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Date Created: 10/20/14
ME 200 Thermodynamics I Division 4 Spring 2012 Exam 1 Review Exam 1 When Tuesday February 7 2012 What time 800 to 900 pm Where For Division 4 gtgt All Students PHYS 114 gtgt Be on time gtgt Don t go to EE 129 or WTHR 200 Closed book and closed notes Equation sheet and unit conversion will be provided Property tables will be provided Exam 1 Contd o Exam 1 covers Lectures 1 through 10 c There are three problems worth 100 points For Exam 1 a Review textbook lecture notessolved examples and assigned homework textbook problems and SP o Bring calculator with fresh batteries pencils eraser a Do not study all night before examination Get enough sleep on the night before not during the day and definitely not during 800 to 900 pm Eat well to keep up energy level during the evening exam a Do not stress yourself over this exam During Exam 1 Read each problem carefully and be sure to answer what is being asked When in doubt seek clarification Show all work neatly to maximize your score Write basic equations and assumptions Budget your time wisely First solve problems which boost your confidence Do not spend time when you get stuck come back to it after you have maximized your score Hand in the exam when time is over Do not force your instructor to wrestle you to the floor to get your exam UnitsDimensions a Do you know your units gtgt Pressure 1 bar 100 kPa gtgt Temperature TK T C 27315 ATK AT C gtgt Energy1J1Pax1m3OR1kJ1kPax1m3 Some Basic Concepts o What are the types of systems a What is property a What are the types of properties a What is state a What is state principle Energy Interactions a Work W and Heat Transfer Q gtgt These are not properties gtgt These are not related to states gtgt These are related to path of the given process a Boundary Work Wb Only applicable for quasi 2 Wb t equilibrium process 1 What is Wb for Constant Volume Constant Pressure PV constant and PV constant polytropic Energy Balance o For a closed system such as piston cylinder device rigid lam Q W AU AKE APE W includes boundary work plus any other work interactions electrical work shaft work etc gtgt Heat added positive heat rejected negative gtgt Work output during expansion positive work input during compression negative o For a cycle Qcycle cycle a Difference between basic cycles power vs refrigeration Power Cycle Heat Engine Hot Reservoir T i Net work output cycle QH QC QH Wcycle 77thermal Desired Effect Work Output H Q Heat Input c Thermal efficiency always Cold Reservoir positive and less than 100 To Refrigeration Cycle Hot Reservoir Heat removed from TH X cold space col C Q QH QC Refrigerator wcyde Work input Q Desired Effect Cooling COP always positive no upper limit C Cold Reservoir Tc Heat Pump Cycle Heat supplied to maintain H t Re5equotV 39 Temperature of hot space TH C0PQ Q HP H Desired Effect Heating Ac ycleQH QC Work input Wcycle COP always positive no upper limit Cold Reservoir To Property Determination for Pure Substance o Always need two independent intensive properties to fix the state of pure substance near phase change region a Determine the phase as saturated liquid SL two phase mixture SLVM saturated vapor SV compressed liquid CLsubcooled liquid or superheated vapor SHV o Always start in saturated temperature or pressure table to fix the phase from two intensive properties a Apply other equations such as boundary work and energy balance as necessary when solving problems Property Determination oontd o Tv or Tu or Th or Ts known OR Pv or Pu or Ph or Ps known a fvgtvgorugtuggthgthgorsgtsgatthegivenTor P then the state is SHV We must use an appropriate table 0 fVfltVltVgOrUfltUltUgOrhflthlthgOrSfltSltSgatthe given T or P then the state is SLVM We must compute quality x and use x to get other properties v v u u h h S S X f X f X r X f vg vf ug uf hg hf sg sf Property Determination contd o PT known a At given T if P gt Psat OR at given P if T lt Tsat then the state is compressed liquid subcooled liquid We can look up properties in an appropriate table if available Otherwise use the following approximations In Lectures 1 to 10 we have not discussed these V Vf T approximations These are included here only for the sake of completeness T On Exam 1 we must use property tables if the M quot M f phase of the given substance is compressed liquid 1 TVfTP PSafT Property Determination oontd PT known a At given T if P Psat OR at given P if T Tsat then the state is SLVM In this case either the quality x must be known or additional information must be available to compute quality and use x to get other properties VVfxVg f hhfC lg lf Itlif39CItg Iif SSfCSg Sf Property Determination oontd PT known a At given T if P lt Psat OR at given P if T gt Tsat then the state is SHV We must use an appropriate table a For a general pure substance near its vapor dome we must follow the above procedure which is summarized in the table given below Property Determination contd Compressed Superheated Given Liquid S LVM Vapor T V V lt VfT VfT lt V lt VgT V gt VgT P V V lt Vfp Vfap lt V lt Vgap V gt Vgp T u u lt um um lt u lt ugT u gt ug P u u lt ufap ufap lt u lt ugap u gt ugap T h h lt hfT hfT lt h lt hgT h gt hgg P h h lt hfap hfap lt h lt hgap h gt hgap T s s lt Sf T SfT lt s lt sgg s gt sgT P s s lt Sfap Sfap lt s lt Sgap s gt Sgap T P T lt Tsatap T Tsatap T gt Tsatap T P P gt Pm P PS3 P lt PS3 Problem 1 A o A refrigerator contains R134a at 8 bar and 40 C and its total volume is 02691 m3 gtgt Show the state of R134a on Pv and Tv diagram gtgt What is the internal energy kJ of R134a at this state Problem 1B o A rigid tank initially contains liquid vapor mixture with 1985 liquid by mass of water at 100 C The mixture is heated until the pressure increases to 3 bar gtgt Show the heating process on Pv and Tv diagram gtgt Calculate the change in internal energy kJkg Problem 1C o 100 kg of water at 10 bar and 1799 C fills up a rigid tank What is the volume of the tank gtgt 01127 m3 gtgt 1944 m3 gtgt Insufficient Information Problem 1 D o For a simple compressible substance pressure and temperature are known How many additional properties are required to fix the state of the substance gtgt 0 gtgt 1 gtgt 2 gtgt Insufficient Information Problem 2 o A piston cylinder contains a gas initially occupying 0028 m3 State 1 undergoing a cycle In a constant volume process internal energy of the gas increases by 264 kJ State 2 Expansion occurs to a pressure of 14 bar State 3 such that internal energy does not change with PV constant The cycle is completed with a constant pressure process which requires 105 kJ of work input gtgt Show the cycle on P V diagram gtgt What is the pressure bar at state 2 gtgt Calculate heat transfer kJ for process 23 and 31 Answers Not attached 515 bar 1878 kJ heat addition to the system 369 kJ heat rejected from the system Problem 3 o A horizontal piston cylinder device fitted with stops initially contains 01 kg of water at 1 MPa and 500 C State 1 Constant pressure cooling occurs such that volume occupied by water is onehalf its initial volume and the piston just rests against the stops State 2 With piston face resting against the stops the water is cooled to 25 C State 3 gtgt Show the entire process on Pv and Tv diagram gtgt Evaluate the heat transfer kJ during each process Answers Not attached Q12 8815 kJ heat rejected from the system Q23 23058 kJ heat rejected from the system Problem 1 Answers 1A Not attached 25213 kJ 1B Not attached 120868 kJkg 1C Insufficient Information 1D Insufficient Information
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