ME 201 Thermodynamics
|
|
- Claribel Gibbs
- 5 years ago
- Views:
Transcription
1 ME 0 Thermodynamics Solutions First Law Practice Problems. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 5 C and inside pressure of 00 kpa. The diameter of the balloon is measured and found to be 0.3 m. The balloon is then taken outside and allowed to come to equilibrium with the outside temperature of -5 C and outside pressure of 00 kpa. Determine the boundary work, heat transfer, and final balloon diameter. System Type: Closed System Substance Type: Ideal gas Process: Isobaric Final State: Fixed Q UNKNOWN W bnd UNKNOWN Conservation of Mass: m m st Law: m(u - u ) Q - W Boundary Work: P(v - v) State State T 5 C 98 K T -5 C 68 K P 00 kpa P 00 kpa D 0.3 m D 0.55 m V.50 x 0 m 3 V.034 x 0 m 3 m.345 x 0 kg m.345 x 0 kg u 0 kj/kg u -.6 kj/kg φ 6.7 kj/(kg K) φ kj/(kg K) v m 3 /kg v m 3 /kg Italicized values are from ideal gas relations or tables. Bold values are calculated. Approach: We begin by evaluating the properties at state and state by using the air tables and the ideal gas law. We can then use our boundary work equation to calculate the boundary work. Finally, we use the conservation of energy to determine the heat transfer. We start by determining our total volume at state. The volume of a sphere is given by V πr.50 x 0 m 3 Next we determine our mass. Using the ideal gas law we have P V (00)(.50 x 0 ) m.345 x 0 kg RT (0.87)(98) (.50 x 0 ) (.345 x 0 ) V v m m 3 /kg bnd
2 ME 0 Thermodynamics We can go to the air tables and find u 0 kj/kg φ 6.7 kj/(kg K) u -.6 kj/kg φ kj/(kg K) The specific volume at state must be given by the ideal gas law, so that RT (0.87)(68) 3 v m /kg P (00) The total final volume is then V m v (.345 x 0 )(0.769).034 x 0 m 3 The final diameter is /3 3V D 0.55 m 4 π The boundary work is calculated Wbnd P(V - V ) (00)(.034 x x 0 ) kj and the heat transfer is Q m(u - u ) + W kj bnd (.345 x 0 )(-.6-0) + (-0.06). Air at 800 K and 800 kpa enters an ideal turbine at.3 kg/s. The power output required of this turbine is 700 kw. Determine the exhaust temperature and pressure. System Type: Control Volume System Substance Type: Ideal gas Process: Isentropic Final State: Unknown Q 0 W sh 700 kw Conservation of Mass: m& m&.3 kg/s m& h - h - W& st Law: ( ) State State T 800 K T 55.7 K P 800 kpa P 44 kpa h 70.5 kj/kg h kj/kg φ kj/(kg K) φ kj/(kg K) Italicized values are from ideal gas relations or tables. Bold values are calculated. Approach: We begin by evaluating the properties at state by using the air tables and the ideal gas law. We can then use the conservation of energy to determine the enthalpy at state. The air tables will give T. Finally, the isentropic relation will give the exit pressure At state we can go to the air tables and find h 70.5 kj/kg φ kj/(kg K) From the first law we solve for h W& 700 h h - sh kj/kg m&.3 sh
3 ME 0 Thermodynamics The air tables can then give T 55.7 K φ kj/(kg K) The isentropic relation is P s 0 φ - φ R ln P Solving for P gives φ P P exp φ (800)exp R kpa 3. Refrigerant- as saturated liquid at 3 C enters a valve and exits at 0.5 MPa. What is the fluid phase at the exit? Solution System Type: Control Volume System Substance Type: phase change Process: Isenthalpic Final State: Unknown Q 0 st Law: h - h 0 State State T 3 C T C P MPa P 0.5 MPa h kj/kg h kj/kg Phase: sat.liq. Phase: phase, x 0.30 Italicized values are fromr4a tables. Bold values are calculated. Approach: We begin by evaluating the properties at state by using the refrigerant tables. We can then use the conservation of energy to determine the enthalpy at state. The refrigerant tables will give the fluid phase. At state we can go to the refrigerant tables and find h kj/kg P MPa From the first law we solve for h h h kj/kg At 0.5 MPa, we find from the tables h f 7.86 kj/kg h g kj/kg Since h is between these two values we have a two phase mixture at C and with quality h - h x f 0.30 h - h g f 3
4 ME 0 Thermodynamics 4. A tank contains Refrigerant- as saturated vapor at 00 kpa. There is heat transfer to the tank of 66 kj/kg. Determine the final temperature and pressure. System Type: Closed System Substance Type: phase change Process: Isotropic Final State: Unknown q 66 kj/kg W bnd 0 Conservation of Mass: m m st Law: u - u q State State T C T 80 C P 00 kpa P 50 kpa u kj/kg u 4.39 kj/kg v m 3 /kg v m 3 /kg phase: sat.vap. phase: sup.vap. Italicized values are from tables. Bold values are calculated. Approach: We begin by evaluating the properties at state by using the refrigerant tables. We can then use isotropic process to determine the specific volume at state. Next, we use the conservation of energy to determine the final internal energy. Finally, we go to the refrigerant tables and find the T and P. We can go to the refrigerant tables and find u kj/kg T C v m 3 /kg Since our process is isotropic, we have v v m 3 /kg The final internal energy is given by the st law or u q + u kj/kg So now we must go to the refrigerant tables and find the T and P that will correspond to these values of u and v. Since we had saturated vapor and we added heat, we will assume that we have gone to superheated vapor. Then scanning the tables we find that at 0.5 MPa and 80 C, we have u 7.67 kj/kg v m 3 /kg which is close enough. 5. Consider a 30 gallon hot water heater which is to provide water a 80 F and 0 psia. If a typical shower consumes 3 gallons/minute, last 5 minutes, and requires that the hot water should stay above 60 F, determine the heat transfer rate required. Water is supplied at 63 F and psia and gallons/minute. 4
