Example 12.5 from College Physics, 12.3 The Most General Applications of Bernoulli’s Equation

Calculating Pressure: A Fire Hose Nozzle

Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of $1 . 62 \times 10^{6} {N/m}^{2}$ . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?

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
Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.
Solution
Bernoulli’s equation states

$P_{1} + \frac{1}{2} {ρv}_{1}^{2} + ρ {gh}_{1} = P_{2} + \frac{1}{2} {ρv}_{2}^{2} + ρ {gh}_{2},$
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where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds $v_{1}$ and $v_{2}$ . Since $Q = A_{1} v_{1}$ , we get
$v_{1} = \frac{Q}{A_{1}} = \frac{40 . 0 \times 10^{- 3} m^{3} /s}{π (3 . 20 \times 10^{- 2} m)^{2}} = 12.434 m/s .$
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Similarly, we find

$v_{2} = 56.588 m/s .$
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(This rather large speed is helpful in reaching the fire.) Now, taking $h_{1}$ to be zero, we solve Bernoulli’s equation for $P_{2}$ :

$P_{2} = P_{1} + \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) - ρ {gh}_{2} .$
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Substituting known values yields

$\begin{matrix} P_{2} = \\ 1.62 \times 10^{6} {N/m}^{2} + \frac{1}{2} (1000 {kg/m}^{3}) [(12.434 m/s)^{2} - (56.588 m/s)^{2}] - (1000 {kg/m}^{3}) (9.80 {m/s}^{2}) (10.0 m) \\ \approx 1800 {N/m}^{2} . \end{matrix}$
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Discussion
This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure is very close to atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.

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