Thanks for your suggestion. We have received similar ideas from other community members.
Based on publicly available information from South East Water, it looks as though the daily volume from the Somers Treatment Plant is significantly less than what is required for our project operating in either open or closed loop modes.
We are contacting South East Water to investigate further and will update you next week.
There is not enough water produced to enable 450ML per day but there are two issues here. First, the plant can be upgraded to produce more water and secondly the vessel should be able to recycle water without having to discharge and recharge 450ML every day.
My original question for AGL is what is the cost to recycle the water as opposed to pumping it in and out every day.
Hi @RobH1
Apologies it has taken so long to get back to you. I have included a response to your queries below.
ROBH1. You say that the plumes can dilute individually until it reaches the seabed or are mixed into the ambient flow. If, as you say the flow is 5.4 metres per second or 10.5 knots. A flow of 10.5 knots will not reach the seabed in a very short time, it is a very fast flow and having a continuous flow from the outlet at 18 million liters an hour the flow will create it's own "tidal" flow and could not reach the seabed. If the normal tidal flow of around the 2.5 knot flow has a 10.5 knot flow added to it then the area north and south of the terminal will be so fast small craft would be in danger, then if there is a northerly wind at 20 knots blowing against this very fast flow the creation of pressure waves would be extremely dangerous for most craft. In a normal fast tide flow on an incoming tide with a strong north quarter wind blowing the pressure waves can be over 2 metres high with a very short pitch {the distance between each wave and the angle of vertical height} is considered dangerous for boating, so what size pressure waves could be expected with this super fast water.
Response. The six ports discharge horizontally at a velocity of 5.4 m/s (or 10.5 knots). As you observe, this is a high velocity and thus, there will be rapid mixing between the high velocity jets of discharged seawater and the ambient seawater. As the discharge is horizontal, at 2 m below the water surface, and not vertical, the diluted jets will not reach the seabed until they have mixed through the 11 or 12 m depth of water between the alignment of the ports and the seabed. This does not occur until the jets have travelled about 50 to 60 m and after considerable dilution.
The current speed in the jets decreases with distance due to dilution, from 5.4 m/s at the port to 0.8 m/s at 10 m distance along the jet, to 0.4 m/s distance at about 30 m distance. As the jet dilutes, it absorbs the momentum of the ambient flow and at 50 m distance, is travelling at only a small angle to the average ambient current. The postulated 2 m high pressure waves do not occur when proper account is taken of conservation of mass and conservation of momentum in the jets.
There is an exclusion zone around vessels in the Port including the FSRU and thus, there should not be any small vessels in the discharge area.
Tidal currents have been measured in the area of Crib Point jetty from 19 March to 16 April 2019 and are plotted below. The spring-neap-spring tidal pattern can be seen in the blue line, which records the north-south current speed at mid-depth. The peak north tidal current at Crib Point is about 0.6 m/s which is 1.2 knots. The peak south tidal current at Crib Point is about 0.8 m/s which is 1.6 knots. The tidal current speeds of 2.5 knots you postulated were not recorded in the month of measurements at Crib Point.
ROBH1. The word Dilute appears in many articles with the added number of total litres for the north arm, which is a very deliberate misleading piece of information as the North arm is from the shore at stoney point across to the middle spit all the way up past Joes Island which is massive, yet the only part of the North arm affected is the area as wide as the terminal which is 49m a miniscule amount compared to the whole of the north arm, so dilution and your data is completely incorrect.
Response. It is obvious that a series of jets discharging at 5 m/s mix very rapidly with seawater moving up and down WPB in the tidal currents at speeds of up to 0.8 m/s (average speed of 0.3 m/s). The jets travel and spread over a greater width than the 49 m quoted. But even if they just mixed with the seawater travelling past the FSRU in a 49 m strip on one side of the vessel, the average dilution is 39:1, calculated as follows.
Average seawater flow = Width x Depth x Average speed
= 49 m x 14 m x 0.3 m/s
= 206 m3/s.
