Water Discharge from a Large Tank A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap (Fig. 5–40). A tap near the bottom of the tank is now opened, and water flows out from the smooth and rounded outlet. Determine the maximum water velocity at the outlet.

Question

Water Discharge from a Large Tank

A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap (Fig. 5–40). A tap near the bottom of the tank is now opened, and water flows out from the smooth and rounded outlet. Determine the maximum water velocity at the outlet.

Accepted Answer

A tap near the bottom of a tank is opened. The maximum exit velocity of water from the tank is to be determined.
Assumptions 1 The flow is incompressible and irrotational (except very close to the walls). 2 The water drains slowly enough that the flow can be approximated as steady (actually quasi-steady when the tank begins to drain). 3 Irreversible losses in the tap region are neglected.
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. We take point 1 to be at the free surface of water so that [latex]P_1 = P_{atm}[/latex] (open to the atmosphere), [latex]V_1[/latex] is negligibly small compared to [latex]V_2[/latex] (the tank diameter is very large relative to the outlet diameter), [latex]z_1 = 5 m[/latex], and [latex]z_2 = 0[/latex] (we take the reference level at the center of the outlet). Also, [latex]P_2 = P_{atm}[/latex] (water discharges into the atmosphere). For flow along a streamline from 1 to 2, the Bernoulli equation simplifies to

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

Solving for [latex]V_2[/latex] and substituting,

[latex] V_2 = \sqrt{2gz_1} = \sqrt{2(9.81 m/s^2)(5 m)} = 9.9 m/s [/latex]

The relation [latex] V = \sqrt{2gz} [/latex] is called the Torricelli equation.
Therefore, the water leaves the tank with an initial maximum velocity of 9.9 m/s. This is the same velocity that would manifest if a solid were dropped a distance of 5 m in the absence of air friction drag. (What would the velocity be if the tap were at the bottom of the tank instead of on the side?)
Discussion If the orifice were sharp-edged instead of rounded, then the flow would be disturbed, and the average exit velocity would be less than 9.9 m/s. Care must be exercised when attempting to apply the Bernoulli equation to situations where abrupt expansions or contractions occur since the friction and flow disturbance in such cases may not be negligible. From conservation of mass, [latex](V_1 /V_2)^2 = (D_2/D_1)^4[/latex]. So, for example, if [latex]D_2/D_1 = 0.1[/latex], then [latex](V_1 /V_2)^2 = 0.0001[/latex], and our approximation that [latex] V^2_1\ll V^2_2[/latex] is justified.

Question 5.6: Water Discharge from a Large Tank A large tank open to the a...

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