Thermodynamics I

Spring 1432/1433H (2011/2012H)

Saturday, Wednesday 8:00am - 10:00am &

Monday 8:00am - 9:00am

MEP 261 Class ZA

Dr. Walid A. AissaDr. Walid A. Aissa

Associate Professor, Mech. Engg. Dept.

Faculty of Engineering at Rabigh, KAU, KSA

Chapter #1

September XX, 2011

Announcements:Dr. Walid’s e-mail and Office Hours

walid_aniss@yahoo.com

Office hours for Thermo 01 will be every Sunday and Tuesday from 9:00 – 12:00 am

Dr. Walid’s office (Room 5-213)in Dr. Walid’s office (Room 5-213).Text book:

Thermodynamics An Engineering Approach

Yunus A. Cengel & Michael A. Boles7th Edition, McGraw-Hill Companies,

ISBN-978-0-07-352932-5, 2008

Chapter 1INTRODUCTION AND BASIC

CONCEPTS

Objectives of CH1: To

• Identify unique vocabulary associated with thermodynamics.

• Review metric SI units.

• Explain basic concepts of thermodynamics (system, state, state postulate, equilibrium, (system, state, state postulate, equilibrium, process, cycle).

• Review concepts of temperature, temperature scales, pressure, and absolute and gage pressure.

Engineeringsciences

1) Thermodynamics

Fluid Mechanics

Heat & mass transfer

…….

Chapter 1

INTRODUCTION AND BASIC CONCEPTS

1–1 ■ THERMODYNAMICS AND ENERGY

The name thermodynamics stems from the Greek

words therme (heat) and dynamis (power),

One of the most fundamental laws is the One of the most fundamental laws is the

conservation of energy principle. It simply states

that during an interaction, energy cannot be

created or destroyed.

Fig. 1–1 ■ Energy cannot be created or destroyed

Change in energy content of a body (or any other

System) is equal to difference between energy input

and energy output, energy balance is expressed as

Ein - Eout = ∆∆∆∆E.

First law of thermodynamics is simply expression of

conservation of energy principle, and it asserts that conservation of energy principle, and it asserts that

energy is a thermodynamic property.

Second law of thermodynamics asserts that

energy has quality as well as quantity, and actual

processes occur in the direction of decreasing quality

of energy. For example, a cup of hot coffee left on a

table eventually cools, but a cup of cool coffee in the

same room never gets hot by itself .

1–2 ■ UnitsPrimary (Fundamental) (Main) (base) quantities: SI Units

S Quantity Designation Unit Symbol

1 mass m kilogram kg

2 length L meter m

3 time t second s

4 temperature T Kelvin K

Secondary (Derived) quantities : SI Units

s Quantity Designation& Equation Unit

1 velocity V = L/t m/s

2 Acceleration a = L/t2 m/s2

3 Volume V = L3 m3

* g (Gravitational acceleration) = 9.807 m/s2 ≈ 9.81 m/s2

SI Unit Prefixes

Factor Prefix Symbol Factor Prefix Symbol

1012 tera T 10-1 deci d

109 giga G 10-2 centi C

106 mega M 10-3 milli m106 mega M 10-3 milli m

103 kilo k 10-6 micro µµµµ

102 hecto h 10-9 nano n

101 deka da 10-12 pico p

Force

1N = Kg m/s2 , Newton

s Quantity Designation& Equation SI Symbol

4 Force F = m a Kg m/s2

Secondary quantities : SI Units: (Continued)

1 kgf = 9.807 N

Weight is gravitational force applied to

a body, its magnitude is determined from Newton’s second law

W = m g (N) (1-2)

Mass of a body; m remains the same

regardless of its location in universe.

However, Its weight; W , changes with change in gravitational acceleration, g.

For a body on sea level ( g = 9.807

m/s2 ) having a mass; m of 1 kg, Its

weight; W = 1 kg× 9.807 m/s2 =9.807 kg m/s2 =9.807 N=1 kgf.

Work, which is a form of energy, can

simply be defined as force times

distance. Therefore, it has the unit

“Newton-meter (N · m),” which is called a joule (J).a joule (J).

