()cos()+cos()sin()sin()cos()
1. A 30m2 room is lit by five luminaires. The room must be lit by the luminaires for eight
hours per day. Each luminaire has two lamps which can be fitted with either incandescent
bulbs or Compact Florescent Light (CFL) bulbs. The luminaires have a CU of 0.80 and
LLF of 0.87. The cost of energy is K0.3/kWh. The details for the bulbs are found in the
following table:
Bulb Type
Incandescent
CFL
Lumen Output
(lm)
800
810
Power
Consumption (W)
60
15
Cost
(K/bulb)
20
70
Lifespan
(years)
2.5
10
a. Assume the incandescent bulbs are used in the luminaires. Compute the resulting
lux in the room. [10 pts]
b. What is the total cost of using incandescent bulbs in the luminaires over a 10-year
period? Include the purchase cost of the bulbs, their replacement cost, and energy
cost. [5 pts]
c. Is it less expensive to use CFL bulbs instead of incandescent bulbs over a 10-year
period? If so, compute the total savings. [5 pts]
A=30; (area)
LLF=0.87;
CU=0.8;
n=2; (bulbs per luminaire)
F=800; (lumen output per bulb)
N=5; (number of luminaires)
h=8; (number of hours lit per day)
r=0.3 (cost of electricity in K/kWh)
cost = 20; (cost per incandescent bulb)
costCFL = 70; (cost per CFL)
y=10 (years to consider)
E = n*N*F*CU*LLF/A = 185.6 lux
Incandescent Cost
P=60 (Watts)
Energy=N*n*P*365*y*h = 17520 kWh/10 years
Energycost=Energy*r/1000 = K5256 per 10 years
Bulb=cost*n*N*4 = K800 (purchase and replacement costs)
Total = Energycost + Bulb = K6056
CFL Cost
PCFL=15;
ECFL = N*n*PCFL*365*y*h = 4380 kwh/10 years
ECFLcost = ECFL * r/1000 = K1314 per 10 years
BulbCFL= n*N*costCFL = K700
CFLTotal = ECFLcost + BulbCFL = K2014
It is less expensive to use CFL bulbs. The savings over 10 years is K4042.
2. Dr. Louie accidently touched an energized 230V wire with his left hand while touching a ground
rod in his right hand. The ground rod is 2 m in length, has a radius of .01 m, and the soil in which it
is buried has a resistivity of 1000 Ohms-meters. Dr. Louie’s hand-to-hand resistance is 1800 Ohms.
Dr. Louie is standing on one foot. The resistance from his left hand to his foot is 1000 Ohms.
Assume the wire has zero resistance and the source has zero ground resistance.
a. Draw a circuit diagram showing the voltage source, and the relevant resistances. Label all
circuit elements. [3 pts]
b. Compute the current flowing through Dr. Louie’s left hand. [10 pts]
c. Compute the current flowing through Dr. Louie’s foot. [5 pts]
d. If Dr. Louie places his other foot on the ground, is he more likely to survive the shock?
Why or why not? [2 pts]
The problem can be interpreted several ways, credit is given based on the circuit drawn.
V=230 (voltage source)
l=2 (ground rod length, in m)
r=0.01 (ground rod radius, in m)
rho=1000 (resistivity in Ohm-m)
Rhh=1800 (hand-to-hand resistance)
Rhf=1000 (hand to foot resistance)
Rf = 3133 (computed by given equation) OR 3000 (approximation)
Rg=(rho/(2*pi*l) )*log((2*l+r)/r) = 476.98 Ohm
Now reduce the circuit
Req1=Rhh+Rg = 2277 Ohm (resistance of ground rod and between the hands)
Req2=Rhf + Rf = 4133 Ohm (from left hand to foot)
Req3=(Req2*Req1)/(Req1+Req2) = 1468 Ohm (equivalent resistance)
Ilh = V/Req3 = 156.7mA (current through his left hand)
If = Ilh*(Req1/(Req1+Req2)) = 55.6 mA (current through the foot)
3. Consider the figure shown below. Assume that the circuit breaker is rated at 10 A and the
source voltage is 230 V. The system parameters are Rn = Rh = 0.5 , Rg-man = 500 , Rman =
1000 , Rg = 10 , and Rg-xfm = 30 .
