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Engineering: environmental fluids
Engineering: environmental fluids

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2.5 Force on a floating tunnel

A fixed rail/road link has been proposed in the North Channel between south-west Scotland and Northern Ireland (see Figure 20).

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Figure 20 UKHO Chart 2198: North channel, southern part

Although Portpatrick in Scotland to Donaghadee in Northern Ireland is only 35  km (for comparison, the Channel Tunnel is 50  km long and the Lake Pontchartrain Causeway viaduct is 38  km long) the location poses a series of problems.

  • Since the whole of the northern half of the Irish Sea has to fill and drain twice a day through the North Channel, the tidal streams are strong: at maximum (spring tides) the tide flows at 3 knots (1.5  m  s−1) mid-channel and at 4.5 knots (2.25  m  s−1) near the Irish coast (see Figure 20). Note that tides are normally expressed in knots: 1 knot is 1 nautical mile per hour, equal to 0.514  m  s−1.
  • The weather in the North Channel is notoriously wild, and the combination of strong winds blowing across fast tidal flow in the opposite direction regularly produces huge waves of up to 20  m peak to trough (10  m amplitude). A breakwater at Portpatrick constructed in 1836 by John Rennie the Younger using techniques developed by his father for building lighthouses lasted less than three years before it was destroyed by a winter storm.
  • There is significant shipping traffic, for which a route must be left clear. Any crossing solution must also be able to withstand a collision with a ship.
  • In the middle of the channel, slightly towards the Scottish side, is Beaufort’s Dyke, a glacial valley 45  km long, 3  km wide and up to 300  m deep. On its own it would pose a significant challenge, but to make matters worse it was used as a dumping ground for hazardous waste after World War II and contains many thousands of tonnes of high explosives, incendiary bombs, poison gas and some nuclear waste, all poorly contained.

One possible solution to the problems is a floating tunnel (see Figure 21). At the time of writing this technology is under development in Norway as a possible solution to the similar problem of fjord crossings needed for the coastal highway project [Tip: hold Ctrl and click a link to open it in a new tab. (Hide tip)] .

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Figure 21 An artist’s impression of a proposed floating tunnel for Norway

A floating tunnel may be held up by its own buoyancy and anchored to the sea bed, or it may have negative buoyancy (i.e. a tendency to sink) and be suspended in position by surface pontoons.

In Activities 3.3a–3.3c you will investigate whether a floating tunnel in the North Channel will be able withstand the wave and tidal forces at this location.

Before you attempt Activity 3, it may be necessary to review your understanding of two dimensionless quantities which often occur when analysing and describing fluids: drag coefficient, Cd, and Reynolds number, Re.

Before you attempt Activity 3, it may be necessary to review your understanding of two dimensionless quantities which often occur when analysing and describing fluids: drag coefficient, Cd, and Reynolds number, Re. Start of Activity

Reynolds number

In a fluid flow situation, the Reynolds number is an important dimensionless parameter which characterises the nature of the flow. It is effectively the ratio of inertial forces to viscous forces in the fluid, both of which are resisting changes to velocity (i.e. accelerations of an object or fluid). For a cylinder of circular cross-section placed at right angles to a fluid flow, the equation for Reynolds number Re can be stated as

Re equals cap u sub infinity times d divided by nu

Where cap u sub infinity is the transverse fluid flow velocity some distance away i.e. not disturbed by the cylinder), d is the cylinder diameter and ν (Greek letter nu) is the kinematic viscosity of the fluid.

Drag coefficient

From the definition of drag force presented in the earlier part of the course on the atmosphere, you know that drag force, Fd can be stated as

cap f sub cap d equals one divided by two times cap c sub cap d times rho times u squared times cap a

Where Cd is the dimensionless parameter, the drag coefficient. The drag coefficient is an experimentally determined value which varies characteristically with Reynolds number (Re) for a given flow situation.

Activity 3

Tidal forces: Question 1

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Assuming that a single tunnel is 10 m in external diameter, find the Reynolds number based on diameter for a maximum tidal stream of 2.5 m s−1. Assume that the density of seawater is rho equals 1020 kg m super negative three and kinematic viscosity is nu equals 1.38 multiplication 10 super negative six m super two s super negative one . Give your answer to 2 significant figures.

