Thermodynamics worked examples

Size: px
Start display at page:

Download "Thermodynamics worked examples"

Transcription

1 An Introduction to Mechanical Engineering Part hermodynamics worked examles. What is the absolute ressure, in SI units, of a fluid at a gauge ressure of. bar if atmosheric ressure is.0 bar? Absolute ressure g + a bar kpa. Conert - o C to a temerature in degrees Kelin. [K] [ o C] K. Calculate the following: (i) the kinetic energy of a body which has a mass of kg and a elocity of 0m/s; (ii) the change in otential energy of a mass of kg when it is raised a height of m; (iii) the strain energy stored in a sring comressed by 8mm from its free length if the sring constant is.0 MN/m; (i) the increase in internal energy of a gas in a closed system during a rocess in which -00 J of heat transfer and 400 J of work transfer take lace. (i) Kinetic energy mc 00 (ii) Change in otential energy mg z 0J J (iii) Strain energy stored kx J (i) Increase in internal energy U U First Law: W+Q U U J Hodder Education 009

2 An Introduction to Mechanical Engineering Part 4. he iston in Figure 4. has no mass and is free to moe. he oint force F acting on the iston is alied gently so that the iston moes slowly to the left. As the gas in the cylinder is comressed, the roduct of ressure and system olume, V, remains constant. he iston area is 0 m and when x is x the system olume is 006m. Atmosheric ressure is bar. Calculate the distance (x -x ) moed by the iston as F increases from 0 to 70N. Equilibrium of forces on iston: A 70 + aa x 0 Pa Gien V constant V V V m. 8 0 V. x x Distance moed by iston: l x x l l l V V V A m or 8.4mm. Calculate the work inut to the closed system undergoing the cycle shown in Figure 4. if the ressure during rocess () to () is raised from 000 kpa to 400 kpa. 400 kpa P 00 kpa 4 m.0m Hodder Education 009

3 An Introduction to Mechanical Engineering Part W net W + W + W 4 + W x 0 (.0) x 0 (.0 ) -7 x 0 + x 0-600kJ 6. (i) (ii) Calculate the efficiency of a reersible heat engine oerating between a hot reseroir at 900K and a cold reseroir at 00K. he temerature of one of the heat reseroirs can be changed by 00 degrees kelin u or down. What is the highest efficiency that can be achieed by making this temerature change? 00 (i) η % 900 (ii) he largest reduction in lower temerature. 400 η.% 900 is roduced by subtracting the temerature change from the 7. A erfect gas at a ressure of 8 bar and a temerature of 40K has a density of 0 kg/m. he ratio of secific heats γ is.48. (i) Calculate the alues of molar mass m ~, secific heat at constant ressure c and secific heat at constant olume c. (ii) Calculate the change in secific entroy of the gas if the ressure is raised to 00 bar and the temerature is lowered to 400K. (i) Perfect gas, R ρ 8 0 R 7. 77J / kgk ρ 0 40 R ~ m ~. kg / kmol R γ R. 48 c J / kgk γ 48 Hodder Education 009

4 An Introduction to Mechanical Engineering Part R c 7. 0J / kgk γ 48 (ii) Secific entroy change s s cln Rln l n 7. 8l n J/kgK 8. Calculate the dryness fraction x of saturated steam if it has a secific enthaly of 800 kj/kg and at the same ressure and temerature, the secific enthaly of saturated liquid is 0 kj/kg and the latent heat of aorization is 0 kj/kg. h h f + xh fg h hf x 4 h 0 fg 9. A erfect gas with a γ alue of.4 undergoes an exansion rocess from a ressure of 600 kpa. he ratio of secific olumes / is. Calculate the ressure at the end of the rocess if this is (i) olytroic with an index n of.6, (ii) isothermal, (iii) isentroic. (i) n ( ) kPa (ii) R constant kPa (iii) γ ( ) kPa x. Hodder Education 009

5 An Introduction to Mechanical Engineering Part A erfect gas with a γ alue of. can be comressed in either a non-flow or a steady flow rocess between the same initial and final states. Neglecting any change in kinetic and otential energy, what is the ratio of the secific work done is the two cases if the rocess is (i) olytroic with an index n of.4; (ii) (iii) isothermal; isentroic? Refer to results gien in able 4.8 w Ratio w (i).4 (ii) (iii). flow non flow is:. Calculate the efficiency of the following ideal cycles when undergone by a erfect gas with a γ alue of.4: (i) a Stirling cycle oerating between a hot reseroir at 600K and a cold reseroir at 00K; (ii) a Brayton cycle with a ressure ratio of 8; (iii) an Otto cycle with a comression ratio of 8; (i) a Diesel cycle with a comression ratio of and a cut-off ratio of. (i) η ηcarnot 0% (ii) η ( ) γ γ 4 /. 4 r % (iii) η ( ) γ 4 c 6.% r 8 (i) η c γ c0 γ r ( ) ( ) γ c0 Hodder Education 009

6 An Introduction to Mechanical Engineering Part. 4 ( ) 4 6.7% (. 4 ). A erfectly insulated, rigid tank with a olume of m contains a erfect gas which has a molar mass of 8 kg/mol and a ration of secific heats of.4. Initially the ressure and temerature in the tank are 9 bar and 0 K resectiely. A fan inside the tank is sun at 600 re/min for 0 seconds. he torque required to turn the fan is 0 Nm. Calculate the following: (a) he R, c and c alues of the gas and the mass of gas in the tank. (b) he work inut (gien by torque x angular rotation in radians) to the gas from the fan. (c) he final temerature of the gas. (Exlain briefly why the temerature continues to rise for a short time after the fan has stoed rotating.) (d) he increase in entroy of the gas. V m γ.4 m ~ 8kg / 0K P 9bar kmol orque 0 Nm 600 N 600 re / min 60re / s 60 (a) ~ R 84. J / kmolk R 46J / kgk m ~ 8kg / kmol R C C C C C γ 4 C 06. 7J / kg / K C γc kg / K Hodder Education 009

