Chapter 4

Energy Analysis of Closed Systems

Course Outcome

• Ability to acquire and explain the basic

concepts in thermodynamics.

• Ability to comprehend and apply the concept

to the actual conditions and problems; i.e. to the actual conditions and problems; i.e.

closed and open systems.

• Ability to apply and correlate the concept with

the appropriate equations and principles to

analyze and solve engineering problems.

Course Learning Outcomes

The student should be able to:

• State the first law of thermodynamics and identify it simply as statement of the conservation of energy.

• Explain the mechanisms of energy transfer for a closed system.

• Write down the energy balances for a closed system.

• Calculate the boundary work for isochoric, isobaric, isothermal and polytropic processes for closed systems.polytropic processes for closed systems.

• Define the specific heat at constant volume and the specific heat at constant pressure.

• Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases.

• Describe incompressible substances and determine the changes in their internal energy and enthalpy.

• Solve energy balance problems for closed systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.

4.1 First Law of Thermodynamics

4.2 General Energy Balance

4.3 Energy Balance for Closed Systems

4.4 Specific Heat4.4 Specific Heat

4.1 THE FIRST LAW OF THERMODYNAMICS• The first law of thermodynamics (the conservation of energy principle) provides

a sound basis for studying the relationships among the various forms of energy

and energy interactions.

• The first law states that energy can be neither created nor destroyed during a

process; it can only change forms.

• The First Law: For all adiabatic processes between two specified states of a closed

system, the net work done is the same regardless of the nature of the closed

system and the details of the process.

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Energy

cannot be

created or

destroyed;

it can only

change

forms.

The increase in the energy of a potato

in an oven is equal to the amount of

heat transferred to it.

The work

(electrical) done

on an adiabatic

system is equal to

the increase in

the energy of the

system.

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In the absence of any

work interactions, the

energy change of a

system is equal to the

net heat transfer.

The work (shaft)

done on an

adiabatic system is

equal to the

increase in the

energy of the

system.

4.2 General Energy Balance

The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process.

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Energy Change of a System, ∆Esystem:

Internal, kinetic, and

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Internal, kinetic, and

potential energy changes

4.3 Energy Balance of Closed System

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The work (boundary) done on an adiabatic

system is equal to the increase in the

energy of the system.

The energy change of

a system during a

process is equal to

the net work and

heat transfer

between the system

and its surroundings.

Energy balance when sign convention is used (i.e., heat input and work output

are positive; heat output and work input are negative).

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Various forms of the first-law relation for

closed systems when sign convention is

used.For a cycle ∆E = 0, thus Q = W.

The first law cannot be proven mathematically, but no process in nature is known to

have violated the first law, and this should be taken as sufficient proof.

Example 2.10 A rigid tank contains a hot

fluid that is cooled while

being stirred by a paddle

wheel. Initially, the internal

energy of the fluid is 800 kJ.

During the cooling process, During the cooling process,

the fluid loses 500 kJ of

heat, and the paddle wheel

does 100 kJ of work on the

fluid. Determine the final

internal energy of the fluid.

Neglect the energy stored in

the paddle wheel. 11

Energy balance for a constant-pressure expansion or

compression process:

HWUb

∆=+∆

For a constant-pressure expansion or

compression process:

An example of constant-pressure process

General analysis for a closed system

undergoing a quasi-equilibrium

constant-pressure process. Q is to

the system and W is from the

system.

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Problem 4.31

A substance is contained in a well-insulated

rigid container that is equipped with a stirring

device. Determine the change in the internal

energy of this substance when 15 kJ of work

is applied to the stirring device.is applied to the stirring device.

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Solution

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Problem 4.28

Saturated water vapor at 200oC is isobarically condensed

to a saturated liquid in a piston-cylinder device. Calculate

the heat transfer and the work done during this process in

kJ/kg.

Solution

4.4 Specific HeatSpecific heat at constant volume, cv: The energy required to raise the

temperature of the unit mass of a substance by one degree as the volume is maintained constant.

Specific heat at constant pressure, cp: The energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant.

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Specific heat is the energy

required to raise the

temperature of a unit mass of

a substance by one degree in a

specified way.

Constant-volume

and constant-

pressure specific

heats cv and cp

(values are for

helium gas).

