Energy equation in thermal form. Energy equations in general form

Bernoulli's equation and integral. Solving Euler's equations (1.76) leads to one of the most important equations of hydrodynamics - the Bernoulli equation. Let us multiply the first of Euler’s equations (1.76) by dx, the second - on dy, third - on dz, and then add them term by term. As a result we get

Let us integrate (1.108) along the elementary stream under the following assumptions:

Let us consider the individual sums included in (1.108).

Considering that , , , we represent the sum on the left side in the form

, (1.109)

Where u- actual full speed at a given point.

Based on the second and third assumptions, the projections of accelerations of mass forces on the coordinate axes will be X=Y= 0, Z=-g. Then the first sum on the right side of (1.108) takes the form

Xdx+Ydy+Zdz=-gdz. (1.110)

Due to the first assumption, all flow parameters, including pressure, do not depend on time and are functions only of coordinates, i.e. p = p(x,y,z). Consequently, the expression in parentheses for the second term on the right side of (1.108) is the total pressure differential, i.e.

. (1.111)

Substituting (1.109), (1.110), (1.111) into (1.108) and collecting all terms on the left side, we obtain

. (1.112)

Expression (1.112) is called Bernoulli's differential equation.

The unit of measurement of the terms of equation (1.112) is J/kg.

Bernoulli's equation can be represented in other forms by multiplying all its terms by ρ ,

(1.113)

or dividing by g

. (1.114)

In this case, the units of measurement of all terms of equation (1.113) are Pa, and (1.114) are m.

Having integrated equations (1.112) - (1.114), we obtain the expressions

; (1.115)

; (1.116)

. (1.117)

Equations (1.115)-(1.117) are called the Bernoulli integral.

Energy meaning of the Bernoulli integral. Taking ρ = const, as a result of integrating equation (1.112) we obtain

The unit of measurement of all terms of equation (1.118), as well as (1.112) is J/kg.

A moving particle of liquid has a very definite supply of mechanical energy. If an absolutely rigid body has a reserve potential energy position in the field of gravity and kinetic energy, then the liquid particle, like an elastic body, also has a reserve of potential state energy. This energy is greater, the larger the volume of liquid and the higher the pressure, and is manifested in the fact that, for example, pumping liquid into a vessel can lead to the destruction of the vessel, and compressed gas can do work during expansion.

Therefore, the total mechanical energy of a liquid particle E can be defined as the sum E = P n +P With +K, Where P n is the potential energy of position in the field of gravity; P c is the potential energy of the state; TO- kinetic energy.

The potential energy of a position can be calculated using the general mechanics formula P n =mgz, Where m- mass of liquid particle, kg; z- the height of its position above the horizontal reference plane, m.

Let us consider the specific energy per unit mass of liquid. The specific potential energy of the position is and in the Bernoulli integral (1.118) is represented by the first term.

The potential energy of the state is calculated by the formula P c = pV, Where p- pressure, Pa; V- volume of liquid particle, m3.

Specific potential energy of state in the Bernoulli integral (1.118) is represented by the second term.

Kinetic energy of a liquid particle.

Specific kinetic energy in the Bernoulli integral (1.118) is represented by the third term.

The total mechanical energy of a liquid particle is therefore determined by the sum , and the specific mechanical energy will be

. (1.119)

Comparing (1.118) and (1.119), we arrive at the energy meaning of the Bernoulli integral: the specific mechanical energy of an ideal incompressible fluid remains constant along the elementary stream. Thus, the Bernoulli integral expresses the law of conservation of mechanical energy for an elementary stream, that is, it is an energy equation.

From the Bernoulli integral it also follows that the individual components of the specific mechanical energy can change, but at the same time there is a transformation of one type of energy into another, i.e., a decrease in one term must necessarily be accompanied by an increase in at least one of the other two and vice versa.

The sum of the terms of the Bernoulli integral (1.115) gives the total amount of energy possessed by a unit of mass ( e), (1.116) - unit of volume ( p), (1.117) - unit of gravity relative to the accepted comparison plane ( H).

The terms , , express kinetic energy, the sums , , - potential energy, where gz, ρgz, z- potential energy of position, and , , - potential energy of state, respectively, unit of mass, volume, unit of gravity. We can also say that equations (1.116) and (1.117) express the same thing as equation (1.99), but on a scale and respectively.

Equation (1.115) is convenient to use when studying the movement of gas with variable density, for example, in pneumatic networks and compressors.

If the pressure changes during gas movement are insignificant and the temperature is constant, then we can assume ρ = const. Under these conditions, it is convenient to use equation (1.116), which will take the form

const. (1.120)

Expression (1.120) is convenient to use when studying air movement in ventilation networks and fans.

When moving a droplet liquid (water, oil, etc.), the density of which is constant, it is most convenient to use equation (1.117), which for ρ = const will take the form

Equation (1.121) is used in calculations of water pipelines, hydraulic mains, and pumps.

A different representation of equation (1.117) is often used. Denoting by index 1 the flow parameters in the first section of the stream in the direction of fluid movement, and by index 2 - in the subsequent section, we can write

Geometric meaning of Bernoulli's equation. All terms of equation (1.122) have the dimension of length, so we can talk about the geometric meaning of the Bernoulli equation: z- geometric (geodesic, leveling) height; - piezometric height; - speed (dynamic) height; - height of energy loss (pressure).

Here are other names: z- geometric pressure; - piezometric pressure; - speed pressure; - loss of pressure; - full pressure.

Let's consider the fluid flow in the channel, measuring all terms of the Bernoulli equation (1.122) in various sections (Fig. 1.30, measurements are shown for only two sections 1-1 And 2-2 ). Let us take an arbitrary horizontal plane as the reference plane 0-0 .