a. Compute the fault current, If. [12 pts]
b. Does the circuit breaker trip? [2 pts]
c. Compute the current through the person. [8 pts]
Vs=230 (source voltage)
Rh=0.5 (hot wire resistance)
Rn=0.5 (neutral wire resistance)
Rgman=500 (ground resistance of the man)
Rman=1000 (resistance of the man)
Rg=10 (ground rod resistance)
Rgs=30 (source ground (xmfr))
Reduce the circuit
Req1=Rg*(Rgman+Rman)/(Rg+Rgman+Rman) = 9.93 Ohms
Req2=Req1+Rgs = 39.93 Ohms
Req3=(Req2*Rn)/(Rn+Req2) = 0.493 Ohms
Req=Req3+Rh = 0.994 Ohms
If=Vs/Req = 231.4 A (the circuit breaker trips)
There are three paths from the chassis to the ground (the person, the equipment ground conductor
and the neutral)
In=If*(Req2/(Rn+Req2)) = 228.6A (neutral current)
Iman= (If-In)*(Rg)/(Rg+Rgman+Rman) = 19mA
4. Consider a surface above Earth’s atmosphere over Cairo, Egypt (latitude: 30.3o N) tilted at
30.3o. The day is 31 January.
a. Compute the angle of incidence, at9:00, in degrees. [5 pts]
b. Compute the irradiance on the surface at 12:00. [10 pts]
c. Compute the irradiance on the surface at 15:00. [5 pts]
phi=30.3 (latitude)
d=31 (day of the year)
B=30.3 (tilt)
(note MATLAB uses “sind” and “cosd” for sine and cosine computations in degrees)
delta= 23.5*sind(360*(284+d)/365) = -17.82 degrees (declination)
w=-3*15 = -45 degrees (hour angle)
costheta = cosd(delta)*cosd(w) = 0.6732 (using the reduced given equation)
theta = acos(costheta)*180/pi = 47.7 degrees
Gon = 1367*(1+0.033*(cos(2*pi*d/365))) = 1405.8 W/m2 (irradiance on plane horizontal to the
sun)
w=0 degrees (hour angle at 12:00)
Gon*cosd(delta)*cosd(w) = 1338.4 W/m2
Gon*costheta = 946.4 W/m2 (using the shortcut that the angle of incidence at 9:00 is the same as
at 15:00)
MULTIPLE CHOICE
Select the best response(s) to the following questions. [3 pts each]
5. A light source has a luminous intensity of 79.6 cd measured 2m from the source. The light is
emitted through a hemisphere. What is the lumen output of the source?
A. 79.6 lm
B. 100 lm
C. 500 lm
D. 1000 lm
E. None of the above.
W = 4*pi/2 (sr for a hemisphere)
F = 79.6
I = W*F = 500 lm
6. How does frequency of the source generally affect the severity of an electric shock?
A. Shocks DC sources are less severe than AC
B. Shocks from AC sources are less severe than DC
C. Shocks from lower frequency (50-100 Hz) sources are more severe than higher
frequency sources, but less severe than shocks from DC sources.
E. None of the above.
7. At solar noon in Kitwe on 21 December, the sun is:
A. To the North
B. To the South
C. Directly overhead
Declination is -23.5 degrees (in the south, since the latitude of Kitwe is only -12.8 degrees, the sun
appears in the south)
8. What are the disadvantages with the “bonded neutral” wiring scheme below. (Circle all that
apply)
A. The circuit breaker will likely trip if there is an internal equipment fault.
B. A person touching the chassis will experience a harmful shock if the person has a large
ground resistance.
C. The equipment chassis may develop an unsafe voltage under high loads.
E. None of the above.
9. The Global Horizontal Irradiance is the sum of the beam irradiance and the direct irradiance.
A. True
B. False
10. What is the maximum luminous efficacy of a 1W source?
A. 1 lm/W
B. 1/683 lm/W
C. 0 lm/W
D. None of the above.