Thermo Lecture Notes CH2

Course Tutor: Jasim M. Mahdi / University of Baghdad

3-1

Energy Engineering Department (Fundamentals of Thermodynamics/ 2nd Year)

Energy and First Law Of Thermodynamics

Background

Some important thermodynamic quantities are called a point function which depends

on initial and final states of the process (not on the process path in which the system got to that

state).

For the point function quantity, the integral from point 1 to point 2 on the path A and the

integral from point 2 to point 1 through the path B are equal (exact differential).

2

Apath 1

1

Bpath 2

0dFdF

Such quantities are pressure, temperature, enthalpy, and entropy. Other thermodynamic

quantities are called a path function which depends on the process path (inexact differential).

2

A 1,

1

B2,

0FF

Such quantities are work and heat.

Note:

By way of notation, we will specify the use of d to denote an exact differential. In other words,

the integral of d will be equal to 2 1. The symbol will be reserved for an inexact

differential, which cannot be integrated without full knowledge of the path. For example W =

PdV will be used to denote an infinitesimal increment of work.

Th

erm

od

yn

amic

qu

anti

ty F

A

B

1

2


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Definition of Work

Work (W) is defined as a force (F) acting through a distance(s). This is a mechanical definition of

work.

2

1

21ds.FW

A more general definition of work is the thermodynamic definition: Work, an interaction

between a system and its surroundings, is done by a system on its surroundings if the sole effect

on everything external to the system could have been the raising of a weight.

Work is a path function, i.e. the value of it depends on the details of the interaction taking

place between the system and the surroundings during a process, e.g. F(s), and not just the

initial and final state. Differential of work is inexact, i.e. the following integral cant be

evaluated without knowing details of the process

2

1

WW not 12

WW 2

1

W

Thus, work is not a property of the system, since property is evaluated at a specific time and is

independent of the process.

Expansion and Compression Work

Consider the expansion of the gas in a piston-cylinder assembly (Pp is average pressure on

piston face)

2

1

2

1

pp

2

1

21dVPdx)AP(ds.FW


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For a slow or quasi-equilibrium process all the states through which the system passes are

considered equilibrium states and thus the intensive properties, i.e., pressure, are uniform

throughout the system Pp = Pgas , so

2

1

21PdVW

Note:

1.Work done BY the system (an increase in volume) is +ve

2.Work done ON the system (a decrease in volume) is ve

Graphical Interpretation:

W = PdV shaded area

2

1

2

1

21PdVW W total area under the curve

Work is a form of energy transfer


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Consider two processes with the same initial and final state

Since the area under each curve is different, the amount of work done for each path is different.

2path

V

V1path

V

V

2

1

2

1

PdVPdV

Work done depends on the path taken and not just the value of the end states.

Work is not a property!

The work done for the cycle (1-2-1) is

1

2

2

1

V

V

V

V

1221 PdVPdVWWW

The work done per unit mass of a system is denoted by w and is expressed as

m

Ww

kg

kJ

The work done per unit time is called power and is denoted by .

The unit of power is kW or hp, (1 hp = 0.746 kW)

Thermodynamic Processes

Thermodynamic process may be one of the following processes:

1. Isobaric process (P = C)

2. Isothermal process (T = C)

3. Isometric process (V = C)

4. Adiabatic process (Q = 0)

5. Polytrophic process ( PVn = C)

6. Isentropic process (S = C)


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a. Constant pressure process (isobaric process)

P = C )VV(PdV12

2

1

2

1

21PPdVW

b. Constant volume process (isometric process)

V = C, i.e. dV = 0 2

1

210PdVW

c. Constant temperature process (isothermal process)

T = C PV= C

2

1

11

1

2

11

2

111

2

1

2

1 P

PlnVP

V

VlnVPVlnVP

V

dVCdV

V

C

2

1

21PdVW

1

21 ln

V

VmRTW21 (for ideal gas)

d. Adiabatic process

Q= 0 PVk = C , v

p

gasesfor C

Ck

k

VVPVVP

k

CVCV

k

VC

V

dVCdV

V

CPdVW

kkkkkkk

k

2

1

21

111

1

111

1

222

1

1

1

2

2

1

12

1

2

1

k

VPVP

1

1122

k

RTRTmW

1

1221 (for ideal gases)

e. Polytrophic process (PVn = C)

n1

VPVPW 1122

21

n

RTRTmW

1

1221 (for ideal gases)


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Other Types of Work

1. Gravitational Work:

Gravitational work (Wg) is the work done by or against the gravitational force field.

