Fluid Mechanics, Lecture 4 PDF

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Fluid Mechanics Ammar Ali Abd Email: [email protected] Fluid Dynamics Under some circumstances the flow will not be as changeable as this. He following terms describe the states which are used to classify fluid flow: • uniform flow: If the flow velocity is the same magnitude and direction at...

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Fluid Mechanics

Ammar Ali Abd Email: [email protected]

Fluid Dynamics Under some circumstances the flow will not be as changeable as this. He following terms describe the states which are used to classify fluid flow: • uniform flow: If the flow velocity is the same magnitude and direction at every point in the fluid it is said to be uniform. • non-uniform: If at a given instant, the velocity is not the same at every point the flow is non-uniform. (In practice, by this definition, every fluid that flows near a solid boundary will be non-uniform - as the fluid at the boundary must take the speed of the boundary, usually zero. However, if the size and shape of the of the cross-section of the stream of fluid is constant the flow is considered uniform.) • steady: A steady flow is one in which the conditions (velocity, pressure and crosssection) may differ from point to point but DO NOT change with time. • unsteady: If at any point in the fluid, the conditions change with time, the flow is described as unsteady. (In practise, there is always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady. Combining the above we can classify any flow in to one of four type: 1. Steady uniform flow. Conditions do not change with position in the stream or with time. An example is the flow of water in a pipe of constant diameter at constant velocity. 2. Steady non-uniform flow. Conditions change from point to point in the stream but do not change with time. An example is flow in a tapering pipe with constant velocity at the inlet - velocity will change as you move along the length of the pipe toward the exit. 3. Unsteady uniform flow. At a given instant in time the conditions at every point are the same, but will change with time. An example is a pipe of constant diameter connected to a pump pumping at a constant rate which is then switched off. 4. Unsteady non-uniform flow. Every condition of the flow may change from point to point and with time at every point. For example, waves in a channel. Compressible or Incompressible All fluids are compressible (even water) their density will change as pressure changes. Under steady conditions, and provided that the changes in pressure are small, it is usually possible to simplify analysis of the flow by assuming it is incompressible and has constant density. As you will appreciate, liquids are quite difficult to compress - so under most steady conditions they are treated as incompressible. In some unsteady conditions, very high pressure differences can occur and it is necessary to take these into account even for liquids. Gasses, on the contrary, are very easily compressed, it is Al-Qasim Green University

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essential in most cases to treat these as compressible, taking changes in pressure into account. Some terms important to know, this will help to understand and solve problems: 1- Mass flow rate: 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒, 𝑚. , =

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑐𝑜𝑙𝑙𝑒𝑐𝑡 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑

2- Volume flow rate – Discharge:

𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒, 𝑄 . , =

Discharge can be written by: 𝑄=

𝑓𝑙𝑢𝑖𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡𝑖𝑚𝑒

, 𝑢𝑛𝑖𝑡𝑠, 𝑘𝑔/𝑠𝑒𝑐

, 𝑢𝑛𝑖𝑡𝑠, 𝑚3 /𝑠𝑒𝑐

𝐴𝑟𝑒𝑎 ∗ 𝑚𝑒𝑎𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ∗ 𝑡𝑖𝑚𝑒 = 𝐴𝑟𝑒𝑎 ∗ 𝑚𝑒𝑎𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒

Continuity Equation:

Continuity equation represents the law of conservation of mass. In general, for unsteady flow the continuity equation is: Matter cannot be created or destroyed (Mass flow rate into the system) - (Mass flow rate out of the system) = Rate of change of storage. For steady state condition: (Mass flow rate into the system) - (Mass flow rate out of the system) = 0 Example: Inflow: The flow that is coming into a system or an elemental volume such as rainfall in y direction, flow entering the river or a channel. Outflow: The flow escaping from the system such as evaporation, seepage, water released from a system.