1 J = 1 (1-3)

A more common unit for energy in SI is the kilojoule (1 kJ = 103 J).

1 Cal = 4.1868 J

Another energy unit is calorie Cal).

Dimensional Homogeneity

In engineering, all equationsmust be dimensionally homogeneous.

EXAMPLE 1–1 Spotting Errors from Unit Inconsistencies.Unit Inconsistencies.

While solving a problem, a person

ended up with the following equation

at some stage:

Where, E is the total energy and has

the unit of kilojoules. Determine how

to correct the error and discuss what

may have caused it.

E = 25 kJ + 7 kJ/kg

may have caused it.

Solution: The two terms on RHS do

not have the same units. Therefore,

they cannot be added to obtain the

total energy.

Multiplying ast term by mass will

eliminate the kilograms in the

denominator. The whole Eq. will

become dimensionally homogeneous;

i.e., every term in the Eq. will have the

same unit.same unit.

Obviously this error was caused by

forgetting to multiply the last term

by mass at an earlier stage.

EXAMPLE 1–3 What is the weight of a mass of 454 grams in N.

Solution

W = m g = (454/1000) kg ∗ 9.807 m/s2 W = m g = (454/1000) kg ∗ 9.807 m/s2

= 4.45 N.

1–3 ■ SYSTEMS AND CONTROL

VOLUMES

A system is defined as a

quantity of matter (or a

region in space) chosenfor study.for study.

The mass (or region)outside the system

is called the surroundings.

The real (or imaginary) surface that

separates the system from itssurroundings is called the boundary.

System

Closed system (Control mass)

Open system (Control volume)

Closed system:

Energy, in the form of

heat or work, can cross

the boundary; and the

volume of a closed system

If, as a special case, even energy is not allowed to cross the boundary, that system

is called an isolated system.

volume of a closed system

does not have to be fixed.

A closed system with a moving boundary.

* A large number of engineering problems

involve mass flow in and out of a system and,

therefore, are modeled as control volumes

(CV).

* A water heater, car radiator, turbine, and compressor all involve mass flow and should

(Open systems)

compressor all involve mass flow and should be analyzed as control volumes (open systems) instead of as control masses(closed systems).

* CV can also involve heat and work

interactions (just as a closed system), in

addition to mass interaction

An open

system (CV)

with one inlet

and one exit.and one exit.

CV with fixed

and moving

boundaries.

Boundaries of CV (called control surfaces) may be fixed, moving, real, and imaginary.

CV with real

and

imaginaryimaginary

boundaries.

Boundaries of CV (called control surfaces) may be fixed, moving, real, and imaginary.

In general, proper choice of the system may

greatly simplify the analysis.

1–4 ■ PROPERTIES OF A SYSTEMAny C/C of a system is called a property

Intensiveproperties {those

independent on

Extensiveproperties {those

dependent on the independent on

the size or extent

of the system}:

such as pressure

p, temperature T,

density; ρρρρ)

dependent on the

size or extent of

the system}: such

as mass; m and

volume; V)

Density; ρ is mass of a unit volume of a substance

ρ = m /V (Kg/m3) (1-4)

1–5 ■ DENSITY, SPECIFIC VOLUME, SPECIFIC GRAVITY & SPECIFIC WEIGHT

(m3/kg) (1-5)

Specific volume; v is volume per unit

mass (reciprocal of density)

ρ = m /V (Kg/m3) (1-4)

v = V /m

Specific gravity; SG (relative density)

defined as the ratio of the density of a

substance to the density of water at 4°C, ( ρH2O = 1000 kg/m3)

(1-6)

)at 4°C

(N/m3) (1-#1)

Specific weight; γγγγ is weight of a unit volume of a substance

γγγγ = W /V= mg/V= ρρρρg

)at 4°C

EXAMPLE 1–2 Obtaining Formulas from Unit Considerations.

A tank is filled with oil whose density is

ρ = 850 kg/m3. If the volume of thetank is V = 2 m3, determine the amount tank is V = 2 m , determine the amount of mass m in the tank..

