# When solving using Bernoulli's principle, is the pressure potential energy per volume of atmospheric pressure water 0 Pa or 100,000 Pa?

I was just given a question using Bernoulli's principle that gives you the pressure of water (e.g. $$1,000,000 \mathrm{Pa}$$) in a hose, and asks for the speed of the water after the water has left the hose if all the pressure potential energy was converted to Kinetic energy, I assume you would solve using:

$$\text{Pressure Potential Energy} = \text{Kinetic Energy}$$

and by diving both sides by volume, get:

$$\text{Pressure} = \frac{1}{2} \cdot\text{density} \cdot \text{(speed)}^2$$

My question is: since this implies that $$\text{Pressure Potential Energy} = 0$$ after being converted to Kinetic Energy, doesn't this mean that the pressure of the water is $$0$$?

(Since $$\text{Pressure Potential Energy} = \text{volume}\cdot \text{pressure})$$

But how can this be? I thought the atmospheric pressure of water is $$\sim 100,000 Pa$$

Hope that question makes sense

## 2 Answers

Potential exists between two points. In this case, a high-pressure region A exists inside the hose and a low-pressure region B exists outside the hose. The water moves from the higher potential (high pressure) region, from the hose, towards the lower potential (lower pressure) region, the atmosphere. As soon as the water reaches the atmosphere the potential difference between regions A and B that once existed is now zero.

In order to find the velocity of water after it left the hose, apply Bernoulli's principle at a point in region A and a second point in region B. Now we obtain, $$P_{hose}+\rho_{water}\frac{v_{hose}^2}{2}+\rho_{water}gh_{hose}=P_{air}+\rho_{air} v_{air}^2+\rho_{air}gh_{air}$$ and usually, we consider $$h_{air}=h_{hose}$$, $$v_{air}=0$$. From this, the velocity of water leaving the hose is found as $$v_{hose}=\sqrt{2(P_{air}-P_{hose})/\rho_{water}}$$.

There is no such thing as "pressure potential energy" in the usual applications of Bernoulli's equation

$$\frac{1}{2}\rho v_1^2 + \rho g h_1 + p_1 = \frac{1}{2}\rho v_2^2 + \rho g h_2 + p_2.$$

The pressure term is just pressure, it is not associated with any additional energy. The only energy there is - is kinetic energy $$\frac{1}{2}\rho v_1^2$$ and potential energy $$\rho g h_1$$. A water element is not carrying any additional energy due to its higher pressure. This is because it takes zero work to create pressure in an incompressible fluid.

To find the final speed of water outside the hose, just use the Bernoulli equation above with $$p_1 = 1~000~000~000~\text{Pa}$$ (inside hose) and $$p_2 = 100~000~\text{Pa}$$ (outside hose). In case the height is the same at both points, you can drop the $$\rho g h$$ terms.

In reality water is somewhat compressible so there is some energy associated with its higher pressure, and this makes the above Bernoulli equation actually somewhat incorrect. But in practice for water this is negligible and the incompressibility is assumed.

• So, in the textbook I was reading which did a sample problem and solution of this concept, they seem to have dropped the pgh term as well as the p_2 term. I assume this is because they set p_2 = 0 rather than = 100,000 Pa. In short, was the textbook incorrect? May 10, 2020 at 0:39
• Somewhat incorrect, as the water usually does not flow into region of zero pressure. But since the difference of pressures in that case is much higher than 100 000 Pa, the error is small and can be neglected. May 10, 2020 at 2:33