The First Law of Thermodynamics

JABATAN KEJURUTERAAN MEKANIKAL POLITEKNIK SEBERANG PERAI FIRST LAW OF RASHIDAH BINTI IBRAHIM ZAILINDA BINTI YASAK THERMODYNAMICS

FIRST LAW OF THERMODYNAMICS Rashidah binti Ibrahim Zailinda binti Yasak 2023 [JABATAN KEJURUTERAAN MEKANIKAL] ©All rights reserved for electronic, mechanical, recording, or otherwise, without prior permission in writing from Politeknik Seberang Perai.

ii eBook PSP | First Law of Thermodynamics All rights reserved No part of this publication may be translated or reproduced in any retrieval system, or transmitted in any form or by any means, electronic, mechanical, recording, or otherwise, without prior permission in writing from Politeknik Seberang Perai. Published by Politeknik Seberang Perai Jalan Permatang Pauh, 13500 Permatang Pauh Pulau Pinang Editor Rashidah binti Ibrahim Zailinda binti Yasak Content Reviewer Mohammad Najib bin Ramli Cover Designer Azilah binti Abd Rahim Tel : 04-538 3322 Fax : 04-538 9266 Email : [email protected] Website : www.psp.edu.my FB : politeknikseberangperai Ig : politeknikseberangperai eISBN : 978-967-2774-43-3 Rashidah Ibrahim, editor Zailinda Yasak, editor FIRST LAW OF THERMODYNAMICS / Editor Rashidah binti Ibrahim, Zailinda binti Yasak 2023 Politeknik Seberang Perai, Electronic books - - Thermodynamics - - Closed systems - - Open system - - Government publications - - Malaysia - -

First Law of Thermodynamics | eBook PSP iii Acknowledgment First of all we’d like to thank Allah almighty, the most merciful and compassionate, for his support, help and generosity. Peace and blessings be upon our beloved Prophet Muhammad PBUH who was never tired in spreading his love and guidance for us to be on the foundation of truth. We would like to thank profusely Mechanical Engineering Department, Seberang Perai Polytechnic and all the colleagues for support, effort and motivation for helping us completing this e-book. Not forgetting to the family members for their endless support and patience throughout the process of writing. Without those support and courage, this e-book might not complete, a big hug. May Allah shower everyone reading this e-book with more success, honour and blessing throughout your life… Rashidah binti Ibrahim Zailinda binti Yasak

iv eBook PSP | First Law of Thermodynamics Preface The topic of First Law of Thermodynamics, which is included in the course of Thermodynamics, is written for all Malaysian Polytechnics students, who are pursuing their Diploma in Mechanical Engineering. This e-book covers Chapter 3: First Law of Thermodynamics. This e-book serves basic understanding and simple notes of The First Law of Thermodynamics which contents Closed System and Open System in Thermodynamics. This e-book also gives various samples of previous Polytechnics Thermodynamics’ final examination questions. In addition, students could cover their studies of question no.2 and no.3 in Thermodynamics final examination. Hopefully this publication can enhance students to understand better as it contains interesting concise summaries and calculation work paths that students can follow easily and effectively.

First Law of Thermodynamics | eBook PSP v Table of Content 01 Introduction To The First Law of Thermodynamics 1.1 Introduction 1.2 Forms of Energy 1.3 Energy Transfer by Heat and Work 1.4 Mechanical Forms of Work 1.5 Heat, Work and Internal Energy for The Gas and Steam/Water 1 2 3 3 5 6 02 The Concept of The First Law of Thermodynamics in Closed System 2.1 Closed System 2.2 Specific Heat at Constant Pressure and Constant Volume, and Enthalpy 2.3 Non-Flow Energy 2.4 Non-Flow Processes (Isobaric, Isochoric, Isothermal, Adiabatic and Polytrophic) Example of Final Examination Questions 7 8 9 10 10 13 03 The Concept of The First Law of Thermodynamics In Open System 3.1 Open System 3.2 Steady -Flow Process 3.3 Energy Analysis of Steady-Flow Systems 3.4 Application of A Steady-Flow Energy Equation Example of Final Examination Questions 42 43 43 44 47 50 References 66

INTRODUCTION TO THE FIRST LAW OF THERMODYNAMICS OBJECTIVES: At the end of this chapter, students should be able to: ▪ Describe forms of energy ▪ Describe energy transfer by heat and energy transfer by work ▪ Classify the mechanical forms of work ▪ Relate heat, work and internal energy for the gas and steam/water (Source: https://cilisos.my/a-malaysian-woman-changed-theworlds-view-on-lightning-but-what-did-she-discover-anyway/)