• The equations in the figure are valid for any substance undergoing any process.

• cv and cp are properties.

• cv is related to the changes in internal energy and cp to the changes in enthalpy.

• A common unit for specific heats is kJ/kg · °C or kJ/kg · K. Are these units identical?

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The specific heat of a substance

changes with temperature.

True or False?

cp is always greater than cv.

Formal definitions of cv and cp.

INTERNAL ENERGY, ENTHALPY,

AND SPECIFIC HEATS OF IDEAL GASES

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Joule showed

using this

experimental

apparatus that

u=u(T)

For ideal gases, u,

h, cv, and cp vary

with temperature

only.

Internal energy and

enthalpy change of an

ideal gas

• At low pressures, all real gases approach ideal-

gas behavior, and therefore their specific heats

depend on temperature only.

• The specific heats of real gases at low pressures

are called ideal-gas specific heats, or zero-

pressure specific heats, and are often denoted

cp0 and cv0.

• u and h data for a number of gases

have been tabulated.

• These tables are obtained by

choosing an arbitrary reference

point and performing the

integrations by treating state 1 as

the reference state.

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Ideal-gas

constant-

pressure

specific heats

for some gases

(see Table A–

2c for cp

equations).

In the preparation of ideal-gas

tables, 0 K is chosen as the

reference temperature.

(kJ/kg)

Internal energy and enthalpy change when

specific heat is taken constant at an average

value

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For small temperature intervals, the specific

heats may be assumed to vary linearly with

temperature.

The relation ∆ u = cv ∆T is

valid for any kind of

process, constant-volume

or not.

1. By using the tabulated u and h data. This is the easiest and most accurateway when tables are readily available.

2. By using the cv or cp relations (Table A-2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate.

Three ways of calculating ∆∆∆∆u and ∆∆∆∆h

very accurate.

3. By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large.

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Three ways of calculating ∆∆∆∆u.

Specific Heat Relations of Ideal Gases

On a molar basis

The relationship between cp, cv and R

Specific heat

ratio

dh = cpdT and du = cvdT

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The cp of an ideal gas can be

determined from a knowledge of cv

and R.

ratio

• The specific ratio varies with

temperature, but this variation is very

mild.

• For monatomic gases (helium, argon,

etc.), its value is essentially constant at

1.667.

• Many diatomic gases, including air,

have a specific heat ratio of about 1.4

at room temperature.

Example 4.7

Air at 300 K and 200 kPa is heated at constant

pressure to 600 K. Determine the change in

internal energy of air per unit mass, using

(a) Data from the air table (Table A-17)(a) Data from the air table (Table A-17)

(b) The functional form of the specific heat (Table A-2c)

(c) The average specific heat value (Table A-2b)

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Example 4.8

An insulated rigid tank initially contains 0.7

kg of helium at 27oC and 350 kPa. A paddle

wheel with a power rating of 0.015 kW is

operated within the tank for 30 min. operated within the tank for 30 min.

Determine

(a) The final temperature

(b) The final pressure of the helium gas

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Example 4.10

A piston-cylinder device initially contains air at 150

kPa and 27oC. At this state, the piston is resting

on a pair of stops, and the enclosed volume is 400 L.

The mass of the piston is such that a 350 kPaThe mass of the piston is such that a 350 kPa

pressure is required to move it. The air is now

heated until its volume has doubled. Determine

(a) The final temperature

(b) The work done by the air

(c) The total heat transferred to the air

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INTERNAL ENERGY, ENTHALPY, AND

SPECIFIC HEATS OF SOLIDS AND LIQUIDS

Incompressible substance: A substance whose specific volume (or

density) is constant. Solids and liquids are incompressible substances.

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The specific volumes of

incompressible substances

remain constant during a

process.

The cv and cp values of incompressible

substances are identical and are

denoted by c.

Internal Energy Changes

Enthalpy Changes

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The enthalpy of a

compressed liquid

A more accurate relation than

Example 4.12

A 50-kg iron block at 80oC is dropped into

an insulated tank that contains 0.5 m3 of

liquid water at 25oC. Determine the

temperature when thermal equilibrium is temperature when thermal equilibrium is

reached.

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