Adiabatic Processes: Energy Analysis In Isolated Systems

Analysis of Adiabatic Processes in the Context of Energy Changes in Isolated Systems Matters relating to energy balance: Adiabatic, isothermal, isobaric, and isochoric are processes used in determining a condition where no change occurs in the process. Concepts such as state functions or variables are important to understand. Intensive variables or commonly referred to as state variables: Indicates the final result obtained, regardless of the trajectory as shown in Figure 1. Example: Temperature and pressure 2 T A A B 1 p Changes in the value of the same state variable in each of the trajectories A and B from 1 to 2. The unit of energy in the SI system: Joule (J) lOMoARcPSD|26370826 Energy units in the AE (American Engineering) system: Btu, (ft) (lb )f , and (kW) (hr).Calorie units are often used on food packages to indicate information about the number of calories, but the calories used are not precise. 1 calorie is the energy to raise the temperature of 1 mL of water by 1 o C. However, the calories used on food wrappers do not correspond to Definition above. Actually, what is often called calories in the information is kilocalories (kcal). 1 kilocalorie (kcal) = 1000 calories (in thermochemical terms) which is the energy to raise the temperature of 1 L of water by 1 C. o So for example, if the information mentioned in the food package is calories, then we can calculate the actual energy in Joules / hour, namely 2000 kcal 1000 cal 4,184 J 1day day kcal cal 24 hours 350,000 J/h Table 1.1 Energy balance terms. WordUnderstandingPicture System Amount of substance or region of a selected Environment part (surroundings) Everything outside the system boundary Boundary A surface that can be a real shape or a surface that is visualized, fixed or movable to separate the system from the environment Open system (open system/flow system) Open system to happen mass, heat, and work transfer with the environment lOMoARcPSD|26370826 (Continued table 1) Word of Understanding Closed system Systems where no mass transfer occurs (close with the environment. However, heat and system/nonflow work can be transferred system) Property A characteristic of a system that can observed or calculated such as pressure, temperature, volume and others. State The condition of a system described Specialized in temperature , pressure, composition , etc. Steady state Accumulation in the system is zero, inflow and outflow are constant, Unsteady state properties in the system are unchanged. state/transient) Accumulation in the system is not zero, Equilibrium inflow and outflow are not constant, properties in the system change. The properties of the system do not vary so that it undergoes equilibrium conditions, such as thermal, mechanical, phase, and chemical equilibrium. Phase A part or whole of the systems that are macroscopically homogeneous at variable compositions such as gas, liquid, or solid Image lOMoARcPSD|26370826 Table 2 Additional energy balance terms WordUnderstandingPicture Adiabatic system A system where no heat transfer with the environment during the process (perfect insulation). Isothermal System Systems where the temperature is fixed during process. system) Isobaric system Systems where the pressure remains fixed throughout process. Isochoric system Systems where the volume remains fixed during process. State variable Any variable or function whose value (State function) depends only on the state of the system (Point function) and not on the previous state, such as internal energy. Pathvari Any variable or function whose value able depends on how the function) The process takes place and can be different from the previous state, such as heat and work. 1.1Forms of energy 1. Work (W) circumstances2 F .ds (1.1) State1 where F is the external force and s is the direction of the acting force vector Work or positive work (+) when the system receives work from the environment Work or negative work (-) when the system performs work on the environment lOMoARcPSD|26370826 2. H eat (Q) Total energy flowing at the system boundary caused by the temperature difference between the system and the environment. In an adiabatic process, Q = 0. Heat will be positive (+) when transferred to the system Q UA(T 2 T1 ) (1.2) where : Q: heat transfer rate (J/s) A: heat transfer area (m )2 (T )1 T ): temperature difference between the environment (T2 ) and the system (T 21 (o C) 3. Kinetic Energy (EK) 1 EK 2 2 mv (1 .3) Where: m : mass (kg) v : velocity (m/s) 4. Potential Energy ( EP ) EP m g h (1.4) Where: m : mass (kg) g: acceleration of gravity (m/s ) 2 h : height (m) 5. Energy In (∆Ȗ) U ∆Ȗ = Ȗ -Ȗ1= 2 2 T2 dU U1 Cv dT T 1 (1.5) lOMoARcPSD|26370826 Where: Ȗ : internal energy (temperature function) Cv: Heat capacity at fixed volume (J/(kg)(K)) T : Temperature 6. Enthalpy ( H) H2 H H2 H1 dH H H: T2 1 Cp dT (1.6) T1 Enthalpy (temperature function) Cp: Heat capacity at fixed pressure (J/(kg)(K)) T : Temperature Example Problem 1.1 Calculation of Mechanical Work by Piston Ideal gases at temperatures of 300 K and 200 K are placed in a closed cylinder with a piston assumed without friction in it, the gas slowly presses on the piston so that the piston moves and the volume of gas in the cylinder expands as shown in Figure C1.1. Figure C1.1a Calculate the amount of work done by the gas on the piston if two ways are used from the start (1) to the end (2). Path A: Expansion occurs at constant pressure (isobaric): p = 200 kPa Path B: Expansion occurs at constant temperature (isothermal): T = 300 K Completion: Assumptions: lOMoARcPSD|26370826 1) No friction on the piston 2) Idealized process (happens slowly) 3) The movement of the piston at the boundary, namely state 1 to state 2 Base: Gas Quantity n 200 kPa 0.1m3 (kg mol) (K) 300 K 8.314 (kPa) (m 3 0. 00802 kg mol ) Mechanical work performed by the system on the piston per unit area s ta te 2 W state1 v2 F A .A ds p dV v1 A negative sign is a sign that the system is doing work Figure C1.1b A way: v2 W dV p p(v v) 2 1 v1 W 200 x103 Pa 0.1m3 1N 1(m 2 ) (Pa) V1 1J 1( N ) (m) 20 kJ lOMoARcPSD|26370826 0.00802 kg mol W 8.314 kJ 300 K (kg mol) (K) ln 2 20 ln 2 86 kJ 13. Example Problem 1.2 Calculation of Internal Energy What is the change in energy when 10 kg moles of air is cooled from 60 o C to 30o C in a constant volume process? Completion: Using Cv because the process occurs at constant volume. The value of Cv for this temperature range is 2.1 x 10 4 J/(kg mol) ( o C), so the energy in : 30o U C 2.1 x10 4 J 5 (kg mol ) C o dT 2.1 x10 (30 60) 60 o C 6.3 x106 J Problem! 1. A cylinder contains N2 at a pressure of 200 kPa and a temperature of 80 o C. As a result of cooling during t h e night, the cylinder pressure reduced to 190 kPa and a temperature of 30 o C. How much work was done by the gas? 2. Nitrogen gas in a cylinder has 4 stages of the ideal process as shown in Figure S1.1. Figure S1.1 Calculate the work done by the gas at each stage (in Btu)