Assessing flow rate from open sea into a container

I'm currently assessing sub sea pumped hydro capabilities, and need to use the flow rate in order to calculate power output. I can calculate the energy storage capacity using E = mgh:

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

However, I don't know what the flow rate from open sea into the pipe would be. In the model below, the top of the pipe has a valve that will open to let seawater in during the power generation stage, 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.

What would the best way to calculate the flow rate to use in theoretical power calculations, since the sea is an open reservoir?

• Pipe Length = 1594m (assuming a straight path between elevations)
• Vessel Volume = 25,000 m3
• Seawater Density = 1027 kg/m3

These variables can all be changed this is just preliminary/theoretical. • What exactly does the sentence pressure is negated through a bag mean? Water flows when there's a pressure difference. with the bag system in place as on pg 8 of the presentation there will be no flow. It appears you idea won't work.
– mart
May 3 '21 at 7:41
• @mart Sorry I just meant that the storage design doesn't need to consider the pressure at the seabed May 3 '21 at 7:51
• i started typing an answer but the fact that you include a pipe in your scheme and that you didn't say anything about what's in the tank before water flows in tells me that youlack basic understanding of the issue. Maybe someone else is more willing to do basic schooling. look at head loss equations on wikipedia but mostly understand what defines the pressure at both ends of your pipe.
– mart
May 3 '21 at 8:24
• Does this answer your question? I want to determine the flow rate and required turbine size for an offshore hydroelectric scheme and need help May 3 '21 at 8:26
• Also, what is the point of the long pipe in the first place? Why not just have the turbine inlet/discharge right at the bottom, making the whole system much more compact and eliminating pressure losses? The water down there is going to be very similar to the water near the surface. May 3 '21 at 14:55

You can't pump water into the ocean and gain the potential energy of it again like this. In this example, if you pump water out, the tank fills with water anyway because it's really just a balloon. If you somehow had a cave you could pump out then it's a simple head calc, using the total pipe equivalent length.

Basically, putting this in the ocean instead of below a lake (like this is normally done) just makes all the materials more expensive and difficult to maintain.

• I think OP has in mind a giant tank ("cave")? it would have to be comparable in size to a resirvoir used with a hydroelectric dam. However, roof of dome would be bigger than wall of conventional dam?! So can't scale big, but maybe could have niche applications... I'm keeping open mind, considering that crazy Swiss concrete block + crane energy storage scheme someone posted a month or two ago seems to've got funded May 5 '21 at 22:39
• @PeteW you're correct! The tank would act as the lower reservoir, with the pipe running the natural shelf to the wind turbine. It'll have to be a large drop in order to justify the length of pipeline that would be needed, so currently looking at coastal Spain in order to try and keep within the 1:12 L/H ratio that's recommended for hydro storage schemes to be viable, like the Avile's Canyon or something. There's only one or two spots in Scotland that could possibly work. It's only a working theory anyway, it's more to balance out wind farm inconsistencies than be it's own standalone system May 5 '21 at 23:43

I've had a go at calculated the pressure loss along the pipe, can someone verify this?

A 5MW Francis turbine has been chosen using this;

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 Qp;

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

Assuming pump velocity (vp) 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 again?