Spraying Water into the Air Water is flowing from a garden hose (Fig. 5–39). A child places his thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. The pressure in the hose just upstream of his thumb is 400 kPa. If the hose is held upward, what is the maximum

Question

Spraying Water into the Air

Water is flowing from a garden hose (Fig. 5–39). A child places his thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. The pressure in the hose just upstream of his thumb is 400 kPa. If the hose is held upward, what is the maximum height that the jet could achieve?

Accepted Answer

Water from a hose attached to the water main is sprayed into the air.
The maximum height the water jet can rise is to be determined.
Assumptions 1 The flow exiting into the air is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The surface tension effects are negligible. 3 The friction between the water and air is negligible. 4 The irreversibilities that occur at the outlet of the hose due to abrupt contraction are not taken into account.
Properties We take the density of water to be 1000 kg/m³.
Analysis This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation. The water height will be maximum under the stated assumptions. The velocity inside the hose is negligibly small compared to that of the jet ([latex]V_1^2 ≪ V_j^2[/latex], see magnified portion of Fig. 5–39) and we take the elevation just below the hose outlet as the reference level ([latex]z_1 = 0[/latex]). At the top of the water trajectory [latex]V_2 = 0[/latex], and atmospheric pressure pertains. Then the Bernoulli equation along a streamline from 1 to 2 simplifies to

[latex] \frac{P_1}{\rho g} + \overset{ignore}{\frac{\cancel{V^2_1}}{2g} } + \overset{0}{\cancel{z_1}} = \frac{P_2}{\rho g} + \overset{0}{\frac{\cancel{V^2_2}}{2g} } + z_2 \rightarrow \frac{P_1}{\rho g} = \frac{P_{atm}}{\rho g} + z_2[/latex]

Solving for [latex]z_2[/latex] and substituting,

[latex] z_2 = \frac{P_1 - P_{atm}}{\rho g} = \frac{P_{1, gage}}{\rho g} = \frac{400 kPa}{(1000 kg/m^3)(9.81 m/s^2)} \left(\frac{1000 N/m^2}{1 kPa} \right) \left(\frac{1 kg.m/s^2}{1 N} \right) [/latex]
= 40.8 m

Therefore, the water jet can rise as high as 40.8 m into the sky in this case.
Discussion The result obtained by the Bernoulli equation represents the upper limit and should be interpreted accordingly. It tells us that the water cannot possibly rise more than 40.8 m, and, in all likelihood, the rise will be much less than 40.8 m due to irreversible losses that we neglected.

Question 5.5: Spraying Water into the Air Water is flowing from a garden h...

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