Figure 16-29 shows liquid discharging from an orifice in a large tank at a distance \(h\) below the liquid surface. The tank is open at the top. (a) Apply Bernoulli's equation to a streamline connecting points 1,2, and 3, and show that the speed of efflux is $$ v=\sqrt{2 g h} $$ This is known as Torricelli's law. (b) If the orifice were curved directly upward, how high would the liquid stream rise? ( \(c\) ) How would viscosity or turbulence affect the analysis?

Short Answer

Expert verified
The speed of liquid efflux from the tank is \(v = \sqrt{2 g h}\). If the orifice were curved upwards, the liquid would rise to the height of the original liquid surface level in the tank. Viscosity and turbulence would decrease the speed and height of the liquid flow.

Step by step solution

01

Apply Bernoulli's equation to points 1 and 2

Bernoulli's equation can be stated as \(P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2\), where, \(P\) represents pressure, \(\rho\) is fluid density, \(v\) is velocity, \(h\) is height and \(g\) is acceleration due to gravity. At the liquid surface (point 1), pressure \(P_1\) is atmospheric pressure \(P_0\), velocity \(v_1\) is 0, and height \(h_1\) is \(h\). At the orifice (point 2), pressure \(P_2\) is also \(P_0\), height \(h_2\) is 0, and velocity \(v_2\) is \(v\). Substituting these conditions into Bernoulli's equation yields \(P_0 + \rho g h = P_0 + \frac{1}{2} \rho v^2\).
02

Solve Bernoulli's equation for velocity

Rearrange the equation derived in step 1 to solve for velocity \(v\). First, cancel the atmospheric pressure term \(P_0\) from both sides because it's same at points 1 and 2. You get \(\rho g h = \frac{1}{2} \rho v^2\). We can cancel out \(\rho\) from both sides, leaving \(gh = \frac{1}{2}v^2\). Rearranging to solve for \(v\) and taking square root of both sides, you get \(v = \sqrt{2 g h}\).
03

Apply Bernoulli's equation to find the rise height when the orifice is curved upwards

To find how high the liquid will rise if the orifice were curved directly upward, we use Bernoulli's equation similarly to step 1, but set \(v = 0\) (the point of highest rise). Solving for \(h\), we find that \(h = \frac{v^2}{2g}\). Substituting the result of Step 2 for \(v\) gives \(h = \frac{((\sqrt {2gh})^2)}{2g} = h\), stating that the liquid would rise to the original liquid surface level
04

Analyze the effect of viscosity and turbulence

Bernoulli's equation assumes ideal fluid flow (inviscid and streamline flow). Introducing viscosity would decrease the speed of efflux due to frictional forces between the fluid layers. Turbulence causes mixing of fluid, deviating the flow from ideal streamline flow and decreasing velocity, thus reducing the height of liquid rise.

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