2

1

2

1

gdZmgdZ.FW

E.PZZmgW12g

2. Accelerational Work:

Accelerational work (Wa) is the work associated with the change in velocity.

dt

dcmmaF ; since

dt

dsc cdtds

2

1

22

1

2

1

a2

cmcdt

dt

dcmds.FW

21

2

2accm

2

1W = E.K

3. Shaft Work:

Shaft Work (Wsh) is the energy transferred by a rotating shaft.

= Fr r

F

; is the torque

s = 2 n r; n is the no. of revolutions per unit time

Wsh = F. s = r

. 2 n r = 2 n

3. Electrical Work:

We = IVt

Note, I is the current & V is the potential difference

Z

m

Z1

Z2

+ve system falls -ve system raises


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5. Spring Work:

F Spring force = kx

k Spring constant (N/m)

x Displacement from relaxed position

2

1

22

1

2

1

sp2

xkdx.kxdx.FW

2122sp xxk2

1W

Example:

A pistoncylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a

linear spring that has a spring constant of 150 kN/m is touching the piston but

exerting no force on it. Now heat is transferred to the gas, causing the piston

to rise and to compress the spring until the volume inside the cylinder

doubles. If the cross-sectional area of the piston is 0.25 m2, determine (a) the

final pressure inside the cylinder, (b) the total work done by the gas, and (c)

the fraction of this work done against the spring to compress it.

Solution:

F

=

x


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Example:

A gas contained within a pistoncylinder assembly

undergoes a thermodynamic cycle consisting of three processes:

Process 12: Compression with PV=constant from P1 =1 bar, V1

= 1.0 m3 to V2 = 0.2 m

3

Process 23: Constant-pressure expansion to V3 = 1.0 m3

Process 31: Constant volume

Sketch the cycle on a PV diagram, and determine the work for each process and the net work

for the cycle, each in kJ.

Solution:

For process 1-2 : PV = const.

P2 = (V1/V2) P1 = 5 bar

1

211

V

V

21V

VlnVPPdVW

2

1

-160.9 kJ

For process 2-3 : P = const.

40232323

2

VVPPdVWV

V

kJ

For process 3-1 : V = const.

013 W

Thus;

9.120409.160133221 WWWWnet kJ

Gas

2 3

1 0.8 0.6 0.4 0.2

2

3

4

5

P (

bar)

V (m3)

1 1


3-9


Heat

Heat (Q) is defined as the form of energy that is transferred between two systems (or a system

and its surroundings) due to a temperature difference. The transfer is always toward the lower

temperature. Like work, heat is not a property and the amount of energy transfer depends on

the process details (path function), therefore

2

1

21 QQ

Heat transfer per unit mass of a system is denoted q and is determined from

m

Qq

kg

kJ

The rate of heat transfer is denoted by .

Q (kW)

The total energy transferred via heat over a period of time is t2

t1

.

21 dtQQ

Note:

1.Heat received by the system is +ve

2.Heat rejected from the system is ve

Types of Heat Transfer

1. conduction heat transfer

2. convection heat transfer

3. radiation heat transfer

System

+Q

-W

+W

-Q


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First Law of Thermodynamics

The first law of thermodynamics state that during any cycle the net heat transfer is

equal to the net work done. This is expressed in equation form by

WQ or WQ

where the symbol implies an integration around a complete cycle.

The first law can be illustrated by considering the following experiment. Let a weight be

attached to a pulley-paddle-wheel setup, such as that shown. Let the weight fall a certain

distance thereby doing work on the system, contained in the tank shown in Fig. a., equal to the

weight multiplied by the distance dropped. The temperature of the system (the fluid in the

tank) will immediately rise an amount T. Now, the system is returned to its initial state (the

completion of the cycle) by transferring heat to the surroundings, as implied by the Q in Fig. b.