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Generally, the mass balance is written in all the three directions namely x, y and z. 𝜕𝜌𝑢

𝜕𝜌𝑣 𝜕𝜌𝑤 = 0 + + 𝜕𝑧 𝜕𝑥 flow𝜕𝑦 When the density constant and the in one dimension the equation will be: 𝜌𝜕𝑢 =0 𝜕𝑥

The continuity equation can be used to calculate the velocity at different points: 𝑈1 𝐴1 = 𝑈2 𝐴2

flow rate in = flow rate out The mass flow rate can be illustrated in the same way: Mass rate in = Mass rate out Now the continuity equation can be shown in different way:

𝑈1 ∗

𝑈1 𝐴1 = 𝑈2 𝐴2

𝜋 𝜋 ∗ 𝑑1 2 = 𝑈2 ∗ ∗ 𝑑2 2 4 4

𝑈1 ∗ 𝑑1 2 = 𝑈2 ∗ 𝑑2 2

Example 1: Figure below shows the branching of a water pipe system. The diameters of the circular cross-section pipes at positions 1, 2 and 3 are, respectively 20, 6 and 3 cm. Assume inviscid flow so that the flow across any cross-section is uniform. The volumetric flow rates past positions 1 and 2 are, respectively 0.01 and 0.004 m3/s.

1- Calculate the mass flow rates through the pipes at positions 1 and 2. 2- Find the mass flow rate at position 3. 3- Find the flow speed at positions 1, 2 and 3. Solution: the fluid here is water, so, the density will be 1000 kg/m3. 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 1 = 0.01 ∗ 1000 = 10 𝑘𝑔/𝑠𝑒𝑐

𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 2 = 0.004 ∗ 1000 = 4 𝑘𝑔/𝑠𝑒𝑐 Al-Qasim Green University

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And we know that: Mass rate in = Mass rate out 𝑚1 = 𝑚2 + 𝑚3 → 10 = 4 + 𝑚3 →= 𝑚3 = 6 𝑘𝑔/𝑠𝑒𝑐

Now to calculate the speed at 1, 2, and 3: 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 1 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 2 =

𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 1 𝐴𝑟𝑒𝑎 1

0.01 = 0.318 𝑚/𝑠𝑒𝑐 = 𝜋 2 4 ∗ 0.2

𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 2 0.004 = 𝜋 = 1.415 𝑚/𝑠𝑒𝑐 𝐴𝑟𝑒𝑎 2 2 ∗ 0.06 4

Now the velocity at point 3 will need some preparations:

And:

𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 3 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 3 =

𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 6 = 0.006 𝑚3 /𝑠𝑒𝑐 = 1000 𝑑𝑒𝑛𝑠𝑖𝑡𝑦

0.006 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 3 = 𝜋 = 0.53 𝑚/𝑠𝑒𝑐 𝐴𝑟𝑒𝑎 3 2 ∗ 0.12 4

Example 2: Figure below shows a sprayer system. At the inlet side, the pipe diameter is 2 cm and flow speed is 2 m/s. The spray rose has 100 holes, each with a diameter of 1 mm. Assuming inviscid flow throughout, calculate the speed of the jets that leave the sprayer.

Solution: the inlet flow rate can be calculated: 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑖𝑛 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ∗ 𝐴𝑟𝑒𝑎 = 2 ∗

𝜋 ∗ 0.022 = 0.00063 𝑚3 /𝑠𝑒𝑐 4

𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑖𝑛 = 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡 = 0.00063 𝑚3 /𝑠𝑒𝑐

The flow rate out is for 100 holes so:

𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑢𝑡 𝑓𝑜𝑟 𝑜𝑛𝑒 ℎ𝑜𝑙𝑒 = Al-Qasim Green University

0.00063 = 0.0000063 𝑚3 /𝑠𝑒𝑐 100

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And the velocity for one hole: 0.0000063 = 8 𝑚/𝑠𝑒𝑐 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝜋 ∗ 0.0012 4

The Bernoulli equation Work and Energy

We know that if we drop a ball it accelerates downward with an acceleration 𝑔 = 9.81 𝑚2 /𝑠𝑒𝑐 (neglecting the frictional resistance due to air). We can calculate the speed of the ball after falling a distance h by the formula: 𝑣 2 = 𝑢 2 + 2𝑎𝑠 (a = g and s = h)

The equation could be applied to a falling droplet of water as the same laws of motion apply. A more general approach to obtaining the parameters of motion (of both solids and fluids) is to apply the principle of conservation of energy. When friction is negligible the sum of kinetic energy and gravitational potential energy is constant. 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 =

