THERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial.

Size: px
Start display at page:

Download "THERMODYNAMICS, FLUID AND PLANT PROCESSES. The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial."

Transcription

1 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 SOLUTIONS SELF ASSESSMENT EXERCISE No kg/s of steam flows in a pipe 40 mm bore at 200 bar pressure and 400oC. i. Look up the specific volume of the steam and determine the mean velocity in the pipe. ii. Determine the kinetic energy being transported per second. iii. Determine the enthalpy being transported per second. A πd 2 /4 π x (0.04) 2 / m 2 p 200bar θ 400 o C V m v m /s u V/A / m/s KE mu 2 /2 1 x /2 1. W h 2819 kj/kg (from the tables) SELF ASSESSMENT EXERCISE No.2 1. The shaft of a steam turbine produces 600 Nm torque at 0 rev/s. Calculate the work transfer rate from the steam. SP 2πNT 2π x 0 x W 2. A car engine produces 0 kw of power at 000 rev/min. Calculate the torque produced. T P/2πN 0000/(2π x 0) 9. Nm SELF ASSESSMENT EXERCISE No. 1. A non-flow system receives 80 kj of heat transfer and loses 20 kj as work transfer. What is the change in the internal energy of the fluid? U Q + W 80 + (-20) 60 kj 2. A non-flow system receives 100 kj of heat transfer and also 40 kj of work is transferred to it. What is the change in the internal energy of the fluid? U Q + W kj. A steady flow system receives 00 kw of heat and loses 200 kw of work. What is the net change in the energy of the fluid flowing through it? Φ + P E/s 00 + (-200) 00 kw E/s 4. A steady flow system loses 2 kw of heat also loses 4 kw of work. What is the net change in the energy of the fluid flowing through it? Φ + P E/s -2 + (-4) -6kW E/s

2 . A steady flow system loses kw of heat also loses 20 kw of work. The fluid flows through the system at a steady rate of 70 kg/s. The velocity at inlet is 20 m/s and at outlet it is 10 m/s. The inlet is 20 m above the outlet. Calculate the following. i. The change in K.E./s (-10. kw) ii. The change in P.E/s (-1.7 kw) iii. The change in enthalpy/s (1.2 kw) Φ - kw KE/s (m/2)( v v 2 1 ) P -20 kw KE/s (70/2)( ) -10. kw m 70 kg/s v 1 20 m/s PE/s mg(z 2 - z 1 ) v 2 10 m/s PE/s 70 x 9.81(0-20) kw z 1 20 m z 2 0 m Φ + P H/s + KE/s + PE/s (-) + (-20) H/s + (-10.) + (-1.74) H/s 1.24 kw SELF ASSESSMENT EXERCISE No.4 1. Gas is contained inside a cylinder fitted with a piston. The gas is at 20oC and has a mass of 20 g. The gas is compressed with a mean force of 80 N which moves the piston 0 mm. At the same time Joules of heat transfer occurs out of the gas. Calculate the following. i. The work done.(4 J) ii. The change in internal energy. (-1 J) iii. The final temperature. (19.9 o C) Take cv as 718 J/kg K Q + W U W F x 80 x J U -1 J U m c v (θ 2 20) 0.02 x 718 x (θ 2 20) θ o C 2. A steady flow air compressor draws in air at 20oC and compresses it to 120oC at outlet. The mass flow rate is 0.7 kg/s. At the same time, kw of heat is transferred into the system. Calculate the following. i. The change in enthalpy per second. (70. kw) ii. The work transfer rate. (6. kw) Take cp as 100 J/kg K. Φ + P H/s (ignore PE and KE) H/s m c p T 0.7 x 1.00 (120 20) 70. kw + P 70. P 6. kw. A steady flow boiler is supplied with water at 1 kg/s, 100 bar pressure and 200oC. The water is heated and turned into steam. This leaves at 1 kg/s, 100 bar and 00oC. Using your steam tables, find the following. i. The specific enthalpy of the water entering. (86 kj/kg) ii. The specific enthalpy of the steam leaving. (7 kj/kg) iii. The heat transfer rate. (7.7 kw) H/s 1 (h 2 h 1 ) 1(7 86) 77 kw Φ + P H/s P 0 Φ 7.7 MW