Average dilution = Seawater flow / Discharge
= 206 m3/s / 5.2 m3/s
= 39:1.
After a dilution of 39:1, the 7 degree temperature difference between the discharge jet and seawater is reduced to 0.18 deg C. The computer model developed for the EES studies uses a more complex 3-D representation of jet movement, mixing and dispersion on a fine scale, and predicts similar results for the average current, with a lower dilution at slack water and higher dilution at times of peak tidal currents. Under no scenario is there any water less than 0.2 deg C below ambient outside the Port area.
RobH1 . Now we look at the most ridiculous comparison between 4 outboard motors flow at 20 knots. Each propeller is about 10" or 25.4cm, your outlet of 6 ports of 45cm has a much greater volume of flow. The statement where you say the 10.3 knot flow will dissipate after 9 m is completely untrue as that is absolutely impossible. If it was on a slack tide with very little tide flow it is defying all physic law, if the tide is flowing then I would say you need a new calculator or glasses, it is physically impossible.
Response. It may be surprising to learn how much water flows through the propeller of a small boat travelling at 20 knots. The boat speed at 2 knots is 10.3 m/s. To maintain the speed of the boat, the propeller is pushing an extra 20 % of water out behind the boat (this is the ‘slippage’ referred to in outboard motor catalogues). Thus the water speed relative to the boat is 12.3 m/s. If the propeller has a diameter of 0.245 m, the corresponding area is 0.047 m2, and the volume of water flowing through the propeller is 0.047 m2 x 12.3 m/s = 0.58 m3/s. So with a small propeller, the discharge is equivalent to 9 outboard motors
The 40 HP Mercury outboards typically use a 13 inch (0.33 m diameter) propeller with a corresponding area of 0.085 m2. The volume of water flowing through the propeller is 0.085 m2 x 12.3 m/s = 1.05 m3/s. So with a larger propeller, the discharge of 5.2 m2/s is equivalent to the flow through 5 outboard motors
In one way (laterally) you are correct in observing that the lateral momentum from a 5.2 m3/s discharge at a velocity of 5.4 m/s (about 10.5 knot) does not dissipate after 9 m. The lateral momentum is conserved and only dissipates slowly due to friction.
But this does not mean that the jets continue at the same velocity across the Bay until they run into the other side of the Channel. The reason is that, as the jets mix with the ambient seawater, the longshore momentum in the tidal flow also must be conserved.
An example will help to illustrate this. The lateral momentum of the jets is 5.2 m3/s at 5.4 m/s giving an eastwards momentum of 28.1 m4/s2 (we can change units but these mixed units make the arithmetic easier to follow). At the average tidal current of 0.3 m/s, the jets mix with 39 volumes of seawater (dilution is 39:1, as calculated above). The longshore momentum is 39 x 5.2 m3/s at 0.3 m/s giving a northwards momentum of 60.8 m4/s2. So from conservation of east momentum of 28.1 units and conservation of north momentum of 60.8 units, the jets are heading 25 degrees east of north (north-north-east) at a velocity of 0.43 m/s. After more mixing, the jets come closer and closer to the ambient current speed and the ambient current direction.
RobH1. On the speed of the flow, I am trying to understand how you came to this figure. Lets say the tide is a slow incoming tide of 1.6 knots and you are telling us the flow from the outlet will reduce to 0.5 or 0.97 knots, or on a fast tide of 3 knots your outlet flow will slow to 0.97 knots, do you actually realise what you are asking us to believe ?
Response. Yes, I am asking you to believe in conservation of mass and conservation of momentum (in each of the three directions of motion).
RobH1. On the temperature. are these figure based on the absurd total number of litres entering the Western entrance or the total number of litres for the North arm or the area around the terminal, I ask this because your figures are rubbery at best. to calculate the 19m from the terminal did you consider the tide phase and at what speed the flow is? what the slack water time on the dead or very slow tides are ? A flow of 18 million litres in one hour of chilled water that is 7 deg colder than the ambient temperature travelling in a plume is going to dissipate within 19m of the terminal! The flow is in a plume so how can it mix if you say the flow almost stops after travelling 9m ?
