Energy Equation in Flow Systems

When energy crosses the boundary of a system, the energy in the system changes by an equal amount. Thus a decrease in the energy of a system equals the increase in energy of the surroundings and vice versa.

Energy can cross a system boundary in the form of heat transfer (to or from the system), in the form of work (done by or on the system), or by mass entering or leaving. A process during which no heat is transferred to or from a system is known as an adiabatic process. The law of conservation of energy states that for a system of particles

E₂ – E₁ = Q͂ - Wʹ

Or

dE = d Q͂ - dWʹ

Where:

• E2 – E1: is the change in energy experienced by the system
• Q͂: is the heat added to the system
• Wʹ: is work done by the system

The total energy is a property of the system and includes internal (U), kinetic (KE), and potential energies (PE):

E = U + KE + PE

If we express this on a per-unit-mass basis, we obtain:

Heat and work are not properties of the system, but they represent energy transfers moving across the system boundaries in specific forms.

Remember the general conservation expression:

If we substitute, N = E and n = e, we obtain:

Equation 1 is generally known as the first law of thermodynamics and is a statement of the conservation of energy.

The Wʹ term consists of all forms of work crossing the boundary. This includes shaft work, electric and magnetic work, viscous shear work, and flow work. The last of these forms represents work done by the fluid system in pushing mass into or out of the control volume. Thus:

where dW_f /dt is flow work per unit time.

To develop an expression for dW_f/dt, we can consider an element of mass dm leaving a control volume through an area dA, as shown below:

The fluid in the control volume must work against the external pressure p to move dm out. Since work = force x distance, we have:

Similarly, for an element of mass entering, work is being done on the system. Therefore, we get:

Combining the previous two equations, we can obtain the flow work as:

If we substitute Equations 3 and 2 into Equation 1 we get:

Combining the integral terms results in:

Enthalpy is a property defined as h = u + p/ρ; the conservation of energy equation now becomes:

For a steady, one-dimensional flow, the energy equation becomes:

For many incompressible flows, Q͂ is usually assumed to be zero, and changes in internal energy are frequently negligible. Thus equation 5 can be simplified into:

For compressible flows, however, the change in internal energy is significant and cannot be neglected. 

Relationship Between Heat Transfer and the First Law of Thermodynamics

In this topic we are interested in the efficiency of heat engines. We are going to build upon the knowledge of thermodynamics and show how heat transfer plays an integral role in managing and promoting the efficiency of a wide range of energy conversion devices. 

Remember that we have defined a heat engine previously as any device that continuously or cyclically operates and converts heat to work. Power plants and thermoelectric devices are examples of heat engine. 

It is extremely important to improve the efficiency of heat engines. For example, an efficient combustion engine consumes less fuel to produce a given amount of work thus reduces emissions of pollutants. More efficient thermoelectric devices can generate more electricity from waste heat. 

The second law of thermodynamics can be represented in a number of distinct but comparable ways and is frequently employed when efficiency is an issue. The KelvinPlanck statement is very relevant to the operation of a heat engine. It states:

"It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of work to its surroundings while receiving energy by heat transfer from a single thermal reservoir"

Remember that a thermodynamic cycle is a process for which the initial and final states of the system are identical. Consequently, the energy stored in the system does not change between the initial and final states, and the first law of thermodynamics reduces to W = Q.

Because of the Kelvin–Planck statement, a heat engine must exchange heat with two or more reservoirs, gaining thermal energy from the higher-temperature reservoir and rejecting thermal energy to the lower-temperature reservoir. Therefore, converting all of the input heat to work is impossible and:

W = Q_in – Q_out,

where Q_in and Q_out are both defined to be positive. i.e.,

Q_in = heat transferred from the high temperature source to the heat engine

Q_out = the heat transferred from the heat engine to the low temperature sink.

The efficiency of a heat engine is the fraction of heat transferred into the system that is converted to work:

For a reversible process the ratio Q_out/Q_in is equal to the ratio of the absolute temperatures of the respective reservoirs (From the 2^nd Law of Thermodynamics. Thus, the efficiency of a heat engine undergoing a reversible process, i.e., Carnot efficiency, η_C, (as previously discussed) is given by:

where T_c and T_h are the absolute temperatures of the low and high temperature reservoirs, respectively. 