7 An Introduction to Mechanical Engineering Part P V 9 0 m. kg R (b) Work inut: W fan θ ( N 0 π) π 6. kj (c) No heat transfer and constant olume roes so. st Law: U + U Q+ W W Wfan where Q0 and W has constant olume, 0 mc ( ) W fan W mc fan K (d) V S S mr ln mc ln V ln 0 6 J / K emerature will continue to increase after the fan stos because the KE generated by the fan takes a (short) time to dissiate from KE into internal energy.. A closed system containing argon undergoes a reersible isothermal rocess from an initial state () where 0 bar, V 0m and 40 K to state (). he work done during the rocess is -00 kj. he system is then heated reersibly at constant olume to final state (). he total heat transferred during the two rocesses is 70 kj. reat argon as a erfect gas with c 0 J/KgK and molar mass m ~ 40 kg/kmol. Calculate the following: Hodder Education 009

8 An Introduction to Mechanical Engineering Part (a) he mass of argon in the system. (b) he heat transferred during the constant olume rocess () (). (c) he final temerature,. (d) Sketch the rocesses on a ressure olume diagram. Argon m ~ 40kg / kmol C 0 J / KgK ~ R m ~R KJ / kgk (a) Mass V m R kg (b) Isothermal () () Q Q + W W U U 00kJ 0 But Q + Q 70kJ Q kJ (c) Constant olume () () Q + W U U where W 0 U U 70kJ mc ( ) Hodder Education 009

9 An Introduction to Mechanical Engineering Part and C 40K (() () Isothermal) C C R J / kgk 70, K (d) isothermal P () () () Constant olume V 4. A closed system containing steam undergoes a reersible constant ressure rocess during which 400 kj/kg of heat transfer takes lace. Initially the steam has a dryness fraction, x, of.0 and a temerature of 7ºC. Using the tables, and using linear interolation where necessary, determine: (a) he secific enthaly, temerature and secific internal energy of the steam at the end of the rocess. (b) he secific work transfer. (c) State whether the rocess is an exansion or comression rocess and gie two reasons to suort your choice. (a) Initial conditions dry saturated (x.0) at 7 C u h g 00bar 7kJ / kg h g 0kJ / kg 007m / kg Constant ressure rocess (reersible) st Law: q+ w u u q ( h ) + u - u KJ / kg h h emerature interolate at 80 bar and 90kJ Hodder Education 009

10 An Introduction to Mechanical Engineering Part u (90 888) (4 400) ( ) C h (. 8. 9) m / kg u 09m kJ/kg / kg ( ) 0 kj / kg (b) st Law: q+ w u u w kJ / kg. 6kJ / kg (c) Exansion. (i) w is negatie (ii) >. Air flows through an oen system at the steady mass flow rate of kg/s. At inlet the air elocity is negligible, the ressure is bar and the temerature is ºC. he air flow is comressed isentroically and leaes the system at a ressure of bar through a ie with an internal diameter of 80 mm. Calculate: (a) the temerature of the air as it leaes the system; (b) the elocity in the exit ie; (c) the ower inut to the air; (d) Exlain why the ower inut is not gien by Assume for air, R87 J/kgK and γ.4. W ~ m& d Hodder Education 009

11 An Introduction to Mechanical Engineering Part m& W & Q & 0 () () C o φ 80mm bar bar K (a) Isentroic comression P P γ γ K (b) mass continuity m& ρ Ac m& c ρ A ρ R 0. 8kg / m πd π A c 60m / s m (c) S.F.E.E Q& + W& m(h & 60 W& (c( 46 88) + ) ( ) x 6kW 0 kw c h+ ) (d) W & m ~ c d neglects changes in And otential energy between () and (). Would underestimate by.8 kw in this examle. Hodder Education 009

12 An Introduction to Mechanical Engineering Part 6. (a) Determine the rate of heat rejection from a reersible heat engine oerating between a hot reseroir at 900K and a cold reseroir at 400K if the engine roduces a ower outut of 400 kw. (b) A ie with an inside diameter of 80 mm, a wall thickness of 0 mm and a thermal conductiity of 09 W/mK is lagged with a 0 mm thick layer of lagging with a thermal conductiity of 0 W/mK. he ambient temerature is 4ºC and the inner surface of the ie wall is maintained at 80ºC. he heat transfer coefficient at the outer surface of the lagging is 8 W/mK. Calculate: (i) the rate of heat transfer er unit length of ie; (ii) the temerature at the inner surface of the lagging. (a) Reersible, and two reseroirs so: η η carnot % 900k Q R? 400k W& Q& S Q& s 40 7kW First law: Q & R Q& S W& 7 40 kw (b) 0 k0 W/mk 0 () 80 o C? () h8 W/m k Hodder Education 009

13 An Introduction to Mechanical Engineering Part q& ΣR H ln ln π. 09 π. 0 π W / m 80 q & ln 40 π C Hodder Education 009

OUTCOME 1. TUTORIAL No. 2 THERMODYNAMIC SYSTEMS

OUTCOME 1. TUTORIAL No. 2 THERMODYNAMIC SYSTEMS UNI 6: ENGINEERING HERMODYNAMICS Unit code: D/60/40 QCF level: 5 Credit value: 5 OUCOME UORIAL No. HERMODYNAMIC SYSEMS. Understand the arameters and characteristics of thermodynamic systems Polytroic rocesses:

More information

ES-7A Thermodynamics HW 5: 5-62, 81, 96, 134; 7-29, 40, 42, 67, 71, 106 Spring 2003 Page 1 of 7