5 ME 0 Thermodynamics System Type: Transient System Substance Type: Incompressible Process: Isotropic Final State: Fixed Inlet State: Fixed Exit State: Fixed Q UNKNOWN W bnd 0 m - m Conserv. of Mass: m& in - m& out t m u - mu st Law: m& h - m& h + & in in out out t State State In State Out State T 80 F 8. C T in 63 F 7. C T out 70 F 76.7 C T 60 F 7. C P 0 psia 37.9 kpa P in psia 5.7 kpa P out 0 psia 37.9 kpa P 0 psia 37.9 kpa u kj/kg h in 7.7 kj/kg h out kj/kg u kj/kg v m 3 /kg v in m 3 /kg v out m 3 /kg v m 3 /k m 0. kg m in 0.3 kg/s m out 0.89 kg/s m kg Note that we have used the average temperature at state and to fix our outlet temperature Italicized values are from steam tables. Bold values are calculated. Approach: We begin by evaluating the properties at all states by using the steam tables. We can then use the conservation of mass to determine the mass at state. Finally the st law is used to determine the heat transfer rate. Even though we will treat water as an incompressible substance in this case, we can use the steam tables to determine the specific volume and internal energy since for an incompressible substance they only depend on temperature, we can take the values for saturated liquid at the given temperature. For the enthalpy, we can use the enthalpy of saturated liquid and then bring it up to the appropriate pressure or h incompressible sub. h f + v f (P-P sat ) Our values are then entered on the table. To obtain the masses and mass flows we convert gallons to m 3 and divide by the specific volume. These values are entered on the table. The final state mass is given by conservation of mass or m t( m& in - m& out ) + m (5)(60)( ) kg Then the heat transfer rate becomes m u - m u Q& m& inhin + m& outhout t (50.80)(97.34) (0.)(343.88) (5)(60) 6.39 kw 90,046 Btu/hr - (0.3)(7.7) Q + (0.89)(30.95) 5
6 ME 0 Thermodynamics 6. A piston cylinder system contains gas at 300 K, 500 kpa, and 0.03 liters. The gas then undergoes a polytropic expansion with a polytropic exponent of.5 to 0.3 liters. Compare the work performed in kj for air as the gas versus hydrogen as the gas. We start this problem by working in air and then in hydrogen System Type: Closed System Substance Type: Ideal gas Process: Polytropic with n.5 Final State: UNKNOWN Q UNKNOWN (but not asked for) W bnd UNKNOWN Conservation of Mass: m m st Law: m(u - u ) Q - W Boundary Work: P v bnd - P v Wbnd (- n) State State T 300 K T NA P 500 kpa P 6.8 kpa V 3 x 0-5 m 3 V 3 x 0-4 m 3 Italicized values are from ideal gas relations. Bold values are calculated. Approach: At state we can determine the pressure from the polytropic relation. We can then use our boundary work equation to calculate the boundary work From the polytropic relationship we have P or solving for P n V P V V 0.03 P P (300) V 0.3 The boundary work is then given by n n kpa (6.8)(3 x 0 ) - (300)(3 x 0 ) Wbnd 0.34 kj (-.5) Since the calculation above never used the fact that air was our ideal gas, we will obtain the same boundary work for hydrogen as the gas. 7. In an open feedwater, subcooled liquid water is heated to saturated liquid by mixing directly with steam. In a given situation water at 0 MPa and 00 C enters an open feedwater heater at kg/s. Steam at 0 MPa and 350 C is available to be added to this water to produce saturated liquid at 0 MPa. How much steam in kg/s must be added? 6
7 ME 0 Thermodynamics System Type: Control Volume System Substance Type: Water (phase change substance) Process: Isobaric Inlet State: Fixed Exit State: Fixed Q 0 Conserv. of Mass: m& in - + m& in- - m& out 0 st Law: m& h + m& h - m& h 0 in - in- in- in- out out State In- State In- State Out T in- 00 C T in- 350 C T out 3 C P in- 0 MPa P in- 0 MPa P out 0 MPa h in kj/kg h in kj/kg h out kj/kg m& in- kg/s m& in kg/s m& out kg/s phase: sub.liq. phase: sup.vap. phase: sat.liq. Italicized values are from steam tables. Bold values are calculated. Approach: We begin by evaluating the properties at all states by using the steam tables. We can then use the conservation of mass to algebraically solve for the exit mass flow rate. Substituting this into the st law steam flow rate can be determined. The enthalpies are read from the steam tables and entered into our table above. Solving for m out from conservation of mass m& out m& in- + m& in- Substituting into the energy equation m& h + m& h - m& + m& h 0 ( ) in - in- in- in- in- in- out Solving for m in- m& ( h - h ) m& ( ) in- in- out in - hout - hin 4.37 kg 8. Most of the time during the winter Dr. Somerton turns down the thermostat to 50 F when he leaves in the morning. When he is in the house he likes to have the temperature at 68 F. The house may be considered to be composed of air, occupying a volume of 0,000 ft 3, and structural material (mostly wood with c P.76 kj/(kg K)) of,000 lb m. Determine the total heat transfer required to bring the house up to 68 F. What fraction of this total goes to heating up the air and what fraction goes to heating up the structural material? 7