Response. The calculations of the increase in temperature from the discharge (7 degrees below ambient) to the near-ambient temperature have been made using two models. The plume model tracks flow conditions along the centreline of each plume, calculating the dimensions, dilution and temperature in the plumes, using equations based on conservation of mass and conservation of momentum. The plume model extends for several hundred metres from the discharge ports.
A separate 3-D model of flow in the whole of Westernport is then used to check the predictions of the plume model. The 3-D model simulates currents over a 30-day period and allows the effects of reversing currents and variations in current speed and direction over the depth to be represented. Both models show that the “worst case” situation for dilution occurs for a short period around slack water, when the plumes will reach the seabed (with a temperature about 0.3 degrees below ambient). As the currents increase (with the following ebb or flood tide), the pool of colder water on the seabed (in the port around the FSRU) will be mixed upward into the passing seawater flow. Within a few hundred metres, the temperature difference will be under 0.1 deg C and within a kilometre, the temperature difference will be around 0.01 deg C."
Note that in Westernport Bay, water temperature varies by an average of 1 deg C over a day. Thus temperature variations of 0.1 or 0.3 deg C are not of ecological significance.
RobH1. The words "for a short period of time" depends on what you consider short, a few minutes an hour 2 hours or more, you do not explain that small piece of critical data.
Response. The current measurements show that slack water at Crib Point occurs over a short period of 10 to 20 minutes.
RobH1. You also claim all the discharged water will go straight to the seabed because it is cold, but being pumped out at 18 million litres an hour 2m below the surface has a couple of problems, the first it is being pumped out at 10.3 knots which at that speed could not go the the bottom, second it has about 17 m to get to the bottom, so how fast does water 7 deg less than the surrounding water take to get to the bottom. an example I can use is with fishing which a lot of people will understand. In a fast flowing tide in 17m of water I sometimes have to use a 10 oz sinker, if I use a one ounce sinker it will not reach the bottom for a very long distance.
Response. The plume does not go straight to the seabed at slack water. It travels laterally away from the FSRU and spreads vertically due to mixing. As the temperature difference is small, the cooler water plume descends gracefully toward the seabed (in the same manner as your light sinker).
RobH1. There is another problem you have that for some reason is missing in all of this, if the discharge is 18 million litres an hour then the intake is the same, you intake is 5m off the bottom the outlet 2m below the surface, so that would create a huge turbulence with the ports only 49m apart, a 10.3 knot suction which will create currents around the area of intake, the same as it will on the outflow so it would not be unreasonable to believe that the area around the Terminal will be extremely turbulent causing some of the discharged chilled water to be sucked in and chilled again. On the "slack" tide flow this water will be chilled over and over, worse if your assumption the outflow water will go to the bottom only to be sucked up by the intake. If there is no tide flow and on certain moon phases this slack tide flow can be for 2 hours creating a massive plume of super chilled water. As the tide flows this super chilled and highly chlorinated water will travel in a massive plume and as it will travel at the same speed as the tide flow, then how does it mix, it can't catch up to the water in front of it nor the water behind can catch it. So your statement of " Within a few hundred metres, the temperature difference will be under 0.1oC and within a kilometre, the temperature difference will be around 0.01oC." is very misleading because you claim it mixes, how can it mix if it travels at the same tide flow speed with all the water in and around travelling at the same speed.? Are these figures based on the misleading total amount of litres of water in westernport, or the north arm or just around the terminal?
Response. We have a computer model to examine the risk of discharge water being re-entrained into the seawater intake. This can occur to a minor extent around the time of reversing tidal flows, but the calculated effects are much smaller that you suppose. There will not be any super-chilled water occurring.