2 eBook PSP | First Law of Thermodynamics 1.1 Introduction 1. First Law of Thermodynamics The first law of thermodynamics stated as: “ When a system goes through a thermodynamic cycle, the net work that the system does on its environment is equal to the net heat that the system receives from its surroundings”. Figure 1.1 : First law of thermodynamics. Energy cannot be generated or destroyed, but it can be converted or transferred, according to the First Law of Thermodynamics, often known as the conservation of energy. 2. Energy Conversion Energy can be found in many different forms and converted. Figure 1.2 : Energy conversion. dQ = dW (Source: https://byjus.com/physics/energy-conversion/) (Source: https://courses.lumenlearning.com/suny-physics/chapter/15-1-the-first-law-of-thermodynamics/)

First Law of Thermodynamics | eBook PSP 3 1.2 Forms of Energy The capacity of a body to function is its energy. There are many different types of energy, including gravitational, kinetic, potential, thermal or heat, mechanical, chemical, electrical, elastic, acoustic, and electrical energy. (Source: mskuksclass.weebly.com) Figure 1.3 : Forms of energy. 1.3 Energy Transfer by Heat and Work ENERGY TRANSFER: Only occurs across the boundaries of the system by interaction with the environment in forms of heat, work and mass flow. It represents the energy that a system gains or loses throughout the process.

4 eBook PSP | First Law of Thermodynamics Table 1.1 : Energy transfer by heat and work. ENERGY TRANSFER BY HEAT, Q The transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings. Example: heat addition and loss ENERGY TRANSFER BY WORK, W Energy interaction between system and its surroundings associated with a force acting through a distance. Example: a rotating shaft or a rising piston. Figure 1.4 : Sign convention for heat transfer. Heat can be transferred in 3 ways: No media transmitter is required. It is produced when the electronic structure of atoms or molecules is altered. Because of this, heat can move because electromagnetic waves are produced. Conduction Convection Radiation Transfer of energy of bulk motion of material containing a different of energy per volume than its surrounding. Due to interactions between the molecules, heat is transferred from the more energetic molecules to the less energetic molecules. Note that: Qin Heat energy flows into the system from the surroundings: +ve (heat addition) Qout Heat energy flows from the system to the surroundings : -ve (heat loss) Wout Work energy is transferred out the system to the surroundings : +ve (Wout ) Win Work energy is transferred from the surroundings to the system : -ve (Win )

First Law of Thermodynamics | eBook PSP 5 (Source: mskuksclass.weebly.com) Figure 1.5 : Mode of energy transfer. 1.4 Mechanical Forms of Work Table 1.2 : Example of some common mechanical forms of work. Shaft Work Transmission of energy using a rotating shaft. Example: ship's propeller. Spring Work An applied force causes a spring to lengthen. Work done on elastic solid bar Elastic solid bar's expansion or contraction caused by pressure. Work associated with the stretching a liquid film Stretching a liquid film using a moveable wire frame component. referred to as surface tension. Work done to raise or to accelerate a body Work is represented by potential energy or kinetic energy used to lift or accelerate a body.

6 eBook PSP | First Law of Thermodynamics 1.5 Heat, Work and Internal Energy for The Gas and Steam/Water Table 1.3 : Heat, Work and Internal Energy for The Gas and Steam/Water HEAT, Q • Heat is the energy in transit whenever temperature differences exist. • Symbol: Heat: Q; unit: kJ Specific heat: q; unit: kJ/kg WORK, W • A force created from any source can do work by moving an object through a displacement. • Symbol: Work: W; unit: kJ Specific work: w; unit: kJ/kg INTERNAL ENERGY, U • The amount of energy that a fluid has because of the microscopic motion of its molecules. • Energy stored in the fluid can be increased (+VE value) or decreased (-VE value). • Symbol: Internal energy: U; unit: kJ Specific internal energy: u; unit: kJ/kg Figure 1.6 : Heat, work and internal energy Heat rejected from the system Heat added to system Work done by system Work done on system U (Internal energy) +Q -Q +W -W

First Law of Thermodynamics | eBook PSP 7 THE CONCEPT OF THE FIRST LAW OF THERMODYNAMICS IN CLOSED SYSTEM OBJECTIVES: At the end of this chapter, students should be able to: ▪ Define a closed system ▪ Explain the specific heat at constant pressure and constant volume, and enthalpy ▪ Explain non-flow energy ▪ Determine the properties for constant pressure, constant volume, isothermal, adiabatic and polytrophic process (Source: Canva in atonergi.com)