Answer

Using the tidal stream velocity of u equals 2.5 m s super negative one , external diameter l equals 10 m and kinematic viscosity of seawater of nu equals 1.38 multiplication 10 super negative six m super two s super negative one , the Reynolds number is

Re equation sequence part 1 equals part 2 u times l divided by nu equals part 3 2.5 m s super negative one multiplication 10 m divided by 1.38 multiplication 10 super negative six m super two s super negative one equals part 4 1.8 multiplication 10 super seven left parenthesis to two s full stop f full stop right parenthesis full stop

Tidal forces: Question 2

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Using the following graph, estimate the drag coefficient at the Reynolds number found in Question 1 and hence the expected lateral force per kilometre on the tunnel. Give your answer to 2 significant figures.

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Figure 22

Answer

Flow is completely turbulent and the drag coefficient will be close to that shown for Re = 107, so from the graph cap c sub cap d equals 0.9 .

A 1 km length of pipe has a transverse area of 1000 m prefix multiplication of 10 m equals 10 000 m super two , so the drag force per km is

equation sequence part 1 cap f sub cap d equals part 2 one divided by two times cap c sub cap d times rho times u squared times cap a equals part 3 one divided by two multiplication 0.9 multiplication 1020 times kg m super negative three multiplication left parenthesis 2.5 times m s super negative one right parenthesis squared multiplication 10 000 times m super two equals 29 MN left parenthesis to two s full stop f full stop right parenthesis full stop

This is a substantial force, to put it mildly. However you may care to compare it to the buoyant force exerted by the sea on the same length of tunnel when the centreline is submerged by only 20 m. The buoyant force cap f sub b on an immersed object is equal to the weight of fluid displaced by it (Archimedes’ principle), so

cap f sub b equals rho times g times cap v
cap f sub b equals 1020 times prefix multiplication of 9.81 prefix multiplication of pi multiplication left parenthesis 10 m right parenthesis squared divided by four multiplication 1000 m
cap f sub d equals 790 italic MN left parenthesis normal t times normal o postfix times two postfix times normal s full stop normal f full stop

This is around 27 times greater than the tidal force, which is therefore not particularly large by the standards of the project.

A winter storm creates deep water waves of amplitude 10 m and wavelength 100 m in the middle of the North Channel. Work through the following questions to find whether the tunnel will be able to withstand the wave forces in winter.

Give your answer to 2 significant figures where appropriate.

Wave forces: Question 3

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How fast will the waves travel?

Answer

Using the formula for deep water waves in Equation 12, the wave speed is

equation sequence part 1 c sub w equals part 2 Square root of cap l times g divided by two pi equals part 3 Square root of 100 m prefix multiplication of 9.81 m s super negative two divided by two times pi equals part 4 12.49 times ellipsis m s super negative one equals 12 times m s super negative one left parenthesis to two s full stop f full stop right parenthesis full stop

Wave forces: Question 4

What is the period of the waves if the centre of the tunnel is at a depth of 50 m?

Answer

Using Equation 13, at 50.0 m depth the period of the waves is

equation sequence part 1 cap t equals part 2 cap l divided by c sub w equals part 3 100 m divided by 12.49 times ellipsis m s super negative one equals part 4 8.0 s left parenthesis to two s full stop f full stop right parenthesis full stop

Wave forces: Question 5

Will the waves cause significant forces?

Answer

According to Section 1.2, deep-water, wave-induced motion at half the wavelength is around 4% of the surface value. That is the case here, so the wave motion at the tunnel centre line will be only 4% × 10 m = 40 cm. Since tidal flow in the centre regularly reaches 150 cm s−1, the extra wave displacement of 40 cm there-and-back every 4 seconds will not add significant additional forces.

Conclusion: Question 5

Is a floating tunnel a viable solution to the problem from the fluid dynamics point of view?

Answer

Conclusion: A floating tunnel should easily be able to withstand both tidal and wave forces in this location.