ThermodynamicsEGR 334

Lecture 01:Introduction to Thermodynamics

Today’s Objectives:• Distribute and understand syllabus• Take quick tour of thermodynamics topics covered in this

course• Understand difference between system and control

volume.• Review units of thermodynamics related quantities

Reading Assignment:

Homework Assignment:

• Read Chap 1. Sections 1 - 9

From Chap 1: Problems 19, 37, 51, 59

Thermodynamics

Definition-- Thermodynamics is the science of energy and entropy-- Thermodynamics: the branch of physics dealing with

the relationships between heat, work, and forms of energy.

From the Greek: - θερμη, therme, meaning “heat“ → Energy- δυναμις, dynamis, meaning “power“→ Transport

To the Engineer…Thermodynamics means:- Understanding the 4 Laws of

Thermodynamics- Learn to work in with four different

temperature scales.- Learn to balance energy, heat, and work

with respect to open and closed systems.- Learn about common thermodynamcis

devices and applications and how the principles can be used to predict system performance and efficiencies.

Thermo….a Quick Survey

-- Properties of systems: Temperature, Pressure, Specific Volume, Phase, Quality, Density, Enthalpy, Entropy

-- Processes and Cycles: State to state transitions. Carnot cycle, Rankine cycle, Otto cycle, Diesel cycle, Brayton cycle, Refrigeration cycle

-- Work-Energy-Heat Balance: Applying the 1st Law of thermodynamics

Thermo…A Quick Survey…continued

-- Entropy Production and Accounting: Applying the 2nd Law of thermodynamics

-- Thermodynamics devices: Turbines, Heat Exchangers, Condensers, Engines Pumps, Cooling Towers, Compressors, Diffusers

-- Psychrometrics: HVAC, heating, ventilation, air conditioning, and humidity

-- Combustion and Power Production: Chemical energy production and balances

A basic concept: System•A System is whatever it is that we want to

study.• In thermodynamics, the first step in defining

any problem is to define exactly what is to be monitored, examined, measured, etc.

System

Environment

Boundary

Interactions

System Definition Radiation

Radiation

System Definition Radiation

Radiation

System Definition

Q

Gas

System Definition

W

Gas Gas

Q

Types of Systems• Isolated Systems – matter and energy may

not cross the boundary.• Adiabatic Systems – heat may not cross the

boundary.• Diathermic Systems - heat may cross

boundary.• Closed Systems – matter may not cross the

boundary.• Open Systems – heat, work, and matter may

cross the boundary (more often called a Control Volume (CV)).

Types of Systems• Isolated Systems – matter and energy may

not cross the boundary.• Adiabatic Systems – heat must not cross the

boundary.• Diathermic Systems - heat may cross

boundary.• Closed Systems – matter may not cross the

boundary.• Open Systems – heat, work, and matter may

cross the boundary (more often called a Control Volume (CV)).

Closed System•Matter (m) may not cross the boundary.•Heat (Q) and Work (W) may cross the boundary

which change the energy (ΔE) of the system.

Po, Vo, To

Pf, Vf, TfCombustion

Q

∆V →WCxHy+A O2 B CO2 + C H2O

Control Volume / Open System•Heat (Q), work (W), and matter (m) may

cross the boundary

Q

Gas

System Properties• The State of a system is defined by its properties

(T, V, P, E, ρ, v, u, h, s)• Extensive properties – Depends on mass• Intensive properties – Does not depend on mass.•

Give some examples of Extensive Properties Intensive properties

Process vs. Cycle.

•A Process is when properties of the system undergo a change. A process moves from one “state” to another “state”.

•If State i = State f then system is said to be Steady State.

•If a Process undergo changes that eventually bring it back to the original state, these transitions together are called a Cycle.