8 ME 0 Thermodynamics System Type: Closed System Substance Type: Ideal gas (air) and Incompressible (wood) Process: Isobaric Final State: Fixed Q UNKNOWN W bnd 0 Conservation of Mass: m m st Law: U - U Q State State T 50 F 0 C T 68 F 0 C V,a 0,000 ft m 3 V,a 0,000 ft m 3 m,a kg m,a kg m,w,000 lb m kg m,w,000 lb m kg Italicized values are from ideal gas relations. Bold values are calculated. Approach: We begin by determining the mass of air in the house. Then we can use the st law to calculate the heat transfer required. We determine our mass of air assuming a pressure of 4.7 psia. Using the ideal gas law we have P V,a (00)(83.) m, a kg RT (0.87)(83) We can calculate the heat transfer from the energy equation U - U m c (T - T ) + m c (T - T ) Q a v,a w P,w (348.7)(0.79)(0-0) + (4989.5)(.76)(0-0) 89,794 kj 85,65 Btu of which 98% goes into the structural material and % into the air. 9. Engine oil enters an ideal pump at 00 F and psia and leaves at 7 psia. The oil flow rate is 0.04 lb m /s. The pump inlet has a diameter of 5 inches and the pump outlet has a diameter of inches. What pumping power is required? 8
9 ME 0 Thermodynamics System Type: Control Volume System Substance Type: Incompressible Process: Isentropic, can't neglect KE Final State: Unknown Q 0 W sh UNKNOWN Conservation of Mass: m& m& 0.04 lbm/s r r v v st Law: m& h + - h - - W& sh State State T 00 F 37.8 C T 37.8 C 00 F P psia 8.7 kpa P 7 psia 7. kpa v r ft/s v r 0.03 ft/s Bold values are calculated. Approach: We begin by using our process description to fix state. Then from the diameter and mass flow information the velocities are obtained. Finally, the st law is used to calculate the power. Since the process is isentropic and we have an incompressible substance, we write T s - s cp,avg ln T which yields T T 37.8 C The velocities can be calculated from one of our continuity relations r m& m& v ρ A c πd ρ 4 Using a density for engine oil of 04 kg/m 3, we find v r 0.08 m/s v r 0.08 m/s The enthalpy change for an incompressible substance is given by - h c (T T ) + v (P P ) h P,avg avg Then the shaft work is calculated (using the density rather than the specific volume) r r P - P v - v W& - m sh & + ρ (0.08) - (0.08) kw.93 Btu/hr - (0.08) 0 3 9
10 ME 0 Thermodynamics 0. The exhaust process for an internal combustion engine may be modeled as transient system undergoing an isobaric process with boundary work. Just before the exhaust valve opens the cylinder of 0.5 liters contains air at 00 kpa and 500 K. At the end of exhaust the volume is liters. Assume that the process is adiabatic. Determine the final temperature and mass and the boundary work. System Type: Transient System Substance Type: Ideal gas Process: Isobaric Outlet State: UNKNOWN Final State: UNKNOWN Q 0 W bnd P(V -V ) Conserv. of Mass: m -m - m out st Law: m u -m u - m out h out - W bnd State State Out State T 500 K T out T P 00 kpa P out 00 kpa P 00 kpa V 5 x 0-4 m 3 V out NA V 8.33 x 0-5 m 3 Bold values are calculated. Approach: We begin by calculating the boundary work. We then use the ideal gas law and constant specific heat to determine the enthalpy and internal energies. The boundary work is then W bnd P(V -V ) (00)[8.33 x x 0-4 ] kj Now we want to solve for the final temperature, but we note that in the first law four different things depend on T : u, m, h out, and m out. We can eliminate m out with the help of our conservation of mass so that m u -m u - (m -m )h out - W bnd Since the exhaust air temperature is changing during the process, we will use our linear average approach to determine h out, or T + T hout hair at The mass at the final state will be given by PV m RT So substituting RT PV T + T u mu - m hair at - Wbnd PV RT When we consider that the air tables provide the relationships among u and T and h and T, we see that we do have a well posed problem, but we can't do the algebra, so we will have to solve the problem by iteration. Our process will be Guess T Evaluate h air and u from the air tables Calculate u from the above equation 0
11 ME 0 Thermodynamics Compare the u at the guessed value of T to the calculated value of u Re-guess T and repeat the process until the difference between the values of u at the guessed T and the value of the calculated u is negligible I set this up in an Excel spreadsheet and the results are shown below T,guess u,guess T out h out m u u,guess -u,calc E E E So it would appear that our final temperature is about 65 K.
