ENERGY BALANCE FOR CLOSED SYSTEMS

The First Law and Other Basic Concepts

2.4 ENERGY BALANCE FOR CLOSED SYSTEMS

If the boundary of a system does not permit the transfer of matter between the system and its surroundings, the system is said to be closed, and its mass is necessarily constant. The devel- opment of basic concepts in thermodynamics is facilitated by a careful examination of closed systems. For this reason they are treated in detail here. Far more important for industrial prac- tice are processes in which matter crosses the system boundary as streams that enter and leave process equipment. Such systems are said to be open, and they are treated later in this chapter, once the necessary foundation material has been presented.

Because no streams enter or leave a closed system, no energy associated with matter is transported across the boundary that divides the system from its surroundings. All energy exchange between a closed system and its surroundings is in the form of heat or work, and the total energy change of the surroundings equals the net energy transferred to or from it as heat and work. The second term of Eq. (2.1) can therefore be replaced by variables representing heat and work, to yield

Δ(Energy of surroundings) = ±Q ± W

Heat Q and work W always refer to the system, and the choice of sign for numerical values of these quantities depends on which direction of energy transfer with respect to the system is regarded as positive. We adopt the convention that makes the numerical values of both quanti- ties positive for transfer into the system from the surroundings. The corresponding quantities

2.4. Energy Balance for Closed Systems 27 taken with reference to the surroundings, Q_surr and W_surr, have the opposite sign, i.e., Q_surr = −Q and W_surr = −W. With this understanding:

Δ(Energy of surroundings) = Q surr + W surr = − Q − W Equation (2.1) now becomes:²

Δ ( Energy of the system ) = Q + W (2.2) This equation states that the total energy change of a closed system equals the net energy transferred into it as heat and work.

Closed systems often undergo processes during which only the internal energy of the system changes. For such processes, Eq. (2.2) reduces to:

Δ U ^t = Q + W (2.3)

where U^t is the total internal energy of the system. Equation (2.3) applies to processes of finite change in the internal energy of the system. For differential changes in U^t:

dU ^t = dQ + dW (2.4)

In Eqs. (2.3) and (2.4) the symbols Q, W, and U^t pertain to the entire system, which may be of any size, but must be clearly defined. All terms require expression in the same energy units. In the SI system the unit is the joule.

Total volume V^t and total internal energy U^t depend on the quantity of material in a system, and are called extensive properties. In contrast, temperature and pressure, the principal thermodynamic coordinates for pure homogeneous substances, are independent of the quantity of material, and are known as intensive properties. For a homogeneous system, an alternative means of expression for the extensive properties, such as V^t and U^t, is:

V ^t = mV or V ^t = nV and U ^t = mU or U ^t = nU

where the plain symbols V and U represent the volume and internal energy of a unit amount of material, either a unit mass or a mole. These are specific or molar properties, respectively, and they are intensive, independent of the quantity of material actually present.

Although V^t and U^t for a homogeneous system of arbitrary size are exten- sive properties, specific and molar volume V  and specific and molar internal energy U are intensive.

Note that the intensive coordinates T and P have no extensive counterparts.

For a closed system of n moles, Eqs. (2.3) and (2.4) can now be written:

Δ(nU ) = n ΔU = Q + W (2.5)

d(nU ) = n dU = dQ + dW (2.6)

2The sign convention used here is recommended by the International Union of Pure and Applied Chemistry. How- ever, the original choice of sign for work and the one used in the first four editions of this text was the opposite, and the right side of Eq. (2.2) was then written Q − W.

In this form, these equations show explicitly the amount of substance comprising the system.

The equations of thermodynamics are often written for a representative unit amount of material, either a unit mass or a mole. Thus, for n = 1, Eqs. (2.5) and (2.6) become:

ΔU = Q + W and dU = dQ + dW

The basis for Q and W is always implied by the mass or number of moles associated with the left side of the energy equation. That is, Q and W may be expressed as total values for a system undergoing a particular change or for that process applied to a unit mass or mole of a sub- stance. The basis of the calculation is set by the number of moles or mass associated with properties of the system. These energy flows are then expressed on that basis.

These equations do not provide a definition of internal energy. Indeed, they presume prior affirmation of the existence of internal energy, as expressed in the following axiom:

Axiom 1: There exists a form of energy, known as internal energy U, which is an intrinsic property of a system, functionally related to the measurable coordinates that characterize the system. For a closed system, not in motion, changes in this property are given by Eqs. (2.5) and (2.6).

Equations (2.5) and (2.6) not only supply the means of calculating changes in internal energy from experimental measurements, but they also enable us to derive further property relations that supply connections to readily measurable characteristics (e.g., temperature and pressure). Moreover, they have a dual purpose, because once internal-energy values are known, they enable the calculation of heat and work quantities for practical processes. Having accepted the preceding axiom and associated definitions of a system and its surroundings, one may state the first law of thermodynamics concisely as a second axiom:

Axiom 2: (The First Law of Thermodynamics) The total energy of any system and its surroundings is conserved.

