Tidal Energy

Question 1

Explain the main mechanisms that lead to the tides.

Answer 1

Tides are caused by the effect of gravity in the Earth-Moon-Sun system, and the movement of those three bodies.

Question 2

What are the main tidal cycles and what are they governed by? What other factors
manifest themselves as a tidal cycle?

Answer 2

Main cycle is the semi-diurnal cycle arising from attraction of Moon and the rotation of the Earth which results in 12 h 25 min cycle.
Next is spring-neap cycle arising from relative position of Sun and Moon: this is approximately 14 day cycle.
Other cycles arise from Moon’s elliptical and out of plane orbit as well as Sun’s elliptical orbit.

Question 3

A typical ocean depth is 4400 m.

(i) What is the speed of a freely travelling tidal wave?
(ii) How long would it take the tidal wave to circle the Earth’s equator?
(iii) How does this compare with the speed of the tidal force from the Moon? (Assume radius at equator is 6400 km)

Answer 3

(i) Speed of tidal wave c = √(gh). For h = 4400 m, c =√(9.81×4400) = 207.8 m/s.
(ii) Earth’s circumference at equator s = 2πR = 2π×6400 = 40217.6 km. Wave covers this distance in time t_wave = s/c = 40217.6×103/207.8 = 193540 s = 54 h.
(iii) The tidal force moves at around the same rate as the Earth rotates, i.e., period of 24 hours (or speed v = 2πR/t = 2π×6400×103 /(24×60×60) = 465 m/s). This means the
tidal wave induced by the Moon cannot form a continuously revolving bulge in line with the Moon. It will effectively lag the Moon.

Question 4

The Severn Estuary has a length of around 200 km and an average depth of 30 m. Demonstrate that it is resonant.

Answer 4

Resonance occurs when the length L is a fraction of the wavelength λ. L=1/4nλ, where n=1, 3, etc.

We can now estimate the dimensions of resonant basins whose period matches the tidal period T:
L/√h=nT/4√g

The dominant semi-diurnal period is 45,000s, which means the first resonance (n=1) occurs at 36,000m^1/2

L/√h ≈ 200 × 103/√(30) ≈ 36400 m^1/2. This is close match to 36000 m^1/2 where the semi-diurnal tidal period matches estuary natural frequency.

Question 5

Real estuaries do not conform to the simple model suggested. How is their resonant
behaviour examined in practice?

Answer 5

Originally this would have been conducted using scale models of the estuary in large hydraulics laboratories. The model would be filled with water and the rise and fall of the tide at the open sea end of the estuary would allow observation of effect further up the estuary. With availability of cheap computing power, this is now largely carried out with software able to simulate the shallow water equations and using detailed bathymetric data, e.g. Telemac, Mike, POL-COMS.

Question 6

The Atlantic and Pacific oceans and English Channel exhibit resonant behaviour – explain the circumstances in each case.

Answer 6

Boyle (2012), pp 246-8 talks a little about this but it needed you to use the
principles of resonance to see if the dimensions of these areas would result in
wavelengths that would create resonant behaviour. Essentially you will get
resonance where the wavelength is a multiple of a quarter wavelength: The Atlantic
is ~4000 m deep and ~4000 km wide (Canada-UK). The depth tells us the speed of
the tidal wave c = √(gh) = √(9.81 ×4000) ~ 200 m/s and its wavelength for the main
semi-diurnal tidal period (T = 12h 25 m = 44700s): λ = c/f = cT = 200×44700 =
8940 km. This is close enough to being twice the width of the Atlantic for weak
resonance to occur. Similarly in the Pacific (~20000km, 4000m) a wavelength of
almost a quarter the width of the ocean is resonant. The English Channel is ~500
km long and ~50m deep: the wave speed and wavelength will be different due to
the depth (~22m; ~1000km) resulting in a wavelength of twice the length..

Question 7

Outline the main components of and the methods for constructing tidal barrages.

Answer 7

Either build cofferdam as for hydro scheme (see Boyle (2012), p254) or use caissons which requires them to be towed into place (discussed in lecture). The 2nd method does not fully block estuary.

Question 8

Outline the differences in operating a tidal barrage in ebb-only, flood-only and 2-way modes. Why is pumping considered valuable with barrages?

Answer 8

Ebb-only:
Sluices closed at high water and head develops on lagoon side.
Flood-only:
Sluices closed at low water and head develops on open sea side.
Two-way:
Generates each way, Peak power lower, smoother. Pumping enhances head.

Question 9

What are the main environmental impacts of tidal barrages?

Answer 9

Flow is blocked. Constructing a gate which allows water level on the open motion side to increase/decrease but on the basin side you keep it closed. Serious environmental impact; no fish movement, biotop stopped cause no flushing of water, nothing to feed on for birds. Long term sedimentation because blocking sediment to move.

Question 10

The mean tidal range for a tidal barrage is 8.45 m and its basin area is 22 km2. If the mean output is 75 MW, what proportion of the theoretical production does this represent?
If roh not given of water use 1025.

Answer 10

Each ebb or flood cycle the potential energy E = ½ρgAR^2 = 8 × 10^12 which equals 2,220 MWh. If operating 2-way this is 4440 MWh over a 12h 25 min period. Mean power of 75 MW over 12 h 25 min period is equal to 75 × 12.416 = 931.25 MWh. Proportion harnessed ≈ 931.25 / 4440 ≈ 21%.

Question 11

What are the relative merits of tidal lagoons over barrages?