ME 201 Thermodynamics
Fall 204 ME 20 Thermodynamics Homework #8 Solutions. A rigid wall, insulated container is divided into two regions by a removable wall. One region contains 0. lb m of CO at 00 F, while the other region
More informationME 201 Thermodynamics
Spring 01 ME 01 Thermodynamics Property Evaluation Practice Problems II Solutions 1. Air at 100 K and 1 MPa goes to MPa isenthapically. Determine the entropy change. Substance Type: Ideal Gas (air) Process:
More informationME 201 Thermodynamics
8 ME 0 Thermodynamics Practice Problems or Property Evaluation For each process described below provide the temperature, pressure, and speciic volume at each state and the change in enthalpy, internal
More informationReadings for this homework assignment and upcoming lectures
Homework #3 (group) Tuesday, February 13 by 4:00 pm 5290 exercises (individual) Thursday, February 15 by 4:00 pm extra credit (individual) Thursday, February 15 by 4:00 pm Readings for this homework assignment
More informationI. (20%) Answer the following True (T) or False (F). If false, explain why for full credit.
I. (20%) Answer the following True (T) or False (F). If false, explain why for full credit. Both the Kelvin and Fahrenheit scales are absolute temperature scales. Specific volume, v, is an intensive property,
More informationChapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 5 Mass and Energy Analysis of Control Volumes by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics:
More informationChapter 5. Mass and Energy Analysis of Control Volumes
Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)
More informationCHAPTER 7 ENTROPY. Copyright Hany A. Al-Ansary and S. I. Abdel-Khalik (2014) 1
CHAPTER 7 ENTROPY S. I. Abdel-Khalik (2014) 1 ENTROPY The Clausius Inequality The Clausius inequality states that for for all cycles, reversible or irreversible, engines or refrigerators: For internally-reversible
More informationKNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.
Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity
More informationThe First Law of Thermodynamics. By: Yidnekachew Messele
The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy
More informationME Thermodynamics I
Homework - Week 01 HW-01 (25 points) Given: 5 Schematic of the solar cell/solar panel Find: 5 Identify the system and the heat/work interactions associated with it. Show the direction of the interactions.
More informationME 2322 Thermodynamics I PRE-LECTURE Lesson 10 Complete the items below Name:
Lesson 10 1. (5 pt) If P > P sat (T), the phase is a subcooled liquid. 2. (5 pt) if P < P sat (T), the phase is superheated vapor. 3. (5 pt) if T > T sat (P), the phase is superheated vapor. 4. (5 pt)
More informationDishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)
HW-1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment
More information5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE
Chapter 6 Using Entropy 1 2 Chapter Objective Means are introduced for analyzing systems from the 2 nd law perspective as they undergo processes that are not necessarily cycles. Objective: introduce entropy
More informationfirst law of ThermodyNamics
first law of ThermodyNamics First law of thermodynamics - Principle of conservation of energy - Energy can be neither created nor destroyed Basic statement When any closed system is taken through a cycle,
More informationCHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi
More informationME 200 Final Exam December 12, :00 a.m. to 10:00 a.m.
CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 1:30 p.m. 3:30 p.m. Mongia Abraham Sojka Bae Naik ME 200 Final Exam December 12, 2011 8:00 a.m. to 10:00 a.m. INSTRUCTIONS
More informationChapter 5: The First Law of Thermodynamics: Closed Systems
Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy
More informationFirst Law of Thermodynamics
CH2303 Chemical Engineering Thermodynamics I Unit II First Law of Thermodynamics Dr. M. Subramanian 07-July-2011 Associate Professor Department of Chemical Engineering Sri Sivasubramaniya Nadar College
More informationME 200 Final Exam December 14, :00 a.m. to 10:00 a.m.