3 4. A pump delivers 0 dm/min of water from an inlet pressure of 100 kpa to an outlet pressure of MPa. There is no measurable rise in temperature. Ignoring K.E. and P.E, calculate the work transfer rate. (2.42 kw) Φ + P H/s + KE/s + PE/s KE/s PE/s 0 Φ 0 P H/s FE + U U 0 P FE p 2 V 2 p 1 V 1 V 1 V 2 0 x 10 - m /s P 0 x 10 - ( x x 10 ) 144 kw. A water pump delivers 10 dm/minute (0.1 m/min) drawing it in at 100 kpa and delivering it at 00 kpa. Assuming that only flow energy changes occur, calculate the power supplied to the pump. (860 W) Φ + P E/s FE/s Φ 0 P FE/s (0.1/60)(00 100) x kw 6. A steam condenser is supplied with 2 kg/s of steam at 0.07 bar and dryness fraction 0.9. The steam is condensed into saturated water at outlet. Determine the following. i. The specific enthalpies at inlet and outlet. (21 kj/kg and 16 kj/kg) ii. The heat transfer rate. (46 kw) h 1 h f + x h fg at 0.07 bar h (2409) 21.1 kj/kg h 2 h f 16 kj/kg Φ m(h 2 - h 1 ) 2( ) kj kg/s of gas is heated at constant pressure in a steady flow system from 10oC to 180oC. Calculate the heat transfer rate Φ. (7.4 kw) C p 1.1 kj/kg K Q 0.2 x 1.1 (180 10) 7.4 kj kg of gas is cooled from 120oC to 0oC at constant volume in a closed system. Calculate the heat transfer. (-16.8 kj) Cv 0.8 kj/kg. Q 0. x 0.8 (0 120) kj

4 SELF ASSESSMENT EXERCISE No. 1. A vapour is expanded from 12 bar and 0 cm to 10 cm and the resulting pressure is 6 bar. Calculate the index of compression n. (0.6) p 1 12 bar p 2 6 bar V 1 0 cm V 2 10 cm 12 x 0 n 6 x 10 n 12/6 2 (10/0) n n 2 n ln2 n ln n ln2/ ln a.A gas is compressed from 200 kpa and 00 cm to 800 kpa by the law pv1.4c. Calculate the new volume. (111.4 cm ) 200 x V V 2 00(200/800) 1/ cm 2.b.The gas was at 0oC before compression. Calculate the new temperature using the gas law pv/t C. (207 o C) T2 p 2 V 2 T 1 /(p 1 V 1 ) 800x x 2/(200x 00) K or 207 o C.a. A gas is expanded from 2 MPa and 0 cm to 10 cm by the law pv1.2 C. Calculate the new pressure. (06 kpa) 2 x p 2 x p 2 06 kpa.b. The temperature was 00oC before expansion. Calculate the final temperature. (14 o C) T2 p 2 V 2 T 1 /(p 1 V 1 ) 0.06 x 10 x 77/(2 x 0) 86.7 K or 14 o C SELF ASSESSMENT EXERCISE No A gas is expanded from 1 MPa and 1000oC to 100 kpa. Calculate the final temperature when the process is i. Isothermal (n1) (1000 o C) ii Polytropic (n1.2) (94 o C) iii. Adiabatic (γ 1.4) (86 o C) iv. Polytropic (n 1.6) (264 o C) p 1 1 MPa p MPa T K ISOTHERMAL POLYTROPIC ADIABATIC POLYTROPIC T 2 T o C or 127 k T 2 T 1 (p 2 / p 1 ) 1-1/n 127(0.1) K or 94 o C T 2 T 1 (p 2 / p 1 ) 1-1/γ 127(0.1) K or 86 o C T 2 T 1 (p 2 / p 1 ) 1-1/n 127(0.1) K or 264 o C

5 2. A gas is compressed from 120 kpa and 1oC to 800 kpa. Calculate the final temperature when the process is i. Isothermal (n1) (1 o C) ii. Polytropic (n1.) (17 o C) iii Adiabatic (γ1.4) (222 o C) iv. Polytropic (n 1.) (269 o C) p kpa p kpa T K ISOTHERMAL POLYTROPIC ADIABATIC POLYTROPIC T 2 T K o C or 1 o C T 2 T 1 (p 2 / p 1 ) 1-1/n 288(800/120) K or 17 o C T 2 T 1 (p 2 / p 1 ) 1-1/γ 288(800/120) K or 222 o C T 2 T 1 (p 2 / p 1 ) 1-1/n 288(800/120) K or 269 o C. A gas is compressed from 200 kpa and 20oC to 1.1 MPa by the law pv1.c. The mass is 0.02 kg. cp100 J/kg K. c v 718 J/kg K. Calculate the following. i. The final temperature. (44 K) ii. The change in internal energy (2.0 kj) iii. The change in enthalpy (2.84 kj) p kpa p MPa T 1 29 K T 2 T 1 (p 2 / p 1 ) 1-1/n 29(1.1/0.2) K U 0.02 x ( ) kj H 0.02 x 1.00( ) 2.88 kj 4. A gas is expanded from 900 kpa and 1200oC to 120 kpa by the law pv1.4 C. The mass is 0.01 kg. cp1100 J/kg K cv 70 J/kg K Calculate the following. i. The final temperature. ii. The change in internal energy iii. The change in enthalpy p kpa p kpa T K T 2 T 1 (p 2 / p 1 ) 1-1/n 147(120/900) K U 0.01 x 0.7 ( ) -7.2 kj H 0.01 x 1.1( ) kj