8 eBook PSP | First Law of Thermodynamics 2.1 Closed System A closed system, on the other hand, can exchange only energy with its surroundings, not matter. Figure 2.1 : Differences between closed and open system. ► CLOSED SYSTEM ▪ control mass ▪ fixed amount of mass, no mass/matter can cross the boundary ▪ energy (heat & work) can cross the boundary ▪ non-flow process ▪ example: piston, hydraulic jack ► OPEN SYSTEM ▪ control volume ▪ involve mass/matter flow ▪ mass and energy (heat & work) can cross the boundary ▪ flow process ▪ example: boiler, compressor, turbine

First Law of Thermodynamics | eBook PSP 9 2.2 Specific Heat at Constant Pressure and Constant Volume, Internal Energy and Enthalpy 1. Specific heat at constant pressure and constant pressure The specific heat capacity of any substance is defined as the amount of heat energy required to raise the unit mass through one-degree temperature raise. Table 2.1 : Specific heat at constant pressure and constant pressure Specific Heat at Constant Pressure, cp ▪ If 1kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of the gas by 1 degree whilst the pressure of the gas remains constant, the amount of heat energy supplied is known as the specific heat capacity at constant pressure. ▪ Unit: cp = J/kgK or kJ/kgK Specific Heat at Constant Volume, cv ▪ If 1kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of the gas by 1 degree whilst the volume of the gas remains constant, the amount of heat energy supplied is known as the specific heat capacity at constant volume. ▪ Unit: cv = J/kgK or kJ/kgK 2. Enthalpy • Symbol: Enthalpy: H; unit: kJ Specific enthalpy: h; unit: kJ/kg Figure 2.2 : Enthalpy

10 eBook PSP | First Law of Thermodynamics 3. Relationship between Specific Heat, Constant Gas and Heat Ratio = = − = = ( − 1) = ( − 1) 2.3 Non-Flow Energy Non-flow energy is energy that flows through a system as heat and work but does not involve any mass moving into or out of the system. When there is no mass flow and the initial and final energies of a closed system are sometimes equal, the energy balance is simplified to Ein = Eout. (Energy in = Energy out). The majority of systems do not change their velocity or elevation during the process (∆KE = 0, ∆PE = 0). So, the First Law of Thermodynamics' principle of heat and work interactions can be used to express the energy balance for a cycle. ∆ = ∆ ∆ = ∆ − ∆ ( − ) +( − ) = 0 = = ∑ = ∑ 2.4 Non-Flow Processes (Isobaric, Isochoric, Isothermal, Adiabatic and Polytropic) Isobaric, isochoric, isothermal, adiabatic, and polytrophic processes are examples of non-flow processes. Where; Ru = Molar gas constant (kJ/kmolK) R = Specific gas constant (kJ/kgK) M = Molar mass (kg/kmol) cp = Specific heat at pressure constant (kJ/kgK) cv = Specific heat at pressure constant (kJ/kgK) = Specific heat ratio

First Law of Thermodynamics | eBook PSP 11 Table 2.2 : Types of non-flow process Process Graph (Source: https://www.sciencedirect.com/topics/engineering) Polytropic process The process that include expansion and compression of heat transfer

12 eBook PSP | First Law of Thermodynamics Table 2.3 : Formulas of non-flow process Process Formula Heat (Q) Work (W) Change of Internal Energy Isobaric -Constant pressure -1 = 2 = 1 1 = 2 2 = (2 − 1 ) = (2 − 1 ) ∆ = − Isochoric -Isometric -Constant volume -1 = 2 = 1 1 = 2 2 = (2 − 1 ) 0 ∆ = Isothermal - Constant temperature - 1 = 2 - = 11 = 22 = = 11 ( 2 1 ) = 11 ( 1 2 ) = ( 2 1 ) = ( 1 2 ) 0 Adiabatic - Isentropic - = 11 = 22 2 1 = [ 2 1 ] −1 2 1 = [ 1 2 ] −1 0 = 11 − 22 − 1 = (1 − 2 ) −1 ∆ = − ∆ = (2 − 1 ) Polytropic - = 11 = 22 2 1 = [ 2 1 ] −1 2 1 = [ 1 2 ] −1 = [ − − 1 ] × = + ∆ = 11 − 22 − 1 = (1 − 2 ) − 1 ∆ = − ∆ = (2 − 1 )

First Law of Thermodynamics | eBook PSP 13 Example of Final Examination Questions December 2015 Session Question List FOUR (4) characteristics of non-flow process. Answer 1. Mass cannot cross the boundary. 2. Energy in the form of heat, Q and work, W can cross the boundary. 3. Known as closed system or control mass. 4. Heat and work cannot enter and leave the system at the same time. December 2016 Session Question Define the following: i. First Law of Thermodynamics. ii. Law of Conservation of Energy. iii. Closed System. Answer i. First Law of Thermodynamics. a) Net heat supplied to the system is equal to the net work done to the surroundings. b) ∑Q=∑W ii. Law of Conservation of Energy. a) Energy cannot be created nor destroyed, but rather transformed from one state to another.