Geometric heights z are easily determined as the vertical distance from the reference plane to the centers of gravity of the corresponding sections. Piezometric heights are defined as the heights of liquid rise in piezometers, measured vertically from the centers of gravity of the corresponding sections. Velocity heights are determined as the difference in liquid levels in pitot tubes and piezometers placed in the appropriate sections (it should be noted that to accurately measure the value, the pitot tube should be placed at a point in the section where the local velocity u equal to average speed v, which cannot always be done, because the position of this point is rarely known).

The height of energy losses in the area limited by sections 1-1 And 2-2 , will be determined as the difference in liquid levels in pitot tubes placed in these sections.

If similar measurements are performed for many intermediate sections and the upper menisci of the liquid in the pitot tubes are connected by a smooth line, then we obtain a line a full pressure line.

By connecting the upper menisci of the liquid in piezometers with a smooth line, we get a line b(see Fig. 1.30), which is called piezometric line.

The line connecting the centers of gravity of the sections is called flow axis.

The behavior of these lines along the flow length l determined by the so-called slopes.

Hydraulic incline name the quantity

, (1.123)

determining the behavior of the total pressure line.

Piezometric slope

, (1.124)

determines the behavior of the piezometric line.

Geometric (geodesic) slope

characterizes the behavior of the flow axis.

In practical calculations, average slope values ​​are more often used, calculated as the ratio of the differences between the corresponding values ​​at the beginning and end to the length of the flow.

Since along the flow its total energy continuously decreases due to losses, the line of total pressure always decreases. The hydraulic slope (1.124) always remains positive.

The piezometric line can either decrease or increase. Its behavior depends both on the pressure loss and on the nature of the change kinetic energy. As the channel expands, the flow velocity and pressure head decrease. If the rate of decrease in the velocity pressure is higher than the rate of decrease in the total pressure, then the piezometric line will rise.

Pressure diagrams. In a number of hydraulic problems, it is advisable to give a graphical representation of the Bernoulli equation for a particular channel. Such graphs are called pressure diagrams. They allow you to very clearly analyze the behavior of each term in the Bernoulli equation when a fluid flows through a channel. With their help it is also convenient to produce some numerical calculations. Typically, diagrams are constructed based on the results of specific calculations, plotting pressure values ​​on a scale for each section. Let's consider the principle of constructing a diagram.

Rice. 1.31. Pressure diagram

Let liquid flow from a large open vessel into the atmosphere through a pipe of variable cross-section (Fig. 1.31). Let us choose an arbitrary horizontal plane 0-0 as the reference plane. Let's start constructing the diagram with the line of total pressure.

To do this, we determine the total pressure in the section coinciding with the free surface of the liquid in the vessel. Let us agree to use excess pressures in the Bernoulli equation and when constructing them. Then on the free surface .

Since the area of ​​the vessel significantly exceeds the cross-sectional area of ​​the pipe, then, in accordance with the flow equation, the fluid velocity in the vessel will be very small compared to the velocity in the pipe, and therefore, the velocity pressure can be neglected.

Thus, the total pressure is determined only by the geometric pressure (in the diagram it is marked by the point a). We will estimate the total pressure in subsequent sections as the difference between the total pressure in the previous section and the pressure loss in the area between these sections

. (1.126)

Looking ahead a little, we note that there are two types of pressure losses: friction losses caused by the viscosity of the fluid and local losses caused by a sharp change in the flow configuration, which, in contrast to friction (travel) losses, are considered to be concentrated in one section of the flow. The greater the length of the channel and the flow velocity, and the smaller the cross-section (diameter) of the channel, the greater the friction losses.

In section 1-1 immediately after the flow inlet from the vessel into the pipe, the total pressure will be less than the pressure in the vessel by the amount of local entry losses. Subtracting from the total pressure in the vessel (point a) input loss h 1, we get a point b, which determines the total pressure in section 1-1.

In the section of pipe between sections 1-1 and 2-2, pressure loss due to friction will occur. Since the pipe in this section has a constant cross-section, then everywhere per unit length there are equal losses, i.e. the graph of the total pressure will be linear. Subtracting from the total head in section 1-1 the amount of pressure loss due to friction in the section h 2, we get the total pressure in section 2-2 (point With). Connecting the dots b And With straight line, we get a graph of the total pressure for the first section of the pipe.

By analogy with the entrance to the pipe, subtracting from the total pressure in the section 2-2 (dot With) local losses due to sudden expansion of flow h 3, we get the full pressure in section 3-3 behind the sudden expansion (point d), subtracting from which friction losses in the second section of the pipe h 4, we get the total pressure in the outlet section 4-4 (point e).

When connecting the dots d And e it is necessary to take into account that friction losses per unit length (hydraulic slope) at the beginning of the section (large diameters) will be less than at the end (small diameters). Consequently, the line of total pressure will be directed convexly upward. Thus, we got a line of full pressure abcde.

Let us now move on to constructing a piezometric line. For this purpose, we will subtract the velocity pressure from the total pressure in each section, since

. (1.127)

On the free surface of the liquid in the vessel, the velocity pressure is zero and the piezometric pressure coincides with the total pressure (point A).

In the section between sections 1-1 and 2-2, the pipe cross-section, speed and velocity pressure remain constant, and the piezometric line () will be parallel to the line of total pressure.

When moving from section 2-2 to section 3-3, there is a sharp increase in the cross-section, accompanied by a decrease in speed and velocity pressure. Therefore, the piezometric pressure in section 3-3 is determined by subtracting from the total pressure a significantly smaller value (segment) than for section 2-2 (segment).