6 SELF ASSESSMENT EXERCISE No.7 1. kg/s of steam is expanded in a turbine from 10 bar and 200oC to 1. bar by the law pv1.2c. Determine the following. i. The initial and final volumes. (0.618 m and m ) ii. The dryness fraction after expansion. (0.86) iii. The initial and final enthalpies. (2829 kj/kg and 288 kj/kg) iv. The change in enthalpy kw) p 1 10 bar p 2 1. bar T 1 47 K θ oc V 1 m v 1 x m 10 x (V 2 ) 1.2 V 2 m v 2 / 1 m /kg V 2 x m v g x /( x 1.19) 0.86 h kj/kg h (2226) 288 kj/kg H ( ) MW kg/s of steam is expanded from 70 bar and 40oC to 0.0 bar by the law pv1. C. Determine the following. i. The initial and final volumes. (0.066 m /kg and 17.4 m /kg) ii. The dryness fraction after expansion. (0.411) iii. The initial and final enthalpies. (287 kj/kg and 11 kj/kg) iv. The change in enthalpy. (-228 kw) p 1 70 bar p bar θ 1 40 oc v m /kg 70 x (V 2 ) 1. v 2 m /kg x 28.2 x h kj/kg h (242) 11.9 kj/kg H 1. ( ) kw. A horizontal cylindrical vessel is divided into two sections each 1m volume, by a nonconducting piston. One section contains steam of dryness fraction 0. at a pressure of 1 bar, while the other contains air at the same pressure and temperature as the steam. Heat is transferred to the steam very slowly until its pressure reaches 2 bar. Assume that the compression of the air is adiabatic (γ1.4) and neglect the effect of friction between the piston and cylinder. Calculate the following. i. The final volume of the steam. ii. The mass of the steam. iii. The initial internal energy of the steam. iv The final dryness fraction of the steam. v. The final internal energy of the steam. vi. The heat added to the steam. t s at 1 bar 99.6 o C so the initial air temperature is 99.6 o C AIR Adiabatic compression T 2 T 1 (p 2 / p 1 ) 1-1/γ 72.6(2/1) K m pv/rt 1 x 10 x 1/(287 x 72.6) 0.9 kg U 0.96 x 718 x( ) 4.86 J Q + W U Q 0 hence W 4.86 J p 1 V 1 /T 1 p 2 V 2 /T 2 V m V 0.9 m p V p V 2x10 x x10 x 1 W J γ 1 0.4

7 STEAM V m W kj Q U W u 1 u f + x u fg at 1 bar ( ) kj/kg V 1 1 m x m v g at 1 bar m 1/(0. x 1.694) 1.97 kg V x m v g at 2 bar x 1.9/(0.886 x 1.97) 0.8 u 2 u f + x u fg at 2 bar ( ) kj/kg Q kj SELF ASSESSMENT EXERCISE No g of steam at 10 bar and 0oC expands reversibly in a closed system to 2 bar by the law pv1.c. Calculate the following. i. The initial volume. ( m ) ii. The final volume. ( m ) iii. The work done. (-2.92 kj) p 1 10 bar p 2 2 bar θ 1 0 oc m 10 g From the tables v m /kg V x m p 1 V 1 p2 V 2 10 x V 2 V m p2v2 p1v1 2x10 x x 10 x W 2921J g of gas at 20oC and 1 bar pressure is compressed to 9 bar by the law pv1.4 C. Taking the gas constant R 287 J/kg K calculate the work done. (Note that for a compression process the work will turn out to be positive if you correctly identify the initial and final conditions). (.67 kj) m 20 g T 1 29 K p 1 1 bar p 2 9 bar p 1 V p2 V /1.4 p2 T2 T p 1 mr( T1 T ) 0.02 x n W () 49 K ( ).68 kj. Gas at 600 kpa and 0.0 dm is expanded reversibly to 100 kpa by the law pv1. C. Calculate the work done. V m p kpa p kpa 1/ x 0.0 p 1 V 1 p2 V 2 V m 100 p V2 p1v1 100 x10 x x 10 x W J

