FE Thermodynamics Review
description
Transcript of FE Thermodynamics Review
+FE FE ThermodynamicThermodynamicssReviewReview
Dr. Omar MezaDr. Omar MezaAssistant ProfessorAssistant ProfessorDepartment of Mechanical Department of Mechanical EngineeringEngineering
+Topics coveredTopics covered
Thermodynamics LawThermodynamics Law 11stst and 2 and 2ndnd law law
Energy , heat and workEnergy , heat and work
Availability and reversibilityAvailability and reversibility
CyclesCycles
Ideal gasesIdeal gases
Mixture of gasesMixture of gases
Phase changePhase change
Heat TransferHeat Transfer
Properties of:Properties of: enthalpyenthalpy entropyentropy
+Tips for taking examTips for taking exam
Use the reference handbookUse the reference handbook Know what it containsKnow what it contains Know what types of problems you can use it forKnow what types of problems you can use it for Know how to use it to solve problemsKnow how to use it to solve problems Refer to it frequentlyRefer to it frequently
Work backwards when possibleWork backwards when possible FE exam is multiple choice with single correct answerFE exam is multiple choice with single correct answer Plug answers into problem when it is convenient to do soPlug answers into problem when it is convenient to do so Try to work backwards to confirm your solution as often as Try to work backwards to confirm your solution as often as
possiblepossible Progress from easiest to hardest problemProgress from easiest to hardest problem
Same number of points per problemSame number of points per problem Calculator tipsCalculator tips
Check the NCEES website to confirm your model is allowedCheck the NCEES website to confirm your model is allowed Avoid using it to save time!Avoid using it to save time! Many answers do not require a calculator (fractions vs. decimals)Many answers do not require a calculator (fractions vs. decimals)
+Properties of Single-Component Properties of Single-Component SystemsSystems
Handbook page:Handbook page:
For a simple substance, For a simple substance, specification of any specification of any two two intensive, independent intensive, independent propertiesproperties is sufficient to is sufficient to fix all the rest.fix all the rest.
+Properties of Single-Component Properties of Single-Component SystemsSystems
Handbook page:Handbook page:
A substance that has a A substance that has a fixed chemical fixed chemical composition throughoutcomposition throughoutis called a is called a pure substancepure substance..
+Properties of Single-Component Properties of Single-Component SystemsSystems
A substance whose properties are uniform throughout is A substance whose properties are uniform throughout is referred to as referred to as A.A.A solid A solid B.B.An ideal substance An ideal substance C.C.A pure substance A pure substance D.D.A standard substance A standard substance
A substance whose properties are uniform throughout is A substance whose properties are uniform throughout is referred to as referred to as A.A.A solid A solid B.B.An ideal substance An ideal substance C.C.A pure substance A pure substance D.D.A standard substanceA standard substance
+Properties of Single-Component Properties of Single-Component SystemsSystems
+Properties of Single-Component Properties of Single-Component SystemsSystems
Given:Given: Steam at 2.0 kPa is saturated at 17.5 Steam at 2.0 kPa is saturated at 17.5 ooC. In what C. In what state will the steam be at 40 state will the steam be at 40 ooC if the pressure is 2.0 kPa?C if the pressure is 2.0 kPa?
Analysis:Analysis:
@ P = 2.0 kPa, T@ P = 2.0 kPa, Tsatsat = 17.5 = 17.5ooCC
TTsatsat < T < T “superheated vapor”“superheated vapor”
Tsat= 17.5Tsat= 17.5ooCCT= 40T= 40ooCC
+Properties of Single-Component Properties of Single-Component SystemsSystems
+Properties of Single-Component Properties of Single-Component SystemsSystems
Find the volume occupied by 20 kg of steam at 0.4 MPa, Find the volume occupied by 20 kg of steam at 0.4 MPa, 400400ooCC
At 0.4 MPa the TAt 0.4 MPa the Tsatsat=142=142ooC approximately. It means that the C approximately. It means that the
steam is in the steam is in the superheated regionsuperheated region
+Properties of Single-Component Properties of Single-Component SystemsSystems
Handbook page:Handbook page:
Real gases exhibit ideal-Real gases exhibit ideal-gas behavior at relatively gas behavior at relatively low pressures and high low pressures and high temperatures.temperatures.