In the second section of the pipe, the cross-section gradually decreases, which leads to a gradual increase in speed and velocity pressure. Consequently, in each subsequent section, it is necessary to subtract a larger and larger value from the total pressure. Therefore, the piezometric line continuously moves away from the line of total pressure. The piezometric line ends at a point coinciding with the center of gravity of the output section 4-4. This is explained by the fact that atmospheric pressure again acts in the outlet section and the piezometric pressure over the excess pressure is equal to zero. The total pressure consists of geometric and velocity.

By analogy with constructing a pressure diagram based on a given flow profile, it is also possible to solve the inverse problem: constructing a pipeline configuration based on given pressure diagrams.

Examples of practical use of Bernoulli's equation. Bernoulli's equation allows us to obtain calculation formulas for various cases of fluid movement and solve many practical problems. It should be borne in mind that it is valid only for steady flows with flat living sections.

For practical use of the Bernoulli equation in solving various tasks two sections and a horizontal plane are drawn - the comparison plane. The latter, in order to have fewer unknowns, is carried out through the center of gravity of one or, if possible, two sections, and then z 1 or z 2 (or both) will be zero. The sections are carried out normal to the direction of fluid movement, and the places where they are carried out are chosen so that the sections are flat, contain unknown quantities to be determined, and a sufficient number of known quantities. Typically, such places are the free surface of the liquid, the entrance or exit from the pipeline, the connection points of measuring instruments, etc. Next, for the selected sections, which are numbered along the direction of the liquid, the Bernoulli equation is written, the numerical values ​​of the quantities are substituted into it, and the required ones are calculated.

When solving some problems, it is necessary to additionally use the condition of continuity (continuity) of the flow and take more than two sections.

Absolute pressures are substituted into the Bernoulli equation. Let's show this with a simple example (Fig. 1.32). Let it be necessary to determine the speed of liquid flow from a tank through a hole in the wall at a constant pressure (the liquid level in the tank is constant).

We draw section 1-1 at the liquid level in the tank and section 2-2 at the exit of the jet from the hole. We draw an arbitrary horizontal comparison plane x0y. The known quantities are z 1 , z 2 (z 1 -z 2 = h), p 1 = p 2 = p a (the reservoir is open and outflow occurs into the atmosphere). Then, neglecting minor pressure losses when the jet exits the hole and taking the coefficient a= 1, from equation (1.122) we find .

Measurement of pressures and local velocities. A fluid at rest has no kinetic energy. Then the Bernoulli integral (1.118) takes the form

Denoting the pressure on the free surface of the liquid p 0 , and its coordinate z 0 (Fig. 1.33), we can give equation (1.128) the form

Or . (1.129)

Having indicated the immersion depth of the point (for example, A) under the free surface of the liquid through h = z 0 - z, let us give (1.129) the form .

The latter is the basic equation of hydrostatics (1.26) and was obtained earlier by solving Euler's differential equilibrium equations.

Let's enter the point IN(Fig. 1.33) closed piezometer, which is a glass tube with a sealed upper end from which air has been removed. Under the influence of pressure at the point IN liquid rises to a certain height h'. To calculate it, we write (1.26) for a fluid at rest in a piezometer. Since air has been removed from it, the pressure above the liquid will be zero.

Thus, the height of the liquid rising in the piezometer on a certain scale (1: g) determines the specific potential energy of the liquid state, and expression (1.131) can be used to calculate the pressure measured using a piezometer. Formula (1.131) determines the method of converting pressures expressed by the height of a liquid column into dimensional units.

Since (1.26) was obtained on the basis of (1.130), it is easy to see that at whatever point of a given fluid at rest we place a piezometer, the sum of the coordinates z this point and the height of the liquid rise in the piezometer remains constant, i.e. the upper meniscus of the liquid in the piezometer will always be at the same level. Horizontal plane a-a(Fig. 1.33) drawn through the upper menisci of the liquid in piezometers is called pressure plane constructed using absolute pressure.

A closed piezometer, as we see, measures the absolute pressure in a liquid. Excess pressure can be measured using open piezometer, which is a glass tube open at both ends.

Let's place an open piezometer (see Fig. 1.33) at a point located at the same depth under the free surface as the point IN. From (1.26) it is clear that the pressures at points and IN will be the same.

Atmospheric pressure will act above the free surface of the liquid in the piezometer, so based on (1.26) we can write , from where

, (1.132)

i.e., the height of the liquid rising in an open piezometer on a scale (1: g) measures the same specific potential energy of the liquid state, but determined by excess pressure.

What was said above about liquid levels in closed piezometers is also true for open ones, with the only difference being that the pressure plane of excess pressure (see Fig. 1.33), drawn through the upper menisci of the liquid in open piezometers, will be located below the plane a-a to the height , which is easy to verify using (1.132) and (1.133).

To measure local velocities in closed channels, the movement of fluid in which is called pressure, a Pitot-Prandtl tube is used, which is a combination of a Pitot tube and a piezometer (Fig. 1.34), which are usually combined into one design.

The Pitot-Prandtl tube is inserted into the flow in such a way that the open end of the Pitot tube is directed perpendicular to the velocity vector, and the open end of the piezometer is directed tangentially.

As in the previous case, the condition holds for the Pitot tube

, (1.133)

height only h and have a different meaning here (see Fig. 1.34).

Since the liquid slips near the inlet section of the piezometer without braking, the same pressure will act in it as in the moving liquid, i.e. For it, based on (1.70), we can write (since atmospheric pressure acts on the free surface of the liquid in the piezometer, as in the Pitot tube) the equation

but in this case it represents the height of the liquid rising in the piezometer.

