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Water flows from a horizontal tube into a large tank. The tube is located \(2.5 \mathrm{m}\) below the free surface of water in the tank. The head loss is \(2 \mathrm{J} / \mathrm{kg}\). Compute the average flow speed in the tube.

Short Answer

Expert verified
After calculation, the average flow speed in the tube comes out to be approximately 6.65 m/s.

Step by step solution

01

Use Bernoulli's equation

Bernoulli's equation states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid is constant along a streamline, assuming no energy is added or removed by pumps or other means. The equation is formulated as \(P + \frac{1}{2}蟻v^2 + 蟻gh = constant\), where \(P\) is pressure, \(蟻\) is density, \(v\) is velocity, and \(h\) is height. But the pressure drops out because the start and end points are both at atmospheric pressure.
02

Set up the energy balance equation

We set the potential energy (in the tank) equal to the sum of the kinetic energy (in the pipe) and the energy loss: \(蟻gh_{tank} = \frac{1}{2}蟻v^2_{pipe} + h_{loss}\). Here, \(h_{tank}\) is the height of the liquid in the tank = 2.5m, \(v_{pipe}\) is the speed in the pipe that we want to find, and \(h_{loss}\) is the energy loss = \(2 \frac{J}{kg}\), which can be considered as another kind of effective height, \(h_{loss} = \frac{2}{g}\) where \(g\) is the gravitational acceleration.
03

Solve the energy balance equation for pipe speed

Rearranging the energy balance equation to solve for the average speed in the pipe \(v_{pipe}\) implies \(v_{pipe} = \sqrt{2 (gh_{tank}-h_{loss})}\). Plug in the given values into the equation: \(g = 9.81 m/s^2\), \(h_{tank} = 2.5m\), and \(h_{loss} = \frac{2}{9.81}\) m = 0.204m to calculate \(v_{pipe}\).

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Most popular questions from this chapter

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