These two axioms cannot be proven, nor can they be expressed in a simpler way. When changes in internal energy are computed in accord with Axiom 1, then Axiom 2 is universally observed to be true. The profound importance of these axioms is that they are the basis for formulation of energy balances applicable to a vast number of processes. Without exception, they predict the behavior of real systems.³

Example 2.1

The Niagara river, separating the United States from Canada, flows from Lake Erie to Lake Ontario. These lakes differ in elevation by about 100 m. Most of this drop occurs over Niagara Falls and in the rapids just above and below the falls, creating a natural opportunity for hydroelectric power generation. The Robert Moses hydroelectric power plant draws water from the river well above the falls and discharges it well below them. It has a peak capacity of 2,300,000 kW at a maximum water flow of 3,100,000 kg·s⁻¹. In the following, take 1 kg of water as the system.

3For a down-to-earth treatment designed to help the student over the very difficult early stages of an introduction to thermodynamics, see a short paperback by H. C. Van Ness, Understanding Thermodynamics; DoverPublications.com.

2.4. Energy Balance for Closed Systems 29 (a) What is the potential energy of the water flowing out of Lake Erie, relative to

the surface of Lake Ontario?

(b) At peak capacity, what fraction of this potential energy is converted to electri- cal energy in the Robert Moses power plant?

(c) If the temperature of the water is unchanged in the overall process, how much heat flows to or from it?

Solution 2.1

(a) Gravitational potential energy is related to height by Eq. (1.8). With g equal to its standard value, this equation yields:

E P = mzg

= 1 kg × 100 m × 9.81 m⋅ s ⁻² = 981 kg⋅ m ² ⋅s ⁻² = 981 N⋅m = 981 J

(b) Recalling that 1 kW = 1000 J⋅s⁻¹, we find the electrical energy generated per kg water is:

2.3 × 10 ⁶  kW

_____________3.1 × 10 ⁶  kg⋅ s ⁻¹ = 0.742 kW⋅s⋅ kg ⁻¹ = 742 J⋅ kg ⁻¹

The fraction of the potential energy converted to electrical energy is 742/981 = 0.76.

This conversion efficiency would be higher but for the dissipation of potential energy in the flow upstream and downstream of the power plant.

(c) If the water leaves the process at the same temperature at which it enters, then its internal energy is unchanged. Neglecting also any change in kinetic energy, we write the first law, in the form of Eq. (2.2), as

Δ ( Energy of the system ) = Δ E p = Q + W For each kilogram of water, W = −742 J and ΔE_P = −981 J. Then Q = Δ Ep − W = − 981 + 742 = − 239 J

This is heat lost from the system.

Example 2.2

A typical industrial-scale wind turbine has a peak efficiency of about 0.44 for a wind speed of 9 m·s⁻¹. That is, it converts about 44% of the kinetic energy of the wind approaching it into usable electrical energy. The total air flow impinging on such a tur- bine with a rotor diameter of 43 m is about 15,000 kg·s⁻¹ for the given wind speed.

(a) How much electrical energy is produced when 1 kg of air passes through the turbine?

(b) What is the power output of the turbine?

(c) If there is no heat transferred to the air, and if its temperature remains unchanged, what is its change in speed upon passing through the turbine?

Solution 2.2

(a) The kinetic energy of the wind on the basis of 1 kg of air is:

E K 1 = 1_{__}

2 mu ² = (1 kg) (9  m·s _____________⁻¹ ) ²

2 = 40.5 kg· m ² ·s ⁻² = 40.5 J Thus, the electrical energy produced per kilogram of air is 0.44 × 40.5 = 17.8 J.

(b) The power output is:

17.8  J·kg ⁻¹ × 15,000 kg· s ⁻¹ = 267,000  J·s ⁻¹ = 267 kW

(c) If the temperature and pressure of the air are unchanged, then its internal energy is unchanged. Changes in gravitational potential energy can also be neglected. Thus, with no heat transfer, the first law becomes

Δ(Energy of the system) = Δ E K = E K ₂ − E K ₁ = W = −17.8 J⋅ kg ⁻¹

E K ₂ = 40.5 − 17.8 = 22.7 J⋅ kg ⁻¹ = 22.7 N⋅m⋅ kg ⁻¹ = 22.7  m ² ⋅s ⁻²

E K ₂ = u_{__ 2}²

2 = 22.7 m ² ⋅s ⁻² and u 2 = 6.74 m ⋅s ⁻¹ The decrease in air speed is: 9.00 − 6.74 = 2.26 m·s⁻¹.