Chapter 1: Introduction to Basic

Concepts of Thermodynamics 1

PTT 201/4 THERMODYNAMICSSEM 1 (2012/2013)

THERMODYNAMICS & ENERGY

Energy:

The ability to cause changes.

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.

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.

The first law of thermodynamics

Heat flows in the direction of decreasing temperature

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APPLICATION AREAS OF

THERMODYNAMICS

Power plant

Human body

Aircraft and spacecraft Car

Refrigeration systems

Wind turbines Air conditioning systems Boats Solar hot water systems3

L of

light

DIMENSIONS& UNITS

lightAny physical quantity can be characterized by dimensions lightThe magnitudes assigned to the

dimension are called units

lightPrimary/

fundamental dimensions

lightSecondary/

derived dimensions

Electric current

L ofAmount of light

Amount of matter

Temperature

lightLength

lightMass

lightTime

lightVelocity

lightEnergy

lightVolume

lightForce

lightEnglish system lightInternational

system (SI)

Primary dimensions and their units in SI

Standard prefixes in SI units

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MEASURE OF AMOUNT OR SIZE

lightMass, m lightNo. of moles, n lightTotal volume, Vt

light

nMm Mass

No. of moles

Molecular weight

lightM

mn

MassNo. of moles

Molecular weight

Specific volume:

m

VV

t

or

VmV t

Molar volume:

n

VV

t

or

Vn V t

lightRepresent the size of a system

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FORCE

light lightm=32.174 lbm

m=1 kga=1 m/s2

a=1 ft/s2

F=1 N

F=1 lbf

maF

Mass

Acceleration

lightDefinition:

Force required to accelerate a mass of 1 kg or 32.174 lbm at a rate of 1 m/s2 or 1 ft/s2.

lightDefinition:

Weight is gravitational force applied to a body

light

mgW

Mass

Local gravitational acceleration

g = 9.807 m/s2

= 32.174 ft/s2

lightEnglish system

Pound-force (lbf)

light

Unit in many European countries

Kilogram-force (kgf)

lightSI

Newton (N)

light1 N = 1 kg. m/s2

1 lbf = 32.174 lbm.ft/s2 = 4.44822 N1 kgf = 9.807 N

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• 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 closed or open.

• Closed system (Control mass): A fixed amount of mass, and no mass can cross its boundary.

SYSTEMS AND CONTROL VOLUMES

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• 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.

SYSTEMS AND CONTROL VOLUMES, CONT’

Commonly measured with liquid-in-glass thermometer, wherein the liquid expands when heated

TEMPERATURE

lightBoiling point of pure water at

standard atmospheric pressure

lightFreezing point of water saturated with air at standard atmospheric

pressure

lightLower limit of temperature

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TEMPERATURE

lightRelations among temperature scales light

Comparison of magnitude of various

temperature units

SI unit system English unit system10

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.

a) Determine the total energyb) Determine the money saved

SOLUTION :

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a) Determine the total energy

Total energy = (Energy per unit time) (Time interval) = (30 kW) (2200 h) = 66, 000 kWh

b) 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 =

SOLUTION :

66,000 kWh 3600 s 1 kJ/s = 2.38 X 108 kJ

1 h 1 kW

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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 = 1.00 lbm 32.174 ft/s2 1 lbf = 1.00 lbf

32.174 lbm.ft/s2

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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

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light

Some basic pressure gages.

A normal force exerted by a fluid per unit area

PRESSURE

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• 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.

PRESSURE

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The pressure of a fluid at restincreases with depth (as a result of added weight).

VARIATION OF PRESSURE WITH DEPTH

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In a room filled with a gas, the variation of

pressure with height is negligible.

Pressure in a liquid at rest increaseslinearly 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.

PRESSURE, CONT’

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light

Pascal’s law: The pressure applied to a confined fluid increases the pressure throughout by the same amount.

Lifting of a large weight by a small force by the application of

Pascal’s law.

PRESSURE, CONT’

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In stacked-up fluid layers, the pressure change across a fluid layer of density and height h is gh.

Measuring the pressure drop across

a flow section or a flow device by a

differential manometer.

The basic 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.

MANOMETER

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• 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.

BAROMETER AND ATMOSPHERIC PRESSURE

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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

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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.

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SOLUTION

33OH kg/m 850 )0kg/m(0.85)(100)SG(ρρ

2

ρghPP atm

= 96 kPa + 850 kg/m3 9.81 m/s2 0.55 m 1 N 1 kPa

1 kg.m/s2 1000 N/m2

= 100.6 kPa

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

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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

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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 kPa26

WORK, ENERGY AND HEAT

Work, energy and heat will be covered in other chapter!

light

Work = Force Distance1 J = 1 N m∙

1 cal = 4.1868 J1 Btu = 1.0551 kJ

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THANK YOU..