RobH1. Is the design taking into account of the huge amount of floating weed that is in the water column, weed in such density that sometimes boats have trouble pulling up their anchors due to the enormous amount of weed on the chain ? The picture of the grill filter would take about 5 minutes to be completely clogged up, so then what happens?
Response: Yes, the problems of weed clogging the seawater intake are being addressed. We expect there will be a seawater intake on each side of the FRSU and so the port intake can be used while the starboard intake is being cleaned, and vice versa.
Could you clarify something for me please?
The plume is the discharge, the ambient temperature is the surrounding seawater. Your temperature estimates are expressed as such " when the plumes will reach the seabed (with a temperature about 0.3 degrees below ambient" Is the 0.3 temperature measurement estimated from the meeting point/edge of the plume and ambient water? Or is the temperature measurement estimated from the centre of the plume?
Hi @KerryRainer
When we refer to the temperature of the plume, it is the temperature along the centreline of the plume
Sorry this has been a long delay in my response and it was not trying to show you how we have to wait it was because I was away and just got back. My response is in Blue
you wrote.
The six ports discharge horizontally at a velocity of 5.4 m/s (or 10.5 knots). As you observe, this is a high velocity and thus, there will be rapid mixing between the high velocity jets of discharged seawater and the ambient seawater. As the discharge is horizontal, at 2 m below the water surface, and not vertical, the diluted jets will not reach the seabed until they have mixed through the 11 or 12 m depth of water between the alignment of the ports and the seabed. This does not occur until the jets have travelled about 50 to 60 m and after considerable dilution.
I am not sure how you come to the conclusion that the water will mix within the 50 to 60 metres distance from the terminal. The seawater is already travelling at a maximum of 3.3knots plus the added speed of 10.5 knots or 5.4m/s which means the flow is completely mixed with the seawater in 9 to 11 seconds and stops !!! so what does your science data say what the speed of the water will be in 100 metres, 500 metres, 1000 metres.
Now to when the water is not moving and the 10.5 knots of jet water is mixing with stationary water and has severely chilled that water within the area around the terminal. Then the tide starts moving north, travelling in a plume and is not mixing anymore as it is travelling at the same speed as the tide heading north in one massive cold plume of water at lets say the fast tidal speed of 3.3 knots in the 3rd and fourth hour, how far will it travel ?
you wrote
The current speed in the jets decreases with distance due to dilution, from 5.4 m/s at the port to 0.8 m/s at 10 m distance along the jet, to 0.4 m/s distance at about 30 m distance. As the jet dilutes, it absorbs the momentum of the ambient flow and at 50 m distance, is travelling at only a small angle to the average ambient current. The postulated 2 m high pressure waves do not occur when proper account is taken of conservation of mass and conservation of momentum in the jets.
You have not taken into account the speed of the tidal flow combined with the speed of the outflow from the jets. The jets pumping out 18 million litres an hour in a slow moving tide cannot slow down to the ambient flow in such a short distance. For example a small vessel cannot sit behind a stationary tug boat while it is pushing a ship into the wharfe, the turbulence is incredible, but you are saying the 18 million litres and hour will have no turbulence after 50 metres ! now lets look at the peak flow of 3.3m being pushed by a jet flow of 10.5 knots, will create a massive area of turbulent water, now add a northerly wind blowing against that flow will cause pressure waves which are dangerous to small craft.
you wrote
There is an exclusion zone around vessels in the Port including the FSRU and thus, there should not be any small vessels in the discharge area.