CIRCLE YOUR LECTURE BELOW: First Name Last Name 7:30 a.m. 8:30 a.m. 10:30 a.m. 11:30 a.m. Boregowda Boregowda Braun Bae 2:30 p.m. 3:30 p.m. 4:30 p.m. Meyer Naik Hess ME 200 Final Exam December 14, 2015
More informationOutline. Example. Solution. Property evaluation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids Examples
Outline Property ealuation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids s A piston-cylinder deice initially contains 0.5m of saturated water apor at 00kPa.
More informationTHERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.
THERMODYNAMICS, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 2 THERMODYNAMIC PRINCIPLES SAE
More informationChapter 4. Energy Analysis of Closed Systems
Chapter 4 Energy Analysis of Closed Systems The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat
More informationChapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn
Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,
More informationWhere F1 is the force and dl1 is the infinitesimal displacement, but F1 = p1a1
In order to force the fluid to flow across the boundary of the system against a pressure p1, work is done on the boundary of the system. The amount of work done is dw = - F1.dl1, Where F1 is the force
More informationCHAPTER INTRODUCTION AND BASIC PRINCIPLES. (Tutorial). Determine if the following properties of the system are intensive or extensive properties: Property Intensive Extensive Volume Density Conductivity
More informationME Thermodynamics I. Lecture Notes and Example Problems
ME 227.3 Thermodynamics I Lecture Notes and Example Problems James D. Bugg September 2018 Department of Mechanical Engineering Introduction Part I: Lecture Notes This part contains handout versions of
More informationNon-Newtonian fluids is the fluids in which shear stress is not directly proportional to deformation rate, such as toothpaste,
CHAPTER1: Basic Definitions, Zeroth, First, and Second Laws of Thermodynamics 1.1. Definitions What does thermodynamic mean? It is a Greeks word which means a motion of the heat. Water is a liquid substance
More informationFirst Law of Thermodynamics Closed Systems
First Law of Thermodynamics Closed Systems Content The First Law of Thermodynamics Energy Balance Energy Change of a System Mechanisms of Energy Transfer First Law of Thermodynamics in Closed Systems Moving
More information1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.
AME 5053 Intermediate Thermodynamics Examination Prof J M Powers 30 September 0 0 Calorically perfect ideal air at 300 K, 00 kpa, 000 m/s, is brought to rest isentropically Determine its final temperature
More information+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1.
5.6 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.6. Room A is at 200 kpa, v = 0.5 m3/kg, VA = m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane
More informationME 2322 Thermodynamics I PRE-LECTURE Lesson 23 Complete the items below Name:
Lesson 23 1. (10 pt) Write the equation for the thermal efficiency of a Carnot heat engine below: 1 L H 2. (10 pt) Can the thermal efficiency of an actual engine ever exceed that of an equivalent Carnot
More informationEVALUATING PROPERTIES FOR A PURE SUBSTANCES. By Ertanto Vetra
EVALUATING PROPERTIES FOR A PURE SUBSTANCES 1 By Ertanto Vetra Outlines - TV, PV, PT, PVT Diagram - Property Tables - Introduction to Enthalpy - Reference State & Reference Values - Ideal Gas Equation
More informationChemical Engineering Thermodynamics Spring 2002
10.213 Chemical Engineering Thermodynamics Spring 2002 Test 2 Solution Problem 1 (35 points) High pressure steam (stream 1) at a rate of 1000 kg/h initially at 3.5 MPa and 350 ºC is expanded in a turbine
More information(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:
Last Name First Name ME 300 Engineering Thermodynamics Exam #2 Spring 2008 March 28, 2008 Form A Note : (i) (ii) (iii) (iv) Closed book, closed notes; one 8.5 x 11 sheet allowed. 60 points total; 60 minutes;
More informationTwo mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW SVCET
Two mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW 1. What do you understand by pure substance? A pure substance is defined as one that is homogeneous and invariable in chemical composition
More informationENT 254: Applied Thermodynamics
ENT 54: Applied Thermodynamics Mr. Azizul bin Mohamad Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) azizul@unimap.edu.my 019-4747351 04-9798679 Chapter
More informationMAE 11. Homework 8: Solutions 11/30/2018
MAE 11 Homework 8: Solutions 11/30/2018 MAE 11 Fall 2018 HW #8 Due: Friday, November 30 (beginning of class at 12:00p) Requirements:: Include T s diagram for all cycles. Also include p v diagrams for Ch
More informationLecture 44: Review Thermodynamics I
ME 00 Thermodynamics I Lecture 44: Review Thermodynamics I Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R. China Email : liyo@sjtu.edu.cn
More informationDr Ali Jawarneh. Hashemite University
Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Examine the moving boundary work or P d work commonly encountered in reciprocating devices such as automotive engines and compressors.
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 20 June 2005
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 20 June 2005 Midterm Examination R. Culham & M. Bahrami This is a 90 minute, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib
More informationChapter 6. Using Entropy
Chapter 6 Using Entropy Learning Outcomes Demonstrate understanding of key concepts related to entropy and the second law... including entropy transfer, entropy production, and the increase in entropy
More informationLecture 35: Vapor power systems, Rankine cycle
ME 00 Thermodynamics I Spring 015 Lecture 35: Vapor power systems, Rankine cycle Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R.
More informationThermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA
Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering
More informationMAE 110A. Homework 3: Solutions 10/20/2017
MAE 110A Homework 3: Solutions 10/20/2017 3.10: For H 2O, determine the specified property at the indicated state. Locate the state on a sketch of the T-v diagram. Given a) T 140 C, v 0.5 m 3 kg b) p 30MPa,
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 13 June 2007
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 13 June 2007 Midterm Examination R. Culham This is a 2 hour, open-book examination. You are permitted to use: course text book calculator There are
More informationAvailability and Irreversibility
Availability and Irreversibility 1.0 Overview A critical application of thermodynamics is finding the maximum amount of work that can be extracted from a given energy resource. This calculation forms the
More informationFINAL EXAM. ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW:
ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW: Div. 5 7:30 am Div. 2 10:30 am Div. 4 12:30 am Prof. Naik Prof. Braun Prof. Bae Div. 3 2:30 pm Div. 1 4:30 pm Div. 6 4:30 pm Prof. Chen Prof.
More informationME 300 Thermodynamics II
ME 300 Thermodynamics II Prof. S. H. Frankel Fall 2006 ME 300 Thermodynamics II 1 Week 1 Introduction/Motivation Review Unsteady analysis NEW! ME 300 Thermodynamics II 2 Today s Outline Introductions/motivations
More informationBME-A PREVIOUS YEAR QUESTIONS
BME-A PREVIOUS YEAR QUESTIONS CREDITS CHANGE ACCHA HAI TEAM UNIT-1 Introduction: Introduction to Thermodynamics, Concepts of systems, control volume, state, properties, equilibrium, quasi-static process,
More informationContent. Entropy and principle of increasing entropy. Change of entropy in an ideal gas.
Entropy Content Entropy and principle of increasing entropy. Change of entropy in an ideal gas. Entropy Entropy can be viewed as a measure of molecular disorder, or molecular randomness. As a system becomes
More informationChapter 1: Basic Definitions, Terminologies and Concepts
Chapter : Basic Definitions, Terminologies and Concepts ---------------------------------------. UThermodynamics:U It is a basic science that deals with: -. Energy transformation from one form to another..
More informationBasic Thermodynamics Module 1
Basic Thermodynamics Module 1 Lecture 9: Thermodynamic Properties of Fluids Thermodynamic Properties of fluids Most useful properties: Properties like pressure, volume and temperature which can be measured
More informationMAE 320 HW 7B. 1e. For an isolated system, please circle the parameter which will change with time. (a) Total energy;
MAE 320 HW 7B his comprehensive homework is due Monday, December 5 th, 206. Each problem is worth the points indicated. Copying of the solution from another is not acceptable. Multi-choice, multi-answer
More informationUNIT I Basic concepts and Work & Heat Transfer
SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code: Engineering Thermodynamics (16ME307) Year & Sem: II-B. Tech & II-Sem
More informationMAHALAKSHMI ENGINEERING COLLEGE
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI-621213. Department: Mechanical Subject Code: ME2202 U N IT - 1 Semester: III Subject Name: ENGG. THERMODYNAMICS 1. 1 kg of gas at 1.1 bar, 27 o C is compressed
More informationDual Program Level 1 Physics Course
Dual Program Level 1 Physics Course Assignment 15 Due: 11/Feb/2012 14:00 Assume that water has a constant specific heat capacity of 4190 J/kg K at all temperatures between its melting point and boiling
More informationCONCEPTS AND DEFINITIONS. Prepared by Engr. John Paul Timola
CONCEPTS AND DEFINITIONS Prepared by Engr. John Paul Timola ENGINEERING THERMODYNAMICS Science that involves design and analysis of devices and systems for energy conversion Deals with heat and work and
More informationChapter 7. Entropy: A Measure of Disorder
Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic
More informationGAS. Outline. Experiments. Device for in-class thought experiments to prove 1 st law. First law of thermodynamics Closed systems (no mass flow)
Outline First law of thermodynamics Closed systems (no mass flow) Device for in-class thought experiments to prove 1 st law Rubber stops GAS Features: Quasi-equlibrium expansion/compression Constant volume
More informationChapter 7. Dr Ali Jawarneh. Department of Mechanical Engineering Hashemite University
Chapter 7 ENTROPY Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Objectives Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify
More informationQUIZZES RIEPJCPIγPJEJJJY
Che 3021 Thermodynamics I QUIZZES RIEPJCPIγPJEJJJY QUIZ 1. Find Molecular Weights: 1 1 CO 2 2 NaCl 3 Aspirin C 9 H 8 O 4 CO2 = NaCl = C9H8O4 = PIgPJC Quiz 1. Temperature conversion 1 Convert 94 o F, to
More informationEng Thermodynamics I conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics.
