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Suppose you have a $5\,\text{m}$ high waterfall. The water will be accelerated downwards by gravity and impart a force, proportional to the cross sectional area of the falling column of water, on the surface at the bottom. So the falling water exerts a constant hydrodynamic pressure (force/area) as a function of the drop height and the fluid density, ignoring atmospheric drag.

Now suppose you are at the bottom of a pool of water $5\,\text{m}$ deep. You will experience the hydrostatic pressure of just the weight of all the water above you. The water at the bottom will be compressed, and under pressure, and exert a constant hydrostatic pressure as a function of depth and density.

The falling water is not being compressed by gravity, since the water is allowed to fall (Falling observer experiences $0g$), so there is no hydrostatic pressure. But the same water is moving at a velocity enough to exert a given hydrodynamic pressure.

My question is, given the same fluid density, height/depth, and gravitational acceleration, is the hydrodynamic pressure at the bottom of a falling stream always equal to the hydrostatic pressure at the bottom of a static column?

Vincent Thacker
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CPlus
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3 Answers3

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This is very similar to the falling chain problem.

Assume a column of water with a cross-sectional area A and column length L, is suspended directly above a scale.

The weight of water is $$Mg = \rho A L g$$

The total weight measured on the scale after all the water has fallen will be:

$$W_T = W_{water} + F_{impact}$$ where $F_{impact}$ is the force due to the momentum change that occurs due to each each falling drop dm that hits the scale.

The velocity the dm achieves after falling a distance L is: $$ v = \sqrt{2gL}$$

EDIT: The water ground collision is not perfectly inelastic like a falling chain. After collision, the water spreads with a velocity that is a fraction of original:

$$v_s = \gamma v$$

If the water column is uniform, then dm can be stated as a function of dx by using the linear density $\frac{M}{L}$

$$dm = \frac{M}{L}dx$$

$$F = \frac {dp}{dt} = \frac{M}{L}dv\frac{dx}{dt}$$

Since $$dv = \sqrt{2gL}(1-\gamma)$$ and $$\frac{dx}{dt} = \sqrt{2gL}$$

$$F_{impact} = 2Mg(1- \gamma)$$ and $$F_T = Mg + 2Mg(1-\gamma)$$

When $\gamma = 1$ then $F_T = Mg$ which is the answer given by Vincent Thacker using Bernoulli's equation. Otherwise, the water imparts a fraction of its momentum to the ground.

My question is, given the same fluid density, height/depth, and gravitational acceleration, is the hydrodynamic pressure at the bottom of a falling stream always equal to the hydrostatic pressure at the bottom of a static column?

Unless the water collision is perfectly elastic, then No. Water is highly incompressible and at 5m any additional pressure will be small compared to the momentum imparted to the ground if the collision is inelastic.

Falling chain problem: https://www.feynmanlectures.caltech.edu/info/solutions/falling_chain_sol_1.pdf

https://canvas.harvard.edu/files/15494/download?download_frd=1&verifier=fB9E3sf0KFGvm3NYSzOhf8tFhvImRiFVhV8e2Yth

Stevan V. Saban
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If the water is assumed to start from rest at height $h$, using Bernoulli's equation, we get $P_0 + \rho gh = P_1$ so the additional pressure at the central point of the bottom is $P_1-P_0$ which is indeed equal to $\rho gh$. The velocity of the cylindrical jet of water just before hitting the bottom is $v=\sqrt{2gh}$.

If the water breaks up into droplets, the pressure can vary and becomes much more difficult to calculate, cf this post.

Vincent Thacker
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At the center of the falling fluid contact, the pressure is going to be equal to the stagnation pressure $p_s=\frac{1}{2}\rho v^2$. For the velocity $v=\sqrt{2gh}$, that means that, at the center of contact, $$p=\rho g h$$ This is the same as the hydrostatic pressure in the static case. Further from the center of contact at the surface, the pressure will be less than this.

Chet Miller
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