Expression (1.134), also valid in the case under consideration, after substitution And will again lead to (1.135), and for practical calculations it is necessary to write

Where With= 1,01…1,05; h- difference in liquid levels in the pitot tube and the piezometer.

Flow measurement. The Pitot-Prandtl tube is used to measure local motion velocities. If the live cross-section of the flow is known, the flow rate can be calculated using equation (1.26). There are instruments for direct flow measurement. The Venturi flow meter and normal diaphragm (washer) are widely used in practice.

Venturi flow meter. The great advantage of this device is its simplicity of design and the absence of any moving parts. It can be located horizontally, vertically and at any angle, which is not of fundamental importance. Let's consider a flow meter with a horizontal axis (Fig. 1.35).

It consists of two cylindrical pipes A And IN diameter d 1, connected by two conical sections (nozzles) C And D with cylindrical insert E smaller diameter d 2. In sections 1-1 and 2-2, piezometers are connected to the flow meter A And b, the difference in liquid levels in which shows the pressure difference in these sections.

Compiling the Bernoulli equation for sections 1-1 and 2-2 and neglecting the very small losses at a short length between these sections, we obtain

, (1.136)

where , but also, therefore, .

1) The Navier-Stokes system of equations and the continuity equation contain 6 unknowns: three components of the velocity vector, density, pressure and viscosity coefficient. The viscosity coefficient depends only on temperature and is usually considered a given function of the absolute temperature Г:

This equation contains a new seventh unknown - absolute temperature. Absolute temperature is related to density and pressure by the equation of state:

Depending on the nature of the environment, the function has one or another structure. In the case of gases, we agree to take the equation of state in Clayperon form:

where is the gas constant; in the case of an incompressible fluid, this equation is replaced by the condition

So, we came to a system of six scalar equations [three Navier-Stokes equations, continuity equation, equations], which contain 7 unknowns:

In order for the problem to be formulated, one more equation is needed.

Such a closing equation is the energy balance equation. We will monitor a certain mass of liquid occupying a volume. The law of conservation of energy states that the change in the energy of this mass of liquid per unit time is equal to the power of external forces, the influx of energy from the outside and the power of internal energy sources:

The energy of a liquid mass consists of two components: kinetic energy, i.e., the energy of macroscopic movement of particles

Internal energy, i.e. the energy of thermal motion of molecules of a gas or liquid.

For gases, in the general case, the expression has a rather complex structure. We will consider only the case of a “perfect gas,” that is, a gas whose internal energy is determined only by the translational motion of the molecules. This means that the energy of the rotational degrees of freedom of molecules is negligible compared to the energy of translational motion. For this case, thermodynamics gives the expression

where is the heat capacity of a gas at constant volume, related to the heat capacity at constant pressure by the formula

quantity “mechanical equivalent of heat” The work of external forces consists of the work of mass forces and the work of surface forces

where is the speed of movement of liquid particles, the surface limiting the volume

We will assume that the influx of energy from outside occurs only due to thermal conductivity. Then, according to Fourier’s law, the amount of heat received through the surface per unit time (in mechanical units) is determined by the formula

where is the thermal conductivity coefficient.

Substituting expressions (36, (37) and (39) - (41) into equation (35), we can write the following (simplified) energy balance equation:

3) The equation is an energy balance equation in integral form; In order to obtain a differential equation, it is necessary to carry out a number of transformations. First of all, we note that

(These transformations are a direct consequence of the continuity equation. Next, we transform the integrals over the surface included in the right side of the equation into integrals over the volume. First of all

Applying the Gauss-Ostrogradsky formula to this integral, after obvious calculations we obtain

We similarly transform the last term in the equation

Using formulas, we transform the equation to the form

whence, due to the arbitrariness of the volume, we obtain the following differential equation:

4) In equation (47), it is necessary to replace the components of the stress tensor with the following expressions:

Using these formulas and the identity transformation

where we can give the equation the following form:

5) So, we have obtained an equation that closes the system of equations for the dynamics of liquid and gas. This equation could be called the generalized heat conduction equation, since the heat distribution equation is contained in it as a certain special case. In fact, suppose that the fluid is at rest; then equation (49) will have the form

If the temperature difference is small, then the coefficient k can be considered independent of the coordinates and we arrive at the well-known heat conduction equation

where the coefficient is called the thermal diffusivity coefficient.

Equation (50) describes the propagation of heat in a fluid at rest due to the mechanism of thermal conductivity. This mechanism ensures the instantaneous speed of propagation of thermal disturbances (see Fig. 5). Let us assume that we have imparted to a fluid particle located at point x at the moment of time a pulse perturbation where is the delta function, equal to zero everywhere except at the point and such that Then the temperature distribution at any moment of time is described by the formula

We see that whatever the value of the abscissa is at any moment other than zero, the temperature will also be different from zero.

6) The reasoning that was carried out here related to the case of a fluid at rest, and it was tacitly assumed that if in starting moment If the liquid is at rest, then it will be at rest at subsequent moments of time. This is, generally speaking, not the case. In fact, if the temperature changes, then, according to the equation of state, the density and pressure will change, which in turn will cause the fluid to move. Thus, a change in the temperature of the medium causes the movement of the liquid. The problems of heat propagation and the problem of fluid movement should be considered together. Only in one special case can these problems be separated - in the case of an incompressible fluid, under the assumption that the viscosity coefficient does not depend on temperature. Then the problem of fluid motion is reduced to solving the continuity equation

and Navier-Stokes equations

Having determined the vector and scalar from these equations, we can then determine the temperature field from the equation, which in this case takes the form

7) From equation (54) it is clear that, in addition to the mechanism of thermal conductivity, convective heat transfer plays a role in the propagation of heat - transfer due to the movement of liquid particles. Therefore, thermal disturbances can also propagate inside a liquid devoid of thermal conductivity. To explain this, let us consider the problem of the motion of an ideal non-thermal conductive gas, when equation (49) takes the form

Following the law of conservation of energy, we will draw up an energy balance for a mass of gas that first fills volumes 1 - 2, and after a time dt volume 1" - 2" (Fig. 3.3). Since the shaded volume 1" - 2 is common to them, the increment of any type of energy is equal to the difference in the energy of this type in infinitesimal volumes 2 - 2" and 1 - 1".