14 eBook PSP | First Law of Thermodynamics iii. Closed System. a) No mass can cross the boundary (control mass). b) Energy can cross the boundary in form of heat and work. c) Closed system is applicable to non-flow process. December 2017 Session Question Define the following terms: i. Heat transfer. ii. Specific heat. Answer i. Heat transfer Form of energy transferred between two system has difference temperature. ii. Specific heat Amount of heat energy required to raise the unit mass through 1oC temperature raise. June 2016 Session Question List FOUR (4) non-flow processes. Answer i. Isobaric. ii. Isothermal.

First Law of Thermodynamics | eBook PSP 15 iii. Isochoric/ isometric. iv. Adiabatic. v. Polytropic. June 2017 Session Question List THREE (3) characteristics of an adiabatic process. Answer i. No heat enters or leaves the gas. ii. The temperature of the gas changes. iii. The change in internal energy is equal to the mechanical work done. June 2019 Session Question Give TWO (2) differences between closed and open system. Answer Close system Open system ▪ control mass (non-flow process) ▪ control volume (flow process) ▪ fixed amount of mass, no mass/matter can cross the boundary ▪ involve mass/matter flow ▪ energy (heat & work) can cross the boundary ▪ mass and energy (heat & work) can cross the boundary ▪ example: piston, hydraulic jack ▪ example: boiler, compressor, turbine

16 eBook PSP | First Law of Thermodynamics 1: 2022/2023 Session Question Briefly describe the Non-Flow Process. Answer 1. Non-flow processes are thermodynamics processes in which there is no exchange of mass. 2. Occurs in a closed system which can exchange energy with surroundings. 3. Involving internal energy changes. 4. Require non-flowing fluid. 1: 2022/2023 Session Question Briefly discuss energy transfer by heat and energy transfer by work. Answer 1. Energy transfer by heat is energy interaction between system and its surroundings associated with temperature difference. 2. Energy transfer by work is energy interaction between system and its surroundings associated with a force acting through a distance.

First Law of Thermodynamics | eBook PSP 17 December 2015 Session Question A mass of 0.5 kg of air, initially at 130 oC, is heated at a constant pressure of 2 bar until the volume occupied is 0.0658 m3 . Calculate the heat supplied and the work done. Answer Given: = 0.5 = 130° +273 = 403 1 = 2 = 2 = 200 2 2 = 0.06583 22 = 2 2 = 22 2 = 200(0.0658) 0.5(0.287) 2 = 91.7 i. Heat supplied, Q = (2 − 1 ) = 0.5(1.005)(91.7− 403) = −. ii. Work done, W = (2 − 1 ) = 0.5(0.287)(91.7 − 403) = −.

18 eBook PSP | First Law of Thermodynamics December 2016 Session Question A tank containing a fluid which is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. If the tank and the fluid is in a control surface, find the change in internal energy of this control mass. State whether the change in internal energy is increased or decreased. Answer Given: = −5090 = −1500 since the process occured in non-flow process therefore the kinetic energy and the potential energy may be neglected. 2 − 1 = − 2 − 1 = −1500 − (−5090) − = (internal energy has increased) December 2017 Session Question An amount of gas inside an insulated vessel contain 45kJ of internal energy. The gas expands until the internal energy is reduced to 23kJ. Calculate the work produced by the gas. Answer Given: , = 0 1 = 45 2 = 23 ∆ = − 2 − 1 = − 23 − 45 = 0 − =

First Law of Thermodynamics | eBook PSP 19 December 2018 Session Question A quantity of gas has a pressure of 3.5 bar when the volume and the temperature are 0.03 m3 and 35 oC respectively. Specific gas constant for the gas is 0.29 kJ/kgK. Determine the mass of gas. Answer Given: 1 = 3.5 = 3.5 × 102 2 1 = 0.033 1 = 35° + 273 = 308 = 0.29 . = = = 3.5 × 102 (0.03) 0.29(308) = . June 2015 Session Question 0.01 kg of an ideal gas is contained in a pressure vessel that has a volume of 0.003 m3 . In this condition, the absolute pressure is 5 bar and the gas temperature 120 oC. Then, the gas is expanded until the pressure dropped to 1 bar and volume is 0.02 m3 . Calculate the molecular mass and final temperature of the gas. Answer Given: = 0.01