The Carnot efficiency is the maximum possible efficiency that any heat engine can achieve operating between those two temperatures. Any real heat engine, which will necessarily undergo an irreversible process, will have a lower efficiency.

From our knowledge of thermodynamics, we know that, for heat transfer to take place reversibly, it must occur through an infinitesimal temperature difference between the reservoir and heat engine. In heat transfer mechanisms, in order for heat transfer to occur, there must be a nonzero temperature difference between the reservoir and the heat engine. This introduces irreversibility and reduces the efficiency. 

Let us now consider a more realistic heat engine model where heat is transferred into the engine through a thermal resistance R_t,h, while heat is extracted from the engine through a second thermal resistance R_t,c where subscripts h and c refer to the hot and cold sides of the heat engine respectively. 

Let us now consider a more realistic heat engine model where heat is transferred into the engine through a thermal resistance R_t,h, while heat is extracted from the engine through a second thermal resistance R_t,c where subscripts h and c refer to the hot and cold sides of the heat engine respectively. This is shown in the following figure:

 

These thermal resistances are associated with heat transfer between the heat engine and the reservoirs across a nonzero temperature difference through mechanisms of conduction, convection and/or radiation. E.g., the resistances could represent conduction through walls separating the heat engine from the two reservoirs

Note that the reservoir temperatures are still T_h and T_c but that the temperatures seen by the heat engine are T_h,i < T_h and T_c,i > T_c, as shown in the diagram above. The heat engine is still assumed to be internally reversible, and its efficiency is still the Carnot efficiency.

However, the Carnot efficiency is now based on the internal temperatures T_h,i and T_c,i . Therefore in order to account for the realistic irreversible process, the efficiency, ηm, is as follows: 

where the ratio Q_out/Q_in, has been replaced by the corresponding ratio of heat rates, q_out/q_in. This replacement is based on applying energy conservation at an instant in time. Utilizing the definition of a thermal resistance, the heat transfer rates into and out of the heat engine are given by:

The above equations can be solved for the internal temperatures to result in:

For the T_c,i equation, q_out has already been related to q_in and η_m, The more realistic, modified efficiency can then be expressed as:

Solving for η_m, results in:

where R_tot = R_t,h + R_t,c.

It is evident that η_m = η_C only if the thermal resistances R_t,h and R_t,c could somehow be made infinitesimally small (or if qin  0). For realistic (nonzero) values of R_tot, η_m < η_C, and η_m further deteriorates as either R_tot or q_in increases. As an extreme case, note that η_m = 0 when T_h = T_c + q_inR_tot , meaning that no power could be produced even though the Carnot efficiency is nonzero.

In addition to the efficiency, another important parameter to consider is the power output of the heat engine, given by:

Maxwell Equation Solved Example

Example

Show that: 

Solution

This example uses the Maxwell Relations Table from the previous blog entry. You can read it here

Lets combine the first and second laws: dU = TdS - pdV

Divide both sides by dV and keep T as a constant:

Note that:

This results in:

Using the Maxwell Relation for A:

This results in:

Since:

This confirms that

Maxwell Relations

Background

Maxwell’s equations are important for electric and magnetic fields, while Maxwell’s relations focus on thermodynamic quantities like temperature, pressure, volume, and entropy and are important when analyzing thermodynamic systems like internal combustion engines.

Like Maxwell’s equations, Maxwell’s relations are based on partial derivatives. The relations describe the symmetry of second-order partial derivatives of thermodynamic quantities. Maxwell’s relations directly and indirectly cover eight thermodynamic quantities including:

• pressure (P),
• volume (V),
• temperature (T),
• entropy (S) measured in Joules per Kelvin (J/K),
• internal energy (U) measured in Joules,
• enthalpy (H),
• Helmholtz free energy (F or A),
• Gibbs free energy (G).

P, V, and T are simple quantities while the other thermodynamic quantities are more complex, thus Maxwell’s relations are very important.

For example, enthalpy (H) measures a system’s total heat content is defined as the sum of its internal energy (U) plus the product of pressure (P) and volume (V): H = U + PV.

Helmholtz free energy (F) measures the useful work obtainable from a closed system at constant temperature and volume and is defined as F = U – TS.

Gibbs free energy (G) is used to predict the spontaneity of a process at a given temperature and is defined as G = H – TS, or ∂G = ∂H – T∂S:

• When ∂G < 0, the process is spontaneous (favored).
• When ∂G > 0, the process is non-spontaneous (not favored).
• If ∂G = 0, the process is at equilibrium.