ES-7A Thermodynamics HW 5: 5-62, 81, 96, 134; 7-29, 40, 42, 67, 71, 106 Spring 2003 Page 1 of 7 ES-7A hermodynamic HW 5: 5-6, 8, 96, 34; 7-9, 4, 4, 67, 7, 6 Sring 3 Page of 7 5-6 Heat Pum Given: A heat um i ued to maintain a houe at 3 C. he houe loe heat to the outide at a rate of 6, kj/h, and the

More information

EDEXCEL HIGHERS ENGINEERING THERMODYNAMICS H2 NQF LEVEL 4. TUTORIAL No. 1 PRE-REQUISITE STUDIES FLUID PROPERTIES

EDEXCEL HIGHERS ENGINEERING THERMODYNAMICS H2 NQF LEVEL 4. TUTORIAL No. 1 PRE-REQUISITE STUDIES FLUID PROPERTIES EDEXCEL HIGHERS ENGINEERING HERMODYNAMICS H NQF LEEL 4 UORIAL No. PRE-REQUISIE SUDIES FLUID PROPERIES INRODUCION Before you study the four outcomes that make u the module, you should be cometent in finding

More information

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases

HEAT UNIT 1.1 KINETIC THEORY OF GASES. 1.1.1 Introduction. 1.1.2 Postulates of Kinetic Theory of Gases UNIT HEAT. KINETIC THEORY OF GASES.. Introduction Molecules have a diameter of the order of Å and the distance between them in a gas is 0 Å while the interaction distance in solids is very small. R. Clausius

More information

First Law, Heat Capacity, Latent Heat and Enthalpy

First Law, Heat Capacity, Latent Heat and Enthalpy First Law, Heat Caacity, Latent Heat and Enthaly Stehen R. Addison January 29, 2003 Introduction In this section, we introduce the first law of thermodynamics and examine sign conentions. Heat and Work

More information

APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES

APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES APPLIED HERMODYNAMICS UORIAL REVISION OF ISENROPIC EFFICIENCY ADVANCED SEAM CYCLES INRODUCION his tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more advanced

More information

APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES

APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES APPLIED THERMODYNAMICS TUTORIAL 1 REVISION OF ISENTROPIC EFFICIENCY ADVANCED STEAM CYCLES INTRODUCTION This tutorial is designed for students wishing to extend their knowledge of thermodynamics to a more

More information

1.3 Saturation vapor pressure. 1.3.1 Vapor pressure

1.3 Saturation vapor pressure. 1.3.1 Vapor pressure 1.3 Saturation vaor ressure Increasing temerature of liquid (or any substance) enhances its evaoration that results in the increase of vaor ressure over the liquid. y lowering temerature of the vaor we

More information

1 Exercise 4.1b pg 153

1 Exercise 4.1b pg 153 In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit,

More information

be the mass flow rate of the system input stream, and m be the mass flow rates of the system output stream, then Vout V in in out out

be the mass flow rate of the system input stream, and m be the mass flow rates of the system output stream, then Vout V in in out out Chater 4 4. Energy Balances on Nonreactive Processes he general energy balance equation has the form Accumulation Inut Outut Heat added = + of Energy of Energy of Energy to System Work by done System Let

More information

ME 24-221 THERMODYNAMICS I

ME 24-221 THERMODYNAMICS I Solution to extra problem in chapter 8 Noember 9, 000 Fall 000 J. Murthy ME 4- HERMODYNAMICS I 8.5 Water i ued a the working fluid in a Carnot cycle heat engine, where it change from aturated liquid to

More information

Sheet 5:Chapter 5 5 1C Name four physical quantities that are conserved and two quantities that are not conserved during a process.

Sheet 5:Chapter 5 5 1C Name four physical quantities that are conserved and two quantities that are not conserved during a process. Thermo 1 (MEP 261) Thermodynamics An Engineering Approach Yunus A. Cengel & Michael A. Boles 7 th Edition, McGraw-Hill Companies, ISBN-978-0-07-352932-5, 2008 Sheet 5:Chapter 5 5 1C Name four physical

More information

Fugacity, Activity, and Standard States

Fugacity, Activity, and Standard States Fugacity, Activity, and Standard States Fugacity of gases: Since dg = VdP SdT, for an isothermal rocess, we have,g = 1 Vd. For ideal gas, we can substitute for V and obtain,g = nrt ln 1, or with reference

More information

STABILITY OF PNEUMATIC and HYDRAULIC VALVES

STABILITY OF PNEUMATIC and HYDRAULIC VALVES STABILITY OF PNEUMATIC and HYDRAULIC VALVES These three tutorials will not be found in any examination syllabus. They have been added to the web site for engineers seeking knowledge on why valve elements

More information

- The value of a state function is independent of the history of the system. - Temperature is an example of a state function.

- The value of a state function is independent of the history of the system. - Temperature is an example of a state function. First Law of hermodynamics 1 State Functions - A State Function is a thermodynamic quantity whose value deends only on the state at the moment, i. e., the temerature, ressure, volume, etc - he value of

More information

CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS. Blank

CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS. Blank CHAPTER 7 THE SECOND LAW OF THERMODYNAMICS Blank SONNTAG/BORGNAKKE STUDY PROBLEM 7-1 7.1 A car engine and its fuel consumption A car engine produces 136 hp on the output shaft with a thermal efficiency

More information

ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6

ES-7A Thermodynamics HW 1: 2-30, 32, 52, 75, 121, 125; 3-18, 24, 29, 88 Spring 2003 Page 1 of 6 Spring 2003 Page 1 of 6 2-30 Steam Tables Given: Property table for H 2 O Find: Complete the table. T ( C) P (kpa) h (kj/kg) x phase description a) 120.23 200 2046.03 0.7 saturated mixture b) 140 361.3