20 eBook PSP | First Law of Thermodynamics 1 = 0.0033 2 = 0.023 = 120° +273 = 393 1 = 5 = 500 2 2 = 1 = 100 2 i. Molecular mass, M 11 = 1 = 11 1 = 500(0.003) 0.01(393) = 0.382 / = = = 8.314 0.382 = . / ii. Final temperature, T2 22 = 2 2 = 22 2 = 100(0.02) 0.01(0.382) = .

First Law of Thermodynamics | eBook PSP 21 June 2016 Session Question 0.046 m3 of gas is contained in a rigid cylinder at a pressure of 300 kN/m2 and a temperature of 45 oC. The gas is compressed until the pressure reaches 1.27 MN/m2 . The gas is assumed to be a perfect gas. Given R=0.29 kJ/kgK, determine the: i. Mass of gas. ii. Final temperature of gas. Answer Given: 1 = 2 = 0.0463 1 = 300 2 2 = 1.27 2 = 1.27 × 103 2 1 = 45° + 273 = 318 = 0.29 . i. Mass of gas 11 = 1 = 11 1 = 300(0.046) 0.29(318) = . ii. Final temperature of gas 11 1 = 22 2 1 1 = 2 2 2 = 2 × 1 1 2 = 1.27 × 103 × 318 300 2 = 1346.2

22 eBook PSP | First Law of Thermodynamics June 2017 Session Question A quantity of air occupied a pressure of 1.2 bar, volume of 0.334 m3 and temperature of 29 oC. Then the air is compressed at constant pressure until the volume becomes 0.18 m3 . Calculate the mass and the final temperature for the air. Answer Given: 1 = 0.3343 , 2 = 0.183 1 = 1.2 = 1.2 × 102 2 1 = 29° + 273 = 302 = 0.287 . i. Mass 11 = 1 = 11 1 = 1.2 × 102 (0.334) 0.287(302) = . ii. Final temperature 1 1 = 2 2 2 = 2 × 1 1 2 = 0.18 × 302 0.334 = .

First Law of Thermodynamics | eBook PSP 23 June 2018 Session Question 0.25 kg of gas is at a temperature of 17 oC, volume 0.19 m3 and pressure 115 kN/m2 . If the gas has a value of Cv = 720 J/kgK, determine: i. Gas constant. ii. Specific heat at constant pressure. iii. Specific heat ratio. Answer Given: = 0.25 = 17° + 273 = 290 = 0.193 = 720 = 0.72 i. Gas constant = = = 115 × 0.19 0.25 × 290 = . / ii. Specific heat at constant pressure = − = + = 0.301+ 0.72 = . / iii. Specific heat ratio = = 1.021 0.72 = .

24 eBook PSP | First Law of Thermodynamics 1: 2022/2023 Session Question The heat transferred to the system is 500kJ/kg and the initial internal energy is 250kJ/kg greater than the final. Calculate the work done by the system. Answer = 500/ 1 > 2 = 250/ 2 − 1 = 0 − 250/ − = 2 − 1 500 − = −250 500 + 250 = = / 2: 2021/2022 Session Question Table 1(a) shows the information about a process of a closed system. Fill in the blanks in the table with the correct answers. PROCESS Q12 (kJ) W12 (kJ) (U2 - U1) (kJ) a. +60 -30 i. b. +110 ii. -40 c. iii. -80 +120 d. -40 +30 iv.

First Law of Thermodynamics | eBook PSP 25 Answer PROCESS Q12 (kJ) W12 (kJ) (U2 - U1) (kJ) a. +60 -30 i. 90 b. +110 ii. 150 -40 c. iii. 40 -80 +120 d. -40 +30 iv. -70 2: 2021/2022 Session Question A mass of 0.23 kg gas at a temperature of 20 oC, pressure 135 kN/m2 and volume 0.22 m3 . If the gas has a value of Cv=720 J/kgK, calculate the gas constant and specific heat at constant pressure. Answer Given: = 0.23 = 20° + 273 = 293 = 0.223 = 720 = 0.720/ i. Gas constant = = = 135(0.22) 0.23(293) = . /