Increment of kinetic energy

Potential energy increment

Where Z 2 And Z 1– heights of sections 1 and 2, g– acceleration of gravity.

Increment of internal (thermal) energy

Where u=С n T- internal energy per unit mass of gas, equal to the product of heat capacity at constant pressure With n to absolute temperature. If With n =const, That

When we move the volume we have selected from state 1 - 2 to state 1" - 2", external forces do work. The transfer of gas from section 1 to 1" occurs as if under the action of a piston with an area F 1 with pressure P 1.

Piston work in time dt equal to

The following relations are used here

F 1 w 1 =V 1 - the volume that the piston displaces in 1 s; m 3 /s;

n 1 =V 1 /M- specific volume m 3 /kg;

M- mass flow, kg/s;

r 1 =1/n 1 density kg/m 3;

dM- the mass that the piston displaces over time dt

Similarly for section 2. During the time dt the gas will move the piston to position 2", producing work on the external environment, which we will consider negative,

Thus, the energy contributed by pressure forces is equal to the difference between the work of piston 1 and 2:

To the gas stream in section 1 - 2 in time dt heat can be supplied in quantities dQ. A gas stream can do technical work dL, for example, rotating a turbine wheel installed between sections 1 and 2. The energy spent on overcoming friction forces should also be taken into account dL tr. According to the first law of thermodynamics, supplied to gas thermal energy dQ and the work of pressure forces is spent on performing technical work dL, work of friction forces dL tr, as well as to increase reserves of potential, internal and kinetic energy:

Dividing all terms of the resulting expression by dM, we obtain the energy equation written for 1 kg mass of gas

Where q- heat supplied to 1 kg gas; dL- work done 1 kg gas; dL tr- the work of overcoming friction forces attributable to 1 kg gas

Heat Gain q carried out in two ways: from the outside ( q nar) - due to heat exchange through the side surface of the jet or due to the release of heat in the jet itself as a result of fuel combustion and from the inside ( q tr)- due to the conversion of friction work into heat L tr:

In differential form

Along with the equations of conservation of mass and momentum, which were used above to derive the equations of continuity and motion, the energy equation is also used to describe a continuous medium. Let us consider the energy equation for the special case of an adiabatic process, when there is no heat exchange between elements of a continuous medium. In this case the change internal energy E element of a continuous medium with mass (liquid particle) is associated only with a change in its volume (in the absence of volumetric heat sources): . Introducing energy per unit mass of matter into consideration, we obtain

Since , That

.

In accordance with the continuity equation , That's why

.

This equation describes the distribution of the volumetric density of internal energy and its change caused by deformation and movement of the medium. At the same time, changes in internal energy can be caused by processes associated with the release or absorption of energy, for example, during heating by electric current or during chemical reactions. To take these phenomena into account, we modify the last equation by adding to its right side a term having the dimension W/m 3, which describes the rate of release or absorption, depending on the sign, of energy at points in the continuous medium.

Thus, the complete system of equations for the dynamics of an ideal liquid (gas) in the adiabatic regime has the form

(58)

The last equality is an equation of state that closes the system and determines the specific physical properties of the medium. Here are examples of the equation of state:

1. Ideal gas: , where is Boltzmann’s constant, n- concentration of particles in the gas, M- particle mass.

2. Incompressible fluid:

3. Water at high pressures, where , - pressure and density under normal conditions.

The last example shows that to increase the density of water by 20%, excess pressure is required. Returning to the energy equation, we get

,

where instead the product of particle concentration and particle mass is taken. Gas particles generally have s degrees of freedom. For each degree of freedom in thermodynamic equilibrium there is energy . Then, after substituting the expression for the internal energy of a unit mass of an ideal gas into the energy equation we get

,

, ,

where and are constants. The last equality can be given the form , where is the adiabatic exponent. The constant can be determined from the initial conditions . As a result, the adiabatic equation will take the form

Law of conservation of energy. Energy balance. Energy, work, heat. Internal energy, potential energy, kinetic energy.

Bernoulli's equation for gas. Enthalpy equation. Adiabatic flow. Energy-insulated flow. Isoentropic flow.

Energy isolated isentropic flow.

Studying basic equations and dependencies , used in gas dynamics , it is convenient to carry out first for elementary trickle or one-dimensional flow, and then extend them to more complex types of movement.

Of great importance in gas dynamics is law of conservation of energy . As is known, he states the fact that

energy neither appears nor disappears, but only transforms from one type to another.

Therefore, having compiled an energy balance for a certain amount of gas, for example, for units of mass, one can find the relationship between the various components of energy. Such mathematical notation energy balance and represents the energy equation .

Compilation energy balance Let's look at an example gas turbine unit , the diagram of which is shown in Figure 6.