Eng3901 - Thermodynamics I 1 1 Introduction 1.1 Thermodynamics Thermodynamics is the study of the relationships between heat transfer, work interactions, kinetic and potential energies, and the properties
More informationEnergy and Energy Balances
Energy and Energy Balances help us account for the total energy required for a process to run Minimizing wasted energy is crucial in Energy, like mass, is. This is the Components of Total Energy energy
More informationEngineering Thermodynamics. Chapter 6. Entropy: a measure of Disorder 6.1 Introduction
Engineering hermodynamics AAi Chapter 6 Entropy: a measure of Disorder 6. Introduction he second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of
More informationUBMCC11 - THERMODYNAMICS. B.E (Marine Engineering) B 16 BASIC CONCEPTS AND FIRST LAW PART- A
UBMCC11 - THERMODYNAMICS B.E (Marine Engineering) B 16 UNIT I BASIC CONCEPTS AND FIRST LAW PART- A 1. What do you understand by pure substance? 2. Define thermodynamic system. 3. Name the different types
More informationWeek 5. Energy Analysis of Closed Systems. GENESYS Laboratory
Week 5. Energy Analysis of Closed Systems Objectives 1. Examine the moving boundary work or PdV work commonly encountered in reciprocating devices such as automotive engines and compressors 2. Identify
More informationTo receive full credit all work must be clearly provided. Please use units in all answers.
Exam is Open Textbook, Open Class Notes, Computers can be used (Computer limited to class notes, lectures, homework, book material, calculator, conversion utilities, etc. No searching for similar problems
More information20 m neon m propane. g 20. Problems with solutions:
Problems with solutions:. A -m tank is filled with a gas at room temperature 0 C and pressure 00 Kpa. How much mass is there if the gas is a) Air b) Neon, or c) Propane? Given: T7K; P00KPa; M air 9; M
More informationEngineering Thermodynamics
David Ng Summer 2017 Contents 1 July 5, 2017 3 1.1 Thermodynamics................................ 3 2 July 7, 2017 3 2.1 Properties.................................... 3 3 July 10, 2017 4 3.1 Systems.....................................
More informationFind: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k.
PROBLEM 6.25 Three m 3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kpa. The air receives 1546 kj of work from the paddle wheel. Assuming the ideal gas model,
More informationc Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 T145 = Q + W cv + i h 2 = h (V2 1 V 2 2)
Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka-1000, Bangladesh zahurul@me.buet.ac.bd
More informationKNOWN: Air undergoes a polytropic process in a piston-cylinder assembly. The work is known.
PROBLEM.7 A hown in Fig. P.7, 0 ft of air at T = 00 o R, 00 lbf/in. undergoe a polytropic expanion to a final preure of 5.4 lbf/in. The proce follow pv. = contant. The work i W = 94.4 Btu. Auming ideal
More informationName: I have observed the honor code and have neither given nor received aid on this exam.
ME 235 FINAL EXAM, ecember 16, 2011 K. Kurabayashi and. Siegel, ME ept. Exam Rules: Open Book and one page of notes allowed. There are 4 problems. Solve each problem on a separate page. Name: I have observed
More informationECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:
ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION June 19, 2015 2:30 pm - 4:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closed-book examination. 2. Permitted
More informationPROBLEM 6.3. Using the appropriate table, determine the indicated property. In each case, locate the state on sketches of the T-v and T-s diagrams.
PROBLEM 63 Using the appropriate table, determine the indicated property In each case, locate the state on sketches of the -v and -s diagrams (a) water at p = 040 bar, h = 147714 kj/kg K Find s, in kj/kg
More informationENGR Thermodynamics
ENGR 224 - hermodynamics W #5 Problem : 7.14 - he Increase of Entropy Principle - 2 pts 11-May-11 Will the entropy of steam increase, decrease or remain the same as it flows through a real adiabatic turbine?
More informationTDEC202 - Energy II (Recitation #3)
TDEC202 - Energy II (Recitation #3) Summer Term, 2007 M. Carchidi Solution to Problem 3-029 We are given that the substance is R-134a and metric units so that Tables A-11, A-12 and A-13 in Appendix 1 will
More informationCHAPTER 8 ENTROPY. Blank
CHAPER 8 ENROPY Blank SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L.
More informationVersion 001 HW 15 Thermodynamics C&J sizemore (21301jtsizemore) 1
Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.
More informationThis follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore. (for reversible process only) (22.
Entropy Clausius inequality can be used to analyze the cyclic process in a quantitative manner. The second law became a law of wider applicability when Clausius introduced the property called entropy.
More informationSOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 SONNTAG BORGNAKKE VAN WYLEN. FUNDAMENTALS of. Thermodynamics. Sixth Edition
SOLUTION MANUAL ENGLISH UNIT PROBLEMS CHAPTER 4 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition CHAPTER 4 SUBSECTION PROB NO. Concept-Study Guide Problems 7- Simple processes 3-8
More informationT098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26
Conservation of Energy for a Closed System Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET Dhaka-, Bangladesh zahurul@me.buet.ac.bd
More informationQUESTION BANK UNIT-1 INTRODUCTION. 2. State zeroth law of thermodynamics? Write its importance in thermodynamics.
