# Chapter 1: Introduction to Basic Concepts of Thermodynamics - PowerPoint PPT Presentation

Chapter 1: Introduction to Basic Concepts of Thermodynamics

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Chapter 1: Introduction to Basic Concepts of Thermodynamics

## Chapter 1: Introduction to Basic Concepts of Thermodynamics

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1. PTT 201/4THERMODYNAMICS SEM 1 (2012/2013) Chapter 1: Introduction to Basic Concepts of Thermodynamics

2. Thermodynamics & energy The name thermodynamics stems from the Greek words therme (heat) and dynamis (power). Conservation of energy principle: During an interaction, energy can change from one form to another but the total amount of energy remains constant. Energy: The ability to cause changes. The first law of thermodynamics: Energy cannot be created or destroyed; it can only change forms. The second law of thermodynamics: It asserts that energy has quality as well as quantity, and actual processes occur in the direction of decreasing quality of energy. Heat flows in the direction of decreasing temperature The first law of thermodynamics

3. Power plant Aircraft and spacecraft Car Application areas of thermodynamics Human body Refrigeration systems Solar hot water systems Wind turbines Air conditioning systems Boats

4. Electric current L of light light Any physical quantity can be characterized by dimensions The magnitudes assigned to the dimension are called units Amount of light L of Amount of matter Primary/ fundamental dimensions light Temperature light Length light Dimensions & units Mass light Time light Secondary/ derived dimensions light Velocity light Energy light English system light light International system (SI) Volume light Standard prefixes in SI units Force light Primary dimensions and their units in SI

5. Measure of amount or size light light Total volume, Vt light No. of moles, n Mass, m light Represent the size of a system light light No. of moles Mass Mass Specific volume: No. of moles Molecular weight Molecular weight Molar volume: or or

6. light m=1 kg light Mass a=1 m/s2 Definition: Force required to accelerate a mass of 1 kg or 32.174 lbm at a rate of 1 m/s2 or 1 ft/s2. light F=1 N m=32.174 lbm a=1 ft/s2 Acceleration F=1 lbf light Mass Definition: Weight is gravitational force applied to a body light FORCE Local gravitational acceleration g = 9.807 m/s2 = 32.174 ft/s2 English system Pound-force (lbf) light Unit in many European countries Kilogram-force (kgf) light SI Newton (N) light light 1 N = 1 kg. m/s2 1 lbf = 32.174 lbm.ft/s2 = 4.44822 N 1 kgf = 9.807 N

7. SYSTEMS AND CONTROL VOLUMES • System: A quantity of matter or a region in space chosen for study. • Surroundings: The mass or region outside the system • Boundary: The real or imaginary surface that separates the system from its surroundings. • The boundary of a system can be fixed or movable. • Systems may be considered to be closedor open. • Closed system (Control mass): A fixed amount of mass, and no mass can cross its boundary.

8. SYSTEMS AND CONTROL VOLUMES, cont’ • Open system (control volume): Both mass and energy can cross the boundary of a control volume. • Device: compressor, turbine, or nozzle. • Control surface: The boundaries of a control volume. It can be real or imaginary. An open system (a control volume) with one inlet and one exit.

9. TEmperature Commonly measured with liquid-in-glass thermometer, wherein the liquid expands when heated Boiling point of pure water at standard atmospheric pressure light Freezing point of water saturated with air at standard atmospheric pressure light Lower limit of temperature light

10. TEmperature light Comparison of magnitude of various temperature units light Relations among temperature scales SI unit system English unit system

11. Example : Electric power generation by a wind turbine A school is paying \$0.09/kWh for electric power. To reduce its power bill, the school install a wind turbines with a rated power of 30 kW. If the turbine operates 2200 hours per year at the rated power, determine the amount of electric power generated by the wind turbine and the money saved by the school per year. SOLUTION : Determine the total energy Determine the money saved

12. SOLUTION : Determine the total energy Total energy = (Energy per unit time) (Time interval) = (30 kW) (2200 h) = 66, 000 kWh Determine the money saved Money saved = (Total energy) (Unit cost of energy) = (66,000 kWh) (\$0.09/kWh ) = \$5940 Convert your answer of total energy in kJ. Total energy =

13. Example : The weight of one pound-mass Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth. SOLUTION : W = mg

14. Example : Expressing Temperature Rise in Different Units During a heating process, the temperature of a system rises by 10 ⁰C. Express this rise in temperature in K, ⁰F and R. SOLUTION : Δ T(K) = Δ T(⁰C) = 10 K Δ T(R) = 1.8 Δ T(K) = (1.8) (10) = 18 R Δ T(⁰F) = Δ T(R) = 18 ⁰F

15. pressure A normal force exerted by a fluid per unit area light Some basic pressure gages.

16. pressure • Absolute pressure: The actual pressure at a given position. It is measured relative to absolute vacuum (i.e., absolute zero pressure). • Gage pressure: The difference between the absolute pressure and the local atmospheric pressure. Most pressure-measuring devices are calibrated to read zero in the atmosphere, and so they indicate gage pressure. • Vacuum pressures: Pressures below atmospheric pressure.

17. Variation of Pressure with Depth The pressure of a fluid at rest increases with depth (as a result of added weight).

18. Pressure, cont’ In a room filled with a gas, the variation of pressure with height is negligible. Pressure in a liquid at rest increases linearly with distance from the free surface. The pressure is the same at all points on a horizontal plane in a given fluid regardless of geometry, provided that the points are interconnected by the same fluid.

19. Pressure, cont’ Pascal’s law: The pressure applied to a confined fluid increases the pressure throughout by the same amount. light Lifting of a large weight by a small force by the application of Pascal’s law.

20. manometer It is commonly used to measure small and moderate pressure differences. A manometer contains one or more fluids such as mercury, water, alcohol, or oil. The basic manometer. Measuring the pressure drop across a flow section or a flow device by a differential manometer. In stacked-up fluid layers, the pressure change across a fluid layer of density  and height h is gh.

21. BAROMETER AND ATMOSPHERIC PRESSURE • Atmospheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is often referred to as the barometric pressure. • A frequently used pressure unit is the standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0°C (Hg = 13,595 kg/m3) under standard gravitational acceleration (g = 9.807 m/s2). The basic barometer.

22. Example : Absolute Pressure of a Vacuum Chamber A vacuum gage connected to a chamber reads 40 kPa at a location where the atmospheric pressure is 100 kPa. Determine the absolute pressure in the chamber. SOLUTION : Pabs = Patm - Pvac = 100 - 40 = 60 kPa

23. Example : 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, as shown in figure. If the local atmospheric pressure is 96 kPa, determine the absolute pressure within the tank.

24. SOLUTION Gage pressure = 4.6 kPa Determine the gage pressure in the tank.

25. Example : Measuring Pressure with a Multifluid Manometer The water in a tank is pressurized by air and the pressure is measured by a multifluid manometer as shown in the figure. The tank is located on a mountain at an altitude of 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 13600 kg/m3, respectively. Ans: P1 = 130 kPa

26. Example : Measuring Atmospheric Pressure with a Barometer Determine the atmospheric pressure at a location where the barometric reading is 740 mmHg and the gravitational acceleration is g = 9.81 m/s2. Assume the temperature of mercury to be 10 ⁰C, at which its density is 13570 kg/m3. Ans: in unit kPa Ans: 98.5 kPa

27. WORk, energy and heat Work, energy and heat will be covered in other chapter! light Work = Force  Distance 1 J = 1 N∙m 1 cal = 4.1868 J 1 Btu = 1.0551 kJ

28. THANK YOU..