Answer 11

Tidal lagoons can be constructed as part of the coast or be constructed away from the coast as a stand alone unit, whereas barrages has to be constructed over the estuary.
Tidal lagoons are like a pocket of water storage, you can use porous structures that allows water to go inside and come out therefore you wouldn’t see rotten water inside because of water flow. Less siltation effects and such.
Barrage are usually very expensive cause you have to use breakwater structure. Lagoons also very expensive?

Question 12

What are the principal mechanisms for the existence of tidal currents?

Answer 12

• Geographical features that promote
tidal currents include
– channels and between islands
– headlands and peninsulas
– areas around the edges of bays
• Although complex, current flows tend
to occur due to differences in the
water level between points as the
tidal wave advances
• The difference in water level (because of moon, earth and sun rotation the bulges on one side will be larger than the other) forces
the water at a higher level to flow
‘down’ the water gradient
• Locations with high tidal current flows
tend to be where there are relatively
small distances between co-tidal lines
(high timing gradient) and/or rapid
changes in tidal range.

Question 13

Tidal current devices can be typically classified in several ways. What are these and what are the key criteria for classification?

Answer 13

Type: Horizontal axis, vertical axis, linear lift, venturi; Support: Fixed or floating
with subdivisions of fixed by structure type; ‘Generation’: 1st generation is for
shallower water – typically fixed installation; 2nd generation for deeper water where
floating looks more credible: Deployment method: tow and sink; barge/boat and
lift or combination of these; Rotors: number of blades per rotor, number of rotors
per device, contra-rotating or not. Etc, etc.

Question 14

Tidal current devices can be typically classified in several ways. What are these and what are the key criteria for classification?

Answer 14

Type: Horizontal axis, vertical axis, linear lift, venturi; Support: Fixed or floating with subdivisions of fixed by structure type; ‘Generation’: 1st generation is for shallower water – typically fixed installation; 2nd generation for deeper water where floating looks more credible: Deployment method: tow and sink; barge/boat and lift or combination of these; Rotors: number of blades per rotor, number of rotors per device, contra-rotating or not. Etc, etc.

Question 15

Discuss and describe 4 tidal current energy conversion concepts, including what
you believe to be the advantages and disadvantages of each.

Answer 15

Horizontal axis TT: (+) Simple and proven; mechanically efficient; low solidity high speed rotors can simplify generator choice; (-) need to be accurately
positioned to ensure flow in line with rotor axis; size limited by channel depth (typical max diameter is half depth)

Vertical axis TT: (+) Generator/gearing in base for stability; size not limited by depth; multi-directional; (-) generally less efficient than HATT; may require
starting assistance; possible higher impact risk with marine life?

Linear lift TT: (+) Not depth limited; (-) complex power conversion technology;original device overly slow – up to date approach uses multiple foils; sensitive to
flow direction.

Venturi TT: (+) Compact and robust; (-) added complexity – do they add anything?

Question 16

A channel has a length of 2 km, approximate depth of 30 m and a width of 200 m. What is the kinetic energy flux (the kinetic energy per second) passing through a section across the channel, if the cross-sectional average velocity is 3 m/s.

Answer 16

The kinetic energy flux (or power) P = ½ρAU^3. The cross sectional area A of channel is 30×200 = 6000 m2. The density of seawater is 1025 kg/m3. The kinetic energy flux P = 0.5×1025×6000×33 = 83,025,000 W = 83 MW.

Question 17

If this channel is to be exploited using sea bed mounted horizontal axis machines, what might the diameter of the devices be? What might be a reasonable expected power output from such a device at 2 m/s? (Assume a reasonable value for the power coefficient Cp)

Answer 17

The channel is relatively shallow so the challenge is putting in rotors that are large enough to capture lots of energy whilst keeping the hub height high enough in the water column to ensure good flow velocity (slower nearer channel bed) and keeping hub height low enough to avoid surface waves and shipping. A hub height of 20 m with 15 m diameter rotors would probably fit well. 20 m rotors at hub height 15 m might also work.

Choose C_p-value. Suitable between 0,4-0,45.
P=0.5C_pnrohA*U^3.

Question 18

If 20 MW of tidal devices are constructed in the channel, what would the impact on flow velocities and overall power available?

Answer 18

This is a matter of using the extraction limits graph (or an equation based on it) in the notes. This relates the percentage reduction in flow velocities to the proportion
of natural energy flux extracted. The proportion of energy flux extracted by 20 MW of devices at rated output will be 20/83 = 24%. The line on the chart stops at 20% so some extrapolation is required which puts the percentage flow reduction at just over 7%. Applying this to the 3 m/s area averaged flow would see an adjusted flow velocity of Uadj = (1-0.07) ×3 = 2.79 m/s. If this was then used in the channel energy flux equation, P = ½ρAU3, the adjusted flux would be Padj = 0.5×1025×6000×2.793 = 66.8 MW (-24%). For the turbines in Q19 this would be a 24% reduction in
available power.

Question 19

A tidal device is to be installed in 50 m depth of water with its hub height at 30 m above the sea bed. If the flow at the surface is 2.8 m/s what is the flow velocity at
hub height if the power law exponent is (i) 1/7 and (ii) 1/10? How realistic is the use of the power law for tidal current?

Answer 19

Straight forward application of power law with coefficient α = 1/7 or 1/10. Rearrange it to U(z) = U(z_r)/U(z/z_r)^α and insert values: z = 30 m, z_r = 50 m, U(50) = 2.8 m/s. Applying these values gives: (i) 2.60 m/s for α = 1/7 and (ii) 2.66 m/s for α =1/10.