26 eBook PSP | First Law of Thermodynamics ii. Specific heat at constant pressure = − = + = 0.441+ 0.720 = . / December 2016 Session Question Air with mass of 0.65 kg at pressure 15 bar and temperature 230 oC is expanded until its final volume is three times greater than its initial volume. The polytropic expansion process is according to the law PVn = C. Calculate: (Assume n = 1.37 and R = 0.287 kJ/kgK) i. Initial volume and final volume, m3 . ii. Final pressure, bar. iii. Final temperature, K. iv. Work done, kJ. Answer Given: = 0.65 1 = 230° + 273 = 503 2 = 31 = 1.37 = 0.287 / i. Initial volume and final volume, m3 11 = 1 1 = 1 1 1 = 0.65 (0.287)(503) 1500 1 = 0.0626 3

First Law of Thermodynamics | eBook PSP 27 2 = 31 2 = 3(0.0626) = . ii. Final pressure, bar = 11 = 22 2 = 1 ( 1 2 ) 2 = 1 ( 1 2 ) 2 = 1500 ( 0.0626 0.1877) 1.37 2 = 333.24 2 = . iii. Final temperature, K 2 1 = ( 1 2 ) −1 2 = 1 ( 1 2 ) −1 2 = 503 ( 0.0626 0.1877) 1.37−1 2 = 503(0.666) = iv. Work done, kJ = 11 − 22 − 1 = 1500(0.0626) − 333.24(0.1877) 1.37− 1 = 31.3959 0.37 = .

28 eBook PSP | First Law of Thermodynamics December 2017 Session Question A closed system containing 2 kg of air undergoes an isothermal process from 600 kN/m2 at 200 oC to 80 kN/m2 . Determine the initial volume of this system, the work done and the heat transferred during this process. Answer Given: = 2 1 = 600 2 2 = 80 2 1 = 200° + 273 = 473 i. Initial volume, v1 11 = 1 1 = 1 1 1 = 2(0.287)(473) 600 = . ii. Work done, W = 11 ( 600 80 ) = . iii. Heat transfer, Q = = .

First Law of Thermodynamics | eBook PSP 29 December 2018 Session Question A gas with 1.8 m3 of volume at initial temperature and pressure of 20 oC and 1.38 bar is compressed through the polytrophic process with index n = 1.3 in a cylinder capacity of 0.244 m3 volume. If the molecular mass of the gas is 29 kg/kmol and Ro = 8.3145 kJ/kmolK, calculate: i. Mass of the gas. ii. Pressure and temperature of gas after the compression. iii. Work done to compress the gas. Answer Given: 1 = 1.83 2 = 0.2443 = = 1.3 1 = 20° + 273 = 293 1 = 1.38 = 1.38 × 102 2 = 29/ = 8.3145/ i. Mass of the gas = = 8.3145 29 = 0.2867 / 11 = 1 = 11 1 = (1.38 × 102 )(1.8) (0.2867)(293) = .

30 eBook PSP | First Law of Thermodynamics ii. Pressure and temperature of gas after the compression 2 1 = [ 1 2 ] 2 = [ 1 2 ] × 1 2 = [ 1.8 0.244] 1.3 × (1.38 × 102 ) 2 = 1854.072/2 2 1 = [ 1 2 ] −1 2 = [ 1 2 ] −1 × 1 2 = [ 1.8 0.244] 1.3−1 × (293) = . iii. Work done to compress the gas = 11−22 −1 = 1.38×102(1.8)−1854(0.244) 1.3−1 = −. June 2016 Session Question An insulated cylinder has initial temperature, pressure and volume of 15 oC, 183 kN/m2 and 0.22 m3 . After a heating process, the temperature is raised to 29 oC. Calculate the final pressure in bar unit and final volume. The specific heat at constant pressure and specific heat at constant volume are 931 J/kgK and 0.611 kJ/kgK.

First Law of Thermodynamics | eBook PSP 31 Answer Given: 1 = 15+ 273 = 288 2 = 29° + 273 = 302 1 = 183/2 = 931/ = 0.931/ = 0.611/ i. Final pressure in bar unit and final volume. = = 0.931 0.611 = . 2 1 = ( 2 1 ) −1 2 1 = ( 2 1 ) −1 2 = 1 ( 2 1 ) −1 2 = 183 × ( 302 288) 1.52 1.52−1 2 = 210.2368 2 = . ii. Final volume 11 = 22 2 1 = ( 1 2 ) 1 2 = 1 ( 1 2 ) 1 2 = 0.22 ( 183 210.2368) 1 1.52 = .