8 4. 1 g of gas is compressed isothermally from 100 kpa and 20oC to 1 MPa pressure. The gas constant is 287 J/kg K. Calculate the work done. (2.9 kj) m 0.01 kg T 1 29 K p kpa p 2 1 MPa p 1 V 1 p 2 V 2 W p V ln(p 2 / p 1 ) m R T ln(p 2 / p 1 ) 0.01 x 287 x 29 x ln(10) 2.9 kj. Steam at 10 bar with a volume of 80 cm is expanded reversibly to 1 bar by the law pvc. Calculate the work done. ( kj) V 1 80 cm p 1 10 bar p 2 1 bar p 1 V 1 p 2 V 2 V 2 p 1 V 1 / p cm W p V ln(v 1 / V 2 ) 10 x 10 x 80 x 10-6 ln(0.1) J 6. Gas fills a cylinder fitted with a frictionless piston. The initial pressure and volume are 40 MPa and 0.0 dm respectively. The gas expands reversibly and polytropically to 0. MPa and 1 dm respectively. Calculate the index of expansion and the work done. (1.46 and -.24 kj) p 1 40 MPa V dm p 2 0. MPa V 2 1 dm p 1 V 1 n p 2 V 2 n (40/0.) (1/0.0) n n n x10 p V2 p1v1 W x 1 x x 10 x 0.0 x KJ 7. An air compressor commences compression when the cylinder contains 12 g at a pressure is 1.01 bar and the temperature is 20 o C. The compression is completed when the pressure is 7 bar and the temperature 90 o C. (1.124 and 1944 J) The characteristic gas constant R is 287 J/kg K. Assuming the process is reversible and polytropic, calculate the index of compression and the work done. p bar T 1 29 K p 2 7 bar T 2 6 K T T 1 1/n 1 1/n 2 p /n p /n ln(1.29)/ln n mr(t T 1) x 287(6 29) W J

9 SELF ASSESSMENT EXERCISE No.9 Take Cv 718 J/kg K and R 287 J/kg K throughout dm of gas at 100 kpa and 20oC is compressed to 1.2 MPa reversibly by the law pv1.2 C. Calculate the following. i. The final volume. (0.126 dm ) ii. The work transfer. (27 J) iii. The final temperature. (170oC) iv. The mass. (1.189 g) v. The change in internal energy. (128 J) vi. The heat transfer. (-128 J) V 1 1 dm T 1 29 K p kpa p MPa r 1/ x 1 p 1 V 1 p2 V 2 V dm x p V p V 1.2 x10 x x10 10 x 10 W J p2v2 1.2 x 10 x x 29 T2 T1 44 K p1v1 10 x 1 m pv/rt 10 x 10 - /(287 x 29) kg U m c v T x 718 x (44-29) 128 J Q + W U Q kj kg of gas at 20 bar and 1100oC is expanded reversibly to 2 bar by the law pv1. C in a closed system. Calculate the following. i. The initial volume. (9.8 dm ) ii. The final volume. (8 dm ) iii. The work transfer. (-27 kj) iv. The change in internal energy. (-20. kj) v. The heat transfer. (6.7 kj) p 1 20 bar T 1 17 K p 2 2 bar m 0.0 kg pv m R T V x 287 x 17/20 x m 1/ x 10 x p 1 V 1 p2 V 2 V m 2 x 10 p V p V 2 x10 x x 10 x W kj 0. p2v2 T2 T1 807 K p1v1 U m c v T 0.0 x 718 x (807-17) -20. kj Q + W U Q kj

10 kg of air at 700 kpa and 800oC is expanded adiabatically to 100 kpa in a closed system. Taking γ 1.4 calculate the following. i. The final temperature. (61.4 K) ii. The work transfer. (26. kj) iii. The change in internal energy. (-26. J) m 0.08 kg T K p kpa p kpa 1 1/ γ p2 100 T2 T K p m R T 0.08 x 287(61 107) W kj γ U m c v T 0.08 x 718(61 107) kj 4. A horizontal cylinder is fitted with a frictionless piston and its movement is restrained by a spring as shown. a. The spring force is directly proportional to movement such that F/ x k Show that the change in pressure is directly proportional to the change in volume such that p/ V k/a 2 b. The air is initially at a pressure and temperature of 100 kpa and 00 K respectively. Calculate the initial volume such that when the air is heated, the pressure volume graph is a straight line that extends to the origin. (0. dm ) c. The air is heated making the volume three times the original value. Calculate the following. i. The mass. (0.8 g) ii. The final pressure. (00 kpa) iii. The final temperature. (2700 K) iv. The work done. (-200 kj) v. The change in internal energy. (917 J) vi. The heat transfer. (1.12 kj) p 1 1 bar T 1 00 K V 1 1 m p 2 bar p 2 bar V 2 x m V m W F x/2 A p x/2 V p/2 2 x 10 x 0.4/2 40 kj out of system W - 40 kj p2v2 x 1.4 x 00 T2 T1 p1v1 1x K m pv/rt 1 x 10 x 1/(287 x 00) kg U m c v T x 718( ) 800kJ Q kj

