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.

Relationship Between Heat Transfer and the First Law of Thermodynamics

• Thermodynamics and heat transfer are complementary engineering subjects, each addressing different aspects of energy behavior in systems.
• In thermodynamics, heat is treated as a form of energy transfer used to analyze system states and energy requirements, but the theory does not address how heat actually flows or at what rate.
• Heat transfer provides the engineering tools and rate equations needed to quantify heat-flow mechanisms (conduction, convection, radiation) and determine heat-exchange rates.
• Practical engineering design problems—such as sizing equipment, components, or entire systems—require heat-transfer analysis in addition to thermodynamic principles.
• Example: Designing and sizing a power plant cannot be accomplished using thermodynamics alone; engineers must apply heat-transfer principles to ensure feasible and efficient operation.

Relationship to the First Law of Thermodynamics (Conservation of Energy)

The first law of thermodynamics states that energy is conserved within a system. A system's energy can only change if energy crosses its boundaries.

For a closed system (fixed mass), there are only two mechanisms for energy transfer:

• Heat transfer across the system boundaries. 
• Work done by or on the system

These concepts form the standard mathematical statement of the first law for closed systems, as introduced in foundational thermodynamics courses:

Where ΔE_st^tot is the change in the total energy stored in the system, Q is the net heat transferred to the system, and W is the net work done by the system. This is schematically illustrated as follows: 

Figure 1

The first law of thermodynamics also applies to a control volume (open system), where mass can cross the system boundaries.

When mass enters or leaves a control volume, it carries energy with it — a process known as energy advection. Energy advection becomes a third mechanism for energy transfer in addition to heat transfer and work. Thus, for both closed systems and control volumes, the first law states that the change in system energy equals the net energy transferred across its boundaries.

First Law of Thermodynamics over a Time Interval (Δt)

A refresher for the 1st law of Thermodynamics:

1. The change in energy within a control volume is equal to the energy flowing into it minus the energy flowing out.

Energy can cross a control volume boundary through heat transfer, work, and energy advection (energy carried by mass flow). The first law of thermodynamics deals with total energy, which includes:

• Mechanical energy: kinetic + potential
• Internal energy: thermal energy plus chemical, nuclear, and other internal forms

In heat-transfer analysis, attention is mainly on thermal and mechanical energy. These two forms are not conserved on their own because they can be produced or consumed through conversions with other energy forms. Examples:

• Chemical reactions can decrease chemical energy and increase thermal energy.
• Electric motors convert electrical energy into mechanical energy.

These conversions can be viewed as thermal or mechanical energy generation (positive or negative). Therefore, a tailored form of the first law is needed for heat-transfer applications, accounting for these energy conversions.

Thermal and Mechanical Energy Equation over a Time Interval (Δt)

2. The rise in thermal and mechanical energy within a control volume equals the energy entering it, minus the energy leaving it, plus any thermal or mechanical energy produced inside the control volume.

This relation is written for a time period Δt, with all energy quantities expressed in joules. Because the first law must hold at every moment, it can also be written in terms of energy rates. In other words, at any instant, the rates of energy transfer must balance, with all terms expressed in joules per second (watts).

Thermal and Mechanical Energy Equation at an Instant (t)

3. The rate at which thermal and mechanical energy accumulates in a control volume equals the rate at which it enters, minus the rate at which it leaves, plus the rate at which it is produced inside the control volume.

If the combined inflow and generation of thermal and mechanical energy are greater than the outflow, the energy stored in the control volume will increase. If the outflow exceeds inflow and generation, the stored energy will decrease. When inflow and generation exactly match outflow, the system reaches a steady state, with no change in stored thermal and mechanical energy.