Solution:

m = ρρρρ V= (850 kg/m3)∗ (2 m3)

= 1700 kg

1–7 ■ PROCESSES AND CYCLESAny change that a system undergoes from

one equilibrium state to another is called a

process.

A process between states 1 and 2 and the process path.

The series of states through which a system passes during a process is called

the path of the process.

To describe a process completely, one

shouldshould

(1) specify the initial and final states of the

process, as well as

(2) the path it follows, and

(3) the interactions with the surroundings.

When a process

proceeds in such a

manner that the

system remains

infinitesimally close

to an equilibrium to an equilibrium

state at all times, it

is called a quasi-static, or quasi-

equilibrium, process.

Quasi-equilibrium and

non quasi-equilibrium

compression processes

Process diagrams plotted by employing

thermodynamic properties as coordinates r

very useful in visualizing the processes.

Some common properties used as

coordinates r temperature; T, pressure ;p, coordinates r temperature; T, pressure ;p,

and volume; V (or specific volume; v).

p-V diagram of a compression process of a

gas is illustrated.

The P-V

diagram of a

compression

process.

The prefix iso- is often used to designate

a process for which a particular property

remains constant.

An isothermal process, for example, is a process during which the temperature; T

remains constant.remains constant.

An isobaric process is a processduring which the pressure; p remainsconstant.

An isochoric (or isometric) process is a

process during which the specific

volume; v remains constant.

A system is said to have undergone a

cycle if it returns to its initial state

at the end of the process. i.e.,

for a cycle initial & final states r identical.

3

cycle

&Final

state

The Steady-Flow Process

The term steady implies no change with

time. The opposite of steady is unsteady,

or transient.

A large number of engineering devices

operate for long periods of time under the

same conditions, and they r classified as

steady-flow devices.

steady-flow (Continued) That is, fluid

properties can change from point to point

within CV, but at any fixed point they

remain the same

during the entire

Process.

#The term uniform, however, implies nochange with location over a specified region.

When a body is brought into contact with

another body that is at a different

temperature, heat is transferred from the

body at higher temperature to the one at

lower temperature until both bodies attain the

1–8 ■ TEMPERATURE AND ZEROTH

LAW OF THERMODYNAMICS

lower temperature until both bodies attain the

same temperature.

At that point, the heat transfer stops, and the

two bodies r said to have reached thermal

equilibrium. The equality of temperature is

the only requirement for thermal equilibrium.

The zeroth law of thermodynamics

states that if two bodies are in thermalequilibrium with a third body, they are also

in thermal equilibrium with each other.

# By replacing the third body with a # By replacing the third body with a

thermometer, the zeroth law can be

restated as:

two bodies r in thermal equilibrium ifboth have the same temperature reading even if they r not in contact.

Temperature Scales

a) absolute thermodynamic scale: K (SI units)

b) common scale: °C (SI units)

( ) ( ) 15.273+°= CTKT

#When we r dealing with temperature

differences ∆T, the temperature interval is

the same.

# Some thermodynamic relations involve the

temperature T (such as a = bT),and often

the question arises of whether it is in K or °C.

∆T (K) = ∆T (°C).

the question arises of whether it is in K or °C.

# If the relation involves temperature

differences (such as a = b∆∆∆∆T), it makes no

difference and either (K or °C) can be used.

EXAMPLE 1–4 Expressing Temperature Rise in Different UnitsDuring a heating process, the temperature of a system rises by 10°C. Expressthis rise in temperature in K.

Solution:

∆∆∆∆T (K) = ∆∆∆∆T (°C) = 10

EXAMPLE 1–4(B) Expressing Temperature value in Different UnitsDuring a heating process, the temperature of a system is 10°C. Express the temperature in K,.

Solution:

T (K) = T (°C) +273.15 =10 + 273.15=283.15T (K) = T (°C) +273.15 =10 + 273.15=283.15

1-9 ■ Pressure

Pressure is defined as a normal

force exerted by a fluid per unit

area. We speak of pressure only when we deal with a gas or a liquid.,

The counterpart of pressure in solids

is normal stress. Since pressure is

defined as force per unit area, it has

the unit of newtons per square meter(N/m2), which is called a pascal (Pa).