11 SELF ASSESSMENT EXERCISE No kg of dry saturated steam at 10 bar pressure is expanded reversibly in a closed system to 1 bar by the law pv1.2 C. Calculate the following. i. The initial volume. (8.9 dm ) ii. The final volume. (264 dm ) iii. The work transfer. (-62 kj) iv. The dryness fraction. (0.779) v. The change in internal energy. (-108 kj) vi. The heat transfer. (-46 kj) m 0.2 kg dry saturated steam p 1 10 bar p 2 1 bar V 1 m v g 0.2 x m x 10 x p 1 V 1 p2 V 2 V m 1x 10 p V p V 1 x10 x x 10 x W kj 0.2 V 2 m x v g at 1 bar x(1.694) x u 1 u g at 10 bar 284 kj/kg K U 1 m u x kj U 2 m u x kj U kj Q kj 2. Steam at 1 bar and 20 o C is expanded reversibly in a closed system to bar. At this pressure the steam is just dry saturated. For a mass of 1 kg calculate the following. i. The final volume. ii. The change in internal energy. iii. The work done. iv. The heat transfer. p 1 1 bar θ 1 20 o C p 2 bar 1/1.2 v m /kg v 2 v g at bar m /kg u kj/kg u 2 u g at bar 262 kj/kg U ( ) -1 kj/kg n n p1 V p 2 V ( 2.466) n n ln()/ln(2.466) p2v2 p1v1 x10 x x 10 x 0.12 W 187 kj Q U - W kj/kg

THERMODYNAMICS, 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, FLUID AND PLANT PROCESSES The tutorials are drawn from other subjects so the solutions are identified by the appropriate tutorial. THERMODYNAMICS TUTORIAL 1 LIQUIDS VAPOURS - GASES SAE

More information

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA. PRINCIPLES AND APPLICATIONS of THERMODYNAMICS NQF LEVEL 3 OUTCOME 2 -ENERGY TRANSFER

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA. PRINCIPLES AND APPLICATIONS of THERMODYNAMICS NQF LEVEL 3 OUTCOME 2 -ENERGY TRANSFER EDEXCEL NATIONAL CERTIFICATE/DIPLOMA PRINCIPLES AND APPLICATIONS of THERMODYNAMICS NQF LEEL OUTCOME -ENERGY TRANSFER TUTORIAL - CLOSED THERMODYNAMIC SYSTEMS CONTENT Be able to quantify energy transfer

More information

I. (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. 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 information

first law of ThermodyNamics

first 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 information

Chapter 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 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 information

Chapter 5. Mass and Energy Analysis of Control Volumes

Chapter 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 information

Chapter 5: The First Law of Thermodynamics: Closed Systems

Chapter 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 information

MAHALAKSHMI ENGINEERING COLLEGE

MAHALAKSHMI 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 information

Where F1 is the force and dl1 is the infinitesimal displacement, but F1 = p1a1

Where 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 information

First Law of Thermodynamics

First 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 information

Engineering Thermodynamics Solutions Manual

Engineering Thermodynamics Solutions Manual Engineering Thermodynamics Solutions Manual Prof. T.T. Al-Shemmeri Download free books at Prof. T.T. Al-Shemmeri Engineering Thermodynamics Solutions Manual 2 2012 Prof. T.T. Al-Shemmeri & bookboon.com

More information

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

CHAPTER 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 information

Dishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)

Dishwasher. 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 information

ME Thermodynamics I. Lecture Notes and Example Problems

ME 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 information

First Law of Thermodynamics Closed Systems

First 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 information

ME Thermodynamics I

ME 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 information

R13 SET - 1 '' ''' '' ' '''' Code No RT21033

R13 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 information

The First Law of Thermodynamics. By: Yidnekachew Messele

The 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 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.

+ 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 information

(1)5. Which of the following equations is always valid for a fixed mass system undergoing an irreversible or reversible process:

(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 information

CHAPTER 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 information

Thermodynamics ENGR360-MEP112 LECTURE 7

Thermodynamics ENGR360-MEP112 LECTURE 7 Thermodynamics ENGR360-MEP11 LECTURE 7 Thermodynamics ENGR360/MEP11 Objectives: 1. Conservation of mass principle.. Conservation of energy principle applied to control volumes (first law of thermodynamics).