This reduces the temperature of the system to its initial temperature. The first law states that

this heat transfer will be exactly equal to the work which was done by the falling weight.

Statement of the First Law for a Closed System

Consider a process A in the figure shown, the system is

returned to its initial state by two arbitrary processes B and C.

From the first law of thermodynamics

WQ

Considering the cycle 1-A-2-B-1, we have

Bpath

1

2

Apath

2

1

Bpath

1

2

Apath

2

1

)W()W()Q()Q( -------- (1)

Considering the cycle 1-A-2-C-1, we have

Cpath

1

2

Apath

2

1

Cpath

1

2

Apath

2

1

)W()W()Q()Q( -------- (2)

Now subtract Eq. (2) from Eq. (1) yields

(a) (b)

Q

W

W

Gas Gas


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Cpath

1

2

Bpath

2

1

)W-Q()W-Q(

Because path B and C are arbitrary between state 1 and 2, we conclude the quantity(Q - W)

depends only on the initial and final states of the system that means the difference in two path

functions Q and W is a point function, and therefore is a property of the system. This property

is called the energy of the system and given by the symbol E

dE = Q W

for a process 1 to 2 : 1E2 = E2-E1= 1Q2 - 1W2

1Q2 and 1W2 represent the heat and work transferred during the process 12.

(E2-E1) represent the stored energy in the system at the beginning and the end of the process

and which consist of system kinetic, potential and internal energy.

If a closed system undergoes a process from state 1 to state 2, then

1E2 = 1KE2 + 1PE2 + 1U2 = 1Q2 1W2

The increase in the KE + increase in PE + increase in internal energy = heat transferred into

the system work done by the system

Most systems encountered in practice are stationary, that is, they do not involve any changes in

their velocity or elevation during a process. Thus, for stationary systems, the changes in

kinetic and potential energies are zero (that is, KE = KE = 0), and the above equation

reduces to

1U2 = U2 U1 = 1Q2 1W2

The above results prove that the internal energy is a property and the change in its value for

closed system should be equal to the difference between the heat supplied and the work done

(Q W) during the change in state. i.e.:

1Q2 1W2 = U2 U1 (kJ)

1q2 1w2 = u2 u1 (kJ/kg)

dt

dUW

2

.

12

.

1 Q (Watt)

All these equations are different forms of First Law of Thermodynamics for a closed system.

They state that as a closed system undergoes a change of state, internal energy may cross the

boundary as either work or heat, and each may be positive or negative.


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Notes on heat and work

Heat and work are energy transfer mechanisms.

System posses energy, they do not contain heat or work !

Heat and work can be identified with a process, but cannot be identified with a state.

Example:

Suppose that 180 kJ of heat is added to a closed system executing a process from 1 to 2

in which the internal energy is increased by 100 kJ. To restore the closed system to its initial

state, 95 kJ of work must be done on the system. What is 2Q 1?

Solution :

1Q 2 = 180 kJ U = U2 U1 = 100 kJ 2W 1 = - 95 kJ

The general statement of the first law for closed system is

WQ

i.e., (1Q 2 + 2Q 1 ) = (1W 2 + 2W 1 )

2Q 1 = 1W 2 + 2W 1 + 1Q 2

For process 12 :

1Q 2 - 1W 2 = U

80 kJW100 W180 2121

2Q 1 = 1W 2 + 2W 1 + 1Q 2 = 80 + (-95) + 180 = -195 kJ


3-13


Energy exchange in a cycle of a closed system

For a closed system undergoing a cycle, the initial and final states are identical, and thus

(U)cycle= 0. Noting that a closed system does not involve any mass flow across its boundaries,

the energy balance for a cycle can be expressed in terms of heat and work interactions as

outincycle

cyclecycle

QQWQW

That is, the net work output during a cycle in a closed system is equal to net heat input.

