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Transcript of 1 PTT 201/4 THERMODYNAMICS SEM 1 (2012/2013). Energy: The ability to cause changes. The name...
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’
18
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’
19
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
21
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
22
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.
23
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..