More information

Thermodynamics - Example Problems Problems and Solutions

Thermodynamics - Example Problems Problems and Solutions Thermodynamics - Example Problems Problems and Solutions 1 Examining a Power Plant Consider a power plant. At point 1 the working gas has a temperature of T = 25 C. The pressure is 1bar and the mass flow

More information

01 The Nature of Fluids

01 The Nature of Fluids 01 The Nature of Fluids WRI 1/17 01 The Nature of Fluids (Water Resources I) Dave Morgan Prepared using Lyx, and the Beamer class in L A TEX 2ε, on September 12, 2007 Recommended Text 01 The Nature of

More information

Fluid Mechanics: Static s Kinematics Dynamics Fluid

Fluid Mechanics: Static s Kinematics Dynamics Fluid Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three

More information

3 Energy. 3.3. Non-Flow Energy Equation (NFEE) Internal Energy. MECH 225 Engineering Science 2

3 Energy. 3.3. Non-Flow Energy Equation (NFEE) Internal Energy. MECH 225 Engineering Science 2 MECH 5 Egieerig Sciece 3 Eergy 3.3. No-Flow Eergy Equatio (NFEE) You may have oticed that the term system kees croig u. It is ecessary, therefore, that before we start ay aalysis we defie the system that

More information

Pressure Drop in Air Piping Systems Series of Technical White Papers from Ohio Medical Corporation

Pressure Drop in Air Piping Systems Series of Technical White Papers from Ohio Medical Corporation Pressure Dro in Air Piing Systems Series of Technical White Paers from Ohio Medical Cororation Ohio Medical Cororation Lakeside Drive Gurnee, IL 600 Phone: (800) 448-0770 Fax: (847) 855-604 info@ohiomedical.com

More information

FUNDAMENTALS OF ENGINEERING THERMODYNAMICS

FUNDAMENTALS OF ENGINEERING THERMODYNAMICS FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant

More information

1. A belt pulley is 3 ft. in diameter and rotates at 250 rpm. The belt which is 5 ins. wide makes an angle of contact of 190 over the pulley.

1. A belt pulley is 3 ft. in diameter and rotates at 250 rpm. The belt which is 5 ins. wide makes an angle of contact of 190 over the pulley. Sample Questions REVISED FIRST CLASS PARTS A1, A2, AND A3 (NOTE: these questions are intended as representations of the style of questions that may appear on examinations. They are not intended as study

More information

Boiler efficiency measurement. Department of Energy Engineering

Boiler efficiency measurement. Department of Energy Engineering Boiler efficiency measurement Department of Energy Engineering Contents Heat balance on boilers Efficiency determination Loss categories Fluegas condensation principals Seasonal efficiency Emission evaluation

More information

Exergy Analysis of a Water Heat Storage Tank

Exergy Analysis of a Water Heat Storage Tank Exergy Analysis of a Water Heat Storage Tank F. Dammel *1, J. Winterling 1, K.-J. Langeheinecke 3, and P. Stephan 1,2 1 Institute of Technical Thermodynamics, Technische Universität Darmstadt, 2 Center

More information

THE BAROMETRIC FALLACY

THE BAROMETRIC FALLACY THE BAROMETRIC FALLACY It is often assumed that the atmosheric ressure at the surface is related to the atmosheric ressure at elevation by a recise mathematical relationshi. This relationshi is that given

More information

18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a

18 Q0 a speed of 45.0 m/s away from a moving car. If the car is 8 Q0 moving towards the ambulance with a speed of 15.0 m/s, what Q0 frequency does a First Major T-042 1 A transverse sinusoidal wave is traveling on a string with a 17 speed of 300 m/s. If the wave has a frequency of 100 Hz, what 9 is the phase difference between two particles on the

More information

MODEL OF THE PNEUMATIC DOUBLE ACTING CYLINDER COMPILED BY RHD RESISTANCES

MODEL OF THE PNEUMATIC DOUBLE ACTING CYLINDER COMPILED BY RHD RESISTANCES Journal of alied science in the thermodynamics and fluid mechanics Vol. 3, No. 1/009, ISSN 180-9388 MODEL OF THE PNEUMATIC DOUBLE ACTING CYLINDER COMPILED BY RHD RESISTANCES *Lukáš DVOŘÁK * Deartment of

More information

Gases. Macroscopic Properties. Petrucci, Harwood and Herring: Chapter 6

Gases. Macroscopic Properties. Petrucci, Harwood and Herring: Chapter 6 Gases Petrucci, Harwood and Herring: Chapter 6 CHEM 1000A 3.0 Gases 1 We will be looking at Macroscopic and Microscopic properties: Macroscopic Properties of bulk gases Observable Pressure, volume, mass,

More information

Answer, Key Homework 6 David McIntyre 1

Answer, Key Homework 6 David McIntyre 1 Answer, Key Homework 6 David McIntyre 1 This print-out should have 0 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

More information

Steady Heat Conduction

Steady Heat Conduction Steady Heat Conduction In thermodynamics, we considered the amount of heat transfer as a system undergoes a process from one equilibrium state to another. hermodynamics gives no indication of how long

More information

ME 201 Thermodynamics

ME 201 Thermodynamics ME 0 Thermodynamics Second Law Practice Problems. Ideally, which fluid can do more work: air at 600 psia and 600 F or steam at 600 psia and 600 F The maximum work a substance can do is given by its availablity.