32 eBook PSP | First Law of Thermodynamics June 2017 Session Question Nitrogen (molar mass 28 kg/kmol) expands reversibly in a perfectly thermally insulated cylinder form 3.5 bar, 200 oC to a volume of 0.09 m3 . If the initial volume occupied was 0.03 m3 and the nitrogen is assumed as a perfect gas with Cv = 0.741 kJ/kgK, calculate: i. The gas constant. ii. The final gas pressure. iii. The work input. Answer Given: 1 = 3. 5 = 3.5 × 102 1 = 200° + 273 = 473 1 = 0.033 , 2 = 0.093 = 0 (ℎ ) = 0.741/ = 28/ i. The gas constant = = 8.314 28 = . / ii. The final gas pressure = − 1 −1 = = + 1 @ = ( 0.297 0.741)+ 1 = 1.4 = − = + = 0.297+ 0.741 = 1.038 / = = 1.038 0.741 = 1.4

First Law of Thermodynamics | eBook PSP 33 2 1 = ( 1 2 ) 2 = 1 × ( 1 2 ) 1.4 2 = 3.5 × 102 ( 0.03 0.09) 1.4 = . iii. Work done on the gas = 11−22 −1 = 3.5×102(0.03)−75.179(0.09) 1.4−1 = . June 2018 Session Question A quantity of air occupies a volume of 0.45 m3 at the pressure 1.5 bar and 23 oC of temperature. The air is compressed isothermally to a pressure of 6.3 bar and then expanded adiabatically to its initial volume. Given = 1.4, R = 0.287 kJ/kgK and Cp = 1.006 kJ/kgK. Calculate: i. Mass of air during the compression. ii. Heat received or rejected (state which) during the compression. iii. Change of internal energy during the expansion. Answer Given: 1 1 = 0.453 1 = 1.5 = 1.5 × 105/2 1 = 230° + 273 = 503

34 eBook PSP | First Law of Thermodynamics 2 1 = 2 = 503 2 = 6.3 = 6.3 × 105/2 = 3 3 = 1 = 0.453 = 1.5 = 0.287 = 287 = 1.006 = 1006 i. Mass of air during the compression 11 = 1 = 11 1 = (1.5 × 105 )(0.45) (287)(296) = . ii. Heat received or rejected (state which) during the compression = = 1 ( 1 2 ) @ = = 11 ( 1 2 ) = (1.5 × 105 )(0.45) ( 1.5 6.3 ) = −. × ( ) iii. Change of internal energy during the expansion 11 1 = 22 2 Since 1 = 2 11 = 22 2 = 11 2

First Law of Thermodynamics | eBook PSP 35 2 = (1.5 × 105 )(0.45) 6.3 × 105 2 = 0.1073 33 = 22 3 2 = ( 2 3 ) 3 = 2 ( 2 3 ) 3 = 6.3 × 105 ( 0.107 0.45 ) 1.4 3 = 84.33 × 103 3 − = 3 − 2 , since (Q=0) − = 3 − 2 3 − 2 = − 3 − 2 = −( 22 − 33 − 1 ) 3 − 2 = ( 33 − 22 − 1 ) 3 − 2 = (84.33 × 103 × 0.15)− (6.3 × 105 × 0.107) 1.4 − 1 − = −. × June 2019 Session Question An air occupies in the tank with pressure of 6.5 bar and volume 0.7 m3 at temperature 50 oC. After an isobaric expansion, the volume increased to 0.95 m3 . Relate the values given to find the value of mass and the final temperature of the air. Answer Given: 1 = 6.5 2 = 6.5 1 = 0.73

36 eBook PSP | First Law of Thermodynamics 1 = 50+ 273 = 323 = 0.287/ i. Mass, m 11 = 1 = 11 1 = (6.5 × 102 )(0.7) (0.287)(323) = . ii. Final temperature, T2 11 1 = 22 2 1 = 2 1 1 = 2 2 2 = 2 × 1 1 2 = 0.95 × 323 0.7 = . Or 22 = 2 2 = 22 2 = 6.5 × 102 (0.95) 4.908(0.287) 2 = 438

First Law of Thermodynamics | eBook PSP 37 June 2019 Session Question A perfect gas is expanded adiabatically from 6 bar, 0.08 m3 at 70 oC to 0.82 m3 . Calculate the mass of the gas, the final pressure, final temperature and work done of the gas. Given Cv= 0.752 kJ/kgK and molecular mass 30 kg/kmol. Answer Given: 1 = 6 = 6 × 102/2 1 = 0.083 1 = 70° = 343 = 0.752/ = 30/ i. The mass of the gas, m = = 8.3145 30 = 0.277 / 11 = 1 = 11 1 = (6 × 102 )(0.08) (0.277)(343) = 0.5052 ii. The final pressure, P2 = − = + = 0.277+ 0.752 = 1.029/

38 eBook PSP | First Law of Thermodynamics = = 1.029 0.752 = . Or = − 1 0.752 = 0.277 − 1 0.752( − 1) = 0.277 0.752 − 0.752 = 0.277 = 1.029 0.752 = . 2 1 = ( 1 2 ) 2 = 1 × ( 1 2 ) 2 = 6 × 102 ( 0.08 0.82) 1.368 = . iii. The final temperature, T2 2 1 = ( 1 2 ) −1 2 = 1 × ( 1 2 ) −1 2 = 343 ( 0.08 0.82) 1.368−1 = . iv. Work done by the gas, W = 11−22 −1 = 6×102(0.08)−24.859(0.82) 1.368−1 = .

First Law of Thermodynamics | eBook PSP 39 2: 2021/2022 Session Question 1.02 kg of gas is compressed from 1 bar, 20 oC according to the law Pv1.3 = constant, until the pressure is 5.5 bar and temperature 161 oC. If the gas is Argon with molar mass 40 kg/kmol which has Cp= 0.520 kJ/kgK, calculate: i. Work done ii. Heat flow of the gas Answer = 1.02 1 = 20° + 273 = 293 2 = 161° + 273 = 434 = 40 / i. Work done = = 8.3145 40 = 0.2079 / = (1 − 2 ) − 1 = 1.02(0.2079)(293 − 434) 1.3− 1 = −. ii. Heat flow of the gas = − 0.2079 = 0.520 − = 0.520− 0.2079 = 0.3121 / = = 0.520 0.3121 = 1.67

40 eBook PSP | First Law of Thermodynamics = − − 1 × = 1.67− 1.3 1.67 − 1 × (−99.667) = −. 1: 2022/2023 Session Question A certain perfect gas has initial pressure of 12 bar and volume 0.024 m3 is expanded adiabatically to a final pressure of 160 kPa. Given Cp and Cv are 1.046 kJ/kgK and 0.752 kJ/kgK respectively. Calculate: i. Final volume of the gas. ii. Work done by the gas. iii. Change of internal energy of the gas. Answer Given 1 = 12 = 12 × 102 /2 1 = 0.0243 2 = 16 = 16/2 = 1.046 . = 0.752 . i. Final volume of the gas = = 1.046 0.752 = 1.39

First Law of Thermodynamics | eBook PSP 41 11 = 22 1 2 = ( 2 1 ) 1200 160 = ( 2 0.024) 2 1.39 = 7.5 × 5.60410−3 2 = √0.04203 1.39 = . ii. Work done by the gas = 11−22 −1 = 1200(0.024)−160(0.1023) 1.39−1 = . iii. Change of internal energy of the gas (2 −1 ) = − = 0 ( − ) = .

42 eBook PSP | First Law of Thermodynamics THE CONCEPT OF THE FIRST LAW OF THERMODYNAMICS IN OPEN SYSTEM OBJECTIVES: At the end of this chapter, students should be able to: ▪ Define an open system ▪ Differentiate between steady and unsteady flow processes ▪ Explain the steady flow energy equation ▪ Determine the energy balance for open systems (Turbine, Compressor, Boiler, Nozzle, Diffuser, Heat Exchanger) (Source: Image By Robb Kendrick, Nat Geo Image Collection in https://www.nationalgeographic.com/science/article/150817- power-plant-pollution-depends-on-the-weather)

First Law of Thermodynamics | eBook PSP 43 3.1 Open System A space under study that deals with mass flow into or out of a system is known as an open system or a control volume. It usually encloses a device such as turbine, compressor or nozzle. During the control volume, both mass and energy can cross the boundary of the system. The boundaries of a control volume is also known as a control surface. A control surface chosen can be fixed or movable depend on type of system being analyzed. Figure 3.1: Open system 3.2 Steady-Flow Process A process where a fluid steadily passes through a control volume. Although the fluid properties may differ from point to point throughout the control volume, but they do not change with time. 1. A fluid is said to be a steady flow if it meets the following conditions: i. The total mass inside the control volume does not change with time. ii. The properties of fluid must remain constant over time at any given point in the system. iii. A steady flow system’s interaction with its surroundings in terms of work energy and heat must remain constant over time. CONTROL VOLUME Inlet Outlet Q W