Assessing flow rate from open sea into a container

Question

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?

• Head = 200 m
• 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.

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Answer 1

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.

Answer 2

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 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 again?