More information

Physics 1114: Unit 6 Homework: Answers

Physics 1114: Unit 6 Homework: Answers Physics 1114: Unit 6 Homework: Answers Problem set 1 1. A rod 4.2 m long and 0.50 cm 2 in cross-sectional area is stretched 0.20 cm under a tension of 12,000 N. a) The stress is the Force (1.2 10 4 N)

More information

OUTCOME 2 INTERNAL COMBUSTION ENGINE PERFORMANCE. TUTORIAL No. 5 PERFORMANCE CHARACTERISTICS

OUTCOME 2 INTERNAL COMBUSTION ENGINE PERFORMANCE. TUTORIAL No. 5 PERFORMANCE CHARACTERISTICS UNIT 61: ENGINEERING THERMODYNAMICS Unit code: D/601/1410 QCF level: 5 Credit value: 15 OUTCOME 2 INTERNAL COMBUSTION ENGINE PERFORMANCE TUTORIAL No. 5 PERFORMANCE CHARACTERISTICS 2 Be able to evaluate

More information

THE PUBLISHING HOUSE PROCEEDINGS OF THE ROMANIAN ACADEMY, Series A, OF THE ROMANIAN ACADEMY Volume 11, Number 2/2010, pp. 148 155

THE PUBLISHING HOUSE PROCEEDINGS OF THE ROMANIAN ACADEMY, Series A, OF THE ROMANIAN ACADEMY Volume 11, Number 2/2010, pp. 148 155 THE PUBLISHING HOUSE PROCEEDINGS OF THE ROMANIAN ACADEMY, Series A, OF THE ROMANIAN ACADEMY Volume, Number /00,. 48 55 NEW CONCEPT FOR THE RUNNING OF ENGINES Nicolae CHIOREANU, Şerban CHIOREANU University

More information

FEASIBILITY OF A BRAYTON CYCLE AUTOMOTIVE AIR CONDITIONING SYSTEM

FEASIBILITY OF A BRAYTON CYCLE AUTOMOTIVE AIR CONDITIONING SYSTEM FEASIBILITY OF A BRAYTON CYCLE AUTOMOTIVE AIR CONDITIONING SYSTEM L. H. M. Beatrice a, and F. A. S. Fiorelli a a Universidade de São Paulo Escola Politécnica Departamento de Engenharia Mecânica Av. Prof.

More information

Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional

Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional Chapter 14 Fluid Mechanics. Solutions of Selected Problems 14.1 Problem 14.18 (In the text book) Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional area

More information

Natural Convection. Buoyancy force

Natural Convection. Buoyancy force Natural Convection In natural convection, the fluid motion occurs by natural means such as buoyancy. Since the fluid velocity associated with natural convection is relatively low, the heat transfer coefficient

More information

OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS

OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS Unit 41: Fluid Mechanics Unit code: T/601/1445 QCF Level: 4 Credit value: 15 OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS 1. Be able to determine the behavioural characteristics and parameters

More information

Pump Formulas Imperial and SI Units

Pump Formulas Imperial and SI Units Pump Formulas Imperial and Pressure to Head H = head, ft P = pressure, psi H = head, m P = pressure, bar Mass Flow to Volumetric Flow ṁ = mass flow, lbm/h ρ = fluid density, lbm/ft 3 ṁ = mass flow, kg/h

More information

p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh

p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1

More information

9. Forced Convection Correlations

9. Forced Convection Correlations Part B: Heat Transfer incials in Electronics Cooling 9. Forced Convection Correlations Our rimary objective is to determine heat transfer coefficients (local and average) for different flow geometries

More information

4. Introduction to Heat & Mass Transfer

4. Introduction to Heat & Mass Transfer 4. Introduction to Heat & Mass Transfer This section will cover the following concepts: A rudimentary introduction to mass transfer. Mass transfer from a molecular point of view. Fundamental similarity

More information

GAS TURBINE PERFORMANCE WHAT MAKES THE MAP?

GAS TURBINE PERFORMANCE WHAT MAKES THE MAP? GAS TURBINE PERFORMANCE WHAT MAKES THE MAP? by Rainer Kurz Manager of Systems Analysis and Field Testing and Klaus Brun Senior Sales Engineer Solar Turbines Incororated San Diego, California Rainer Kurz

More information

Fluids and Solids: Fundamentals

Fluids and Solids: Fundamentals Fluids and Solids: Fundamentals We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation.

More information

THE KINETIC THEORY OF GASES

THE KINETIC THEORY OF GASES Chapter 19: THE KINETIC THEORY OF GASES 1. Evidence that a gas consists mostly of empty space is the fact that: A. the density of a gas becomes much greater when it is liquefied B. gases exert pressure

More information

DET: Mechanical Engineering Thermofluids (Higher)

DET: Mechanical Engineering Thermofluids (Higher) DET: Mechanical Engineering Thermofluids (Higher) 6485 Spring 000 HIGHER STILL DET: Mechanical Engineering Thermofluids Higher Support Materials *+,-./ CONTENTS Section : Thermofluids (Higher) Student

More information

ENGINEERING INFORMATION Hot water and steam service

ENGINEERING INFORMATION Hot water and steam service ENGINEERING INFORMTION Hot water and steam service WHT IS STEM? Like other substances, water can exist in the form of a solid, when we call it ice; as a liquid when we call it water or as a gas when we

More information

oil liquid water water liquid Answer, Key Homework 2 David McIntyre 1

oil liquid water water liquid Answer, Key Homework 2 David McIntyre 1 Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

More information

CO 2 41.2 MPa (abs) 20 C

CO 2 41.2 MPa (abs) 20 C comp_02 A CO 2 cartridge is used to propel a small rocket cart. Compressed CO 2, stored at a pressure of 41.2 MPa (abs) and a temperature of 20 C, is expanded through a smoothly contoured converging nozzle

More information

One-Chip Linear Control IPS, F5106H

One-Chip Linear Control IPS, F5106H One-Chi Linear Control IPS, F5106H NAKAGAWA Sho OE Takatoshi IWAMOTO Motomitsu ABSTRACT In the fi eld of vehicle electrical comonents, the increasing demands for miniaturization, reliability imrovement

More information

Center of Gravity. We touched on this briefly in chapter 7! x 2

Center of Gravity. We touched on this briefly in chapter 7! x 2 Center of Gravity We touched on this briefly in chapter 7! x 1 x 2 cm m 1 m 2 This was for what is known as discrete objects. Discrete refers to the fact that the two objects separated and individual.