Process: Cycle:

S1 S4

S2

S1

S2

S3

P P

v v

Unit Review: Weight and MassWeight Mass

U.S. Customary pound , lbf pound of mass, lbm or

slug

Metric Newton, N kilogram, kg

Relationships: W =m g where g = 9.81 m/s2 g = 32.2 ft/s2

Conversions: 1 N = 0.2248 lbf

1 lbf = 4.4482 N

1 kg = 2.205 lbm 1 lbm = 0.4536 kg

1 slug = 32.2 lbm = 71.0 kg

Unit review: Density and Specific Volume

Density ρ

Specific Volume v or v

U.S. Customary

mass basis [lbm / ft3 ] [ ft3 /lbm]

molar basis [ ft3 /mol]

Metricmass basis [kg/m3] [ m3/kg]

molar basis [m3/mol]

Relationships m = mass V = Volume

M = molecular weight

n = number of moles

m

V 1

M mn

M

Unit review: Pressure Pressure = Presure head

U.S. Customary [psi ] or [lbf /in2 ]

[psf] or [lbf /ft2 ]

[in of Hg ] [ in of H20 ] [ft of H20]

Metric [N/m2]

[ Pa][bar]

[mm of Hg ] [ mm of H20 ] [cm of H20]

Other atm

Relationships: 1 atm = 14.69 psi = 1.01325 bar = 100 324 Pa = 760 mm of Hg = 29.92 in of Hg = 33.96 ft of H2O

Gage vs. Absolute pressure

pabs = patm + pgage

Vacuum vs. absolute pressurepabs = patm - pvac

Unit review: TemperatureMetric U.S.

Relative Temperature

Celcius [ oC ]

Fahrenheit [ o F]

Absolute Temperature

Kelvin [ K ]

Rankine[ oR ]

Relationships

K = oC + 273.15 oR = oF + 459.67 oC =( oF - 32)/1.8 oF= 1.8 oC + 32

oR = 1.8 K

Example 1: An object whose mass is 35 lb is subjected to an applied upward force of 15 lbf. The only other force acting on the object is the force of gravity. Determine the net acceleration of the object in ft/s2 assuming the acceleration of gravity is constant (g = 32.2 ft/s2). Is the net acceleration up or down?

FAF= 15 lbf

Fg= 15 lbf

mass= 35 lbm

Example 2: A system consists of N2 in a piston-cylinder assembly, initially at P1 = 20 psi, and occupying a volume of 2.5 ft3. The N2 is compressed to P2 = 100 psi and a final volume of 1.5 ft3. During the process, the relationship between P and V is linear. Determine the P in psi at an intermediate state where the volume is 2.1 ft3 and sketch the process on a graph of P vs V.

P1= 20 psi V1 =2.5 ft3

P2= 100 psi V2 =1.5 ft3

Example 3: A monometer is attached to a tank of gas in which the pressure is 104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3. If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate(a) The difference in mercury levels in the manometer, in cm.(b) The gage pressure of the gas in kPa and bar(c) The absolute pressure of the gas kPa, atm, and psi

Example 1: An object whose mass is 35 lbm is subjected to an applied upward force of 15 lbf. The only other force acting on the object is the force of gravity. Determine the net acceleration of the object in ft/s2 assuming the acceleration of gravity is constant (g = 32.2 ft/s2). Is the net acceleration up or down?

FAF= 15 lbf

Fg= ?

m= 35 lbm

32.21532.2

35

18.4

AF g

AF g AF AF

f m m

m f

F ma

F F ma

F F F mg Fa g

m m mlb lb ft s lb ft

alb lb s

fta

s

Example 2: A system consists of N2 in a piston-cylinder assembly, initially at P1 = 20 psi, and occupying a volume of 2.5 ft3. The N2 is compressed to P2 = 100 psi and a final volume of 1.5 ft3. During the process, the relationship between P and V is linear. Determine the P in psi at an intermediate state where the volume is 2.1 ft3 and sketch the process on a graph of P vs V.

312

12 805.25.1

20100

ft

psi

VV

PPm

Since the relationship is linear (y=mx+b)

psipsiftft

psiP

PVVmPVVmPP

52205.21.280 33

1111

Example 3: A monometer is attached to a tank of gas in which the pressure is 104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3. If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate(a) The difference in mercury levels in the manometer, in cm.(b) The gage pressure of the gas in kPa and bar(c) The absolute pressure of the gas atm, and psi

g

PPLgLPP atm

atm

2.0L cm104.0 101.33 2.67

0.0267

guage atmP P P kPa

bar

b)

Solution: a)

b)

3 2 3 2 2

3 2 2

104.0 101.33 10 / 10 1 / (1 )

(13.59 / )(9.81 / ) 1 1 1 (100 )

kPa N m g kg m s mL

g cm m s kPa kg N cm

c)1

104.0 1.026101.33abs

atmP kPa atm

kPa

14.71.026 15.09

1

psiatm psi

atm