Chapter 5

The First Law of Thermodynamics

THE FIRST LAW OF THERMODYNAMICS– FOR A CONTROL MASS UNDERGOING A CYCLE

– FOR A CHANGE IN STATE OF A CONTROL MASS

INTERNAL ENERGY– A THERMODYNAMIC PROPERTY

ENTHALPY – THE THERMODYNAMIC PROPERTY

SPECIFIC HEATS – CONSTANT-VOLUME

– CONSTANT-PRESSURE

INTERNAL ENERGY,ENTHALPY, AND SPECIFIC HEAT OF IDEAL GASES

THE FIRST LAW AS A RATE EQUATION CONSERVATION OF MASS

Introduction

THE FIRST LAW OF THERMODYNAMICS FOR A CONTROL MASS UNDERGOING A CYCLE

– The First Law is often called the

conservation of energy law.

– The first law of thermodynamics states

that during any cycle a system (control

mass) undergoes, the cyclic integral of

the heat is proportional to the cyclic

integral of the work.

THE FIRST LAW OF THERMODYNAMICS FORA CHANGE IN STATE OF A CONTROL MASS

Three observations should be made regarding this equation The first is that the property E, the

energy of the control mass, was found to exist.

– However, rather than deal with this property E, we find it more convenient to consider the internal energy and the kinetic and potential energies of the mass.

The second is that Eqs. 5.10 and 5.11 are in effect a statement of the conservation of energy.

– The net change of the energy of the control mass is always equal to the net transfer of energy across the boundary as heat and work.

– There are two ways in which energy can cross the boundary of a control mass—either as heat or as work.

The third is that Eqs. 5.10 and 5.11 can give only changes in energy.

We can learn nothing about absolute values of these quantities from these equations.– If we wish to assign values to internal energy, kinetic energy,

and potential energy, we must assume reference states and assign a value to the quantity in this reference state.

• The kinetic energy of a body with zero velocity relative to the earth is assumed to be zero.

• Similarly, the value of the potential energy is assumed to be zero when the body is at some reference elevation.

• With internal energy, therefore, we must also have a reference state if we wish to assign values of this property.

Internal energy, kinetic and potential energies are extensive properties, because they depends on the mass of the system.

U designates the internal energy of a given mass of a substance, and u as the specific internal energy.

The values are given in relation to an arbitrarily assumed reference state, which, for water in the steam tables, uf is taken as zero for saturated liquid at the triple-point temperature, 0.010C.

u = ( 1 –x ) uf + x ug u = uf + x ufg

5.3 INTERNAL ENERGY

Example 5.3

What is the control mass or control volume?

What do we know about the initial state (i.e., which properties are known)?

What do we know about the final state? What do we know about the process that

takes place? Is anything constant or zero? Is there some known functional relation between two properties?

5.4 PROBLEM ANALYSIS AND SOLUTION TECHNIQUE

Is it helpful to draw a diagram of the information in steps 2 to 4 (for example, a T .or P–v diagram)?

What is our thermodynamic model for the behavior of the substance (for example, steam tables, ideal gas, and so on)?

What is our analysis of the problem (i.e., do we examine control surfaces for various work modes, use the first law or conservation of mass)?

What is our solution technique? In other words, from what we have done so far in steps 1–7, how do we proceed to find whatever it is that is desired? Is a trial-and-error solution necessary?

Example 5.4

5.5 ENTHALPYLet us consider a control mass undergoing a quasi-equilibrium constant-pressure process, as shown in Fig. 5.6.

[ specific enthalpy, h, and total enthalpy, H. ]

(The significance and use of enthalpy is not restricted to the special process just described. Other cases in which this same combination of properties u +Pv appear will be developed later, notably in Chapter 6 in which we discuss control volume analyses.)

Students often become confused about the validity of this calculation when analyzing system processes that do not occur at constant pressure, for which enthalpy has no physical significance.

We must keep in mind that enthalpy, being a property, is a state or point function, and its use in calculating internal energy at the same state is not related to, or dependent on, any process that may be taking place.

Example 5.5

(Table at page 398, P2=400kPa,T 2=300oC)

5.6 THE CONSTANT-VOLUME AND CONSTANT-PRESSURE 5.6 THE CONSTANT-VOLUME AND CONSTANT-PRESSURE SPECIFIC HEATS SPECIFIC HEATS

5.7 THE INTERNAL ENERGY, ENTHALPY, and SPECIFIC HEAT of IDEAL GASES

For a low-density gas, however, u depends primarily on T and much less on the second property, P or v.

For example,

As gas density becomes so low that the ideal-gas model is appropriate, internal energy does not depend on pressure at all, but is a function only of temperature.That is, for an ideal gas,

P v = R T and u =f (T ) only

where the subscript 0 denotes the specific heat of an ideal gas.

(1)

Calculate the change of Calculate the change of enthalpyenthalpy ------ as 1 kg of ideal gas is heated from T1 to T2 K

Empirical Equation

(2)

(3)

Ideal-gas EntropyIdeal-gas Entropy (to integrate at constant pressure of 0.1 MPa)

Example 5.6

Calculate the change of enthalpy as 1 kg of oxygen is heated from 300 to 1500 K. Assume ideal-gas behavior.

Let us solve this problem in these ways and compare the answers.

(3)

Our most accurate answer for the ideal-gas enthalpy change for oxygen between 300 and 1500 K would be from the ideal-gas tables, Table A.8. This result is, using Eq. 5.29,

h2 - h1 = 1540.2 – 273.2 = 1267.0 kJ/ kg

Example 5.7

5.8 The First Law as a Rate Equation( Dividing the first law equation by δt )

5.9 Conservation of Mass

E = mc2

Ex.

2900 KJ = m x ( 2.9979x10-8 )2

m = 3.23 x 10 –11Kg