1 ∗ 𝑚 ∗ 𝑣2 2

𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑚𝑔ℎ

(m is the mass, v is the velocity and h is the height above the datum). To apply this to a falling droplet we have an initial velocity of zero, and it falls through a height of h. Now: 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 = 0, 1

𝐹𝑖𝑛𝑎𝑙 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 = 2 ∗ 𝑚 ∗ 𝑣 2 ,

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑚𝑔ℎ 𝐹𝑖𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 0

We know that:

kinetic energy + potential energy = constant And: Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy 𝑚∗𝑔∗ℎ =

1 ∗ 𝑚 ∗ 𝑣2 2

Although this is applied to a drop of liquid, a similar method can be applied to a Hence: continuous jet of liquid.

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𝑣 = √2𝑔ℎ

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We can consider the situation as in the figure above, a continuous jet of water coming from a pipe with velocity 𝑢1 . One particle of the liquid with mass 𝑚 travels with the jet and falls from height 𝑧1 to 𝑧2 . The velocity also changes𝑢1 to 𝑢2 . The jet is travelling in air where the pressure is everywhere atmospheric so there is no force due to pressure acting on the fluid. The only force which is acting is that due to gravity. The sum of the kinetic and potential energies remains constant (as we neglect energy losses due to friction). Therefore, 1 1 𝑚𝑔𝑧1 + 𝑚𝑢1 2 = 𝑚𝑔𝑧2 + 𝑚𝑢2 2 2 2

The mass of fluid is constant and the equation will be:

1 1 𝑔𝑧1 + 𝑢1 2 = 𝑔𝑧2 + 𝑢2 2 2 2

This will give a reasonably accurate result as long as the weight of the jet is large compared to the frictional forces. It is only applicable while the jet is whole - before it breaks up into droplets. Case study: We can use a very similar application of the energy conservation concept to determine the velocity of flow along a pipe from a reservoir. Consider the 'idealised reservoir' in the figure below.

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Themovement level of theofwater reservoir is 𝑧is Considering the energy situation is no waterinsothe kinetic energy but the gravitational potentialthere energy 1 .zero

is 𝑚𝑔𝑧1 . If a pipe is attached at the bottom water flows along this pipe out of the tank to a level 𝑧2 . A mass 𝑚 has flowed from the top of the reservoir to the nozzle and it has gained a 1

velocity 𝑢2 . The kinetic energy is now 2 𝑚𝑢2 2 and the potential energy 𝑚𝑔𝑧2 .

Summarising:

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 = 0, 1

𝐹𝑖𝑛𝑎𝑙 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 = 2 ∗ 𝑚 ∗ 𝑢2 2 , We know that:

𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑚𝑔𝑧1

𝐹𝑖𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑚𝑔𝑧2

kinetic energy + potential energy = constant 1 𝑚𝑔𝑧1 = 𝑚𝑔𝑧2 + 𝑚𝑢2 2 2 𝑔(𝑧1 − 𝑧2 ) =

1 2 𝑢 2 2

𝑢2 = √2𝑔(𝑧1 − 𝑧2 )

We now have a expression for the velocity of the water as it flows from of a pipe nozzle at a height (𝑧1 − 𝑧2 ) below the surface of the reservoir. (Neglecting friction losses in the pipe and the nozzle). Case Study:

The pipe is filled with stationary fluid of density 𝜌 has pressures 𝑝1 and 𝑝2 at levels 𝑧1 and 𝑧2 respectively. What is the pressure difference in terms of these levels? 𝑝1 − 𝑝2 = 𝜌𝑔(𝑧1 − 𝑧2 )

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This applies when the pressure varies but the fluid is stationary. Compare this to the equation derived for a moving fluid but constant pressure: 𝑔𝑧1 +

𝑝1

= 𝑔𝑧2 +

𝑝𝜌1

1 𝜌2 1 𝑔𝑧1 + 𝑢1 = 𝑔𝑧2 + 𝑢2 2 2 2 You can see that these are similar form. What would happen if both pressure and velocity varies? Bernoulli's equation has some restrictions in its applicability, they are: 1234-

Flow is steady; Density is constant (which also means the fluid is incompressible); Friction losses are negligible. The equation relates the states at two points along a single streamline, (not conditions on two different streamlines).

