I'm creating a report on the possible implementation of a sub hydro system and need some help. I've attached the image of what I'm proposing, where I've essentially created a head via a riser B with a water inlet just below sea level, which will then flow down into the hydraulic turbine and fill up the pipeline and Riser A to create electrical energy.

In it's initial state, the entire system is full of water. When power isn't in demand I've incorporated an ESP to then evacuate the pipeline and Riser A back out of Riser B (Riser B is still filled after). Initially I calculated the required energy needed to evacuate Riser A and the pipeline which would act as potential energy using the simple energy equation E=mgh. That potential energy would then be converted into electrical energy upon subsequent flooding of the system again.

I'm having trouble in determining the ideal size of the turbine though because I'm not sure how to calculate the flow rate that will be coming into Riser B and subsequently into the turbine itself since Riser B is never completely empty. I assume that I won't need that large a one due to the the large head, but I'm not sure of the appropriate governing equations or if they're even needed given my potential energy calculations.

I'll be testing multiple values of pipe diameter and pipeline length but original values were:

Riser A = 500m

Riser B = 50m

Pipeline L = 10000m

Pipeline, Riser A/B Diameter = 500mm

Any help would be appreciated, because I'm stuck in a bit of a rut finding the connection between the two as I'm using two different ideas and joining them.

Also for reference the ESP will be powered using excess energy from a wind turbine (not shown).

enter image description here

  • $\begingroup$ Check out wells turbines with wave energy - much simpler implementations compared to what you suggest. $\endgroup$
    – Solar Mike
    May 1, 2021 at 7:03
  • $\begingroup$ @SolarMike Thanks for the suggestion I'll be sure to have a look atleast for research, I'm quite deep into this already though hahah with regards to installation etc, it's just the actual calculations I'm confusing myself over because I did the energy calc as a potential justification, then moved on to hydropower equations $\endgroup$
    – Kuriosity
    May 1, 2021 at 7:16
  • $\begingroup$ You may be deep into this, but you will deep, deep pockets with minimal return. Energy recovery is 14 dollars (using transistors numbers). Under sea storage you link to is cost effective for oil company because 2000m3 * 50 dollars/ton * 0.8 = 125k dollars. I'm guessing 50 dollarston. 0.8 is specific gravity of oil. So it is cost effective for them, not you. $\endgroup$ May 2, 2021 at 0:57

1 Answer 1


I can't easily help you with the flow rate calculations but I offer the following observations:

  • This is an undersea pumped storage system.
  • The capacity of the sloped pipe is $\pi r^2 l = \pi\cdot 0.25 \cdot 10000 = 1963 \ \mathrm {m^3} $. It will hold 2001 T of sea water.
  • The empty pipe anchors will have to counter a buoyancy of 2000 T.
  • The pressure on the empty pipe will be 50 bar at 500 m.
  • Riser A has a capacity of 500/10000 of that, so 0.5% and I'll ignore it in the calculations.
  • You've put the hydraulic turbine at the top of the pipe. Your working head, therefore is 50 m no matter how deep the second riser is. The 10 km of pipe and 500 m depth buys you nothing other than storage and pain. You would get the same benefit from a 2000 m3 low-profile storage at the bottom of your 50 m Riser B. These would have to withstand 5 bar external pressure when empty so I'm picturing cylindrical tanks with domed tops and bottoms - maybe 1000 L (1 T) capacity each. Each would need a water inlet/outlet at the bottom manifolded together and an air inlet/outlet at the top also manifolded and connected to the surface by a 50 m Riser A.
  • Most land-based pumped storage systems use the turbine as a pump to return the water to the reservoir. Your system has a second pump for this. This is duplication of equipment with associated cost.
  • Your evacuation pump is only 50 m deep. You can't evacuate the sloped pipe. To evacuate the sloped pipe the ESP would have to be at the bottom of Riser A.
  • The energy available to your turbine is $mgh = 2001000 \times 9.81 \times 50 = 491 \ \mathrm{MJ} = \frac {491\text{M}}{3600} = 136 \ \mathrm {kWh}$. If you can get 10c per kWh that's worth 13.60 in your favorite currency.
  • To pump the water out from 500 m will also be determined by $mgh$ and this will work out to ten times the 50 m drop, $1360 \ \text{kWh}$. Cost 136.00 of your favorite currency - and that assumes that you can purchase off-peak at the same cost as your feed-in tarriff.

I'm not investing in your company for now!

  • $\begingroup$ Thanks so much, I knew I'd need to go through more iterations but hadn't thought of a lot of this! Yeah so it's a variation of pumped storage hydro, negating the need for mountains/rivers/dams etc since the ocean is an infinite supply. I've been getting advice on subsea systems from a relative who's in oil and gas hence the inclusion of the ESP. Do you think that if I remove Riser A and move the turbine to the end of the pipeline (while shortening it to reduce buoyancy issues), therefore creating a bigger head would work? No worries, I'm a long way away from anything like that yet hahah $\endgroup$
    – Kuriosity
    May 1, 2021 at 9:34
  • $\begingroup$ Moving the turbine to the end of the pipeline means it can only pass water until Riser A is filled so you have poor capacity and a head that's decreasing as the riser fills. Back to the drawing board. You need to come up with a scheme that gives fairly constant head. If you look at terrestrial pumped storage the lakes cover large areas and are shallow relative to the head. Therefore they behave consistently through the discharge. You need the same with the added large headache that your under water storage is at high pressure. $\endgroup$
    – Transistor
    May 1, 2021 at 9:48
  • $\begingroup$ An alternative is to use the sea as the lower lake and elevate the storage. Here's a study document for one < 1 hour from where I live. $\endgroup$
    – Transistor
    May 1, 2021 at 9:51
  • $\begingroup$ Yeah the pressure is the big annoyance, originally I'd not even considered the risers and wanted to actually utilise the pressure to flood a storage tank placed in deep water with a turbine inside then emptying it; turns out it's already been done so here I am with this trainwreck haha $\endgroup$
    – Kuriosity
    May 1, 2021 at 10:56
  • $\begingroup$ I've just had a thought, what is I utilised this idea and joined it up to the riser, and pumped the water out through it? Suppose it'd then just be potential energy being wasted as the water would be pumped from below anyway $\endgroup$
    – Kuriosity
    May 1, 2021 at 11:20

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