+Properties of Single-Component Properties of Single-Component SystemsSystems
All real gases deviate somewhat from ideal gas behavior: All real gases deviate somewhat from ideal gas behavior: PV= mRT. For which of the following conditions is the PV= mRT. For which of the following conditions is the deviation the smallest? deviation the smallest? A.A.High temperature and low volume High temperature and low volume B.B.High temperature and low pressures High temperature and low pressures C.C.High pressures and low volumes High pressures and low volumes D.D.High pressure and low temperatures High pressure and low temperatures
When the volume of an ideal gas is doubled while the When the volume of an ideal gas is doubled while the temperature is halved, what happens to the pressure? temperature is halved, what happens to the pressure? A.A.Pressure is doubled Pressure is doubled B.B.Pressure is halved Pressure is halved C.C.Pressure is quartered Pressure is quartered D.D.Pressure is quadrupledPressure is quadrupled
4P
=X
TXv4
=)2/T()v2(X
=TPv
+Properties of Single-Component Properties of Single-Component SystemsSystems
Handbook page:Handbook page:
+Properties of Single-Component Properties of Single-Component SystemsSystems
+Properties of Single-Component Properties of Single-Component SystemsSystems
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
( ) ( ) PEΔ+KEΔ+UΔ=Wout-Win+Qout-Qin
Formal sign convention: Heat transfer to a system and work done by a system are positive; heat transfer from a system and work done on a system are negative.
Wb is positive for expansionWb is negative for compression
+First Law of ThermodynamicsFirst Law of Thermodynamics
+First Law of ThermodynamicsFirst Law of Thermodynamics
During a process, 30J of work are done by a closed During a process, 30J of work are done by a closed stationary system on its surroundings. The internal energy stationary system on its surroundings. The internal energy of the system decreases by 40 j. What is the heat transfer?of the system decreases by 40 j. What is the heat transfer?
PEKEU Wout-Win+Qout-Qin
UWout-Qnet
WoutQnet U
JJJ 103040Qnet
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
Calculate the work done by a piston contained within a Calculate the work done by a piston contained within a cylinder with air if 2mcylinder with air if 2m33 is tripled while the temperature is is tripled while the temperature is maintained at a constant T = 30maintained at a constant T = 30ooC. The initial pressure is C. The initial pressure is PP11=400 kPa absolute.=400 kPa absolute.
+First Law of ThermodynamicsFirst Law of Thermodynamics
Polytropic process in a closed system
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
+First Law of ThermodynamicsFirst Law of Thermodynamics
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
Handbook page:Handbook page:
+First Law of ThermodynamicsFirst Law of Thermodynamics
A steam coil operating at steady state receives 30 kg/min of A steam coil operating at steady state receives 30 kg/min of steam with an enthalpy of 2900 kJ/kg. if the steam leaves steam with an enthalpy of 2900 kJ/kg. if the steam leaves with an enthalpy of 1600 kJ/min, what is the rate of heat with an enthalpy of 1600 kJ/min, what is the rate of heat transfer from the coil?transfer from the coil?
+First Law of ThermodynamicsFirst Law of Thermodynamics
+First Law of ThermodynamicsFirst Law of Thermodynamics
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
Handbook page:Handbook page:
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Basic CyclesBasic Cycles
+Ideal Gas MixtureIdeal Gas Mixture
Handbook page:Handbook page:
+Ideal Gas MixtureIdeal Gas Mixture
+Ideal Gas MixtureIdeal Gas Mixture
+Ideal Gas MixtureIdeal Gas Mixture
+PsychrometricsPsychrometrics
Handbook page:Handbook page:
+PsychrometricsPsychrometrics
Handbook page:Handbook page:
Mollier Diagram Mollier Diagram
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+PsychrometricsPsychrometrics
+Combustion ProcessesCombustion Processes
+Combustion ProcessesCombustion Processes
+Combustion ProcessesCombustion Processes
+Combustion ProcessesCombustion Processes
+Combustion ProcessesCombustion Processes
+Second Law of ThermodynamicsSecond Law of Thermodynamics
Handbook page:Handbook page:
+Second Law of ThermodynamicsSecond Law of Thermodynamics
Part of the heat received by a heat engine is converted to work, while the
rest is rejected to a sink.
This is a law. 1.It is always observed in real heat engines. 2.One cannot derive it from first principles. 3.No exceptions are known.
It is not just that we haven’t looked hard enough and that future discoveries will makeit possible to convert heat completely to work.
+Second Law of ThermodynamicsSecond Law of Thermodynamics
+Second Law of ThermodynamicsSecond Law of Thermodynamics
The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP).
The objective of a refrigerator is to remove heat (QL) from the refrigerated space.
Can the value of COPR be greater than unity?
+Second Law of ThermodynamicsSecond Law of Thermodynamics
The work supplied to a heat pump is
used to extract energy
from the cold outdoors and
carry it into the warm indoors.
for fixed values of QL and QH
+Second Law of ThermodynamicsSecond Law of Thermodynamics
A) 1500-MWB) 600-MWC) 900-MWD) 2100-MW
+Second Law of ThermodynamicsSecond Law of Thermodynamics
A) YesB) NoC) Not clearD) NA
+EntropyEntropy
+EntropyEntropy
A) 2.82 kJ/KB) 6.86 kJ/KC) -8.10 kJ/KD) 8.10 kJ/K
+
Preguntas? Preguntas? Comentarios?Comentarios?
+
Muchas Muchas Gracias !Gracias !