Animated Solution for Physics - Properties of Solids and Liquids: Water flows in a horizontal tube (see figure). The pressure of water changes by $700{\text{ Nm}}^{-2}$ between $A$ and $B$, where the area of cross-section are $40{\text{ cm}}^2$ and $20{\text{ cm}}^2$, respectively. Find the rate of flow of water through the tube. (Take, density of water = $1000{\text{ kgm}}^{-3}$)

Visualizing the Flow

Imagine you are watching water flow through a horizontal pipe that suddenly narrows. As the water moves from the wider section $A$ to the narrower section $B$, it has to squeeze through a smaller space. Because water is incompressible, it can't just pile up; it has to speed up! This increase in speed comes at a cost: a drop in pressure.

In our problem, we are given the cross-sectional areas of both sections: $A_A=40{\text{ cm}}^2$ and $A_B=20{\text{ cm}}^2$. We are also told that the pressure drops by $700{\text{ Nm}}^{-2}$ as the water moves from $A$ to $B$. Our ultimate goal is to find the rate of flow, which is the volume of water passing through any cross-section per second.

What Goes In Must Come Out

Our first powerful tool is the Equation of Continuity. It simply states that for an incompressible fluid, the volume flow rate must be constant throughout the pipe.

Let's plug in the areas we know:

By dividing both sides by $20$, we find a beautiful, simple relationship between the velocities:

This makes perfect intuitive sense. Since the area at $B$ is exactly half the area at $A$, the water must travel exactly twice as fast to get the same amount of volume through in the same amount of time.

The Energy Balance

Now, how do we connect these velocities to the pressure difference? Enter Bernoulli's Principle. This principle is essentially the conservation of energy for flowing fluids. For a horizontal tube, the potential energy due to height doesn't change, so the equation simplifies to balancing pressure energy and kinetic energy per unit volume:

Let's rearrange this to isolate the pressure difference, $p_A-p_B$, which we know is $700{\text{ Nm}}^{-2}$:

Bringing It All Together

Now we substitute all our known values into this rearranged Bernoulli's equation. We know $\rho =1000{\text{ kg m}}^{-3}$ and $v_B=2v_A$:

Be very careful here to square the entire term $(2v_A)$ to get $4v_A^2$. Let's simplify:

Solving for $v_A^2$ gives us:

Taking the square root, we find the velocity at section $A$:

Since our areas are given in ${\text{cm}}^2$ and the options are in ${\text{cm}}^3\text{/s}$, we must convert this velocity to $\text{cm/s}$ by multiplying by $100$:

Finally, the rate of flow $Q$ is the area multiplied by the velocity at any section. Let's use section $A$:

And there we have it! The water flows at a rate of $2720{\text{ cm}}^3\text{/s}$.