What does your data say that this turbulent flow will stop ? If a tug boat can cause turbulence at the stern and for a good 100 metres plus, then how far will your 18 million litres and hour at 10.5 knots of turbulent water travel and how far are you going to put no go zones ? In the reply it seems it will only travel 50 to 60 metres, yet in practical terms it will be much greater plus the added wind factor could see and area will be much greater than the small 50 metres.
you wrote
Tidal currents have been measured in the area of Crib Point jetty from 19 March to 16 April 2019 and are plotted below. The spring-neap-spring tidal pattern can be seen in the blue line, which records the north-south current speed at mid-depth. The peak north tidal current at Crib Point is about 0.6 m/s which is 1.2 knots. The peak south tidal current at Crib Point is about 0.8 m/s which is 1.6 knots. The tidal current speeds of 2.5 knots you postulated were not recorded in the month of measurements at Crib Point.
I am not sure how you gathered those figures but I and thousands of anglers and Yachties will tell you the speed during peak tides will travel at over 3 knots. The graph seems to be in conflict with the height of the tide to the bottom of the tide. The reason I say this is because the flow of the water by your graph would not match the height of the water: for example the tide on the 27th of March is a 6 hour tide with a low of 0.21m at 11.25 am rising to 3.14m at 6.22pm a rise of 2.93m, so you want us to believe the figures you have provided can cause the amount of water to raise the level from 0.21m to 3.14m a total of 2.93m, what does your data say the flow of water is needed to make that happen.
you wrote
It is obvious that a series of jets discharging at 5 m/s mix very rapidly with seawater moving up and down WPB in the tidal currents at speeds of up to 0.8 m/s (average speed of 0.3 m/s). The jets travel and spread over a greater width than the 49 m quoted. But even if they just mixed with the seawater travelling past the FSRU in a 49 m strip on one side of the vessel, the average dilution is 39:1, calculated as follows.
Average seawater flow = Width x Depth x Average speed
= 49 m x 14 m x 0.3 m/s
= 206 m3/s.
I dispute the 0.3 m/s as I have pointed out in previous comments. It will depend on moon phases as to the flow speed, around full moons slow around half moon tides exceptionally fast, then the consideration of the moons affect on the tides month by month as each month will vary depending on the distance the moon is from earth, therefore the strength of the gravitation pull can be high to low.
you wrote
Average dilution = Seawater flow / Discharge
= 206 m3/s / 5.2 m3/s
= 39:1.
After a dilution of 39:1, the 7 degree temperature difference between the discharge jet and seawater is reduced to 0.18 deg C. The computer model developed for the EES studies uses a more complex 3-D representation of jet movement, mixing and dispersion on a fine scale, and predicts similar results for the average current, with a lower dilution at slack water and higher dilution at times of peak tidal currents. Under no scenario is there any water less than 0.2 deg C below ambient outside the Port area.
No consideration has been used in your calculations of dead tides when the water is not moving on a full moon and can stay still with very little movement for a minimum of 2 hours and sometimes up to 4 hours when the mixing is being concentrated, so with the continuous recycling of the same water, chilling it down a lot more than your .2 deg, in fact I would say it would be super chilled water and an area of exceptional size. Then I see no reference to the movement of this large body of super chilled water moving north on an incoming tide or south on an outgoing tide. Will this super chilled water enter any mangrove or Marine park areas ? Of course it will.
you wrote
It may be surprising to learn how much water flows through the propeller of a small boat travelling at 20 knots. The boat speed at 2 knots is 10.3 m/s. To maintain the speed of the boat, the propeller is pushing an extra 20 % of water out behind the boat (this is the ‘slippage’ referred to in outboard motor catalogues). Thus the water speed relative to the boat is 12.3 m/s. If the propeller has a diameter of 0.245 m, the corresponding area is 0.047 m2, and the volume of water flowing through the propeller is 0.047 m2 x 12.3 m/s = 0.58 m3/s. So with a small propeller, the discharge is equivalent to 9 outboard motors.
You talk about a boat being propelled, I think in this case we need to only talk about a stationary boat.
you wrote
The 40 HP Mercury outboards typically use a 13 inch (0.33 m diameter) propeller with a corresponding area of 0.085 m2. The volume of water flowing through the propeller is 0.085 m2 x 12.3 m/s = 1.05 m3/s. So with a larger propeller, the discharge of 5.2 m2/s is equivalent to the flow through 5 outboard motors
In one way (laterally) you are correct in observing that the lateral momentum from a 5.2 m3/s discharge at a velocity of 5.4 m/s (about 10.5 knot) does not dissipate after 9 m. The lateral momentum is conserved and only dissipates slowly due to friction.
