TEE 3108 POWER PLANT ENGINEERING
Worked examples
Date Given: 17th October 2017
1.
Give a detailed account of energy conversion process (both qualitatively and quantitatively but no numerical
calculations) of the following energy conversion systems:
[15 marks]
a) An incandescent electric bulb
Solution
An incandescent electric bulb converts electrical energy into a very high temperature (thermal) then light.
The gas (e.g. argon, nitrogen) enclosed in the bulb prevents the filament from glowing or burning.
Conversion: Electric energy  thermal energy  Light
The thermal energy is given using the equation P = I2R (I is the current and R the resistance of the filament)
[4 marks]
b) A standby diesel fuelled electric generator (energy conversion from fuel to electrical)
Solution
A standby generator uses diesel as a fuel, the energy conversion is given as: The diesel fuel contains
chemical energy due to the hydrocarbon composition, the fuel undergoes combustion in the engine (internal
combustion), and then the chemical energy is converted into thermal energy. The thermal energy as a result
of combustion leads to expansion in the piston driving the calm shaft connected to the driving shaft
(thermal energy converted into mechanical). The mechanical rotational energy from the shaft drives the
generator and then generator converts the mechanical energy of the shaft and the magnetic energy in the
exercitation into electrical energy.
Chemical
energy in
diesel
Internal
combustion
Thermal energy
in engine
cylinder
Electrical
energy from
the generator
Expansion in
the pistons
Mechanical
energy in
Shaft
Electromechanical
conversion
Electromagnetic
conversion
Magnetic
energy in the
excitation fields
[4 marks]
c) A coal fuelled thermal power plant for generating electricity
Solution
A coal fuelled thermal power plant uses coal as the fuel, the energy conversion is given as: The coal fuel
contains chemical energy due to the hydrocarbon composition, the fuel undergoes combustion in the
burner, and then the chemical energy is converted into thermal energy. The thermal energy is used the
1
boil water to produce steam. The thermal energy in the steam is converted into mechanical energy in the
steam turbine (expansion of steam in the turbine), i.e. thermal energy converted into mechanical. The
mechanical rotational energy from the turbine through the shaft drives the generator and then generator
converts the mechanical energy of the shaft and the magnetic energy in the exercitation into electrical
energy.
Chemical
energy in
coal
Combustion
in the burner
Thermal energy
in the steam
Electrical
energy from
the generator
Mechanical
energy in
turbine
Expansion in
steam turbine
Electromechanical
conversion
Magnetic
energy in the
excitation fields
Electromagnetic
conversion
[4 marks]
d) Electrical geyser (electrical to thermal)
Solution
An electrical geyser uses a heating element with resistance. The heater converts electrical energy into
thermal energy with the use of the relationship (P = I2R). The thermal energy is transferred to the water in
the geyser. At all stages there are losses. Conversion: Electric energy  thermal energy
[3 marks]
2.
Calculate the energy, in kJ, a 65 Watt light bulb consumes when turned on for 30 minutes.
[4 marks]
Solution
Energy  Power  time  65W  0.3hr  32.5 Wh  0.0325kWh
[2 marks]
but , 1kWh  3600 kJ, then we have
[2 marks]
Energy  0.0325 kWh  3600kJ/kWh  117 kJ
3.
The kilowatt-hour, or kWh, is not a unit of power but of energy. Notice that kilowatt is a unit of power and
hour is a unit of time. A kilowatt hour is equal to 1 kW delivered continuously for 1 hour (3600 sec). Assume
your electric bill showed you used 1155 kWh over a 30-day period.
[8 marks]
a) Find the energy used, in (kJ, BTU, kcal, toe, tce) for the 30 day period.
Solution
The energy used in 30 days is 1155kWh, expressed in other units as:
1kWh  3,600kJ

Energy  1,155kWh  3,600kJ/kW h  4,158,000kJ
1kWh  3,414BTU

Energy  1,155kWh  3,414BTU/k Wh  3,943,170BTU
[1 mark]
[1 mark]
2
1kWh  860.4kcal

Energy  1,155kWh  860.4kcal/ kWh  993,762kcal
[1 mark]
1kWh  85.98x10 -6 toe

Energy  1,155kWh  85.98x10 -6 toe/kWh  9.92838x10 -2 toe
1kWh  122.8x10 -6 tce

Energy  1,155kWh  122.8x10 -6 tce/kWh  1.41834x10 -1 tce
[1 mark]
[1 mark]
b) Find the average energy used in J/day.
Solution
The average energy used in 30 days is obtained as:
Energy average  Energy
 4,158,000,000kJ
 1.386  108 J/day
30days
30days
[2 marks]
c) At the rate of N$1.52/kWh, what will your electric bill be for this month?
Solution
The electricity bill in the given in 30 days is obtained as:
Bill  1155kWh 1.52 N$/kWh  N$1 755.6
4.
A 100 Watt incandescent light bulb is 5% efficient.
a) How much energy in kWh does the bulb convert to light during 12 hours?
[1 mark]
[7 marks]
Solution
The energy the bulb converts into light in 12 hours is obtained as:
Energy  0.05  100
kW  12hrs  0.06kWh
1000
[2 marks]
b) How much energy in kWh does it use in 12 hours of operation?
Solution
The energy the bulb uses or consumes in 12 hours is obtained as:
Energy  100
kW  12hrs  1.2kWh
1000
[2 marks]
c) Convert total energy use to MJ, toe, tco
Solution
The total energy use or consumption is obtained as:
3
1kWh  3.6MJ

