Conservation of Energy in Fluids

If you ever want to do the dishes or run a bath, it's simple - just turn on the faucet and let the hot or cold water run. A very complicated plumbing system ensures that all houses in a neighborhood have access to water and the gas necessary to heat it. It seems like a self-explanatory part of every household, however, a lot of planning goes into creating functional and efficient pipe systems. 

Conservation of Energy in Fluids Conservation of Energy in Fluids

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    Conservation of Energy in Fluids A brick wall with a pipe system in front of it StudySmarterFig. 1 - Pipes are a real-life example of a closed system utilizing the conservation of energy to move fluids.

    Nowadays, any system that relies on the relocation of fluids, whether it's water, gas, or oil, will utilize fluid mechanics and the law of conservation of energy. Bernoulli's principle describes the fluid mechanics of ideal fluids, so it's a great place to start when trying to understand the complicated nature of fluids in motion. As the pipes of alternating diameters and elevations carry fluids of varying densities, it's important to remember that the total energy and mass of this closed system will remain constant. In this article, we'll derive the expression for the conservation of energy in fluids, and establish a better understanding of the nature of flowing fluids!

    The Law of Conservation of Energy in Fluids

    The law of conservation of energy is one of the fundamental laws of physics, explaining the nature of energy.

    The law of conservation of energy states that energy can be neither created nor destroyed, only converted from one kind of energy into another.

    In other words, the total energy of an isolated system remains constant. The same principle applies to fluid mechanics. The conservation of energy in ideal fluids is described by Bernoulli's principle.

    The Bernoulli principle states that the pressure exerted by a moving fluid is inversely proportional to its velocity in a horizontal flow.

    Bernoulli proved that the total mechanical energy of a flowing liquid or gas remains unchanged at any point along a streamline, assuming that the fluid in question is incompressible and has zero viscosity.

    Derivation of Conservation of Energy in Fluids

    Let's derive the Bernoulli equation, which is simply the more generic mathematical form of the Bernoulli principle! To accomplish that, imagine a pipe system with two different cross-sectional areas at two different heights, as visualized in Figure 2 below.

    Conservation of Energy in Fluids A pipe of different diameters and elevations has a fluid flowing through it. Two distinct regions are defined with distinct pressure, velocity, cross-sectional area, displacement and force StudySmarterFig. 2 - A pipe system used to derive the equation for conservation of energy in fluids.

    The work-energy theorem states that the work done \(W\) is equal to the sum of the change in kinetic energy \(K\) and the change in potential energy \(U\). Mathematically, it can be expressed as

    $$\Delta W = \Delta K + \Delta U,$$

    where the sign for work depends on the direction of the force relative to displacement, meaning it's positive when it's pointing in the same direction as the flow, and negative when it's opposing it. From the diagram above, we can figure out the respective signs and obtain

    $$W_1-W_2 = K_2-K_1+U_2-U_1,$$

    which can be rearranged into

    $$W_1 + U_1 + K_1 = W_2 + U_2 + K_2.$$

    The equations for kinetic and potential energies are the same as usual:

    \begin{align} K&=\frac{1}{2}mv \\ U&=mgh\end{align}

    where \(m\) is the mass, \(v\) is velocity, \(g\) is the acceleration due to gravity, and \(h\) is the height (in our case, height is represented by \(y\)). The work done in this case can be represented as some force \(F\) exerted over a distance \(x\). We know that

    $$ P=\frac{F}{A},$$

    so the work can be expressed as

    $$ W = Fx = PAx.$$

    All of these equations can be plugged into the work-energy relation for each respective region in the pipe:

    $$ P_1A_1x_1+mgy_1+\frac{1}{2}mv_1^2=P_2A_2x_2+mgy_2+ \frac{1}{2}mv_2^2.$$

    We can use

    $$ \rho=\frac{m}{V}$$

    to reexpress mass in terms of density \(\rho\) and volume \(V\). In addition, we can rewrite the product of cross-sectional area \(A\) and distance \(x\) as volume, considering the volume of a cylinder is

    $$V_\text{cylinder}=\pi r^2 \ell = A_\text{base}\ell.$$

    All of this leads to the cancellation of the volume terms

    $$ P_1\bcancel{V}+\rho \bcancel{V} gy_1+\frac{1}{2}\rho \bcancel{V} v_1^2=P_2\bcancel{V}+ \rho \bcancel{V} gy_2+\frac{1}{2}\rho \bcancel{V} v_2^2$$

    and reveals the final version of the Bernoulli equation:

    $$ P_1 + \rho g y_1 + \frac{1}{2} \rho v^2_1 = P_2 + \rho g y_2 + \frac{1}{2} \rho v^2_2.$$

