Bernoulli’s Principle—Bernoulli’s Equation at Constant Depth
Bernoulli’s Principle—Bernoulli’s Equation at Constant Depth
Another important situation is one in which the fluid moves but its depth is constant—that is, \({h}_{1}={h}_{2}\). Under that condition, Bernoulli’s equation becomes
\({P}_{1}+\cfrac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\cfrac{1}{2}{\mathrm{\rho v}}_{2}^{2}\text{.}\)
Situations in which fluid flows at a constant depth are so important that this equation is often called Bernoulli’s principle. It is Bernoulli’s equation for fluids at constant depth. (Note again that this applies to a small volume of fluid as we follow it along its path.) As we have just discussed, pressure drops as speed increases in a moving fluid. We can see this from Bernoulli’s principle. For example, if \({v}_{2}\) is greater than \({v}_{1}\) in the equation, then \({P}_{2}\) must be less than \({P}_{1}\) for the equality to hold.
Example: Calculating Pressure: Pressure Drops as a Fluid Speeds Up
In this figure, we found that the speed of water in a hose increased from 1.96 m/s to 25.5 m/s going from the hose to the nozzle. Calculate the pressure in the hose, given that the absolute pressure in the nozzle is \(1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\) (atmospheric, as it must be) and assuming level, frictionless flow.
Strategy
Level flow means constant depth, so Bernoulli’s principle applies. We use the subscript 1 for values in the hose and 2 for those in the nozzle. We are thus asked to find \({P}_{1}\).
Solution
Solving Bernoulli’s principle for \({P}_{1}\) yields
\({P}_{1}={P}_{2}+\cfrac{1}{2}{\mathrm{\rho v}}_{2}^{2}-\cfrac{1}{2}{\mathrm{\rho v}}_{1}^{2}={P}_{2}+\cfrac{1}{2}\rho ({v}_{2}^{2}-{v}_{1}^{2})\text{.}\)
Substituting known values,
\(\begin{array}{lll}{P}_{1}& =& 1\text{.}\text{01}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\\ & & \text{}+\cfrac{1}{2}({\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3})[(\text{25.5 m/s}{)}^{2}-(\text{1.96 m/s}{)}^{2}]\\ & =& 4.24×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\text{.}\end{array}\)
Discussion
This absolute pressure in the hose is greater than in the nozzle, as expected since \(v\) is greater in the nozzle. The pressure \({P}_{2}\) in the nozzle must be atmospheric since it emerges into the atmosphere without other changes in conditions.
This lesson is part of:
Fluid Dynamics and Applications