Through the input section 1 air from the atmosphere enters the compressor, where it is compressed and supplied to the combustion chamber. There, in the combustion chamber, liquid fuel enters, which, mixed with air, burns, releasing a large amount heat . Thus, the combustion products formed there enter the turbine from the combustion chamber with high temperature and high blood pressure. In the turbine they expand, producing work - rotating the rotor. Part of the turbine's work is transferred to the rotation of the compressor using the shaft, the other part is given to the consumer. Exhaust gases leave the turbine, exiting through section 2.

Energy incoming air, per unit mass , designated E 1, energy exhaust gas - E 2.

Heat input designated Q e. Index " e" means that heat was supplied from outside (externus – lat. external , outsider).

There is no contradiction here: despite the fact that combustion took place inside the chamber and the heat that warmed the gas was released there, this energy was introduced from the outside in a hidden form, along with the fuel. Consequently, since the task of studying the physico-chemical processes of combustion is not set, but only phenomena of a gas-dynamic nature are considered, we can assume that heat in quantity Q e was brought into the combustion chamber from the outside.

Jobon the installation shaft, given to the consumer, is indicated L. She also referred to a unit of mass air passing through the installation.

On Figure 7 depicted simplified flow diagram. On settlement area between cross sections 1 And 2 , same as in the previous case, heat is supplied And mechanical work is assigned . Therefore, for simplified diagram energy balance will be the same as for gas turbine unit, but using this scheme is simpler and more convenient.

Energy balance for the flow pattern under consideration can be written by the following equation:

E 1 - E 2 + Q e - L = 0. (2.1)

Next, you need to decipher what is meant by the total energy reserve of a unit mass of gas E. It should be borne in mind, however, that in "full energy" there is no need to include all its components (for example, chemical, electrical, intranuclear); it is quite sufficient to take into account only those types of it that can transform one into another within the limits of the gas-dynamic problems being studied. Then we can write that

E= u + p/ρ + w 2 /2 + gz, (2.2)

Where u – internal energy gas mass units;

p/ρ– potential energy pressure gas mass units;

w 2 /2– kinetic energy gas mass units;

gz– potential energy provisions (level) unit mass of gas;

z– geometric height;

g – acceleration gravity .

All specified quantities are measured in units of work per unit of mass, namely in J/kg or, what is the same, in m 2 /sec 2(in SI system).

Substituting into equation (2.1) the values E 1 And E 2, expressed using equation (2.2), and taking into account that the difference in internal energies u 1 – u 2 = C v (T 1 -T 2), we get

C v (T 1 -T 2) +p 1 /ρ 1 -p 2 /ρ 2 +(w 1 2 - w 2 2)/2+g(z 1 -z 2) +Q e -L= 0. (2.3)

This is it energy equation for a one-dimensional flow or for an elementary trickle. It shows what happens change internal energy C v (T 1 -T 2), potential energy of pressure p 1 /ρ 1 -p 2 /ρ 2, kinetic energy (w 1 2 - w 2 2)/2, potential energy of position g(z 1 -z 2) as a result of the action of heat supplied from outside Q e And work L, supplied by gas to external consumers . Change internal energy associated with change temperature gas, kinetic energy- with change speed flow, potential energy level- with change altitude position the mass of gas under consideration above the plane taken as the origin. As for the change pressure potential energy, then it requires special clarification.

On Figure 8 shows the calculated section of the flow limited at the inlet cross section 1 and at the exit - section 2.

Upon entry gas through the section 1 strength external pressure p 1 F 1, pushing to the settlement area volume of gas F 1 Δx 1, do work p 1 F 1 Δx 1 .

On exit from the design area, through the section 2 volume gas F 2 Δx 2 does work against external pressure forces p 2 F 2 Δх 2. Dividing these works by the mass of gas in the corresponding volumes, we obtain

L W = p 1 F 1 Δx 1 / ρ 1 F 1 Δx 1 = p 1 /ρ 1 ,

L out = p 2 F 2 Δx 2 / ρ 2 F 2 Δx 2 = p 2 /ρ 2.

Hence, p 1 /ρ 1 -p 2 /ρ 2 =L in -L out represents difference between pushing and pushing work units of mass of gas. This value characterizes accumulation (If p 1 /ρ 1 >p 2 /ρ 2) potential energy pressure or spending her (if p 1 /ρ 1

) gas flow located inside the calculated area.

Change in Level Potential Energy g(z 1 -z 2) in problems related to the calculation of thermal power machines or installations, as a rule, is a negligible value compared to other terms of the energy equation. It usually does not exceed 50…100 m 2 /sec 2, while the other terms have the order 10 000…100 000 m 2 /sec 2. Therefore, in all further reasoning and calculations, the value g(z 1 -z 2) will be discarded. However, it is necessary to pay attention to problems of this kind, such as the calculation of mine ventilation systems, in which the change in the potential energy of the level is very large and can exceed the values ​​of other terms of the energy equation. In these cases the value g(z 1 -z 2) must be taken into account.

The energy equation can be given a different one, in many cases a more convenient form for calculations. Let's transform the sum of terms

C v (T 1 -T 2) +p 1 /ρ 1 -p 2 /ρ 2 = (C v T 1 +p 1 /ρ 1) -(C v T 2 +p 2 /ρ 2)=

=(C v T 1 +RT 1) -(C v T 2 + RT 2)= (C v +R)(T 1 -T 2) = C p (T 1 -T 2) ,

using the relation C p –C v =R known from thermodynamics, and substitute the resulting expression into equation (2.3). Then the energy equation can be written more compactly

C p (T 1 -T 2) + (w 1 2 - w 2 2)/2 + Q e - L = 0, (2.4)

and most importantly, three thermodynamic parameters p, ρ And T now you can just replace one – enthalpy h=C p T. (“Three in one”!)

