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The continuity equation states that if the cross-sectional area decreases (ex. a tube getting narrower), speed has to increase. If the tube is horizontal, this speed increase has to be provided by a force exerted by the surrounding fluid. In symbols, $pA > p'A'$, where $pA$ is the force that is causing the acceleration, and $p'A'$ the force due to pressure on the other side of the fluid element. We know that the tube has become narrower, so $\frac{A'}{A} < 1$. But then, couldn't $p'$ be larger than $p$? For example, if $A = 2, A' = 1$, we need that $2p > p'$. Something like $p = 1, p' = 1.5$ would be a valid solution.

Edit: the second answer of the suggested question assumes the cross-section is constant.

Qmechanic
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2 Answers2

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The correct approach is to analyse the forces in a volume element, instead of 2 different locations of the pipe.

For an horizontal pipe in the direction x, if the forces are not balanced, we have: $$-f(x+\Delta x) + f(x) = ma_x \implies -p(x+\Delta x)\Delta y \Delta z + p(x)\Delta y \Delta z = m\frac{dv_x}{dt}$$

Dividing by the volume $(\Delta x \Delta y \Delta z)$: $$-\frac{p(x+\Delta x)}{\Delta x} + \frac{p(x)}{\Delta x} = \rho\frac{dv_x}{dt}$$ Letting $\Delta$'s go to zero leads to the differential equation: $$\frac{dp}{dx} + \rho \frac{dv_x}{dt} = 0\implies dp + \rho dv_x\frac{dx}{dt}=0\implies dp + \rho v_xdv_x=0 \implies dp + d(\frac{1}{2}\rho v_x^2)=0$$

Integrating this equation, we get the relation between pressure drop and change of kinetic energy: $$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$

So. if the pipe is narrowing downstream, and there is a flow, the velocity must be increasing by the continuity equation. But that means an increase in the kinetic energy, what requires a higher pressure upstream.

Claudio Saspinski
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I think this is how it might be justified, but I'm not sure: $F_1 - F_2 = p_1A_1 - p_2A_2 > 0$. Let $A_2$ be "infinitesimally" larger than $A_1$, so that $A_2 = A_1 + dA$. Then $p_1A_1 > p_2(A_1 - dA)$ is equivalent to $p_1 > p_2(1 - \frac{dA}{A_1})$. In this expression $\frac{dA}{A_1}$ tends to zero and all that's left is $p_1 > p_2$.