QUESTION BANK UNIT-1 INTRODUCTION 1. What do you mean by thermodynamic equilibrium? How does it differ from thermal equilibrium? [05 Marks, June-2015] 2. State zeroth law of thermodynamics? Write its importance
More informationTHE FIRST LAW APPLIED TO STEADY FLOW PROCESSES
Chapter 10 THE FIRST LAW APPLIED TO STEADY FLOW PROCESSES It is not the sun to overtake the moon, nor doth the night outstrip theday.theyfloateachinanorbit. The Holy Qur-ān In many engineering applications,
More informationKNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.
4.3 Steam enters a.6-cm-diameter pipe at 80 bar and 600 o C with a velocity of 50 m/s. Determine the mass flow rate, in kg/s. KNOWN: Pressure, temperature, and velocity of steam entering a.6-cm-diameter
More informationSHRI RAMSWAROOP MEMORIAL COLLEGE OF ENGG. & MANAGEMENT
B.Tech. [SEM III (ME-31, 32, 33,34,35 & 36)] QUIZ TEST-1 Time: 1 Hour THERMODYNAMICS Max. Marks: 30 (EME-303) Note: Attempt All Questions. Q1) 2 kg of an ideal gas is compressed adiabatically from pressure
More informationR13 SET - 1 '' ''' '' ' '''' Code No RT21033
SET - 1 II B. Tech I Semester Supplementary Examinations, June - 2015 THERMODYNAMICS (Com. to ME, AE, AME) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (Part-A and Part-B)
More informationTeaching schedule *15 18
Teaching schedule Session *15 18 19 21 22 24 Topics 5. Gas power cycles Basic considerations in the analysis of power cycle; Carnot cycle; Air standard cycle; Reciprocating engines; Otto cycle; Diesel
More informationChapter 2: The Physical Properties of Pure Compounds
Chapter 2: The Physical Properties of Pure Compounds 2-10. The boiler is an important unit operation in the Rankine cycle. This problem further explores the phenomenon of boiling. A. When you are heating
More informationESO201A: Thermodynamics
ESO201A: Thermodynamics First Semester 2015-2016 Mid-Semester Examination Instructor: Sameer Khandekar Time: 120 mins Marks: 250 Solve sub-parts of a question serially. Question #1 (60 marks): One kmol
More information1 st Law Analysis of Control Volume (open system) Chapter 6
1 st Law Analysis of Control Volume (open system) Chapter 6 In chapter 5, we did 1st law analysis for a control mass (closed system). In this chapter the analysis of the 1st law will be on a control volume
More informationR13. II B. Tech I Semester Regular Examinations, Jan THERMODYNAMICS (Com. to ME, AE, AME) PART- A
SET - 1 II B. Tech I Semester Regular Examinations, Jan - 2015 THERMODYNAMICS (Com. to ME, AE, AME) Time: 3 hours Max. Marks: 70 Note 1. Question Paper consists of two parts (Part-A and Part-B) 2. Answer
More informationExergy and the Dead State
EXERGY The energy content of the universe is constant, just as its mass content is. Yet at times of crisis we are bombarded with speeches and articles on how to conserve energy. As engineers, we know that
More informationECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 12 June 2006
ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER 1 June 006 Midterm Examination R. Culham This is a hour, closed-book examination. You are permitted to use one 8.5 in. 11 in. crib sheet (one side
More informationSpring_#8. Thermodynamics. Youngsuk Nam
Spring_#8 Thermodynamics Youngsuk Nam ysnam1@khu.ac.krac kr Ch.7: Entropy Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify the secondlaw effects. Establish
More information2-21. for gage pressure, the high and low pressures are expressed as. Noting that 1 psi = kpa,
- -58E The systolic and diastolic pressures of a healthy person are given in mmhg. These pressures are to be expressed in kpa, psi, and meter water column. Assumptions Both mercury and water are incompressible
More informationToday lecture. 1. Entropy change in an isolated system 2. Exergy
Today lecture 1. Entropy change in an isolated system. Exergy - What is exergy? - Reversible Work & Irreversibility - Second-Law Efficiency - Exergy change of a system For a fixed mass For a flow stream
More informationDepartment of Mechanical Engineering Indian Institute of Technology New Delhi II Semester MEL 140 ENGINEERING THERMODYNAMICS
PROBLEM SET 1: Review of Basics Problem 1: Define Work. Explain how the force is generated in an automobile. Problem 2: Define and classify Energy and explain the relation between a body and energy. Problem
More informationEngineering Thermodynamics. Chapter 1. Introductory Concepts and Definition
1.1 Introduction Chapter 1 Introductory Concepts and Definition Thermodynamics may be defined as follows : Thermodynamics is an axiomatic science which deals with the relations among heat, work and properties
More information