Multiples of Pa, r:

kilopascal (1 kPa = 103 Pa) and

megapascal (1 Mpa = 106 Pa)

r commonly used.

Three other pressure units commonlyThree other pressure units commonly

used in practice, r :

- bar (bar), - standard atmosphere (atm),

& kilogram-force per square centimeter

(kgf/cm2).

1 bar = 105 Pa = 0.1 MPa = 100 kPa

1 atm =101,325 Pa =101.325 kPa =1.01325 bars

1 kgf /cm2 = 9.807 N/cm2 = 9.807 ×104 N/m2

= 9.807 ×104 Pa = 0.9807 bar = (0.9807/ 1.01325 ) atm = 0.9679 atm

×

×

1.01325 ) atm = 0.9679 atm

EXAMPLE 1–#1

Calculate p)fat man & p)thin man

If m)fat man =136 kg & m)thin man=68 kg, A)feet= 0.0326 m2

Solution:

68 kg 136 kg68 kg 136 kg

p)fat man = 40.93 kPa

p)thin man =

20.46 kPa

# W)fat man = m g = 136 kg ∗ 9.81

m/s2 = 1334.2 N.

p)fat man = W)fat man / A = 1334.2 N /0.0326 m2 = 40.93 3 Pa= 40.93 kPa.

# W)thin man = m g = 68 kg ∗ 9.81 m/s2 = 667.1 N.

p)thin man = W)thin man / A = 667.1 N /0.0326 m2 = 20.46 3 Pa= 20.46

kPa.

Absolute, gage, and vacuum pressures

Movable datum

(zero gage)

Pgage

Pvac

Fixed datum

(zero absolute)

PatmPabsPabs

Pabs = Patm + Pgage

= Patm - Pvac

1 atm = 101,325 Pa = 101.325 kPa =1.01325 bars

=14.7 psi.

EXAMPLE 1–5 Absolute Pressure of a Vacuum Chamber

A vacuum gage connected to a chamber

reads 5.8 psi at a location where the

atmospheric pressure is 14.5 psi.

Determine the absolute pressure in theDetermine the absolute pressure in the

Chamber in psi, atm and bar.

Solution:

Pabs = Patm - Pvac = 14.5 psi - 5.8 psi = 8.7 psi

= (8.7/14.7) atm = 0.5918 atm =

8.7*(1.01325 / 14.7) = 0.6 bar

Variation of Pressure with DepthPressure in a fluid at rest does not change inthe horizontal direction.

This is not the case in

the vertical direction in the vertical direction in

a gravity field.

Pressure in a fluid increases with depthbecause more fluid rests on deeper layers,

and the effect of this “extra weight” on a

deeper layer is balanced by an increase in

pressure.

To obtain a relation

for the variation of

pressure with depth,

consider a rectangular

fluid element of height

∆z, length ∆x, and unit depth (into the page)

F1p1

depth (into the page) in equilibrium, as shown

in Figure:

Free-body diagram of a rectangular

fluid element in equilibrium

F2

p2

∆x

∆z

1

Assuming the density

of the fluid ρ to be

constant, HenceW

F1

F2

W = m g = [ρ (∆z × ∆x × 1)] g = ρ g ∆x ∆z

F2

F2= p2 (∆x × 1)= p2 *∆x

F1 = p1 (∆x × 1)= p1 *∆x

From Newton’s 2nd law (force balance) in

vertical z-direction gives

+ve

Hence, F2 – (F1 + W)= 0Hence, F2 – (F1 + W)= 0

p2 *∆x – ( p1 *∆x + ρ g ∆x ∆z )= 0

Dividing both sides of Eq. by ∆x, we get

p2 – ( p1 + ρ g ∆z )= 0 Hence,

∆∆∆∆ p = p2 – p1 = ρρρρ g ∆∆∆∆z = γγγγ ∆∆∆∆z

(1-17)

(1-18)

From Eq. (1-18), we can conclude:

1) Pressure in a fluid increases linearly with

depth.

2) For a given fluid, the vertical distance ∆z issometimes used as a measure of pressure, and it is called the pressure head.and it is called the pressure head.

