Presentation is loading. Please wait.

Presentation is loading. Please wait.

Thermodynamics EGR 334 Lecture 01: Introduction to Thermodynamics.

Similar presentations


Presentation on theme: "Thermodynamics EGR 334 Lecture 01: Introduction to Thermodynamics."— Presentation transcript:

1 Thermodynamics EGR 334 Lecture 01: Introduction to Thermodynamics

2 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

3 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

4 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.

5 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 1 st Law of thermodynamics

6 Thermo…A Quick Survey…continued -- Entropy Production and Accounting: Applying the 2 nd 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: C hemical energy production and balances

7 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

8 System Definition Radiation

9 System Definition Radiation

10 System Definition Q Gas

11 System Definition W Gas Q

12 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)).

13 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)).

14 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. P o, V o, T o P f, V f, T f Combustion Q ∆V → W C x H y +A O 2  B CO 2 + C H 2 O

15 Control Volume / Open System Heat (Q), work (W), and matter (m) may cross the boundary Q Ga s

16 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

17 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.

18 Process: Cycle: S1S1 S4S4 S2S2 S1S1 S2S2 S3S3 P P vv

19 Unit Review: Weight and Mass WeightMass U.S. Customarypound, lb f pound of mass, lb m or slug MetricNewton, Nkilogram, kg Relationships:W =m gwhere g = 9.81 m/s 2 g = 32.2 ft/s 2 Conversions:1 N = 0.2248 lb f 1 lb f = 4.4482 N 1 kg = 2.205 lb m 1 lb m = 0.4536 kg 1 slug = 32.2 lb m = 71.0 kg

20 Unit review: Density and Specific Volume Density ρ Specific Volume v or v U.S. Customary mass basis[lb m / ft 3 ][ ft 3 /lb m ] molar basis [ ft 3 /mol] Metric mass basis [kg/m 3 ] [ m 3 /kg] molar basis [m 3 /mol] Relationships m = mass V = Volume M = molecular weight n = number of moles

21 Unit review: Pressure Pressure = Presure head U.S. Customary[psi ] or [lb f /in 2 ] [psf] or [lb f /ft 2 ] [in of Hg ] [ in of H 2 0 ] [ft of H 2 0] Metric[N/m 2] [ Pa] [bar] [mm of Hg ] [ mm of H 2 0 ] [cm of H 2 0] 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 H 2 O Gage vs. Absolute pressure p abs = p atm + p gage Vacuum vs. absolute pressure p abs = p atm - p vac

22 Unit review: Temperature MetricU.S. Relative Temperature Celcius [ o C ] Fahrenheit [ o F] Absolute Temperature Kelvin [ K ] Rankine [ o R ] Relationships K = o C + 273.15 o R = o F + 459.67 o C =( o F - 32)/1.8 o F= 1.8 o C + 32 o R = 1.8 K

23 Example 1: An object whose mass is 35 lb is subjected to an applied upward force of 15 lb f. The only other force acting on the object is the force of gravity. Determine the net acceleration of the object in ft/s 2 assuming the acceleration of gravity is constant (g = 32.2 ft/s 2 ). Is the net acceleration up or down? F AF = 15 lb f F g = 15 lb f mass= 35 lb m

24 Example 2: A system consists of N 2 in a piston-cylinder assembly, initially at P 1 = 20 psi, and occupying a volume of 2.5 ft 3. The N 2 is compressed to P 2 = 100 psi and a final volume of 1.5 ft 3. 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 ft 3 and sketch the process on a graph of P vs V. P 1 = 20 psi V 1 =2.5 ft 3 P 2 = 100 psi V 2 =1.5 ft 3

25 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/cm 3. If g = 9.81 m/s 2 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

26 Example 1: An object whose mass is 35 lb m is subjected to an applied upward force of 15 lb f. The only other force acting on the object is the force of gravity. Determine the net acceleration of the object in ft/s 2 assuming the acceleration of gravity is constant (g = 32.2 ft/s 2 ). Is the net acceleration up or down? F AF = 15 lb f F g = ? m= 35 lb m

27 Example 2: A system consists of N 2 in a piston-cylinder assembly, initially at P 1 = 20 psi, and occupying a volume of 2.5 ft 3. The N 2 is compressed to P 2 = 100 psi and a final volume of 1.5 ft 3. 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 ft 3 and sketch the process on a graph of P vs V. Since the relationship is linear (y=mx+b)

28 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/cm 3. If g = 9.81 m/s 2 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 b) Solution: a) b) c)


Download ppt "Thermodynamics EGR 334 Lecture 01: Introduction to Thermodynamics."

Similar presentations


Ads by Google