Putt-putt boats and the reverse sprinkler — Matthew Lee

In the first FOTW,(here: https://quantapublication.wordpress.com/2011/01/16/putt-putt-boats-matthew-lee/) we took a look at the mechanism by which the humble putt-putt boat propels itself through water, and tried to explain how the boat manages to move forward at all despite sucking and blowing water in and out from the rear. Then, I quoted the Flying Circus of Physics and claimed that the boat does not just oscillate back and forth on the spot because the force exerted during the blowing-out stage is greater than the force exerted during the sucking-in stage, since “when water is sucked into the engine, it is drawn into the pipe from a hemisphere around the pipe opening”.

Putt-putt -oops

However, as it turns out, that explanation leaves out another reason which is simpler and probably more significant! This is based on the conservation of momentum in the first stage when water is being sucked in and the second stage when the water is being blown out.

In the first stage, the boat sucks in water from the rear. In the lab frame, the velocity of the boat is v0 (taking the left as the positive direction) and the velocity of the water outside the boat is 0. When the boat has sucked in a volume of water and its pipes are full, a perfectly inelastic collision has taken place, as the water is now contained inside the boat, and moving at the same speed as the boat. Depending on the relative masses of the water and the boat, the final velocity of the boat filled with water is a fraction of its initial velocity.


Fig. 1: Conservation of momentum during the sucking-in stage. The mass M1 represents the boat and the mass M2 represents the water that is sucked into the boat.

 In the second stage, the boat expels the mass of water with an exhaust velocity of vd, which propels it forward (vd has a negative value since it is in the direction of the right). As shown in the last equation in Fig. 2, the final velocity v2, can be very high, as long as vd is negative enough. Since the water is expelled from the boat very rapidly as the water in the diaphragm boils, it is quite likely that vd is negative enough for the boat to accelerate forward each cycle.

Fig. 2: Conservation of momentum in the blowing-out stage. Vd is negative.

In fact, as long as -vd > v0, the final velocity will be greater than the initial velocity v0 and the boat will accelerate in each cycle.

This is what we should expect anyway, because if one looks at the first and last stages only, ignoring the intermediate stage when the two masses are joined, the condition -vd > v0 is necessary for the mass m2 to have a more negative velocity after the whole cycle is over.

Reverse sprinkler

It turns out that this situation is related to a problem discussed by Feynman and other physicists in the 1940s, dubbed the ‘reverse sprinkler’. A normal s-shaped garden sprinkler expels water from its two ends and hence experiences a torque opposite to the direction of water flow. But what happens if the sprinkler is submerged in water and made to suck in water instead?

The answer is that the reverse sprinkler does not accelerate when water is flowing into it! This phenomenon, which may be counterintuitive, is because of two effects that oppose each other.

(By the way, the reverse sprinkler is different from the putt-putt boat in that the sprinkler continuously sucks in water, whereas the boat only sucks in a finite volume of water then stops.)

The first effect is due to the difference in pressure between the inside of the sprinkler and the outside. The pressure inside the sprinkler is lower than the external pressure. If the sprinkler’s openings were capped, as in Fig. 3a, there is no net torque or force on the sprinkler as all the forces on the surfaces of the sprinkler due to the external pressures cancel out. This makes sense: a body lying passively in a fluid should not experience a net force or torque due to hydrostatic pressure alone. But if the caps are removed and water is allowed to flow into the sprinkler, then there is a torque on the sprinkler due to the pressures exerted at the areas marked out by coloured arrows shown in Fig. 3b.

Fig. 3a: Forces exerted by external pressure (blue arrows) cancel out when the sprinkler is capped at both ends.

Fig. 3b: Due to the pressure difference between the inside of the sprinkler and the outside, there is a net clockwise torque when water is allowed to flow into the sprinkler.

This pressure effect would cause the sprinkler to accelerate in the clockwise direction, towards the incoming water. However, this does not actually occur due to an opposing effect.

The incoming water is made of many tiny volume elements that were initally at rest, somewhere out there in the surrounding water where the pressure is p2. These little chunks of water were accelerated along the pressure gradient until they entered the mouth of the sprinkler, where the pressure is p1 (Fig.4). Along this path, the difference in pressure between one side of the volume element and the other side is dP, so the total force exerted on the group of volume elements is AdP, where A is the total area of the group of volume elements when they occupy a thin disk across the cross-section of the sprinkler. Since the force is the rate of change of momentum, at each moment in time during the acceleration of all the small volume elements, the total rate of momentum change experienced by the whole group of volume elements is AdP. Hence, the total rate of change of momentum of the group of volume elements when they travel from far away to the opening of the sprinkler is A(P2-P1).


Fig. 4: The tiny volume elements are accelerated down the pressure gradient into the mouth of the sprinkler. Note that since the cross-sectional area of the sprinkler is taken to be uniform, by Bernoulli’s law, the internal pressure is a constant.

Now, when the group of water volume elements collides with the bend in the sprinkler and makes a right angled turn, their momentum in the horizontal direction becomes zero. This change in momentum is due to a force exerted by the sprinkler on the water, which has reduced its momentum to zero. The magnitude of this force is P2-P1.

This nicely means that the total force exerted on the inner surface of the sprinkler wall is the hydrostatic pressure P1 plus the additional force required to change the momentum of the incoming water P2-P1. As a result there is no net force exerted on the sprinkler walls and no acceleration of the sprinkler occurs.



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