More information

UNIT I Basic concepts and Work & Heat Transfer

UNIT 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 information

UBMCC11 - THERMODYNAMICS. B.E (Marine Engineering) B 16 BASIC CONCEPTS AND FIRST LAW PART- A

UBMCC11 - 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 information

R13. II B. Tech I Semester Regular Examinations, Jan THERMODYNAMICS (Com. to ME, AE, AME) PART- A

R13. 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 information

SECOND ENGINEER REG. III/2 APPLIED HEAT

SECOND ENGINEER REG. III/2 APPLIED HEAT SECOND ENGINEER REG. III/2 APPLIED HEAT LIST OF TOPICS A B C D E F G H I J K Pressure, Temperature, Energy Heat Transfer Internal Energy, Thermodynamic systems. First Law of Thermodynamics Gas Laws, Displacement

More information

THE FIRST LAW APPLIED TO STEADY FLOW PROCESSES

THE 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 information

BME-A PREVIOUS YEAR QUESTIONS

BME-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 information

ENT 254: Applied Thermodynamics

ENT 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 information

5/6/ :41 PM. Chapter 6. Using Entropy. Dr. Mohammad Abuhaiba, PE

5/6/ :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 information

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 13 June 2007

ECE309 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 information

Two mark questions and answers UNIT I BASIC CONCEPT AND FIRST LAW SVCET

Two 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 information

Chapter 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 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 information

III. Evaluating Properties. III. Evaluating Properties

III. Evaluating Properties. III. Evaluating Properties F. Property Tables 1. What s in the tables and why specific volumes, v (m /kg) (as v, v i, v f, v g ) pressure, P (kpa) temperature, T (C) internal energy, u (kj/kg) (as u, u i, u f, u g, u ig, u fg )

More information

SHRI RAMSWAROOP MEMORIAL COLLEGE OF ENGG. & MANAGEMENT

SHRI 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 information

To receive full credit all work must be clearly provided. Please use units in all answers.

To 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 information

Unit code: H/ QCF level: 5 Credit value: 15 OUTCOME 1 - THERMODYNAMIC SYSTEMS TUTORIAL 2

Unit code: H/ QCF level: 5 Credit value: 15 OUTCOME 1 - THERMODYNAMIC SYSTEMS TUTORIAL 2 Unit 43: Plant and Process Princiles Unit code: H/60 44 QCF level: 5 Credit value: 5 OUCOME - HERMODYNAMIC SYSEMS UORIAL Understand thermodynamic systems as alied to lant engineering rocesses hermodynamic

More information

ME6301- ENGINEERING THERMODYNAMICS UNIT I BASIC CONCEPT AND FIRST LAW PART-A

ME6301- ENGINEERING THERMODYNAMICS UNIT I BASIC CONCEPT AND FIRST LAW PART-A ME6301- ENGINEERING THERMODYNAMICS UNIT I BASIC CONCEPT AND FIRST LAW PART-A 1. What is meant by thermodynamics system? (A/M 2006) Thermodynamics system is defined as any space or matter or group of matter

More information

Exergy and the Dead State

Exergy 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 information

c 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)

c 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 information

Two mark questions and answers UNIT II SECOND LAW 1. Define Clausius statement. It is impossible for a self-acting machine working in a cyclic process, to transfer heat from a body at lower temperature

More information

Chapter 7. Entropy: A Measure of Disorder

Chapter 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 information

Lecture 35: Vapor power systems, Rankine cycle

Lecture 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 information

Dr Ali Jawarneh. Hashemite University

Dr 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 information

MAE 11. Homework 8: Solutions 11/30/2018

MAE 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 information

PTT 277/3 APPLIED THERMODYNAMICS SEM 1 (2013/2014)

PTT 277/3 APPLIED THERMODYNAMICS SEM 1 (2013/2014) PTT 77/3 APPLIED THERMODYNAMICS SEM 1 (013/014) 1 Energy can exist in numerous forms: Thermal Mechanical Kinetic Potential Electric Magnetic Chemical Nuclear The total energy of a system on a unit mass:

More information

Energy and Energy Balances

Energy 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 information

Chapter 4. Energy Analysis of Closed Systems

Chapter 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 information

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 04 AERONAUTICAL ENGINEERING TUTORIAL QUESTION BANK Course Name : THERMODYNAMICS Course Code : AME00 Regulation : IARE - R1 Year

More information

QUESTION BANK UNIT-1 INTRODUCTION. 2. State zeroth law of thermodynamics? Write its importance in thermodynamics.

QUESTION 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 information

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.

KNOWN: 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 information

ME 201 Thermodynamics

ME 201 Thermodynamics 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

More information

20 m neon m propane. g 20. Problems with solutions:

20 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 information

FINAL EXAM. ME 200 Thermodynamics I, Spring 2013 CIRCLE YOUR LECTURE BELOW:

FINAL 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 information

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad -500 043 MECHANICAL ENGINEERING TUTORIAL QUESTION BANK Name : THERMODYNAMICS Code : A30306 Class : II B. Tech I Semester Branch :

More information

Chapter 1: Basic Definitions, Terminologies and Concepts

Chapter 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 information

ME Thermodynamics I

ME Thermodynamics I HW-22 (25 points) Given: 1 A gas power cycle with initial properties as listed on the EFD. The compressor pressure ratio is 25:1 Find: 1 Sketch all the processes on a p-h diagram and calculate the enthalpy,