Mathematical Derivations of Maxwell's Relations

To derive Maxwell’s relations, it is useful to combine the First and Second Laws into a single mathematical statement. In order to do that, one notes that since:

for a reversible process: dq = TdS

Since: dw = TdS – pdV

for a reversible expansion in which only p-V works is done, and since dU = dq + dw

dU = TdS – pdV         (1)

This differential for dU can be used to simplify the differentials for H, A (Helmholtz energy) and G. But it is even more useful due to the constraints it places on the variables T, S, p and V due to the mathematics of exact differentials. 

Equation 1 suggests that the natural variables of internal energy are S and V or simply U (S, V). Thus, the total differential (dU) in equation 1 can be expressed as follows:

By studying equations 1 and 2, we can see that:

Since dU is an exact differential, the Euler relation must hold that:

substituting equations 3 and 4 into the above expression:

This is an example of a Maxwell Relation.

A similar result can be derived based on the definition of H:

H = U + pV

Using chain rule and differentiating d(pV) results in:

dH = dU + pdV + Vdp

Making the substitution using the combined first and second laws (dU = TdS - pdV) for a reversible change involving on expansion (p-V) work:

dH = TdS – pdV + pdV + Vdp

dH = TdS + Vdp         (5)

As in the case of internal energy, this suggests that the natural variables of H are S and p. Thus, 

If we compare equation 5 and 6 we get:

We can now see that equation 3 and equation 7 are equal to T hence are equal to each other:

Furthermore, the Euler Relation must also hold true:

This is the Maxwell Relation of H. The same can be done for G and A.

Summary

Maxwell Relations are summarized below:

Fundamentals of Partial Differentiation of the Exact Differential Equation

Carrying out partial differentiation of the Exact Differential Equation of M and N with respect to z and y, respectively, results in:

dx is a perfect differential when the above equation is satisfied for any function x.

Similarly, if y = y(x, z) and z = z(x, y), then from these two relations, we have:

In terms of p, v and T, the following relation holds true:

Thermodynamic Relations

Some properties like temperature, pressure, volume, and mass can be measured directly while other properties like density and specific volume can be determined from these using some simple relations.

However, properties like internal energy, enthalpy, and entropy are not so easy to determine as they cannot be measured directly or related to easily measurable properties through some simple relations.

Thus, it is important to develop some fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties.

Partial Derivatives and Associated Relations

Most basic thermodynamic relations involve differentials. Let us consider a function f that depends on a single variable x, i.e., f = f(x) as shown below:

The derivative of the function at a point defined as Slope expressed as:

Therefore, the derivative of a function f(x) with respect to x represents the rate of change of f with x.

Let us now consider a function that depends on two (or more) variables, such as z = z(x, y). This time the value of z depends on both x and y.

It is sometimes necessary to examine the dependence of z on only one of the variables by allowing one variable to change while holding the others constant and observing the change in the function.

The variation of z(x, y) with x when y is held constant is called the partial derivative of z with respect to x, and it is expressed as:

To obtain a relation for the total differential change in z(x, y) for simultaneous changes in x and y would be:

The above equation is the fundamental relation for the total differential of a dependent variable in terms of its partial derivatives with respect to the independent variables. This relation can easily be extended to include more independent variables.

An important relation for partial derivatives is used in calculus to test whether a differential dz is exact or inexact. In thermodynamics, this relation forms the basis for the development of the Maxwell relations which was discussed in a previous blog entry here where dz is replaced by dφ.

We can now develop two important relations for partial derivatives – the reciprocity and the cyclic relations.

The function z = z(x, y) can also be expressed as x = x(y, z) if y and z are taken to be the independent variables. Then the total differential of x becomes:

Eliminating dx from the above equation by combining the dz and dx equations we get:

The variables y and z are independent of each other and thus can be varied independently. For example, y can be held constant (dy = 0), and z can be varied over a range of values (dz ≠ 0). Therefore, for this equation to be valid at all times, the terms in the brackets must equal zero, regardless of the values of y and z. 

Setting the terms in each bracket of the above equation equal to zero gives two equations:

This equation is called the reciprocity relation, and it shows that the inverse of a partial derivative is equal to its reciprocal.

This equation is called the cyclic relation, and it is frequently used in thermodynamics.