You talk about laterally, why? it is not going from west to East or East to West it is travelling North/South along the length of the terminal.
you wrote
But this does not mean that the jets continue at the same velocity across the Bay until they run into the other side of the Channel. The reason is that, as the jets mix with the ambient seawater, the longshore momentum in the tidal flow also must be conserved.
Again you use the term "across the bay" the flow is North South not West East which baffles me as to why you would even say that. There maybe a very miniscule amount of West East flow if you are saying "until they run into the other side of the channel" yet the massive amount of flow travels North South and is travelling with the tide not across it. With the Jets pumping out water in a West East direction which means the jets are on the starboard side of the terminal if it is docked in a North South direction.
you wrote
An example will help to illustrate this. The lateral momentum of the jets is 5.2 m3/s at 5.4 m/s giving an eastwards momentum of 28.1 m4/s2 (we can change units but these mixed units make the arithmetic easier to follow). At the average tidal current of 0.3 m/s, the jets mix with 39 volumes of seawater (dilution is 39:1, as calculated above). The longshore momentum is 39 x 5.2 m3/s at 0.3 m/s giving a northwards momentum of 60.8 m4/s2. So from conservation of east momentum of 28.1 units and conservation of north momentum of 60.8 units, the jets are heading 25 degrees east of north (north-north-east) at a velocity of 0.43 m/s. After more mixing, the jets come closer and closer to the ambient current speed and the ambient current direction.
If the jets are pumping out the chilled water at a 25 deg angle, the tide will cause that to move more to the North than the 25 deg as that is the direction of the tide flow , same as the tide flow in a Southerly direction, except the flow will be slightly restricted due to the speed of the tidal flow on an outgoing tide against the outflow direction of 25 deg.
you wrote
The calculations of the increase in temperature from the discharge (7 degrees below ambient) to the near-ambient temperature have been made using two models. The plume model tracks flow conditions along the centreline of each plume, calculating the dimensions, dilution and temperature in the plumes, using equations based on conservation of mass and conservation of momentum. The plume model extends for several hundred metres from the discharge ports.
In the calculation what data was used. Did you use the data from the tide flow which no matter what evidence provided is incorrect as I can drift in my boat at 3 knots yet your data says I cannot. Did you use the total amount of water entering Westernport or the small amount around the terminal. Anyone can use a model but unless all relevant data and correct data is used then it will be flawed.
you wrote
A separate 3-D model of flow in the whole of Westernport is then used to check the predictions of the plume model. The 3-D model simulates currents over a 30-day period and allows the effects of reversing currents and variations in current speed and direction over the depth to be represented. Both models show that the “worst case” situation for dilution occurs for a short period around slack water, when the plumes will reach the seabed (with a temperature about 0.3 degrees below ambient). As the currents increase (with the following ebb or flood tide), the pool of colder water on the seabed (in the port around the FSRU) will be mixed upward into the passing seawater flow. Within a few hundred metres, the temperature difference will be under 0.1 deg C and within a kilometre, the temperature difference will be around 0.01 deg C."
This is using incorrect data and words to strengthen your argument. "the “worst case” situation for dilution occurs for a short period around slack water" The use of the words "short period" shows that you do not understand tides. In practical terms there is only a few days of each month the water is stationary for a short period. So lets look at around a full moon, some months the tides are at a stand still for several hours, while the chilled water is drawing water in and pumping out at 18 million litres an hour, super chilling the water around the terminal, then when the tides starts moving it does not dilute as the whole chilled plume area moves at the same speed, it does not catch up or slows down, it travels in one massive super chilled water plume.
you wrote
Note that in Westernport Bay, water temperature varies by an average of 1 deg C over a day. Thus temperature variations of 0.1 or 0.3 deg C are not of ecological significance
It is if your figures on the temperature of the mixing is wrong. When the tide is stationary the ambien temperature will be decrease a huge amount. You are recycling the already chilled water over and over, some tides can stay stationary or near stationary for several hours. This is from practical experience not a computer model.
you wrote
The current measurements show that slack water at Crib Point occurs over a short period of 10 to 20 minutes.