More information

STEAM TURBINE 1 CONTENT. Chapter Description Page. V. Steam Process in Steam Turbine 6. VI. Exhaust Steam Conditions, Extraction and Admission 7

STEAM TURBINE 1 CONTENT. Chapter Description Page. V. Steam Process in Steam Turbine 6. VI. Exhaust Steam Conditions, Extraction and Admission 7 STEAM TURBINE 1 CONTENT Chapter Description Page I Purpose 2 II Steam Turbine Types 2 2.1. Impulse Turbine 2 2.2. Reaction Turbine 2 III Steam Turbine Operating Range 2 3.1. Curtis 2 3.2. Rateau 2 3.3.

More information

The Second Law of Thermodynamics

The Second Law of Thermodynamics The Second aw of Thermodynamics The second law of thermodynamics asserts that processes occur in a certain direction and that the energy has quality as well as quantity. The first law places no restriction

More information

VISCOSITY OF A LIQUID. To determine the viscosity of a lubricating oil. Time permitting, the temperature variation of viscosity can also be studied.

VISCOSITY OF A LIQUID. To determine the viscosity of a lubricating oil. Time permitting, the temperature variation of viscosity can also be studied. VISCOSITY OF A LIQUID August 19, 004 OBJECTIVE: EQUIPMENT: To determine the viscosity of a lubricating oil. Time permitting, the temperature variation of viscosity can also be studied. Viscosity apparatus

More information

Hydraulic Formulary. Author: Houman Hatami Tel.: +49-9352-18-1225 Fax: +49-9352-18-1293 houman.hatami@boschrexroth.de 10.01.2013 1

Hydraulic Formulary. Author: Houman Hatami Tel.: +49-9352-18-1225 Fax: +49-9352-18-1293 houman.hatami@boschrexroth.de 10.01.2013 1 Hydraulic Formulary Author: Houman Hatami Tel.: +49-95-8-5 Fax: +49-95-8-9 houman.hatami@boschrexroth.de 0.0.0 CONTENTS RELATIONS BETWEEN UNITS... 4 IMPORTANT CHARACTERISTIC VALUES OF HYDRAULIC FLUIDS...

More information

MOS Transistors as Switches

MOS Transistors as Switches MOS Transistors as Switches G (gate) nmos transistor: Closed (conducting) when Gate = 1 (V DD ) D (drain) S (source) Oen (non-conducting) when Gate = 0 (ground, 0V) G MOS transistor: Closed (conducting)

More information

CFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER

CFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER International Journal of Advancements in Research & Technology, Volume 1, Issue2, July-2012 1 CFD SIMULATION OF SDHW STORAGE TANK WITH AND WITHOUT HEATER ABSTRACT (1) Mr. Mainak Bhaumik M.E. (Thermal Engg.)

More information

International Journal of Latest Research in Science and Technology Volume 4, Issue 2: Page No.161-166, March-April 2015

International Journal of Latest Research in Science and Technology Volume 4, Issue 2: Page No.161-166, March-April 2015 International Journal of Latest Research in Science and Technology Volume 4, Issue 2: Page No.161-166, March-April 2015 http://www.mnkjournals.com/ijlrst.htm ISSN (Online):2278-5299 EXPERIMENTAL STUDY

More information

Torsion Tests. Subjects of interest

Torsion Tests. Subjects of interest Chapter 10 Torsion Tests Subjects of interest Introduction/Objectives Mechanical properties in torsion Torsional stresses for large plastic strains Type of torsion failures Torsion test vs.tension test

More information

Chapter 18 Temperature, Heat, and the First Law of Thermodynamics. Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57

Chapter 18 Temperature, Heat, and the First Law of Thermodynamics. Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Thermodynamics study and application of thermal energy temperature quantity

More information

Performance Estimation of Water Thermal Energy Storage Tank by Using Simplified Simulation

Performance Estimation of Water Thermal Energy Storage Tank by Using Simplified Simulation Performance Estimation of Water Thermal Energy Storage Tank by Using Simplified Simulation M. Kubota, K. Sagara, T. Yamanaka, H. Kotani Division of Global Architecture, Graduate School of Engineering,

More information

Module 1 : Conduction. Lecture 5 : 1D conduction example problems. 2D conduction

Module 1 : Conduction. Lecture 5 : 1D conduction example problems. 2D conduction Module 1 : Conduction Lecture 5 : 1D conduction example problems. 2D conduction Objectives In this class: An example of optimization for insulation thickness is solved. The 1D conduction is considered

More information

Use the correct thermodynamic symbols. Determine the properties of a gas. Determine the properties of vapours. Determine the properties of liquids.

Use the correct thermodynamic symbols. Determine the properties of a gas. Determine the properties of vapours. Determine the properties of liquids. UNIT 61: ENGINEERING THERMODYNAMICS Unit code: D/601/1410 QCF level: 5 Credit value: 15 Students studying this unit should be sure that they can determine the various properties of fluids. If you are already

More information

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004

Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004 Statistical Mechanics, Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 Statistical Mechanics and Thermodynamics Thermodynamics Old & Fundamental Understanding of Heat (I.e. Steam) Engines Part of Physics Einstein

More information

FREESTUDY HEAT TRANSFER TUTORIAL 3 ADVANCED STUDIES

FREESTUDY HEAT TRANSFER TUTORIAL 3 ADVANCED STUDIES FREESTUDY HEAT TRANSFER TUTORIAL ADVANCED STUDIES This is the third tutorial in the series on heat transfer and covers some of the advanced theory of convection. The tutorials are designed to bring the

More information

Vapor Chambers. Figure 1: Example of vapor chamber. Benefits of Using Vapor Chambers

Vapor Chambers. Figure 1: Example of vapor chamber. Benefits of Using Vapor Chambers Vapor Chambers A vapor chamber is a high-end thermal management device that can evenly dissipate heat from a small source to a large platform of area (see Figure 1). It has a similar construction and mechanism

More information

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics.