[1 mark]
Energy  1.2kWh  3.6MJ/kWh  4.32MJ
1kWh  85.98x10 -6 toe

[1 mark]
Energy  1.2kWh  85.98x10 -6 toe/kWh  1.0314x10 -4 toe
1kWh  122.8x10 -6 tce

5.
[1 mark]
Energy  1.2kWh  122.8x10 -6 tce/kWh  1.4736x10 -4 tce
A diesel powered electric generator is rated 10kW. The generator uses an engine with average efficiency of
38% (chemical to mechanical) and mechanical to electrical efficiency of 85%. The generator is operated at 75%
of its rated power for 5 hours. The calorific value of diesel fuel used is 43.4 MJ/kg and density 0.832 litres per
kg.
[15 marks]
a) Calculate overall efficiency of the generator
Solution
The overall efficiency of the generator is obtained as:
  cm   me  0.38  0.85  0.323  32.3%
[2 marks]
b) The calorific value of diesel in kWh/kg and kWh/litre
Solution
The calorific value in kWh/kg and kWh/litre:
1kWh  3.6MJ
 1MJ  1 kWh
3.6
 calorific value  43.4MJ/kg  1 kWh / MJ  12.06 kWh/kg
3.6
[2 marks]
1kg  0.832 litre

calorific
value
 43.4MJ/kg  1
3.6
kWh / MJ  1
0.832
kg / litre
[2 marks]
 14.49kWh/l itre
c) Estimate the amount (in litres) of diesel fuel used in 5 hours
Solution
The amount of diesel used in 5 hours obtained by first determining the electrical energy supplied by the
generator as follows:
Energyelectrical  0.75  10 kW  5hrs  37.5kWh
[2 marks]
The energy content in the fuel is obtained as follows:
Energy fuel 
Energyelectricity
37.5kWh
 116.10kWh
 
0.323
[1.5 marks]
Therefore the amount of fuel is obtained as follows:
4
Fuellitres 
Energy fuel
calorific
 116.10kWh
value
14.49kWh/l itre
 8.01 litre
[1.5 marks]
d) If the generator is used for 5 hours daily, estimate the cost of fuel used to operate the generator in 30 days
(assume cost of diesel is UGSH 2850 per litre)
Solution
The total cost of fuel to be used is obtained as follows:
Fuelcos t  8.01 litre /day  UGSHG 2 850/litre  30days  UGSH 684 855
6.
[2 marks]
A coal powered thermal power plant is rated 100 MW (net electric power output). The plant is operated daily
by a utility company as follows: 100% of rated power for 6 hours/day (at overall efficiency of 32%); 80% or
rated power for 6 hours per day (at overall efficiency of 30%); and 60% of its rated power for 12 hours per day
(at overall efficiency of 28%). The fuel used in the plant is hard coal with calorific value of 34.2MJ/kg.
[16 marks]
a) Estimate the daily electric energy (kWh, kJ, BTU, kcal, toe, tce) generated at the plant
Solution
The daily electric energy generated is obtained as follows:
3
Energydaily  100MW   rk  t k hr
[2 marks]
k 1
Energydaily  100MW  100%  6hr   80%  6hr   60%  12hr 
1800MWh  1.8  106 kWh
[2 marks]
The daily electric energy generated expressed in other units as:
1kWh  3,600kJ

Energy  1.8  106 kWh  3,600kJ/kWh  6.48  109 kJ
1kWh  3,414BTU

Energy  1.8  10 6 kWh  3,414BTU/k Wh  6.1452  109 BTU
1kWh  860.4kcal

Energy  1.8  106 kWh  860.4kcal/ kWh  1.54872  109 kcal
1kWh  85.98x10 -6 toe

Energy  1.8  106 kWh  85.98x10 -6 toe/kWh  154.764 toe
1kWh  122.8x10 -6 tce

Energy  1.8  106 kWh  122.8x10 -6 tce/kWh  221.04 tce
[1 mark]
[1 mark]
[1 mark]
[1 mark]
[1 mark]
b) Estimate the amount (in tons) of coal required to operate the plant (daily, for 30 days and annually)
5
Solution
To estimate the amount of coal required to operate the plant daily is obtained by determining the energy
content of the fuel required:
3
 r t 
[2 marks]
Energycoal  100MW    k k 
k 1   k

 100%  6hr   80%  6hr   60%  12hr 
Energycoal  100MW  



[2 marks]
32%
28% 
  30%  

 6046.43 MWh  6.04643  106 kWh
The energy content of the fuel in MJ expressed in other units as:
1kWh  3,600kJ

Energy  6.04643  10 6 kWh  3,600kJ/kW h  2.1767142857  1010 kJ
[1 mark]
 2.1767142857  10 7 MJ
The amount of fuel is obtained as:
1kg  34.2MJ

Coaltons 
2.1767142857  10 7 MJ
 636466.17 tons  6.365  10 5 tons
34.2MJ/kg 1000kg/ton
[2 marks]
6