# What force accelerates a liquid moving in a narrowing pipe?

Say there is a liquid which behaves like an incompressible fluid and is flowing steadily through a pipe which is moving from a cross section of area $$A_1$$ to the cross section of area $$A_2$$, where $$A_2$$ is less than $$A_1$$. As per the continuity equation, $$v_2>v_1$$ and so the liquid seems to be accelerating. What force is causing this acceleration?

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Jun 14 '20 at 22:01
• Bernoulli’s Principle on Atomic Scale from "Physics Videos by Eugene Khutoryansky". Jun 16 '20 at 1:40

You are right. From continuity of the incompressible fluid you have $$A_1 v_1 = A_2 v_2.$$ So obviously the velocity is changing. Thus the fluid is accelerated, and therefore there must be a force causing this acceleration. In this case the force comes from the pressure difference between the wide and the narrow part of the pipe.

This can be described by Bernoulli's equation ($$p$$ is pressure, $$\rho$$ is density) $$\frac{1}{2}\rho v_1^2 + p_1 = \frac{1}{2}\rho v_2^2 + p_2$$

• Nice answer too. Bernoulli's equation helps to understand how pressure changes. Jun 14 '20 at 9:52
• @Bernhard actually, I think this is the cleanest explanation for me of why the pressure changes with velocity. Jun 14 '20 at 16:33
• a tapered pipe is sometimes called a bernoulli transformer. Jun 14 '20 at 16:36
• @JohnDvorak As long as you are aware what the limitations are of applying Bernoulli's equation. Jun 14 '20 at 17:27

There is more mass per area behind than ahead of the constriction so since ppressure is force divided by area there develops a pressure difference

• Oh c'est chouette. Jun 15 '20 at 3:37

As you state, from the continuity equation you can see that the velocity $$v_2>v_1$$. The next step is a momentum balance (like any balance in fluid dynamics: $$\frac{d}{dt}=in-out+production$$). The momentum flowing into the system is smaller than the momentum flowing out of the system ($$\rho A_1 v_1^2 < \rho A_2 v_2^2 = \rho A_1 v_1^2 \frac{A_1}{A_2}$$, $$\frac{A_1}{A_2}>1$$).
The actual force is not the pressure itself, but the pressure difference, or, actually, the difference in force, because on the left the force is $$p_1 A_1$$ and on the right $$p_2 A_2$$.