# Upthrust Reaction - Water Injection @ 5bar -8msw Subsea

• jakeward46
In summary, the problem is to flush a damaged underwater pipeline with a containment box and a surface pump. The box will be placed over the pipeline and connected to an 8" pipeline for flushing. The weight and upthrust of the box and pipeline have been calculated, and the question is how to relate injecting water pressure into the box and if the box will stay down or lift with the pipeline. It has been determined that without a solid seal, the maximum pressure that can be achieved is 2m head of overpressure.
jakeward46
Hello,

I have a problem

Basically, in theory -- a damaged 1220mm dia pipeline sub-sea (-8msw) with a 500mm hole in the 12oclock position 600mm. Pipeline resting on the seabed

The goal of the problem is to flush the pipeline of its contents in both directions.

The idea - have a containment box (open bottom) fabricated to suit the pipe, as shown in my drawing attached - which would go over the pipe, and then have side plates attached. An 8"pipeline would then be quick connected to the inlet 8" at the top. The idea is to flush water into the box from a surface pump at 5bar pressure (flow rate not known) from a height of +10m flushed down a 8" dia, 35m length rubber (smooth internal, 0.1mm roughness) to the box @-8msw, (1.8bar atm) - which would push the contents of the pipe in each direction to a sludge tanker 20km either side of the break.

I used bernoulli equation to try and figure out the flow rate but was coming up with 550m^3/hr.. not too sure on this figure.

So data.

Weight Air = 16,000kg
Upthrust = 2,575kg
Weight Seawater = 13,425kg
Volume of steel = 2.5m^3
Box dimensions (open bottom, 2m of the 3.3m sides go into hard seabed - 2m below the 6oclock of the pipe) box dimensions 5m length, 2m width, 3.3m height.

I worked it as though the seabed would act as a bottom to the box (6oclock of the pipeline) and used the new volume (5mx2mx1.3m = V = 13.2m^3) to work the upthrust if air was to be injected (i know that is not the solution here) and worked that volume of 13.2m^3 ...

upthrust = 13,596kg + upthrust steel volume 2,575kg
total upthrust 16,171 - 13,425 (weight seawater) = 2,746kg of lift.

However i know injecting water wouldn't produce an action like injecting air, so my question is, how can i relate injecting 5bar of water pressure into the box? to see if the box would stay down or if it would lift with the pipeline.

Thanks for any help towards this in advance.

#### Attachments

• Dimensions Rough.png
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Unless the box is sealed to the seabed, you will only get about 2m head of overpressure into the pipe. Any higher pressure will force the water out the bottom of the box.

If you are pumping sea water in, the lift would be due only to the overpressure but still limited to the 2m head of overpressure.

To get any significant flow rate you will need a solid seal around the bottom, whether it is the seabed or a bottom plate on the box.

## 1. What is "Upthrust Reaction" in the context of subsea water injection?

Upthrust Reaction refers to the upward force exerted by a fluid, in this case water, when it is injected into a subsea environment at a pressure of 5 bar and a speed of 8 meters per second. This force is a result of the principles of buoyancy and is important to understand when designing and operating subsea water injection systems.

## 2. Why is water injection used in subsea operations?

Water injection is commonly used in subsea operations to increase the pressure in oil or gas reservoirs. This helps to maintain or increase the production rate of these resources. Additionally, water injection can be used to displace oil or gas towards production wells, improving the overall efficiency of the production process.

## 3. What is the significance of a pressure of 5 bar and a speed of 8 meters per second in subsea water injection?

These values are commonly used in subsea water injection operations as they are considered to be within safe and effective ranges. A pressure of 5 bar is high enough to create the necessary upthrust reaction, but not so high that it risks damaging equipment. Similarly, a speed of 8 meters per second is fast enough to achieve the desired results, but not so fast that it causes excessive turbulence or other complications.

## 4. How is the upthrust reaction of subsea water injection calculated?

The upthrust reaction of subsea water injection can be calculated using the formula P = ρgh, where P is the pressure in bar, ρ is the density of water in kg/m³, g is the acceleration due to gravity (9.8 m/s²), and h is the depth at which the water is injected in meters. This formula takes into account the principles of buoyancy and allows for the calculation of the upward force exerted by the injected water.

## 5. What are the potential challenges associated with subsea water injection at 5 bar and 8 meters per second?

Some potential challenges of subsea water injection at these conditions may include the risk of equipment damage due to high pressure or turbulence, the need for proper design and maintenance to ensure the safety and efficiency of the injection system, and potential environmental impacts such as displacement of marine life. It is important for scientists and engineers to carefully consider and address these challenges in order to effectively utilize subsea water injection for oil and gas production.

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