3) For small to moderate distances, the

variation of pressure with height is negligible

for gases because of their low density.

4) The pressure in a tank containing a gase.g., can be considered to be uniform since

the weight of the gas is too small to make a

significant difference. Also, the pressure in a

room filled with air can be assumed to be

constant.constant.

patm

patm γh

p

h

p

p= patm + γh, Hence, pgage =p- patm = γh(1-19)

p

dp= -ρ g dz

dz

p+dp

dp/dz= - ρ g

or

(1-20) p

The -ve sign is due to our taking the +ve z

direction to be upward so that dp is -ve

when dz is +ve (since pressure decreases in an upward direction).

dp/dz= - ρ g (1-20)

When the variation of density

with elevation is known the

pressure difference between

points 1 and 2 can be

determined by integration to be.

z

z1z2

From Eq. (1-20) dp/dz= - ρ gFrom Eq. (1-20) dp/dz= - ρ g

i.e.

Pascal’s law:

Lifting of a large weight by

a small force by the

application of Pascal’s

law.

Hence,

Hence,Hence,

But,

Hence,

1–10 ■ THE MANOMETER

patmOwing to Pascal’ law

p1=p2=patm+ρgh

Fluid of density; density; density; density; ρρρρA

p2)gage=p2-patm=ρgh

The basic manometermanometermanometermanometer

Fluid of density; density; density; density; ρρρρ

h

W = mg =(ρV) g =(ρAh) g

p2)gage= W /A= ρAh) g /A= ρghHence; ; ; ; Fluid of

density; density; density; density; ρρρρ

EXAMPLE 1–6 Measuring Pressure with a Manometer

A manometer is used to measure the

pressure in a tank. The fluid used has a

specific gravity of 0.85, and the manometer

column height is 55 cm. If the local column height is 55 cm. If the local

atmospheric pressure is 96 kPa, determine

the absolute pressure within the tank

Solution:

2

p=p2=p1=patm+ρgh

ρ =SG* ρ water=0.85*1000 kg/m31

Schematic for Example 1–6.

ρ =850 kg/m3

Hence;

P=96 kPa+{[850*9.81*(55/100)]/1000} kPa

P=96 kPa+4.586 kPa = 100.586 kPa

×1

×2

3×

In stacked-up fluid layers

3×p1=patm+ρ1gh1

p2=p1+ρ2gh2 =(patm+ρ1gh1)+ ρ2gh2=

patm+ρ1gh1 +ρ2gh2

p3=p2+ρ3gh3=(patm+ρ1gh1)+ρ2gh2 )+ρ3gh3

= patm+ρ1gh1 +ρ2gh2 +ρ3gh3

Differential manometer

pA=pB

Used to measurepressure differential

P1+ρ1g(h+a) =p2+ρ1ga +ρ2gh

Hence,

p1- p2 =(ρ1ga +ρ2gh)- ρ1g(h+a)

i.e., ∆p=p1- p2 =ρ2gh- ρ1gh=(ρ2-ρ1) gh

manometer reading

Multifluid ManometerEXAMPLE 1–7 Measuring pressure with a Multifluid ManometerThe water in a tank is pressurized by

air, and the pressure is measured by

a multifluid manometer. The tank is

located on a mountain at an altitude

of 1400 m where the atmospheric 45××××6666of 1400 m where the atmospheric

pressure is 85.6 kPa. Determine the

air pressure in the tank if h1 = 0.1 m,

h2 =0.2 m, and h3 = 0.35 m.

Take the densities of water, oil, and

mercury to be 1000 kg/m3,850 kg/m3,

and 13,600 kg/m3, respectively.