More information

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 3 - ENERGY TUTORIAL 2 HEAT

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 3 - ENERGY TUTORIAL 2 HEAT EDEXCEL NATIONAL CERTIFICATE/DIPLOMA SCIENCE FOR TECHNICIANS OUTCOME 3 - ENERGY TUTORIAL 2 HEAT 3. Energy Mechanical work, energy and power: work - energy relationship, gravitational potential energy,

More information

Chapter 3 First Law of Thermodynamics and Energy Equation

Chapter 3 First Law of Thermodynamics and Energy Equation Fundamentals of Thermodynamics Chapter 3 First Law of Thermodynamics and Energy Equation Prof. Siyoung Jeong Thermodynamics I MEE0-0 Spring 04 Thermal Engineering Lab. 3. The energy equation Thermal Engineering

More information

Engineering Thermodynamics

Engineering 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 information

KNOWN: Pressure, temperature, and velocity of steam entering a 1.6-cm-diameter pipe.

KNOWN: 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 information

12/21/2014 7:39 PM. Chapter 2. Energy and the 1st Law of Thermodynamics. Dr. Mohammad Suliman Abuhaiba, PE

12/21/2014 7:39 PM. Chapter 2. Energy and the 1st Law of Thermodynamics. Dr. Mohammad Suliman Abuhaiba, PE Chapter 2 Energy and the 1st Law of Thermodynamics 1 2 Homework Assignment # 2 Problems: 1, 7, 14, 20, 30, 36, 42, 49, 56 Design and open end problem: 2.1D Due Monday 22/12/2014 3 Work and Kinetic Energy

More information

ME 200 Final Exam December 14, :00 a.m. to 10:00 a.m.

ME 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 information

CHAPTER 8 ENTROPY. Blank

CHAPTER 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 information

MAE 320 HW 7B. 1e. For an isolated system, please circle the parameter which will change with time. (a) Total energy;

MAE 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 information

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd

More information

SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER 2 EXAMINATIONS 2014/2015 ME258. Thermodynamics

SCHOOL OF COMPUTING, ENGINEERING AND MATHEMATICS SEMESTER 2 EXAMINATIONS 2014/2015 ME258. Thermodynamics s SCHOOL OF COMPUING, ENGINEERING AND MAHEMAICS SEMESER EXAMINAIONS 04/05 ME58 hermodynamics ime allowed: WO hours Answer: Any FOUR Questions Items permitted: Any approved calculator Items supplied: Steam

More information

1. (10) Calorically perfect ideal air at 300 K, 100 kpa, 1000 m/s, is brought to rest isentropically. Determine its final temperature.

1. (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

ECE309 INTRODUCTION TO THERMODYNAMICS & HEAT TRANSFER. 20 June 2005

ECE309 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 information

ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:

ECE309 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 information

ME 200 Final Exam December 12, :00 a.m. to 10:00 a.m.

ME 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 information

Unit Workbook 2 - Level 5 ENG U64 Thermofluids 2018 UniCourse Ltd. All Rights Reserved. Sample

Unit Workbook 2 - Level 5 ENG U64 Thermofluids 2018 UniCourse Ltd. All Rights Reserved. Sample Pearson BTEC Level 5 Higher Nationals in Engineering (RQF) Unit 64: Thermofluids Unit Workbook 2 in a series of 4 for this unit Learning Outcome 2 Vapour Power Cycles Page 1 of 26 2.1 Power Cycles Unit

More information

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity

Consequences of Second Law of Thermodynamics. Entropy. Clausius Inequity onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd

More information

Thermodynamics 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 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 information

Answer Key THERMODYNAMICS TEST (a) 33. (d) 17. (c) 1. (a) 25. (a) 2. (b) 10. (d) 34. (b) 26. (c) 18. (d) 11. (c) 3. (d) 35. (c) 4. (d) 19.

Answer Key THERMODYNAMICS TEST (a) 33. (d) 17. (c) 1. (a) 25. (a) 2. (b) 10. (d) 34. (b) 26. (c) 18. (d) 11. (c) 3. (d) 35. (c) 4. (d) 19. HERMODYNAMICS ES Answer Key. (a) 9. (a) 7. (c) 5. (a). (d). (b) 0. (d) 8. (d) 6. (c) 4. (b). (d). (c) 9. (b) 7. (c) 5. (c) 4. (d). (a) 0. (b) 8. (b) 6. (b) 5. (b). (d). (a) 9. (a) 7. (b) 6. (a) 4. (d).

