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Wave Power Formula

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The energy supplied by the wind waves is called the wave power. It is the energy carried by wind waves and is utilized to accomplish beneficial activities, such as electricity production, water pumping into reservoirs, or water desalination. It is denoted by the symbol P. Its standard unit of measurement is watt (W). Its dimensional formula is given by [M1L2T-3]. Wave power, in general, is energy conveyance across the ocean’s wave surface. A wave energy converter is a mechanism that can harness wave power. The diurnal flux of tidal power and the constant gyre of ocean currents are not the same as wave power.

Formula

P = ρg2Th2l / 32π

where,

P is the wave power,

T is the period of wave,

h is the height of wave,

l is the wavefront length,

ρ is a constant, that is, water density with a value of 1.025 kg/m3,

g is the acceleration due to gravity with a value of 9.8 m/s2,

π is a constant with the value of 3.14.

Sample Problems

Problem 1. Find the wave power of a wave of length 8 m at a height of 10 m in 3 seconds.

Solution:

We have,

l = 8

h = 10

T = 3

Using the formula we have,

P = ρg2Th2l / 32π

= (1.025 × 9.8 × 9.8 × 3 × 10 × 10 × 8) / (32 × 3.14)

= 236258.4/100.48

= 2351.29 W

Problem 2. Find the wave power of a wave of length 5 m at a height of 7 m in 4.5 seconds.

Solution:

We have,

l = 5

h = 7

T = 4.5

Using the formula we have,

P = ρg2Th2l / 32π

= (1.025 × 9.8 × 9.8 × 4.5 × 7 × 7 × 5) / (32 × 3.14)

= 108531.20/100.48

= 1080.127 W

Problem 3. Find the length of a wave of power 2000 W at a height of 3 m in 10 seconds.

Solution:

We have,

P = 2000

h = 3

T = 10

Using the formula we have,

P = ρg2Th2l / 32π

=> 2000 = (1.025 × 9.8 × 9.8 × 10 × 3 × 3 × l) / (32 × 3.14)

=> 8859.69 l/100.48 = 2000

=> 88.17 l = 2000

=> l = 22.6 m

Problem 4. Find the length of a wave of power 3450 W at a height of 4 m in 12 seconds.

Solution:

We have,

P = 3450

h = 4

T = 12

Using the formula we have,

P = ρg2Th2l / 32π

=> 3450 = (1.025 × 9.8 × 9.8 × 12 × 4 × 4 × l) / (32 × 3.14)

=> 18900.67 l/100.48 = 3450

=> 188.10 l = 3450

=> l = 18.34 m

Problem 5. Find the height of a wave of power 4561 W if its length is 9 m in 4 seconds.

Solution:

We have,

P = 4561

l = 9

T = 4

Using the formula we have,

P = ρg2Th2l / 32π

=> 4561 = (1.025 × 9.8 × 9.8 × 4 × h2 × 9) / (32 × 3.14)

=> 3543.876 h2/100.48 = 4561

=> 35.26 h2 = 4561

=> h2 = 129.35

=> h = 11.37 m

Problem 6. Find the height of a wave of power 7631 W if its length is 2 m in 7 seconds.

Solution:

We have,

P = 7631

l = 2

T = 7

Using the formula we have,

P = ρg2Th2l / 32π

=> 7631 = (1.025 × 9.8 × 9.8 × 2 × h2 × 7) / (32 × 3.14)

=> 1378.174 h2/100.48 = 7631

=> 13.71 h2 = 7631

=> h2 = 556.60

=> h = 23.59 m

Problem 7. Find the time taken by the wave of length 4.5 m to generate power of 3265 W at a height 14 m.

Solution:

We have,

P = 3265

l = 4.5

h = 14

Using the formula we have,

P = ρg2Th2l / 32π

=> 3265 = (1.025 × 9.8 × 9.8 × T × 14 × 14 × 4.5) / (32 × 3.14)

=> 86824.962 T/100.48 = 3265

=> 864.101 T = 3265

=> T = 3.77 s


Last Updated : 04 Feb, 2024
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