    Conservation of Energy in Fluids Meaning

    Now that we have derived the necessary equation to explain the conservation of energy in fluids, we can explain the actual meaning behind each term.

    Another way to express the Bernoulli equation is

    $$ P + \rho g y + \frac{1}{2} \rho v^2 = \text{constant}.$$

    Here, we can clearly identify the three components contributing to the total energy of a system, which remains unchanged along the streamline. The first term is simply the static pressure of the fluid. The second term represents the hydrostatic pressure exerted by the fluid due to gravity, and the final term accounts for the dynamic pressure.

    Relationship Between the Conservation of Mass and Conservation of Energy in Fluids

    So far, we have established that the energy of an ideal fluid within a closed system is conserved. The same principle applies to the mass of the flowing fluid and can be explained using the law of conservation of mass.

    The law of conservation of mass states that in a closed system matter can neither be created nor destroyed, it can only change forms.

    This law can be applied to a pipe system of varying diameters, to obtain the continuity equation. Mathematically, it can be expressed as

    $$ A_1v_1 = A_2v_2, $$

    where \(A_1\) and \(A_2\) are the cross-sectional areas of the pipe in two different locations, corresponding to the velocity of the fluid at that point. A diagram visualizing this relation is visible in Figure 3.

    In other words, it means that for a pipe with two different cross-sectional areas, the fluid will have greater velocity in the narrow section, and smaller velocity in the wider section.

    Quite often, both of the conservation laws have to be applied to a system for it to be solvable. For instance, if we know the cross-sectional area of a pipe, we can figure out the velocity of the fluid, which can then be used to calculate the kinetic energy term in the Bernoulli equation. An example problem demonstrating that is solved in the next section.

    Conservation of Energy in Fluids Examples

    Let's look at an example problem involving the conservation of energy and mass in fluids!

    Crude oil is commonly transferred through pipes. So, a curved pipe has oil flowing through it from point \(1\) to point \(2\) as pictured in Figure 4 below, has cross-sectional areas of \(4.00\times10^{-2}\,\mathrm{m^2}\) and \(3.00\times10^{-3} \,\mathrm{m^2}\) respectively. The height difference between the two points is equal to \(3.50\,\mathrm{m}\). The pressure exerted by the oil at point \(1\) is equal to \(160 \, \mathrm{MPa}\) with a velocity of \(1.85 \, \frac{\mathrm{m}}{\mathrm{s}}\).

    Conservation of Energy in Fluids A pipe of different diameters and elevations has a fluid flowing through it. The values for pressure, velocity, cross-sectional area and elevation are indicated on the diagram StudySmarterFig. 4 - The pipe setup of the conservation of energy problem.

    Calculate the pressure of the oil at point \(2\), if the density of the crude oil is \(800 \, \, \frac{\mathrm{kg}}{\mathrm{m^3}} \). Is the pressure difference between the two points significant in this situation?

    Answer:

    First, we must calculate the velocity of the oil at point \(2\). That can be done by rearranging the continuity equation

    $$ v_2= \frac{A_1v_1}{A_2} $$

    and plugging in our given values

    \begin{align} v_2& = \frac{(4.00\times10^{-2}\,\mathrm{m^2})(1.85 \, \frac{\mathrm{m}}{\mathrm{s}})}{(3.00\times10^{-3} \,\mathrm{m^2})} \\ v_2& = 24.7 \, \frac{\mathrm{m}}{\mathrm{s}}. \end{align}

    Now, we can use the Bernoulli equation to find pressure in point \(2\):

    $$ P_1 + \rho g y_1 + \frac{1}{2} \rho v^2_1 = P_2 + \rho g y_2 + \frac{1}{2} \rho v^2_2.$$