13

4××××6666

Solution:

p3=p1=p2+ρ mercury gh3=patm+ρ mercury gh3

p3=p4+ρ oil gh2 Hence,

p4=p3-ρ oil gh2

(1-#2)

(1-#3)

Combining Eqs. (1-#2) & (1-#3) to get

p4=[patm+ρ mercury gh3]-ρ oil gh2 (1-#4)

p5=p4 =[patm+ρ mercury gh3]-ρ oil gh2 (1-#5)

p6=p5 =[patm+ρ mercury gh3]-ρ oil gh2 (1-#6)

But, p6=pair +ρ water gh1 (1-#7)

Hence, pair =p6 - ρ water gh1 (1-#8)

Combining Eqs. (1-#6) & (1-#8) to get

pair ={[patm+ρ mercury gh3]-ρ oil gh2}-ρwater gh1

Hence,Hence,pair ={[85.6 kPa + (13,600×9.81×0.35/1000)]-(850× 9.81 × 0.2) /1000}-[(1000× 9.81 × 0.1)/1000]

pair =85.6 kPa + 46.696 kPa -1.668 kPa -

0.981 kPa = 129.65 kPa ≈ 130 kPa

Other Pressure Measurement Devices

# Bourdon tube to measure gage pressure

# Pressure transducers r used to measure gage, absolute, and differential pressures

- Strain-gage pressure transducer.

- Piezoelectric transducers, alsocalled solid-state pressure transducers.

1–11 ■ THE BAROMETER AND

ATMOSPHERIC PRESSURE

patm=pD=pB=0+ρ Mercury gh

W/A=mg/A=

××××DDDD

i.e.,

W/A=mg/A=

(ρMercuryV) g/A =

(ρMercuryAh) g/A=

ρ Mercury gh

Patm=ρ Mercury gh

EXAMPLE 1–8 Measuring Atmospheric Pressure with a Barometer

Determine the atmospheric pressure at a location where the barometric reading is 740 mm Hg and the gravitational acceleration is g = 9.81 m/s2.

Assume the temperature of mercury to be 10°C, Assume the temperature of mercury to be 10°C, at which its density is 13,570 kg/m3..

Solution:

Patm=[(13,570× 9.81×(740mm/1000)]/1000

=98.51 kPa

EXAMPLE 1–9 Effect of Piston Weight on Pressure in a CylinderThe piston of a vertical piston–

cylinder device containing a gas

has a mass of 60 kg and a

cross-sectional area of 0.04 m2.

The local atmospheric pressure

is 0.97 bar, and the gravitational

∑Fz = 0+ve

is 0.97 bar, and the gravitational

acceleration is 9.81 m/s2.

(a) Determine the pressure

inside the cylinder.

(b) If some heat is transferred to

the gas and its volume is

doubled, do you expect the

pressure

inside the cylinder to change?

∑Fz = 0

Solution:

+ve

pA- patmA –W = 0

Hence,

(a)

Hence,

pA= patmA +W i.e., p = patm +(W /A)

patm = 0.97 bar

1 bar = (101.325/ 1.01325) kPa = 100 kPa

From Slide 69:

1 atm = 101,325 Pa = 101.325 kPa =1.01325 bars

(1-#7)

patm = 0.97 bar= (0.97 * 100) kPa= 97 kPa

W = m g = (60 × 9.81)/1000 kN= 0.5886 KN

From Eq. (1-#7),

p = patm+(W /A)= 97 kPa+(0.5886 KN/0.04 m2)

p = 97 kPa+14.715 kPa = 111.715 kPa= i.e., p = 97 kPa+14.715 kPa = 111.715 kPa=

(111.715 /100)=1.11 bar

(b) If the piston moved , then the piston

balance will be the same. Hence, p is the

same.

Homework

Group 1: Thermodynamics : 1–3C.

Group 2: Mass, Force, and Units : 1–5C, 1–7 ,1–8, 1–9, 1–12.

Group 3: Systems, Properties, State, and Processes: 1–15C , 1–16C, 1–18C, 1–20C, 1–23C, 1–24C.

Group 4: Temperature

1–26C , 1–27C, 1–29, 1–31.

Group 5: Pressure, Manometer, and Barometer1–34C , 1–39C, 1–40, 1–42, 1–43, 1–44, 1–45, 1–48, 1–49, 1–51, 1–52, 1–53, 1–55 , 1–57, 1–59, 1–61, 1–62, 1–66, 1–67, 1–73, 1–75.59, 1–61, 1–62, 1–66, 1–67, 1–73, 1–75.

Group 6: Review Problems1–85 , 1–86, 1–106, 1–107, 1–110.