What can I say except I would get your equipment checked and inform everyone to dismiss your findings. In my 55 years of fishing Westernport I have only seen the 20 minute change of tides when there is a very low low tide and a very high high tide meaning the rise in the tide height over 6 hours will cause a very quick turn around. Many times around the full or new moon I and thousands of others have seen the tide stationary for many hours, yet your measuring devices say that is not correct, well I hate to tell you your equipment is 100% faulty.
you wrote
The plume does not go straight to the seabed at slack water. It travels laterally away from the FSRU and spreads vertically due to mixing. As the temperature difference is small, the cooler water plume descends gracefully toward the seabed (in the same manner as your light sinker).
At slack water the water travels laterally, West to East if the jets are on the starboard side pumping the water out in that direction and gradually mixes and then travels vertically. Now that depends on the length of time the slack water is being slack, but I see there is no reference to the current caused by the intake jets, so at 18 million litres an hour I am sure there has to be a current created all around the terminal which could cause a percentage of the outflow to be caught in this current and be recycled. Then I see you wrote that it gracefully descends to the seabed, but if your figures are correct does it have time to that as you claim slack water only lasts 10 to 20 minutes !!
you wrote
We have a computer model to examine the risk of discharge water being re-entrained into the seawater intake. This can occur to a minor extent around the time of reversing tidal flows, but the calculated effects are much smaller that you suppose. There will not be any super-chilled water occurring.
If your computer model uses the same figures you have provided here then it has to be considered another flawed model. You say 10 or 20 minute slack water, yet in practice this can be 2 to 4 hours. The calculated effects using your slack water time would say there is no or very little super chilled water, but supplying incorrect data can only give that result. If we use practical data the water is very very super chilled.
you wrote
yes, the problems of weed clogging the seawater intake are being addressed. We expect there will be a seawater intake on each side of the FRSU and so the port intake can be used while the starboard intake is being cleaned, and vice versa.
Again I am using practical data and would say that what you have proposal is impractical as the weed in Westernport is so great that you would be cleaning on a round the clock basis.
Then I had another read of you statement and discovered something you failed to mention in all the above. You say you will have intake ports on both sides which when in operation will create a massive turbulent current. Pumping in 18 million litres on either side and at the same time pump out the same will create a current that will have the intake pumps drawing in already super chilled water.
Dear @RobH1,
Sorry for the delay in responding to your messages. Unfortunately, due to the level of detail needed to fully respond to your questions, we are required to forward your messages to the technical specialists who are working on the studies for the Environment Effect Statement. In addition, their focus is to complete the marine environment research and reports required for the EES.
However, if you are interested, the environmental scientists and marine technical specialists from CEE will be available to answer your detailed questions at the upcoming community sessions, so you don’t have to wait for written answers on this forum. They will also be able to speak to you about the work carried out to date.
The locations of each CEE attended community sessions are:
Hastings
Saturday 24 August 2019
10.30am-12.30pm
Hastings Community Hub
1973 Frankston-Flinders Road, Hastings
Crib Point
Tuesday 27 August 2019
5pm – 7pm
Crib Point Community Hall
7 Park Road, Crib Point
Pearcedale
Wednesday 28 August 2019
5pm-7pm
Pearcedale Community Centre
710 Baxter-Tooradin Road, Pearcedale
Grantville
Tuesday 3 September 2019
5pm-7pm
Grantville Hall
1470 Bass Highway, Grantville