Example 12.4 from College Physics, 12.2 Bernoulli’s Equation

Calculating Pressure: Pressure Drops as a Fluid Speeds Up

In Example 12.2, 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 $1 . 01 \times 10^{5} {N/m}^{2}$ (atmospheric, as it must be) and assuming level, frictionless flow.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

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

$P_{1} = P_{2} + \frac{1}{2} {ρv}_{2}^{2} - \frac{1}{2} {ρv}_{1}^{2} = P_{2} + \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) .$
12.24

Substituting known values,

$\begin{array}{l} P_{1} & = & 1 . 01 \times 10^{5} {N/m}^{2} \\ +\frac{1}{2} (10^{3} {kg/m}^{3}) [(25.5 m/s)^{2} - (1.96 m/s)^{2}] \\ = & 4.24 \times 10^{5} {N/m}^{2} . \end{array}$
12.25
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.

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Solution

Discussion

Strategy

Solution

Discussion