Let use now define the statement in italics and try to express it as an equation. let E stand for the sum of thermal and mechanical energy. Using the subscript st to represent energy stored in the control volume, the change in thermal and mechanical energy stored over the time interval Δt is then ΔE_st. The subscripts in and out refer to energy entering and leaving the control volume. Finally, thermal and mechanical energy generation is given the symbol E_g. Thus statement 1 is:

Statement 2 is represented as figure (b) in Figure 1 and is expressed as:

The above two equations are essential tools for solving heat transfer problems. Applying the first law begins with identifying an appropriate control volume and its control surface. The analysis typically follows a series of steps:

• the control surface is indicated, often by a dashed line;
• a decision is made whether to perform the analysis over a time interval Δt or on a rate basis, depending on the problem’s objectives and the form of the given data; finally,
• the relevant energy terms for the specific problem are identified.

The remainder of the section focuses on clarifying these energy terms to help develop confidence in applying them.

• Stored thermal and mechanical energy, E_st.
• Thermal and mechanical energy generation, E_g.
• Thermal and mechanical energy transport across the control surfaces, that is, the inflow and outflow terms, E_in and E_out.

the stored thermal and mechanical energy is given by:

E_st = Kinetic Energy (KE) + Potential Energy (PE) + thermal energy (Ut)

where Ut = Sensible Energy (U_sens) + Latent Enegry (U_lat) In many problems, the only relevant energy term will be the sensible energy, that is, E_st = U_sens.

The energy generation term is associated with conversion from some other form of internal energy (chemical, electrical, electromagnetic, or nuclear) to thermal or mechanical energy.

The inflow and outflow terms are surface phenomena. That is, they are associated exclusively with processes occurring at the control surface and are generally proportional to the surface area.

When the first law is applied to a control volume with fluid crossing its boundary, the work term is typically divided into two components. The first, called flow work, arises from pressure forces moving the fluid through the boundary and, for a unit mass, is equal to the product of the pressure and the fluid’s specific volume (pv). Under steady-state conditions (dEst/dt = 0), with no thermal or mechanical energy generation, the first law simplifies to the steady-flow energy equation. The equation for statement 2 becomes:

In most open system applications, changes in latent energy between the inflow and outflow conditions of the above equation may be neglected, thus the thermal energy reduces to only the sensible component.

If the fluid is approximated as an ideal gas with constant specific heats, the difference in enthalpies (per unit mass) between the inlet and outlet flows may then be expressed as:

(i_in  i_out) = c_p(T_in - T_out)

Where:

• c_p is the specific heat at constant pressure
• T_in and T_out are the inlet and outlet temperatures, respectively.
• If the fluid is an incompressible liquid, its specific heats at constant pressure and volume are equal, c_p = c_v = c,

Thus, for the above equation, the change in sensible energy (per unit mass) reduces to

(u_t,in - u_t,out) = c(T_in - T_out).

Unless the pressure drop is extremely large, the difference in flow work terms, (pv)_in = (pv)_out, is negligible for a liquid.

Having already assumed steady-state conditions, no changes in latent energy, and no thermal or mechanical energy generation, the simplified steady-flow thermal energy equation is obtained as follows:

SPECIAL MATHEMATICAL FUNCTIONS

Legendre’s Differential Equation

The differential equation, 

where n is a real constant known as Legendre’s equation of order n. When n is a nonnegative integer, i.e., n = 0, 1, 2, …., one solution of the above equation is called the Legendre polynomial (or Legendre function of the first kind) of degree n and is represented by P_n(x). The functions P_n(x) are expressed as follows:

Legendre polynomials are orthogonal with each other, i.e.,

The second solution of the Legendre equation is denoted by Q_n(x) and is called the Legendre function of the second kind of order n. The functions Q_n(x) are expressed as follows:

where A_n and B_n are suitably chosen constants. In particular:

Therefore, the general solution of Legendre’s differential equation is expressed as follows:

The substitution x = cos θ transforms Legendre’s equation into the following form:

The solution of Eq. (11) is given by

Legendre’s Associated Differential Equation

The differential equation:

where m and n are nonnegative integers, is known as Legendre’s associated equation of order n. Solutions of this equation are called associated Legendre functions. The general solution is:

where P_n^m (x) and Q_n^m (x) are called the associated Legendre function of the first kind and associated Legendre function of the second kind, respectively. Associated Legendre functions of the first kind are defined by:

Associated Legendre functions of the first kind are orthogonal with each other, i.e.,