All these conditions are impossible to satisfy at any instant in time! Fortunately for many real situations where the conditions are approximately satisfied, the equation gives very good results. The derivation of Bernoulli's Equation:

An element of fluid, as that in the figure above, has potential energy due to its height z above a datum and kinetic energy due to its velocity u. If the element has weight mg then: 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑚𝑔𝑧

𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑍 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 =

1 ∗ 𝑚 ∗ 𝑢2 2

𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 =

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1 ∗ 𝑢2 2𝑔

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At any cross-section, the pressure generates a force, the fluid will flow, moving the cross-section, so work will be done. If the pressure at cross section AB is p and the area of the cross-section is a then 𝐹𝑜𝑟𝑐𝑒 𝑜𝑛 𝐴𝐵 = 𝑝𝑎

when the mass mg of fluid has passed AB, cross-section AB will have moved to AA' 𝑣𝑜𝑙𝑢𝑚𝑒 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝐴𝐵 =

So, the distance:

𝐴𝐴′ =

𝑚𝑔 𝜌𝑔

𝑚 𝜌𝑎

Work done = force distance AA’ = 𝑝𝑎 ∗

𝑚 𝑝𝑚 = 𝜌𝑎 𝜌

Work done per unite weight =

𝑝

𝜌𝑔

This term is known as the pressure energy of the flowing stream. Summing all of these energy terms gives: 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 + Or,

𝑝 𝑢2 +𝑍 =𝐻 + 𝜌𝑔 2𝑔

Where: 𝑝 , 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝜌𝑔

𝑢2

ℎ𝑒𝑎𝑑,

2𝑔

𝑍, 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 ℎ𝑒𝑎𝑑,

, 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑

𝐻, 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑

By the principle of conservation of energy, the total energy in the system does not change, Thus the total head does not change. So, the Bernoulli equation can be written: 𝑢2 𝑝 + + 𝑍 = 𝐻 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜌𝑔 2𝑔

As stated above, the Bernoulli equation applies to conditions along a streamline. We can apply it between two points, 1 and 2, on the streamline in the figure below:

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total energy per unit weight at 1 = total energy per unit weight at 2 Or total head at 1 = total head at 2 Or 𝑝1 𝑢1 2 𝑝2 𝑢 2 2 + 𝑍2 + + 𝑍1 = + 2𝑔 𝜌𝑔 2𝑔 𝜌𝑔

Example 1: A U tube manometer containing water is connected to a nozzle of an air tunnel that discharges to the atmosphere as shown in figure below. The area ratio is 𝐴1 𝐴2

= 0.25. For given operational conditions the level difference in the manometer is

ℎ = 94𝑚𝑚. Take the water density 𝜌 = 1000𝑘𝑔/𝑚 3 and the air density 𝜌 = 1.23 𝑘𝑔/𝑚3 .

What is the average air velocity at the nozzle exit, 𝑣2 ?

Solution: The average air velocity at the nozzle exit is: 2𝜌 𝑣2 = √ ∗ 𝜌𝑎𝑖𝑟

𝑔ℎ

𝐴 (1 − ( 1)2 𝐴2

= 40 𝑚/𝑠𝑒𝑐

Example 2: Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000 Pa. At the nozzle, the pressure decreases to atmospheric pressure (101300 Pa), there is no change in height. Use the Bernoulli equation to calculate the velocity of the water exiting the nozzle. Solution: The velocity at the nozzle exit can be calculated direct by using Bernoulli equation:

Al-Qasim Green University

Water Resources Engineering

Fluid Mechanics

Ammar Ali Abd Email: [email protected] 𝑝1

𝑢1 2 𝑝2 𝑢2 2 + 𝑍2 + + 𝑍1 = + 2𝑔 From the question that there change in 𝜌𝑔 height which is 𝑍1 = 𝑍2 . 𝜌𝑔is no2𝑔 𝑝1

𝑢1 2 𝑝2 𝑢22 2 + = + 𝜌 2 𝜌

Now the exist velocity:

