Cfrgtkky

I am interested in building a very basic water jet propulsion system for a toy boat. Before I build it, I would like to know from a conceptual standpoint if this design of a water jet propulsion system will actually produce forward thrust.

Please refer to the conceptual drawing below of this water jet design. This drawing shows a top-down view of the toy boat.

An embedded motor-propeller will pull water into the pipe section coming from the stern of the toy boat and will at the same time be forcing water out of the pipe section going back to the stern of the toy boat. The force which should propel the toy boat forward should come from the dynamic pressure of the rushing water pushing against the inner walls of the two 90 degree elbow sections of the pipe. Will this particular design of a water jet produce forward thrust as expected?

Although I know that a traditional inline water jet system would be the more ideal thing to construct, I am very interested in finding out if a boat can be propelled using just the dynamic pressure(s) generated within a pipe.

No, it would just create a torque couple and rotate the boat anticlockwise.

Let's say the small propeller has an output volume, q grams/s, and the distance between inlet pipe and outlet is 5cm.

The thrust and suction of each end of pipe.
$$ F = ρ q (v2 - v1) $$
say density of water is =1, and V1 is initially zero, for simplicity, even though it wouldn't affect the outcome either way.

$ F = qV2 and V2 = q/a :a is pipe's area$

And you have a torque,
$T = 5*q^2/a. $,

anticlockwise direction

This will turn the boat in place.

Edit

After some comments I added a bit more detail:

The OP's sketch has 6 nodes or bends that we calculate. Note we analyse the sketch as it is, not introducing any changes or recommendations.

The entrance and exit momentums and thrusts have been done above and remain the same, including the torque they cause.

let's call the four bends from left to right c1, c2, c3 , c4. these corners each experience reaction force $ F= ρ q (v2 - v1) = ρ q (v1sin(45) - v1) + ρ q (v1 cos(45) - v1) $

And if we project these two vector component on the x and y axis at C1 we have $$ F_{c1} = q^2/a$$

and its direction is 135 degrees at c1. And it's reaction is pushing the boat back at 135 degrees out.

At c2 we end up with the same reaction as Fc1 and same 135 degrees.

At c3 we have same reaction but pushing the ship at 45 dgrees.

At c4 we have the same reaction again pushing the boat out at 45 degrees.

The horizontal components of these vectors cancel out and the vertical components add up to $$ q^2/a*4* sqrt{2}/2 = 4* 0.707 = 2.82 q^2/a $$

This is forward thrust pushing the boat forward.

However the boat is still rotating under the combined torque and thrust.

This is actually one of the methods cruise ships use as a docking maneuver to turn in place in ports with limited space.

The water starts out stationary. It ends up being jetted out the stern. Yep, that makes thrust. There will be some pressure interaction over the entire hull of the ship. That is geometry dependent. It can be engineered to minimize the parasitic drag from collecting the inlet from the stern.

This is basically a thrust reverser in reverse.

Amrican Airlines jet powerback video - https://www.youtube.com/watch?v=-Zkxh903s_w

Based on the other answers to this question, this simple water jet design should produce forward thrust. Yet for the toy boat to travel forward in a straight line it needs an additional part added to it. To cancel out the anticlockwise rotation which will be caused by the combined thrust and torque, I believe adding a small rudder into the inlet pipe section will provide the means of generating a clockwise torque to cancel out the anticlockwise torque. See modified picture below. I will probably need to keep readjusting the rudder towards the Port side and fixing it in place until I find the ideal position to make the toy boat travel in a straight line. I will then use the toy boat's main rudder to steer the boat.