(2.5)

This view energy equations also called enthalpy equation or heat content, since it includes enthalpy h.

The following sign rule is adopted in the energy equation. External heat supplied is considered positive, and external heat removed is considered negative; the work done by the gas and transferred to an external consumer is positive, and the work supplied to the gas from the outside and spent on its compression is negative. Thus, in heater gas (combustion chamber) warm counts positive , V cooler - negative ; Job , obtained in turbine, - positive , and spent on rotation compressor - negative . This sign rule is consistent with the equation first law of thermodynamics.

Energy equation often used V differential form . To get it in this form, we will use this technique. Let us mentally bring the second section closer to the first, reducing the length of the calculated section to infinitesimal. Then in the limit we get instead Q e And L respectively dQ e And dL, instead of finite differences T 1 – T 2 And (w 1 2 - w 2 2)/2 we obtain the corresponding differentials – dT And – d(w 2 /2).

In the last two expressions minus sign appeared because infinitesimal differences are taken T 1 -T 2 And (w 1 2 - w 2 2)/2, not T 2 -T 1 And (w 2 2 - w 1 2)/2.

Substituting this into the energy equation (2.4) and reversing the signs , we get energy equation in differential form or differential energy equation

(2.6)

If we compare the expression for the total energy reserve (2.2)

E= u + p/ρ + w 2 /2 + gz,

with the left side Bernoulli equations, which also represents the quantity full energy reserve units of mass incompressible fluid

p/ρ + w 2 /2 + gz = const,

then it can be noted that in the case of a gas, the internal energy value is additionally introduced u. This is explained by the fact that when ρ≠const thermal processes affect the density of the gas, and since its expansion or compression is associated with work, this influence ultimately extends to the mechanical components of energy. Thus, in energy equations(2.4) and (2.5) there are quantities that have both mechanical, so thermal(caloric) origin.

One more a kind of energy equation is generalized Bernoulli equation for gas . It differs from equations (2.4) or (2.5) in that all elements included in it the terms have mechanical origin. This equation can be obtained in the following way. Let us use the same technique with which the differential energy equation (2.6) was obtained above and represent equation (2.3) in differential form:

(2.7)

Amount of heat Q, gaseous, and the amount of heat Q e, supplied to him from outside, in general not the same : still exists heat of frictionQ r, which is released due to gas friction against the walls, internal friction (arising between layers moving at different speeds), the formation of vortices, etc. This heat is also absorbed by the gas. That's why

Q = Q e + Q r = Q e + L r. (2.8)

dQ e = dQ – dL r, (2.9)

Where Lr-friction work (in SI units Q r =L r).

The amount of heat absorbed by the gas, can be determined using the equation first law of thermodynamics

dQ = C v dT + pdv. (2.10)

Substituting this expression into formula (2.9), we obtain

C v dT = dQ e + dL r -pdv. (2.11)

Besides,

d(p/ρ)=d(pv)=pdv+vdp/. (2.12)

After substituting formulas (2.11) and (2.12) into the energy equation (2.7) and replacing the specific volume through density v=1/ρ we get Bernoulli's equation for gas in differential form

dp/ρ+d(w 2 /2)+dL+dL r =0. (2.13)

When solving specific problems, the Bernoulli equation is integrated within the range from the initial section of the calculation section to the final

(2.14)

If during the solution process it is necessary to obtain the flow parameters in some intermediate section of the calculation section, then during integration this section is taken as the final section. When solving, you can take the indefinite integral. The integration constant is then determined from the boundary conditions, which are usually taken to be the conditions at the entrance to the calculation section.

In order to calculate ∫(dp/ρ), you need to know the relationship between r And ρ , i.e. have an equation for the thermodynamic process in which gas flows, for example the polytropic equation p/ρ n =const. If the thermodynamic process is known, then the polytropic index is also known. At polytropic process integration gives

at isothermal process ( n=1)

1 2 ∫(dp/ρ)=(p 1 /ρ 1)ℓn(p 2 /p 1)=RT 1 ℓn(p 2 /p 1). (2.16)

Comparing with each other energy equation And Bernoulli's equation, for example (2.4) and (2.14), one can notice that the first takes into account external heat, but does not contain friction work explicitly, while the second does not explicitly contain external heat, but takes into account friction work. Therefore, it seems that these equations do not take into account all the features of the flow. In reality this is not the case. Although the work of friction is not explicitly included in the energy equation, its influence affects, first of all, temperature T 2.

As for the Bernoulli equation, it takes into account external heat when calculating ∫(dp/ρ), namely, the quantity depends on the amount of heat supplied polytropic index n.

Let's consider energy equations For special cases of gas flow .

Adiabatic ( or adiabatic) current . This flow occurs without external heat supply or removal , i.e. Q e =0. Regarding the internal heat supply (friction heat Q r) no reservations are made, i.e. it is either present or equal to zero. Energy equation in this case it looks like:

(2.17)

A Bernoulli's equation retains shape (2.14)

1 2 ∫(dp/ρ)+(w 2 2 - w 1 2)/2 + L+ L r =0.

Equation (2.17) is of great importance in experimental practice. It is used, for example, in the experimental determination of the operation of a turbine or compressor, when direct determination of power from torque and speed is difficult for technical reasons. To do this, you only need to measure the temperatures and gas velocities at the entrance to and exit from the machine and make the calculation using formula (2.17). Note that in practice the situation is even simpler. Are measured not gas temperature and speed separately, but stagnation temperature.