When the net work for a cycle is positive ( Wcycle > 0 Qin > Qout ) ,it is referred to as a

Power Cycle

Wcycle = Qin - Qout (power cycle)

When the net work for the cycle is negative ( Wcycle < 0 Qin < Qout ) ,it is referred to as a

Refrigeration or Heat Pump Cycle

Wcycle = Qout Qin (refrigeration and heat pump cycles)

Schematic diagrams (a) power cycle & (b) refrigeration and heat pump cycles


3-14


Tutorial # (4)

Problem1

A gas is combined in a closed rigid tank. An electric resistor in the tank transfers

energy to the gas at a constant rate of 1000W. Heat transfer between the gas and the

surroundings occurs at a rate of , where is in Watts, and t is time in minutes.

a) Plot the time rate of change of energy of the gas for 0 t 20 min, in watts.

b) Determine the net change in energy of the gas after 20 min, in kJ.

c) If electricity is valued at $0.08 per kWh, what is the cost of the electrical input to the

resistor for 20 min of operation?

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Problem 2

Nitrogen is contained within a piston-cylinder assembly is shown in the following

figure. Initially, the piston face is at x = 0, the volume is 0.0006 m3 and the spring exerts no

force on the piston. As a result of heat transfer, the gas

expands, raising the piston until it hits the stops. At this

point, the piston face is located at x = 0.06 m and he

heat transfer ceases. The force exerted by the spring on

the piston as the gas expands varies linearly with x

according to:

F spring = kx

where k = 9 000 N/m. Friction between the piston and

the cylinder wall can be neglected. The acceleration of

gravity is g = 9.81 m/s2. Additional information can be

found on the figure. Use ideal gas model, find

(a) What is the initial pressure of the gas, in kPa?

(b) Determine the work done by the gas on the piston, in J.

(c) If the specific internal energies of the gas at the initial and final states are 210 and 335 J/kg,

respectively, calculate the heat transfer, in J.


3-15


H.W # (4)

Q1: Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 kJ of heat is transferred to

the water, and 5 kJ of heat is lost to the surrounding air. The paddle-wheel

work amounts to 500 N.m .Determine the final energy of the system if its

initial energy is 10 kJ.

Ans.: 35.5 kJ

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q2: A gas undergoes a thermodynamic cycle consisting of three processes Process 1-2 compression with pV= const, from p1= 1 bar, V1= 1.6 m

3 to V2= 0.2 m

3, U2 - U1= 0

Process 2-3 constant pressure to V3=V1

Process 3-1 constant volume, U1 - U3 = -3549 kJ

There are no significant changes in KE and PE.

a) Determine the heat transfer and work for process 2-3, in kJ.

b) Is this a power cycle or a refrigeration cycle, why?

Ans.: 1120 kJ, 4669 kJ ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q3: A closed system undergoes a process during which there is energy transfer from the system by heat at a constant rate of 10 kW, and the power varies with time according to

{

where t is time, in h, and is in kW. (a) What is the time rate of change of system energy at t = 0.6 h, in kW?

(b) Determine the change in system energy after 2 h, in kJ.

Ans.: -5.2 kW, -28800 kJ ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q4: Warm air is contained in a piston-cylinder assembly oriented horizontally as shown in the figure.

The air cools slowly from an initial volume of 0.003 m3

to a final volume of 0.002 m3. During the process, the

spring exerts a force that varies linearly from an initial

value of 900 N to a final value of zero. The atmospheric

pressure is 100 kPa and the area of the piston face is

0.018 m2. Friction between the piston and cylinder wall can be neglected. For the air, determine the

initial and final pressures, in kPa, and the work, in kJ

Ans.: 150 kPa, 100 kPa, -0.125 kJ ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Q5: 5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energy is u1=2709.9 kJ/kg,

to state 2, where u2= 2659.6 kJ/kg. During the process, there

is heat transfer to the steam with a magnitude of 80 kJ. Also,

a paddle wheel transfers energy to the steam by work in the

amount of 18.5 kJ. There is no significant change in the

kinetic or potential energy of the steam. Determine the

energy transfer by work from the steam to the piston during

the process, in kJ.


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Ans.: 350 kJ

Thermo Lecture Notes CH2

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