Heat and Work. First Law of Thermodynamics 9.1. Heat is a form of energy. Calorimetry. Work. First Law of Thermodynamics. Heat and First Law of Thermodynamics 9. Heat Heat and Thermodynamic rocesses Thermodynamics is the science of heat and work Heat is a form of energy Calorimetry Mechanical equivalent of heat Mechanical

More information

Theoretical and Numerical Analysis of Heat Transfer in Pipeline System

Theoretical and Numerical Analysis of Heat Transfer in Pipeline System APCOM & ISCM -4 th December, 20, Singapore Theoretical and Numerical Analysis of Heat Transfer in Pipeline System Xiaowei Zhu, Hui Tang, *Hua Li, Jiahua Hong, Songyuan Yang School of Mechanical & Aerospace

More information

AN EXPERIMENTAL STUDY OF EXERGY IN A CORRUGATED PLATE HEAT EXCHANGER

AN EXPERIMENTAL STUDY OF EXERGY IN A CORRUGATED PLATE HEAT EXCHANGER International Journal of Mechanical Engineering and Technology (IJMET) Volume 6, Issue 11, Nov 2015, pp. 16-22, Article ID: IJMET_06_11_002 Available online at http://www.iaeme.com/ijmet/issues.asp?jtype=ijmet&vtype=6&itype=11

More information

Lecture 9, Thermal Notes, 3.054

Lecture 9, Thermal Notes, 3.054 Lecture 9, Thermal Notes, 3.054 Thermal Properties of Foams Closed cell foams widely used for thermal insulation Only materials with lower conductivity are aerogels (tend to be brittle and weak) and vacuum

More information

UNIT 2 REFRIGERATION CYCLE

UNIT 2 REFRIGERATION CYCLE UNIT 2 REFRIGERATION CYCLE Refrigeration Cycle Structure 2. Introduction Objectives 2.2 Vapour Compression Cycle 2.2. Simple Vapour Compression Refrigeration Cycle 2.2.2 Theoretical Vapour Compression

More information

PROPERTIES OF NATURAL GAS

PROPERTIES OF NATURAL GAS 1 PROPERIES OF NAURAL GAS Jón Steinar Guðmundsson PG4140 Naturgass Setember 16, 2008 Introduction Equations and roerties used in course Real gas law, z-factor, density and FVF Corresonding states Viscosity

More information

State Newton's second law of motion for a particle, defining carefully each term used.

State Newton's second law of motion for a particle, defining carefully each term used. 5 Question 1. [Marks 20] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding

More information

Energy Transport. Focus on heat transfer. Heat Transfer Mechanisms: Conduction Radiation Convection (mass movement of fluids)

Energy Transport. Focus on heat transfer. Heat Transfer Mechanisms: Conduction Radiation Convection (mass movement of fluids) Energy Transport Focus on heat transfer Heat Transfer Mechanisms: Conduction Radiation Convection (mass movement of fluids) Conduction Conduction heat transfer occurs only when there is physical contact

More information

Thermodynamics AP Physics B. Multiple Choice Questions

Thermodynamics AP Physics B. Multiple Choice Questions Thermodynamics AP Physics B Name Multiple Choice Questions 1. What is the name of the following statement: When two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium

More information

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME 2 ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS ENGINEERING COMPONENTS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS Structural members: struts and ties; direct stress and strain,

More information

So T decreases. 1.- Does the temperature increase or decrease? For 1 mole of the vdw N2 gas:

So T decreases. 1.- Does the temperature increase or decrease? For 1 mole of the vdw N2 gas: 1.- One mole of Nitrogen (N2) has been compressed at T0=273 K to the volume V0=1liter. The gas goes through the free expansion process (Q = 0, W = 0), in which the pressure drops down to the atmospheric

More information

Practice Problems on Pumps. Answer(s): Q 2 = 1850 gpm H 2 = 41.7 ft W = 24.1 hp. C. Wassgren, Purdue University Page 1 of 16 Last Updated: 2010 Oct 29

Practice Problems on Pumps. Answer(s): Q 2 = 1850 gpm H 2 = 41.7 ft W = 24.1 hp. C. Wassgren, Purdue University Page 1 of 16 Last Updated: 2010 Oct 29 _02 A centrifugal with a 12 in. diameter impeller requires a power input of 60 hp when the flowrate is 3200 gpm against a 60 ft head. The impeller is changed to one with a 10 in. diameter. Determine the

More information

Phys222 W11 Quiz 1: Chapters 19-21 Keys. Name:

Phys222 W11 Quiz 1: Chapters 19-21 Keys. Name: Name:. In order for two objects to have the same temperature, they must a. be in thermal equilibrium.

More information

Chapter 8: Heat Exchangers

Chapter 8: Heat Exchangers Chapter 8: Heat Exchangers Section 8.1: Introduction to Heat Exchangers 8.1-1 (8-1 in text) Dry air at T a,in = 30 C, and atmospheric pressure is blown at V a = 1.0 m 3 /s through a cross-flow heat exchanger

More information

APPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction.

APPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction. APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas expansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend

More information

A drop forms when liquid is forced out of a small tube. The shape of the drop is determined by a balance of pressure, gravity, and surface tension

A drop forms when liquid is forced out of a small tube. The shape of the drop is determined by a balance of pressure, gravity, and surface tension A drop forms when liquid is forced out of a small tube. The shape of the drop is determined by a balance of pressure, gravity, and surface tension forces. 2 Objectives Have a working knowledge of the basic

More information

This paper is also taken for the relevant Examination for the Associateship. For Second Year Physics Students Wednesday, 4th June 2008: 14:00 to 16:00

This paper is also taken for the relevant Examination for the Associateship. For Second Year Physics Students Wednesday, 4th June 2008: 14:00 to 16:00 Imperial College London BSc/MSci EXAMINATION June 2008 This paper is also taken for the relevant Examination for the Associateship SUN, STARS, PLANETS For Second Year Physics Students Wednesday, 4th June

More information

Ch 2 Properties of Fluids - II. Ideal Fluids. Real Fluids. Viscosity (1) Viscosity (3) Viscosity (2)

Ch 2 Properties of Fluids - II. Ideal Fluids. Real Fluids. Viscosity (1) Viscosity (3) Viscosity (2) Ch 2 Properties of Fluids - II Ideal Fluids 1 Prepared for CEE 3500 CEE Fluid Mechanics by Gilberto E. Urroz, August 2005 2 Ideal fluid: a fluid with no friction Also referred to as an inviscid (zero viscosity)

More information

Toolbox 6 THERMODYNAMIC AND TRANSPORT PROPERTIES OF MOIST AIR

Toolbox 6 THERMODYNAMIC AND TRANSPORT PROPERTIES OF MOIST AIR PPLIED INDURIL ENERGY ND ENIRONMENL MNGEMEN Z. K. Moray, D. D. Gozdenac Part III: FUNDMENL FOR NLYI ND CLCULION OF ENERGY ND ENIRONMENL PERFORMNCE lied Industrial Energy and Enironental Manageent Zoran

More information

Lecture 24 - Surface tension, viscous flow, thermodynamics

Lecture 24 - Surface tension, viscous flow, thermodynamics Lecture 24 - Surface tension, viscous flow, thermodynamics Surface tension, surface energy The atoms at the surface of a solid or liquid are not happy. Their bonding is less ideal than the bonding of atoms

More information

Sound pr. (db(a)) Phone: +90 216 394 12 82 Pressure hpa 960.075. Fax: +90 216 394 12 87 Capacity kw 1200.000

Sound pr. (db(a)) Phone: +90 216 394 12 82 Pressure hpa 960.075. Fax: +90 216 394 12 87 Capacity kw 1200.000 Hybrid-Re-Cooler V-2x8 - Offer no. 1328 Sound power (db(a)) Sound pr. (db(a)) Company Schenkel & Partner GmbH 1 Branch Ingenieurbüro Street Bahnhofstrasse 44 Country/ZIP/City 8 Zürich Installation Credit

More information

Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation

Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of

More information

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790. CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,

More information

Vector Algebra. Addition: (A + B) + C = A + (B + C) (associative) Subtraction: A B = A + (-B)

Vector Algebra. Addition: (A + B) + C = A + (B + C) (associative) Subtraction: A B = A + (-B) Vector Algebra When dealing with scalars, the usual math operations (+, -, ) are sufficient to obtain any information needed. When dealing with ectors, the magnitudes can be operated on as scalars, but

More information

THE PSEUDO SINGLE ROW RADIATOR DESIGN

THE PSEUDO SINGLE ROW RADIATOR DESIGN International Journal of Mechanical Engineering and Technology (IJMET) Volume 7, Issue 1, Jan-Feb 2016, pp. 146-153, Article ID: IJMET_07_01_015 Available online at http://www.iaeme.com/ijmet/issues.asp?jtype=ijmet&vtype=7&itype=1

More information

FLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference.

FLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference. FLUID MECHANICS TUTORIAL No.7 FLUID FORCES When you have completed this tutorial you should be able to Solve forces due to pressure difference. Solve problems due to momentum changes. Solve problems involving

More information

The First Law of Thermodynamics

The First Law of Thermodynamics Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot

More information

XI / PHYSICS FLUIDS IN MOTION 11/PA

XI / PHYSICS FLUIDS IN MOTION 11/PA Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A

More information

Direct Conversion of Wind Energy into Heat Using Joule Machine

Direct Conversion of Wind Energy into Heat Using Joule Machine International Conference on Environmental and Computer Science ICBEE vol.9() () IACSI ress, Singapore Direct Conversion of Wind Energy into Heat Using Joule Machine Roustiam Chakirov, and Yuriy Vagapov

More information

The paper addresses the boil-off in the cryogenic industry and details the specifics of it when applied to vehicle LNG tanks.

The paper addresses the boil-off in the cryogenic industry and details the specifics of it when applied to vehicle LNG tanks. What is Boil-off? Scope... 1 Boil-off in the cryogenic industry... 1... 1 Measures for boil-off... 2 LNG vehicle tanks... 2 Boil-off for the vehicle LNG Tank... 2 Heat management and types of fuel delivery

More information

Solution for Homework #1

Solution for Homework #1 Solution for Homework #1 Chapter 2: Multiple Choice Questions (2.5, 2.6, 2.8, 2.11) 2.5 Which of the following bond types are classified as primary bonds (more than one)? (a) covalent bonding, (b) hydrogen

More information

a cannonball = (P cannon P atmosphere )A cannon m cannonball a cannonball = (P cannon P atmosphere ) πd 2 a cannonball = 5.00 kg

a cannonball = (P cannon P atmosphere )A cannon m cannonball a cannonball = (P cannon P atmosphere ) πd 2 a cannonball = 5.00 kg 2.46 A piston/cylinder with a cross-sectional area of 0.01 m 3 has a mass of 100 resting on the stops as shown in the figure. With an outside atmospheric pressure of 100 kpa what should the water pressure

More information