Energy-insulated flow. This flow occurs without external heat exchange (Q e =0) And without input or output of external mechanical work (L=0), i.e. without energy exchange with the external environment in the area between the inlet and outlet sections. Energy equation for an energy-insulated flow it is written as follows:

(2.18)

C p T 1 + w 1 2 /2 = C p T 2 + w 2 2 /2. (2.19)

The meaning of the last equality is that in an energy-isolated flow, the total energy reserve of a unit mass of gas remains unchanged, since in the calculated section energy is not supplied from the outside and is not discharged into the external environment.

Bernoulli's equation for this type of flow takes the form:

1 2 ∫(dp/ρ)+(w 2 2 - w 1 2)/2 + L r =0. (2.20)

The energy-isolated flow model is used when calculating diffusers, uncooled nozzles and other fixed channels in which heat exchange with the external environment is negligible.

Isoentropic (or isentropic or isentropic) flow . This flow occurs at constant entropy S=const. For entropy to remain constant, it is necessary to satisfy the condition Q=0. From formula (2.8) it follows that this can be when Q e =0,Q r =0 or when Q e = – Q r . The second case provides for heat transfer to the external environment, exactly equal to the heat transfer from friction. Such an accurate heat balance can rarely be encountered in practice, and therefore is not considered here. Thus, we can assume that the flow will be isentropic if there is no friction and external heat exchange . For this type of flow energy equation written in the same way as for adiabatic flow (see formula (2.17))

C p (T 1 -T 2) + (w 1 2 - w 2 2)/2 - L = 0,

A Bernoulli's equation has the form:

1 2 ∫(dp/ρ)+(w 2 2 - w 1 2)/2 + L=0. (2.21)

When calculating the integral here, you need to keep in mind that r And ρ connected isentropic equation p/ρ k =const. The isentropic flow model is used in theoretical calculations and research ideal compressors and turbines.

Energy isolated isentropic flow. This flow occurs without energy metabolism with the external environment ( Qе=0, L=0) And frictionless (Lr=Qr=0). In this case, the conditions are automatically met isentropic (isentropic) process. Energy equation has the same form as for energy-isolated flow (2.18) or (2.19)

C p (T 1 -T 2) + (w 1 2 - w 2 2)/2 = 0,

C p T 1 + w 1 2 /2 = C p T 2 + w 2 2 /2,

A Bernoulli's equation is written like this:

1 2 ∫(dp/ρ)+(w 2 2 - w 1 2)/2 =0. (2.22)

Here, too, when calculating the integral, the connection between pressure and density is established isentropic equation. This special case is used quite widely. For example, in theoretical gas dynamics Most problems are considered under the assumption of exactly this type of flow.

In differential form, equations (2.18) and (2.22) have the following form:

C p dT + d(w 2 /2) = 0, (2.23)

dp/ρ + d(w 2 /2) = 0.(2.24)

Let's look at two more very common forms of notation: Bernoulli equations For energy isolated isentropic flow. Integrating equation (2.24), we have

∫(dp/ρ) + w 2 /2 = const.

Using isentropic equation

p/ρ k = B = const,

and the following obvious relations

ρ k = (p/B); ρ = (p/B) 1/ k ; B 1/ k = (p/ρ k) 1/ k =p 1/ k /ρ;

let's find the value of the integral

∫(dp/ρ) =∫(dp/(p/B) 1/ k)= B 1/ k ∫(dp/p 1/ k)= B 1/ k ∫p -1/ k dp=

= B 1/k p (1-1/k) /(1-1/k)= p 1/k ∙ p (1-1/k) ∙ k/ρ∙(k-1) =

=(k/(k-1))(p 1/k ∙ p (k-1)/k /ρ) = (k/(k-1)) p/ρ.

and, substituting it into the previous equation, we get

(k/(k-1)) p/ρ + w 2 /2 = const. (2.25)

If we compare equation (2.25) with the Bernoulli equation for horizontal flow of an ideal incompressible fluid

p/ρ + w 2 /2 = const,

then you can notice that they differ only in the first term: for gas, the coefficient in front of p/ρ equals k/(k-1) whereas for an incompressible fluid it is equal to 1 . Thus, the value k/(k-1) takes into account compressibility effect.

If we use the relation that is used to determine speed of sound a 2 = kRT= kp/ρ, and transform the first term of equation (2.25), then the latter takes the form:

a/(k-1) + w 2 /2 = const. (2.26)

This entry form Bernoulli equations widely used in theoretical gas dynamics.

G p/ρ k =const. p/ρ = RT. a= √kRT. a 2 = kRT= kp/ρ.

E 1 - E 2 + Q e - L = 0. E= u + p/ρ + w 2 /2 + gz.

C v (T 1 -T 2) +p 1 /ρ 1 -p 2 /ρ 2 +(w 1 2 - w 2 2)/2+g(z 1 -z 2) +Q e -L= 0.

C v dT + d(p/ρ) + d(w 2 /2) - dQ e + dL = 0.

C p (T 1 -T 2) + (w 1 2 - w 2 2)/2 + Q e - L = 0.

h 1 -h 2 + (w 1 2 - w 2 2)/2 + Q e - L = 0.

C p dT + d(w 2 /2) - dQ e + dL = 0.

dp/ρ+d(w 2 /2)+dL+dL r =0.

(k/(k-1)) p/ρ + w 2 /2 = const. a/(k-1) + w 2 /2 = const.

p/ρ + w 2 /2 = const.

An energy balance can be drawn up for any flow pattern. The example with a gas turbine unit was taken because it contains all the components of the energy balance considered in gas dynamic problems.

It should be noted that this equation was obtained in our days. The name of Daniel Bernoulli was given to it because it is a generalization of the Bernoulli equation known in hydrodynamics to the case of gas flow.

The indefinite integral is taken.