Analysis Calculations
April 30, 2025
Team 10105
1
The Coefficient Of Power
The coefficient of power quantifies the turbine’s efficiency by comparing the
theoretical energy of the cross sectional area of air subjected to the turbine and
the real energy output. Moreover, It reaches average 0.35 in Savinous VAWTs,
so it is essential to be high to satisfy the design requirements. The power
coefficient is defined by equation 1 as:
CP =
Poutput
1/2ρAV 3
(1)
The prototype has a height of 40cm and width of 20cm, then
A = 0.2 × 0.4 = 0.08m2
(2)
ρistheair′ sdensityequals = 1.146kg/m3
V 3 is wind’s speed cubed, V = 4.5m/s
V 3 = 4.53 = 91.125(m/s)3
(3)
T heactualenergyproduced(Poutput ) = 1.15W
The maximum Power Coefficient Of Power (Cp) is equal to
CP =
1.15
= 0.275
1/2 × 91.125 × 1.116 × 0.8
(4)
Which is the maximum Cp achieved, and is relatively high compared to lost
energy due to gear system, low wind seeds, and the gear system. Indicating the
efficiency of the prototype in the generation of electricity.
1
2
The Design’s Aerodynamics
Savinous turbines operate by pressure differences between the concave and
convex sides of the turbine. Four modifications were implemented to the design
to enhance the aerodynamic performance and increase the Cp significantly. The
1st modification was stablizing two plates at the ends of the turbine. They retain
the pressure difference between concave and convex sides through capturing
airflow in the concave side.
Leading to average 20% of power coefficient
(Cp)(Akwa et al., 2012b)
2nd modification was 3 blades instead of 2, which score higher Cp values
compared to 2 bladed turbines, as the 3 blades excel when projected to high
rotational speeds, compared to double-bladed turbines. Where they score a
higher coefficient of power when subjected to high rotational speeds by about
47% due to their special ability to decrease negative torque. Thirdly, an aspect
ratio of 2, which is the ratio between the turbine’s diameter and its height, as
in Eq 5:
H
(5)
D
Where H is the turbine’s height = 40cm and D is the turbine’s diameter equal
to 20cm.
40
A=
=2
(6)
20
Which is relatively high, aiding in increasing self-starting and the power coefficient. However, increasing it too much breaks the turbine, so 2 was optimal
for the turbine. Thus, 2 amplifies performance up to 19.37% compared to the
conventional 0.8, while retaining endurance and stability
Lastly, two wings were stabilized on the wooden border’s corners, which
collect air from different directions and direct it towards the blade’s rotation.
This also decreases the negative torque acting on returning blades, while keeping
the multidirectional advantage of the turbine and increasing the Cp, where
before stablizing the wings, the Cp1 was =
A=
1
= 0.240
1/2 × 91.125 × 1.116 × 0.8
And after stablizing them, the Cp2 was =
Cp1 =
Cp2 =
1.15
= 0.275
1/2 × 91.125 × 1.116 × 0.8
(7)
(8)
Which increases the overall power coefficient by 15%, as in eq (9)
Ef f iciency =
Cp2
0.275
=
= 1.145833 ≈ 1.15
Cp1
0.240
(9)
And increases the energy output by 15% as in eq (10)
Ef f iciency(output) = 1.15W − 1W = 0.15 × 100 = 15%(Increase)
2
(10)
3
The Torque Coefficient
The torque coefficient CT is a dimensionless number that measures how efficiently the turbine converts the kinetic energy into rotational energy. Where,
Higher Ct = Higher efficiency of converting wind into spinning motion. It is
defined as:
CT =
CP
T SR
(11)
Where Cp is the power coefficient= 0.275
And TSR is the tip speed ratio, which is the ratio between the tangential
velocity of the blade’s tips and the actual velocity of the turbine, V. It is defined
in eq (11) as:
T SR =
ω×R
Vwind
(12)
R is the blade’s radius = 0.11m
V is the wind’s speed = 4 m/s
ω is the blade rotation speed in radians/sec. The turbine made 3 revolutions
per second, then ω =
3 × 2π
radians
=
= 6π/sec ≈ 18.85rad/s
sec
1
Then the TSR is equal to
18.85 × 0.11
= 0.5183 ≈ 0.52
4
(13)
(14)
Then, theCT =
0.275
CP
=
= 0.5288
(15)
T SR
0.52
Which is relatively close to the idea 0.7 value, which expresses the turbine’s
unique utilization of wind energy, generating more electricity effectively. This
high Ct value significantly improves the self-start of the turbine.
=
4
Hospital’s Electricity Needs
The hospital has recently closed due to its electricity shortage, 2,220,000
kWh/year of energy worth 3,083,580 EGP (50% of its operational costs). Applying the wind’s speed in Tanta (10m/s), and the coefficient power of the turbine
(0.275), it was found that two turbines, one of 25 meters in width and 50 meters
in height and another turbine of half its size, would generate about 2,308,260
kWh per year, as in eq (16), the large turbine alone generates:
3
Cp =
Poutput
1/2ρAV 3
(16)
The power coefficient (Cp ) = 0.275
Cross-sectional area (A) = 50 × 25 = 1250m2
V is the wind’s speed in Tanta =10m/s
V 3 = 103 = 1, 000(m/s)3
(17)
POutput = 1/2ρAV 3 × (CP )
(18)
1/2 × 1.225 × 1250 × 1000 = 765, 625
(19)
0.275 × 765, 625 = 210, 546.875W
(20)
Then, Poutput =
Then, 1/2 ρ A V 3 =
Then, Poutput
Then, multiplying by the number of hours in a year = 8,760 hours, then
dividing by 1,000:
210, 546.875 × 8, 760
= 1, 844, 416.25kW h/year
1, 000
(21)
Then, the large turbine alone generates 1,844,416.25Kwh yearly.
The, the small turbine has 1/4 the cross-sectional area (A2) of the larger
turbine’s area (A2), then:
A(2) =
(A1)
1250
=
= 312.5m2
4
4
(22)
Then, 1/2 × ρ × A × V 3 =
1/2 × 1.225 × 312.5 × 1000 = 191, 406.25.
(23)
3
Then, Poutput = Cp × 1/2 × ρ × A × V =
0.275 × 191, 406.25. = 52, 585.9375W
(24)
Then, it generates yearly=
52, 585.9375 × 8, 760
= 460, 455KW h/year
1, 000
(25)
Then, the 2 turbines combined generate about
1, 844, 416.25 + 460, 455 = 2, 304, 871.25KW h
4
(26)
Which meets the hospital’s electricity needs of 2,220,000KWh annually with
over backup energy of about:
2, 304, 871.25 − 2, 220, 000 = 84, 871.25KW h
5
(27)
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