Power Plant Engineering

3. Maximum possible power
max
P
A
= (8/27)
ρ
V
i
3
= (8/27 × 1) × 1.226 × 10
3
= 363 W/m
2
.
4. Assuming 
ηηηηη
 = 42%
P/A = 0.42 (P
tot
/A) = 0.42 × 613 = 257 W/m
2
5.  P = 0.257 × 
π
D
2
/4 = 0.257 × 
π
120
2
/4 = 2906 kW.
6. T
max
= 
2
27
(
ρ
DV
i
3
/N
s
) = 
2
27
[1.20 × 1.226 × 10
3
/(40/60)]
= 16,347 N.

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NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
117
Axial thrust (F
x
) in newtons F
x
max = (
π
/9) 
ρ
D
2
V
i
2
= 
π
/9 (1.226 × 120
2
× 10
2
)
 = 616,255 N.
Example 12. A 10 m/s wind is at 1 standard atmosphere and 15°C. Calculate:
1. The total power density in the wind stream.
2. The maximum obtainable power density.
3. A reasonably obtainable power density.
4. Total power produced if the turbine diameter is 120 m.
Solution. The air density,
ρ
= 
P
RT
= (1.01325 × 10
5
)/[287 (15 + 273)] = 1.226 kg/m
3
.
1. Total power density
total
ρ
A
= 
3
2
i
V
ρ
= 1.226 × (10)
3
/2 = 613 W/m
3
2. Maximum power density
max
A
ρ
= 
8
27
ρ
V
i
3
= (8/27) × 1.226 × (10)
3
= 363 W/m
3
.
3. Assuming 
ηηηηη
 = 40%
Actual power density,
P
A
= 0.4 
wt
P
A






= 0.4 × 613 = 245 W/m
3
.
4. Total power produced,
P = 
P
A
 
 
 
2
4
D
π
= 0.245 × 
π
(120)
2
/4 = 2770 kW.
Example 13. A 10 metre diameter rotor has 30 blades, each 0.25 metre wide. Calculate its
solidity.
Solution. Solidity = [(No. of blade × width)/(
π
× dia of rotor)] × 100 in %
Given that
Number of blade = 30
Width = 0.25 m
Diameter of rotor = 10 m

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118
POWER PLANT ENGINEERING
Putting all the values, we get
Solidity = [(30 × 0.25)/( 
π
× 10)] × 100 in %
= 24%.
Example 14. A 5 metre diameter rotor is rotating at 15 revolutions per minute (rpm) and the
wind speed is 3 m/s, calculate tip speed ratio of the rotor.
Solution. Tip speed ratio = [(
π
× dia of rotor × revolution per sec)/ Wind speed]
Given that;
Diameter of rotor = 5 m
Revolution = 15 RPM = 15/60 = 0.25 revolution per second
Wind speed = 3 m/sec
Tip speed ratio = [(
π
× 5 × 0.25)/ 3] = 1.3
Tip speed ratio = 1.3.
Example 15. Ocean wave on the coast of Tamil Nadu, India were with following data :
Amplitude 1 m, Period 6 s. Calculate the following:
Wavelength, velocity, energy density, density, power extracted from a wave of 10.0 m with a
power density, energy in 100 m wide wave. Assume density of ocean water as 1000 kg/m
3
.
Solution.  For ocean wave;
Wavelength (
λ
) = 1.56 T
2
m
In this example 
λ
= 1.56 T
2
= 1.56 × 62 = 56.16 m
Wave velocity, c = Wavelength(
λ
)/Period (T) =56.6/6 = 9.36 m/s
Frequency, f = 1/T = 1/6 s
–1
= 0.1667
Surface energy density
E/A = 1/2. D. a
2
.g J/m
2
= 1.2 × 1000 × 1 × 9.81 J/m
2
= 4.905 × 1000 J/m
2
.
Energy in 100 m wide wave
Area A = Wavelength (
λ
) × Width (W) = 56.16 × 100 m.
Energy = E/A × A
= (4.905 × 1000) × (56.16 × 100)
= 275.46 × 100000 J = 27.546 J.
Power = E × f W
= 27.546 × 10
3
× (1/6) = 4.56 × 10 W.
Power density = P/A
= (4.56 × 10
3
)/(56.16 × 100)
= 256.08 × 10
5
W/m
= 25.60 kW/m
2
.

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NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
119
THEORETICAL QUESTIONS
1. List various type of source of energy.
2. List the advantage of liquid fuel.
3. Write short notes on non-conventional energy.
4. Write short notes on petroleum.
5. Describe biofuels technology.
6. Write short notes on geothermal energy.
7. What is tidal energy and write the source of tidal energy ?
8. Briefly describe about ocean energy.
9. What is Wind energy and wind mill system ?
10. Write short note on biogas plant.
11. What are the features of biomass energy and describe what is the source of biomass?
12. What is solar cell technologies ?
13. Write down working of solar cells.
14. Write short notes on water heater.
15. Write short notes on box type solar cooker.
16. Write short notes on nuclear energy generation.
17. Write down properties and formation of coal.

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Chapter 3
Chapter 3
Chapter 3
Chapter 3
Chapter 3
Power Plant Economics and V
Power Plant Economics and V
Power Plant Economics and V
Power Plant Economics and V
Power Plant Economics and Variable
ariable
ariable
ariable
ariable
Load Pr
Load Pr
Load Pr
Load Pr
Load Problem
oblem
oblem
oblem
oblem
3.1 TERMS AND FACTORS
The main terms and factors are as follows:
1. Load Factor
It is defined as the ratio of the average load to the peak load during a certain prescribed period of
time. The load factor of a power plant should be high so that the total capacity of the plant is utilized for
the maximum period that will result in lower cost of the electricity being generated. It is always less than
unity.
High load factor is a desirable quality. Higher load factor means greater average load, resulting in
greater number of power units generated for a given maximum demand. Thus, the fixed cost, which is
proportional to the maximum demand, can be distributed over a greater number of units (kWh) supplied.
This will lower the overall cost of the supply of electric energy.