    The equation can be rearranged to find \(P_2\). Before we plug in all our values in SI units, we must decide on the reference line for the elevation. Here, it makes sense to do it at \(y_1\), as that will allow us to get rid of the \(\rho g y_1\) term as follows:

    \begin{align} P_2&= P_1 + \bcancel{\rho g y_1} + \frac{1}{2} \rho v^2_1-\rho g y_2 - \frac{1}{2} \rho v^2_2 \\ P_2&= (1.60\times 10^8 \, \mathrm{Pa})+\frac{1}{2}\left(800 \, \, \frac{\mathrm{kg}}{\mathrm{m^3}}\left( 1.85\,\frac{\mathrm{m}}{\mathrm{s}}\right )^2\right)\\ & \qquad- \left(800 \, \, \frac{\mathrm{kg}}{\mathrm{m^3}}\times9.8 \, \frac{\mathrm{m}}{\mathrm{s^2}}\times 3.50 \, \mathrm{m}\right)-\frac{1}{2}\left(800 \, \, \frac{\mathrm{kg}}{\mathrm{m^3}}\left( 24.7\, \frac{\mathrm{m}}{\mathrm{s}}\right )^2\right) \\ P_2&= 159\,729\,893 \, \mathrm{Pa} \\P_2&= 1.60\times10^8\, \mathrm{Pa}. \end{align}

    Based on this result, we can conclude that despite the varying diameters and the height difference, the pressure in both sections of the pipe is almost constant.

    Conservation of Energy in Fluids - Key takeaways

    • The law of conservation of energy states energy can be neither created nor destroyed, only converted from one kind of energy into another.
    • The Bernoulli principle states that the pressure exerted by a moving fluid is inversely proportional to its velocity in a horizontal flow.
    • The total mechanical energy of a flowing fluid is unchanged at any point along a streamline, assuming that the fluid in question is incompressible and has zero viscosity.
    • The mathematical expression for the conservation of energy in fluids is \(P_1 + \rho g y_1 + \frac{1}{2} \rho v^2_1 = P_2 + \rho g y_2 + \frac{1}{2} \rho v^2_2\).
    • The law of conservation of mass states that in a closed system matter can neither be created nor destroyed, it can only change forms.
    • Mathematically, the law of conservation of mass is represented by the continuity equation \(A_1v_1 = A_2v_2 \).



    References

    1. Fig. 1 - Pipes (https://unsplash.com/photos/4CNNH2KEjhc) by Sigmund (https://unsplash.com/@sigmund) on Unsplash is licensed by Public Domain.
    2. Fig. 2 - Conservation of energy in fluids in a pipe, StudySmarter Originals.
    3. Fig. 3 - The continuity equation applied to a pipe system, StudySmarter Originals.
    4. Fig. 4 - Conservation of energy in fluids in a pipe example, StudySmarter Originals.
    Frequently Asked Questions about Conservation of Energy in Fluids

    What is the meaning of the conservation of energy in fluids?

    The total energy of an isolated system of fluids remains constant as a result of the law of conservation of energy.

    How is energy conserved in a fluid?

    In a fluid, the total energy is conserved, as the energy within that system can neither be created nor destroyed, only converted from one kind of energy into another. 

    What is an example of conservation of energy in fluids?

    An example of conservation of energy in fluids is a pipe system with a gas or liquid flowing through it. 

    Which principle is most associated with the law of conservation of energy in fluids?

    The Bernoulli principle describes the conservation of energy in ideal fluids. 

    What does the law of conservation of energy in fluid states?

    The law of conservation of energy in fluid states that the total mechanical energy of a flowing liquid or gas remains unchanged at any point along a streamline, assuming that the fluid in question is incompressible and has zero viscosity. 

    Test your knowledge with multiple choice flashcards

    What happens to the velocity of a fluid if the cross-sectional area of a pipe is increased three times?

    In the Bernoulli equation (\(P + \rho g y + \frac{1}{2} \rho v^2 = \text{constant}\))which term represents the hydrostatic pressure exerted by the fluid due to gravity?

    What does the Bernoulli principle state? 

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