Concept subsea PHES system, /advice with calculations

[kirkmcg97]

TL;DR Summary

I've designed a subsea pumped storage hydro system and want to know if I'm on the right track with my flow rate and head loss calculations

I've created a concept design of a subsea pumped hydro storage system, and need to use the flow rate in order to calculate power output. I can calculate the energy storage capacity using $E = mgh$:

$$E = mgh = 25675000×9.81×200 = 50374 MJ = \frac {50374MJ}{3600} = 13992 kWh$$

I've attempted to determine the flow rate by choosing a turbine (5MW Francis from https://www.researchgate.net/publication/302969536_The_use_of_LINGO_programming_language_to_develop_a_computer_tool_to_provide_a_technical_and_economic_analysis_of_a_hydraulic_potential_that_allows_the_application_of_Francis_turbines_and_Kaplan_turbin/figures?lo=1) and working backwards, to then find the resulting pipe diameter and head loss. In the model below, the top of the pipe has a valve that will open to let seawater in during the power generation stage (high energy demand), down through a 90/10 Cu/Ni pipe, through a turbine and subsequently fill up the vessel, which is located 200 m below the inlet (Head = 200 m). Vessel design doesn't take into account the pressure at this depth through use of a bag seen here. In times of low demand, using excess energy produced from the wind turbine, the water will be pumped back up to the surface to be stored as potential energy.

• Head = 200 m
• Pipe Length = 1594m (assuming a straight path between elevations)
• Vessel Volume = 25,000 m³
• Seawater Density = 1027 kg/m³
• Turbine Power = 5MW
• Generator Efficiency ($\eta_g$) = 0.9

So in generator mode, the rated discharge will be:

$$P_g=\rho \times g \times h \times \eta_g \times Q_g$$

Rearranging for Q while assuming generator efficiency is 0.9 gives:

$$Q_g = \frac{P_g}{\rho \times g \times h \times \eta_g} = 2.7599m^3/s$$
(Can generate rated capacity for 5.03 hours)

Rearranging the formula for pump mode to find Q_p;

$$Q_p = \frac{P_g \times \eta_p}{\rho \times g \times h \times} = 2.2355m^3/s$$
(Can be operated for 6 hours)

Assuming pump velocity (v_p) to be 5m/s, required diameter (d) is found using:

$$A=\frac{Q}{v}$$ then $$ A=\pi \times \frac{d^2}{4}$$ therefore d = 0.754m.

Reynolds Number:

$$Re = \frac{u \times d_h}{v} = 362500$$

So flow is turbulent.

Reading from Moody chart and using Colebrook equation to check, friction coefficient (f) = 0.0147.

Therefore via Darcy Weisbach, Pressure Loss = 399 kPa and Head Loss = 39m.

Have I done this correctly or is there something glaringly obvious that I've missed? Any thoughts or suggestions are welcome! And please note this doesn't (and probably won't) be economically feasible, it's purely a concept to see if something like this could work.

 

[Baluncore]

Welcome to PF.

Any thoughts or suggestions are welcome!

If the pipeline is also filled with water, then I do not understand where in the system there can be a pressure difference or head. When would the pipeline be filled with air ?

 

[kirkmcg97]

Welcome to PF.

If the pipeline is also filled with water, then I do not understand where in the system there can be a pressure difference or head. When would the pipeline be filled with air ?

So the system would be originally empty (including the vessel), enough water would be allowed to pass through to fill up the vessel to generate power leaving the pipe empty/filled with air, then would be pumped back out, leaving the pipe empty once again. Similar to how pumped hydro works with penstocks etc, the vessel will act as the lower reservoir with the sea as the upper. So while empty the pipe would need an operating pressure of 20bar which I'm aware of and have been calculating the required wall thickness.

 

[Baluncore]

Vessel design doesn't take into account the pressure at this depth through use of a bag seen here.

So your underwater 25,000 m³ reservoir is not an exposed bag but a rigid shell that operates with up to 20 bar of external hydrostatic pressure.

 

[kirkmcg97]

So your underwater 25,000 m³ reservoir is not an exposed bag but a rigid shell that operates with up to 20 bar of external hydrostatic pressure.

Do the seawater intakes not mean that there isn't a need to design against the hydrostatic pressure? As when the bag is empty due to pumping, the chamber will fill up with seawater. Then during generation the bag will fill up, replacing the water in the chamber? There's a diagram in that brochure explaining it and with seawater tests too, maybe I've just got the wrong idea and should consider a large concrete chamber instead

 

[Baluncore]

A bag is employed to isolate oil from water, the pressure inside the bag is the same as outside the bag. If there is no pressure difference between the bag and the external seawater in contact then there is no energy storage benefit from the bag. The storage volume then becomes that of the tube that may be filled with air or water.

Consider the pipline and an air pump/turbine. You could pump air into the bag reservoir, raising the world wide sea level very slightly during the process. Then you could release the compressed air to recover the energy. The system would operate at approximately 20 bar pressure difference. The pump/turbine could be at the surface.

But I can't see how you could do it by pumping water through the pipeline since the hydrostatic pressure will then be the same in the pipeline and the ocean, so there is no pressure difference.

 

[Baluncore]

Do the seawater intakes not mean that there isn't a need to design against the hydrostatic pressure?

Maybe you need to draw four diagrams of your system. Discharged, half charged, fully charged, and then half discharged. Mark on it the presence of air or water. Start at sea level and mark the hydrostatic pressure, inside and outside the system. Look for where there is a pressure difference across the wall, between the inside and the outside. When is there any pressure across the turbine?

 

[kirkmcg97]

This

Maybe you need to draw four diagrams of your system.

I created these, I see what you mean about the pressure, since the vessel is always water it'll be in equilibrium; if I was to disregard the bag altogether and create say, a concrete vessel of similar size would it then work? I guess it'd be very similar to the StEnSea project but worth asking.

 

[kirkmcg97]

Also under consideration is to add a riser purely as an air inlet in order to not restrict the water flow