𝑢2 2 𝑝1 𝑢1 2 𝑝2 𝑝1 𝑢1 2 𝑝2 = + − → 𝑢2 = √2 ( + − ) 𝜌 𝜌 2 𝜌 2 𝜌 2 𝑢2 = √2 (

200000 12 101300 + − ) = 14.08 𝑚/𝑠𝑒𝑐 1000 2 1000

Example 3: Through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 Pa. The refinery needs the ethanol to be at a pressure of 2 atm (202600 Pa) on a lower level. How far must the pipe drop in height in order to achieve this pressure? Assume the velocity does not change. (Hint: The density of ethanol is 789 kg/m3) Solution: from Bernoulli Equation: 𝑝1 𝑢1 2 𝑝2 𝑢 2 2 + 𝑍2 + + 𝑍1 = + 2𝑔 𝜌𝑔 2𝑔 𝜌𝑔

From the question that there is no change in height which is 𝑢1 = 𝑢2 . 𝑝1 𝑝2 𝑝2 𝑝1 + 𝑍1 = + 𝑍2 → 𝑍1 − 𝑍2 = − 𝜌𝑔 𝜌𝑔 𝜌𝑔 𝜌𝑔

Case Study:

∆𝑍 =

202600 101300 = 13.08 𝑚 − 789 ∗ 9.81 789 ∗ 9.81

1- Make a prediction. In the pipe shown in figure above, is the pressure higher at point 2, where the fluid flows fastest, or at point 1? The fluid in the pipe flows Al-Qasim Green University

Water Resources Engineering

Fluid Mechanics

Ammar Ali Abd Email: [email protected]

from left to right. Many people predict that the pressure is higher at point 2, where the fluid is moving faster. 2- Apply the continuity equation, and Bernoulli’s equation, to rank points 1, 2, and 3 according to pressure, from largest to smallest. Let’s see if the common prediction, that the pressure is highest at point 2, is correct. First, apply the continuity equation: 𝐴1 𝑣1 = 𝐴2 𝑣2 Looking at the tube, we know that: 𝐴1 = 𝐴3 > 𝐴2 which tells us that 𝑣2 > 𝑣1 = 𝑣3 . Now, let’s apply Bernoulli’s Equation. Comparing points 1 and 2, we start with: 𝑝1 𝑢1 2 𝑝2 𝑢 2 2 + 𝑍2 + + 𝑍1 = + 2𝑔 𝜌𝑔 2𝑔 𝜌𝑔

The vertical positions of these two points are equal so the terms Z cancel out:

Let’s re-write this as: 𝑝1 − 𝑝2 =

𝑢1 2 𝑝2 𝑢2 2 𝑝1 + = + 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔

𝑢2 2 2

𝜌−

𝑢1 2 2

𝜌

The continuity equation told us that 𝑣2 > 𝑣1 , so the right-hand side of the above equation is positive. This means the left-hand side must also be positive, implying that 𝑝1 > 𝑝2 , Thus, the pressure at point 2, where the fluid speed is highest, is less than the pressure at point 1. For points at the same height, higher speed corresponds to lower pressure. We can make sense of this by considering a parcel of fluid that moves from point 1 to point 2. Because this parcel of fluid speeds up as it travels from point 1 to point 2, there must be a net force acting on it that is directed right. This force must come from a difference in pressure between points 1 and 2. For the force to be directed right, the pressure must be larger on the left, at point 1. We can also use Bernoulli’s equation to show that the pressure at point 3 is equal to that at point 1. Thus, we can conclude that 𝑝1 = 𝑝3 > 𝑝2 .

Al-Qasim Green University

Water Resources Engineering

Fluid Mechanics

Ammar Ali Abd Email: [email protected]

Example 4: A Styrofoam cylinder, filled with water, sits on a table. You then poke a small hole through the side of the cylinder, 20 cm below the top of the water surface. What is the speed of the fluid emerging from the hole?

Solution: We’re going to apply Bernoulli’s equation, which means identifying two points that we can relate via the equation. Point 2 is outside the container where the hole is, because that is the place where we’re trying to find the speed. Point 1 needs to be somewhere inside the container. Any point inside will do, although the most sensible places are either at the top of the container, where we know the pressure, or inside the container at the level of the hole. Let’s choose a point at the very top, and apply Bernoulli’s equation: 𝑝1 𝑢1 2 𝑝2 𝑢 2 2 + 𝑍2 + + 𝑍1 = + 2𝑔 𝜌𝑔 2𝑔 𝜌𝑔

First, we should recognize that, because both...