Associated Legendre functions of the second kind are defined by:

The substitution x = cos θ transforms Legendre’s associated differential equation into the equation:

which is satisfied by P_n^m (cos θ) and Q_n^m (cos θ)

Hermite's Differential Equation

The differential equation:

where n is a real constant, is known as Hermite’s equation of order n. If n is a nonnegative integer, i.e., n = 0, 1, 2, ….. then solutions of Hermite’s equation are Hermite polynomials H_n(x) given by:

Hermite polynomials are orthogonal with each other, i.e.,

Laguerre's Differential Equation

The differential equation:

where n is a real constant, is known as Laguerre’s equation of order n. If n is a nonnegative integer, i.e., n = 0, 1, 2, ….., then solutions of Laguerre’s equation are Laguerre polynomials L_n(x) given by:

Laguerre polynomials are orthogonal with each other, i.e.,

Chebyshev's Differential Equation

The differential equation:

is known as Chebyshev’s equation of order n. The general solution of Chebyshev’s differential equation is:

where T_n(x) and S_n(x) are called the Chebyshev polynomials of the first kind and Chebyshev polynomials of the second kind, respectively. Chebyshev polynomial of the first kind is defined by:

Chebyshev polynomials of the first kind are orthogonal with each other, i.e.,

Chebyshev polynomial of the second kind S_n(x) is defined by:

Chebyshev polynomials of the second kind are orthogonal with each other, i.e.,

Solved Examples For Vapor Compression Refrigeration Cycles

Note: For example 1, values are obtained from the following book:

G. van Wylen, R. Sonntag, C. Borgnakke, Fundamentals of Classical Thermodynamics, 4th Ed., Wiley, 1994

Example 1

A conventional refrigerator uses HFC-134a as its working fluid. Saturated vapor at − 20°C leaves the evaporator and enters the insulated compressor whose compression ratio is 9:1, i.e., the outlet pressure is 9 times the inlet pressure. The compressor is 85% efficient based on an adiabatic reversible compression over the same pressure range. The gas leaving the compressor is cooled and condensed isobaricly to a saturated liquid. The saturated liquid is passed through an insulated throttle valve whose downstream pressure corresponds to a saturation temperature of − 20°C. The mixture from the throttle valve goes to the evaporator, where it absorbs just enough heat from the ice trays and the interior of the refrigerator box to become saturated vapor, which enters the compressor and repeats the cycle. Under summertime conditions (greatest load) it is expected that 120 kJ/min will have to be absorbed in the evaporator.

a) What is the rate of circulation of the refrigerant in kg/min?

b) Determine the power of the compressor.

Solution

The cycle can be schematically represented as follows:

a) Since the rate of heat that must be removed from the low temperature region, Q_c is given, then:

State: 1

At the exit of the evaporator, we have saturated vapor at − 20°C. From HFC-134a table we obtain:

State: 3

At the exit of the condenser, we have saturated liquid. The pressure is

P₃ = P₂ = (9)(0.1337) = 1.2 MPa

Hence,

State: 4

The energy balance around the throttling valve gives:

                                                 

Thus, the circulation rate can be determined from Eq. (1) as follows:

b) The pressure at state 2 is 1.2 MPa. If the compression takes place reversibly and adiabatically:

Example 2

A heat pump is being used to maintain a room at 21°C by removing heat from groundwater and discharging heat to the room. For the cycle shown in the figure below:

a) What is the quality of the fluid leaving the evaporator?

b) What is the maximum quantity of heat that can be delivered to the room per J removed from the groundwater?

c) If the room is to receive 1.7 kW, what is the minimum horsepower rating of the motor driving the compressor?

d) What is the minimum energy that would have to be supplied to provide 1.7 kW into the room using the groundwater as a source?

The fluid properties are as follows:

Solution

a) Since compressor operates reversibly and adiabatically, then:

b) The entropy at state 4 is:

c) The circulation rate of the refrigerant can be calculated from:

d) Power required will be minimum when a Carnot refrigeration cycle is to be operated between evaporator and condenser temperatures. The coefficient of performance of such a heat pump is: