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DICKY DERMAWAN www.dickydermawan.net78.net dickydermawan@gmail.com ITK-233 Termodinamika Teknik Kimia I 3 SKS Basic Principles & First Law

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Score & Grading 20 4 all homework & quiz 25 4 1 st midterm exam 25 4 2 nd midterm exam 30 4 final term examination A 4 74.5 ++ B 4 59.5 ++ C 4 49.5 ++ D 4 39.5 ++

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The Scope of Thermodynamics Thermodynamics deals with transformations of energy of all kinds from one form to another. The general restrictions within which all such transformations are observed to occur are known as the first and second law of thermodynamics. These laws cannot be proved in the mathematical sense. Rather, their validity rests upon experience. The universal applicability of thermodynamics is shown by the fact that it is employed alike by physicists, chemists, and engineer. The basic principles are always the same, but the applications differ.

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Chemical Engineering Thermodynamics The chemical engineer must be able to cope a wide variety of problems viz. the determination of heat & work requirements for physical & chemical processes & the determination of equilibrium conditions for chemical reactions and for the transfer of chemical species between phases. Thermodynamic consideration alone are not sufficient to allow calculation of the rates of chemical or physical processes, because rates depend on both driving force & resistance. Driving forces are thermodynamic variables, resistances are not.

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Chemical Engineering Thermodynamics Thermodynamics is a macroscopic-property formulation. It cannot reveal the microscopic (molecular) mechanisms of physical or chemical processes. But, on the other hand, knowledge of the microscopic behavior of matter can be useful in the calculation of the thermodynamic properties. Such property values are essential to the practical application of thermodynamics; numerical results of thermodynamic are accurate only to the extent that the required data are accurate. The chemical engineer must deal with many chemical species & their mixtures, and experimental data are often unavailable.

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Applying Thermodynamics The application of thermodynamics to any real problem starts with the identification of a particular body of matters as the focus of attention. This quantity of matter is called the system, and its thermodynamic state is defined by a few measureable macroscopic properties: Force Temperature Volume: specific volume, molar volume, density Pressure: gauge pressure, absolute pressure

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Pressure The reading on a mercury manometer at 70 o F (open to the atmosphere at one end) is 25,62 in. The local acceleration of gravity is 32,243 ft/s 2. Atmospheric pressure is 19,86 inHg. What is the absolute pressure in (psia) being measured? The density of mercury at 70 o F is 13,543 g/cm 3.

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Unit Conversion Convert of physical parameters below: 500 o F = ….. K = ….. o C = ….. R R = 8,314 J/mol.K = ….. Btu/lbmol.R = ….. cmHg.cm 3 /mol.K = ….. kPa.m 3 /mol.K 4500 kW = ….. Btu/jam = …..lb f.ft/detik = ….. kgf.m/detik

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Thermodynamics Variable: Work, Energy & Heat Work W is done whenever a force acts through a distance. The quantity of work done is defined by the equation: F is the component of the force acting in the direction of the displacement dl. In engineering thermodynamics an important type of work is that which accompanies a change of volume of a fluid, i.e. compression or expansion.

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2 A gas is confined in a 0.47-m-diameter cylinder by piston, on which rest a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m s -2, and atmospheric pressure is 101.57 kPa. a. What is the force in newton exerted on the gas by atmosphere, the piston and the weight, assuming no friction between the piston and cylinder b. What is the pressure of the gas in kPa c. If the gas in the cylinder is heated, it expands, pushing the piston & weight upward. If the piston & weight are raised 15 cm, what is the work done by the gas in kJ d. What is the change in potential energy in kJ of the piston & weight?

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Work & Energy Work done on a body in accelerating it from an initial velocity u 1 to a final velocity of u 2 is equal to the change of kinetic energy of the body mu 2 /2 Work done on a body in raising it through the distance z 2 -z 1 is equal to the change in the quantity of potential energy mgz If the work done on a body in accelerating it or in elevating it can subsequently be recovered, then the body by virtue of its velocity or elevation must contain the ability or capacity to do this work. Energy is the capacity of a body for doing work

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Work An automobile having a mass of 1,250 kg is traveling at 40 m s -1. What is its kinetic energy in kJ? How much work must be done to bring it to a stop? Liquid water at 0 o C and atmospheric pressure has a density of 1.000 g/cm 3. At this condition, ice has a density of 0.917g/cm 3.How much work is done at these conditions by 1 kg of ice as it melts to liquid water?

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Work: Energy in Transit When work is done, it is done by the surroundings on the system, or vice versa, and energy is transferred from the surroundings to the system, or vice versa. It is only during this transfer that the form of energy known as work exist. In contrast, kinetic & potential energy reside with the system. Their value, however, are measured with reference to the surroundings, i.e. kinetic energy depends on velocity with respect to the surroundings, and potential energy depends on eleveation with respect to a datum level. Change in kinetic and potential energy do not depend on these reference conditions, provided they are fixed.

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Heat & Energy Heat is another form of energy that always flows from a higher temperature to a lower temperature. The rate of heat transfer is proportional to the temperature difference between the two bodies. In the thermodynamic sense, heat is never regard as being stored within a body. Like work, it exist only as energy in transit between a system and its surroundings. When energy in the form of heat is added to a body, it is stored as kinetic & potential energy of the atoms and molecules making up the body.

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Internal Energy U Kinetic & potential energy are energy which the substance may posses as a result of its macroscopic position or motion. They can be regarded as external form of energy because their numerical values refer to the surroundings. In contrast, internal energy of a substance refers to the energy of the molecule making up the substance. Although absolute values of internal energy are unknown, this is not a disadvantage in thermodynamic analysis, because only changes in internal energy is required.

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First Law of Thermodynamics Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form it appears simultaneously in other forms. The first law applies to the system and surroundings, NOT to the system alone. For closed system when only heat and/or work is transferred: When no external energy change:

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Thermodynamic State and State Function State functions are quantities which depend only on present conditions, however reached. State functions can be expressed mathematically as functions of thermodynamic coordinates such as temperature & pressure, their value can always be identified with points on a graph. Viz.Nitrogen at 300 K & 1 bar has a definite set properties: specific volume, a definite viscosity, a definite thermal conductivity, etc. Internal energy is a state function, and is therefore, a property of the system. Work & heat are not state functions because they depend on path/process. They cannot be identified with points on a graph, but rather are represented by areas.

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State Functions Not State Functions Thermodynamic State and State Function Represent a property of a system and always has a value Appear only when changes are caused in a system by a process, which requires time Work & heat ARE NOT state functions but the diiference: Q – W IS a state function

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Example A gas is confined in a cylinder by a piston. The initialpressure of the gas is 7 bar, and the volume is 0.1 m 3. The piston is held in place by latches in the cylinder wall. The whole apparatus is placed in air at standard atmospheric pressure of 101.33 kPa. What is the energy change of the apparatus if the retaining latches are removed so that the gas suddenly expands to double its initial value? The piston is again held by latches at the end of the process. Assume the rate of heat exchange between the apparatus and the surrounding air slow compared with the rate at which the process occur.

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Extensive Intensive Depend on the quantity involved Double when the quantity of material is doubled V, U,…. Independent of the quantity involved Fix even though the quatity of material changed P, T Intensive vs Extensive Property Volume is an extensive property BUT specific volume and molar volume are intensive properties

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Entalphy H Entalphy is another thermodynamic function defined as: Since U, P & V are all state functions, H must also be a state function. Like V & U, H is an extensive property Specific entalphy, h, is of course, an intensive property

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Example Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the constant temperature of 100 o C and the constant pressure of 101.33 kPa. The specific volumes of liquid & vapor water at these conditions are 0.00104 & 1.673 m 3 /kg. For this change, heat in the amount of 2256.9 kJ is added to the water.

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Problem 2.1 (1) An insulated and nonconducting container filled with 10 kg of water at 20 o C is fitted with a stirrer. The stirrer is made to turn by gravity acting on a weight of mass 25 kg. The weight falls slowly through a distance of 10 m in driving the stirrer. Assuming that all work done on the weight is transferred to the water and that the local acceleration of gravity is 9.8 ms -2, determine: a. The amount of work done on the water b. The internal-energy change of the water c. The final temperature of the water d. The amount of heat that must be removed from water to return it to its initial temperature

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Problem 2.1 (2) An insulated and nonconducting container filled with 10 kg of water at 20 o C is fitted with a stirrer. The stirrer is made to turn by gravity acting on a weight of mass 25 kg. The weight falls slowly through a distance of 10 m in driving the stirrer. Assuming that all work done on the weight is transferred to the water and that the local acceleration of gravity is 9.8 ms -2, determine the total energy change if the universe because of: a. The process of lowering the weight b. The process of cooling the water back to its initial temperature c. Both processes together

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Problem 2.4 Liquid water at 100 o C and 1 bar has an internal energy (on an arbitrary scale) of 419.0 kJ/kg & a specific volume of 1.044 cm 3 /g. a.What is its enthalpy? b. The water is brought to the vapor state at 200 o C and 800 kPa, where its entalphy is 2838.6 kJ/kg and its specific volume is 260.79 cm 3 /g. Calculate ΔH & ΔU for the process.

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First Law Expression for Steady-state Flow Process When no external energy change:

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Example Air at 1 bar & 25 o C enters a compressor at low velocity, discharges at 3 bar, & enters a nozzle in which it expands to a final velocity of 600 ms -1 at the initial conditions of pressure & temperature. If the work of compression is 240 kJ/kg of air, how much heat must be removed during compression?

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Problem 2.10 Liquid water at 70 o F flows in a straight horizontal pipe in which there is no exchange of either work or heat with the surroundings. Its velocity is 30 ft/s in a pipe with an internal diameter of 1 in until it flows into a section where the pipe diameter abruptly increases. a. What is the entalphy change of water if the downstream diameter is 1.5 in? b. What is the entalphy change of water if the downstream diameter is 3 in? c. What is the maximum change in entalphy for an enlargement in the pipe?

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Problem 2.11 Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil the water changes state from 2 atm & 180 o F to 1 atm & 250 o F. Its entering velocity is 10 ft/s & its its exit velocity is 600 ft/s. Determine the heat transferred through the coil perunit mass of water. Entalphy of the inlet and outlet water streams are: Inlet: 148 Btu/lbm Outlet: 1168.8 Btu/lbm

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Steam TableSteam Table & Unit Conversion Unit Conversion Complete of steam table as following: P, kPaT, o Cx, %, m 3 /kg h, kJ/kgs, kJ/kg.K 30070 ………… … 200 …… 1500 … 1200 …… 0.142 …… 800 … 0 ……… … 4000 ……… 6.0689

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Example To operate a generator, the utility unit generates amount mechanic energy by operating a turbine. The turbine uses steam as working fluid. Assume that turbine works adiabatically and reversibly. Steam enters the turbine at 1.2 MPa and 250 o C and leave the turbine at saturated condition with pressure and vapor quality of 300 kPa and 96.92%, respectively. Electrical power produced by generator is 10 MW at mechanical efficiency of 80%. a.Calculate the change of enthalpy and internal energy of steam (in kJ/kg)! b. Calculate the required mass flow rate of steam (in ton/hour)! c. Determine the steam’s temperature leave a turbine (in o C)!

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Example Water at 200 o F is pumped from a storage tank at the rate of 50 gpm. The motor for the pump supplies work at the rate of 2 hp. The water passes through a heat exchanger where it gives up heat at the rate of 40,000 Btu/min, & is delivered to a second storage tank at an elevation 50 ft above the first tank. What is the temperature of the water delivered to the second tank?

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Concept of Process Reversibility A process is reversible when its direction can be reversed at any point by an infinitesimal change in external conditions. The reversible process is ideal in that it can never be fully realized; it represents a limit to the performance of actual processes. Results for reversible processes in combination with appropriate efficiencies yield reasonable approximations of the work for actual processes. Vocabulary: Irreversible process, dissipative process

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Irreversible Process An irreversible process is a process that cannot return both the system and the surroundings to their original conditions. That is, the system & the surroundings would not return to their original conditions if the process was reversed. For example, an automobile engine does not give back the fuel it took to drive up a hill as it coasts back down the hill. There are many factors that make a process irreversible. Four of the most common causes of irreversibility are friction, unrestrained expansion of a fluid, heat transfer through a finite temperature difference, and mixing of two different substances. These factors are present in real, irreversible processes and prevent these processes from being reversible.

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Constant Volume Constant Pressure Constant Volume Process & Constant Pressure Process Heat capacity:

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Problem 2.17 The internal energy U t of an amount of gas is given by the equation: Where P is in (psia) and V t is in (ft) 3. The gas undergoes a mechanically reversible process from an initial sate at 1500 psia and 500 R. During the processV t is constant and equal to 10 ft 3 and P increases by 50 percent. Determine the value for Q and H t in Btu for the process

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Heat Capacity Five moles of nitrogen at 80 o C is contained in a rigid vessel. How much heat must be added to the system to raise its temperature to 300 o C if the vessel has a negligible heat capacity? If the mass of the vessel is 100 kg and if its heat capacity is 0.5 J/g o C, how much heat is required? Three moles of nitrogen at 230 o C is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 80 o C if the heat capacity of the piston & cylinder is neglected? Take C V = 20.8 and C P = 29.1 J/mol o C for nitrogen gas.

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Illustration: Calculation of Work A cylinder with a piston is used to compress carbon dioxide from 1 to 20 bar. Assuming that the process can be carried out reversibly and isothermally at 300 K, calculate the work required per mole, given that at 300 K the P-V-T properties of carbon dioxide follow the equation P( - b) = RT with b = -0,00011 m 3 /mol

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Heat Capacity Five moles of nitrogen at 100 o F is contained in a rigid vessel. How much heat must be added to the system to raise its temperature to 400 o F if the vessel has a negligible heat capacity? If the mass of the vessel is 250 lb m and has a heat capacity of 0.12 Btu/lb m o F how much heat is required? Three moles of nitrogen at 450 o F is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 100 o F if the heat capacity of the piston & cylinder is neglected? Take C V = 5 and C P = 7 Btu/lbmol o F for nitrogen gas.

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Problem 2.19 The path followed by a gas during a particular mechanically reversible process is described by the equation: Where a & c are constants. In the initial state, P 1 = 60 bar and V t 1 =0.002 m 3. In the final state, P 2 = 20 bar and V t 2 = 0.004 m 3. During the process, heat in the amount of 5000 J is transferred to the gas. Determine W and ΔU t for the process.

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U Heat in the amount of 7.5 kJ is added to a closed system while its internal energy decreases by 12 kJ. How much energy is transferred as work? For a process causing the same change of state but for which the work is zero, how much heat is transferred?

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H 1.5 kg of saturated steam at 2 bars is confined in a cylinder by piston. Assume no mechanical friction. Amount work 750 kJ given to that steam, so pressure and specific volume of steam to be 10 bars and 0.042. Determine: Change of enthalpy and internal energy of steam (in kJ/kg) Temperature at final compression (in oC) Amount of heat transferred during process (kJ)

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