More information

Lecture 38: Vapor-compression refrigeration systems

Lecture 38: Vapor-compression refrigeration systems ME 200 Termodynamics I Lecture 38: Vapor-compression refrigeration systems Yong Li Sangai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Cuan Road Sangai, 200240, P. R. Cina Email

More information

Brown Hills College of Engineering & Technology

Brown Hills College of Engineering & Technology UNIT 4 Flow Through Nozzles Velocity and heat drop, Mass discharge through a nozzle, Critical pressure ratio and its significance, Effect of friction, Nozzle efficiency, Supersaturated flow, Design pressure

More information

CHAPTER 2: THE NATURE OF ENERGY

CHAPTER 2: THE NATURE OF ENERGY Principles of Engineering Thermodynamics st Edition Reisel Solutions Manual Full Download: http://testbanklive.com/download/principles-of-engineering-thermodynamics-st-edition-reisel-solutions-manual/

More information

Chapter Four fluid flow mass, energy, Bernoulli and momentum

Chapter Four fluid flow mass, energy, Bernoulli and momentum 4-1Conservation of Mass Principle Consider a control volume of arbitrary shape, as shown in Fig (4-1). Figure (4-1): the differential control volume and differential control volume (Total mass entering

More information

Spring_#8. Thermodynamics. Youngsuk Nam

Spring_#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 information

Readings for this homework assignment and upcoming lectures

Readings 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 information

c Dr. Md. Zahurul Haq (BUET) Entropy ME 203 (2017) 2 / 27 T037

c Dr. Md. Zahurul Haq (BUET) Entropy ME 203 (2017) 2 / 27 T037 onsequences of Second Law of hermodynamics Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & echnology BUE Dhaka-000, Bangladesh zahurul@me.buet.ac.bd

More information

1 st Law Analysis of Control Volume (open system) Chapter 6

1 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 information

ME 2322 Thermodynamics I PRE-LECTURE Lesson 23 Complete the items below Name:

ME 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 information

CHAPTER 2 ENERGY INTERACTION (HEAT AND WORK)

CHAPTER 2 ENERGY INTERACTION (HEAT AND WORK) CHATER ENERGY INTERACTION (HEAT AND WORK) Energy can cross the boundary of a closed system in two ways: Heat and Work. WORK The work is done by a force as it acts upon a body moving in direction of force.

More information

First Law of Thermodynamics

First Law of Thermodynamics First Law of Thermodynamics During an interaction between a system and its surroundings, the amount of energy gained by the system must be exactly equal to the amount of energy lost by the surroundings.

More information

ESO201A: Thermodynamics

ESO201A: 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 information

PROBLEM 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 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 information

Introduction CHAPTER Prime Movers. 1.2 Sources of Energy

Introduction CHAPTER Prime Movers. 1.2 Sources of Energy Introduction CHAPTER 1 1.1 Prime Movers Prime mover is a device which converts natural source of energy into mechanical work to drive machines for various applications. In olden days, man had to depend

More information

Introduction to Chemical Engineering Thermodynamics. Chapter 7. KFUPM Housam Binous CHE 303

Introduction to Chemical Engineering Thermodynamics. Chapter 7. KFUPM Housam Binous CHE 303 Introduction to Chemical Engineering Thermodynamics Chapter 7 1 Thermodynamics of flow is based on mass, energy and entropy balances Fluid mechanics encompasses the above balances and conservation of momentum

More information

CHAPTER 7 ENTROPY. Copyright Hany A. Al-Ansary and S. I. Abdel-Khalik (2014) 1

CHAPTER 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 information

Chapter One Reviews of Thermodynamics Update on 2013/9/13

Chapter One Reviews of Thermodynamics Update on 2013/9/13 Chapter One Reviews of Thermodynamics Update on 2013/9/13 (1.1). Thermodynamic system An isolated system is a system that exchanges neither mass nor energy with its environment. An insulated rigid tank

More information

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 8 Introduction to Vapour Power Cycle Today, we will continue

More information

T098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26

T098. 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 information

Availability and Irreversibility

Availability 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 information

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Objectives Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Conservation of Mass Conservation of Mass Mass, like energy, is a conserved

More information

Department of Mechanical Engineering Indian Institute of Technology New Delhi II Semester MEL 140 ENGINEERING THERMODYNAMICS

Department 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 information

Today lecture. 1. Entropy change in an isolated system 2. Exergy

Today 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 information

MAE 320 Thermodynamics HW 4 Assignment

MAE 320 Thermodynamics HW 4 Assignment MAE 0 Thermodynamics HW 4 Assignment The homework is de Friday, October 7 th, 06. Each problem is worth the points indicated. Copying of the soltion from any sorce is not acceptable. (). Mltiple choice

More information

Answers to questions in each section should be tied together and handed in separately.

Answers to questions in each section should be tied together and handed in separately. EGT0 ENGINEERING TRIPOS PART IA Wednesday 4 June 014 9 to 1 Paper 1 MECHANICAL ENGINEERING Answer all questions. The approximate number of marks allocated to each part of a question is indicated in the

More information