thermodynamics formulas and notes typed, Lecture notes of Thermodynamics

Download thermodynamics formulas and notes typed and more Lecture notes Thermodynamics in PDF only on Docsity! 1-1 Chapter 1 Introduction 1.1 Basic Definitions Thermodynamics is the science that seeks to predict the amount of energy needed to bring about a change of state of a system from one equilibrium state to another. While thermodynamics tells us nothing about the mechanisms of energy transfer, rates of change, and time associated with a system changing from one equilibrium state to another, it is still the lynch-pin that allow us to answer these questions. • Definition of 'heat': Heat is energy in transit solely as a result of a temperature difference. • Definition of 'work': Work is energy exchange between system and surroundings due to any phenomenon except a temperature difference. • Definition of 'temperature': Temperature is a measure of the mean kinetic energy of molecules. Absolute zero (0oK) is a state of complete motionless of molecules. • 'Rate': 'Rate' implies an element of speed, how fast an event happens, and time. • 'System': In thermodynamics, the universe can be divided into two parts. One part is the system, the other part is the rest of the universe called the surroundings. System can be classified as (1) isolated system where no mass or energy is transferred across the system boundaries, (2) closed system (system) where only energy is transferred across the system boundaries, or (3) open system (control volume) where mass and energy can be transferred across the system boundaries. A system is any designated region of a continuum of fixed mass. The boundaries of a system may be deformable but they always enclose the same mass. ` Surroundings Boundary System Figure 1.1 Schematic diagram of the "universe", showing a system and the surroundings. 1-2 • 'Control volume': A 'control volume' is also any designated region of a continuum except that it may permit matter to cross its boundaries. If the boundaries of a control volume are such that matter may not enter or leave the control volume, the control volume is identical to a system. In these respects, a 'system' is a subset of a 'control volume'. • 'Equilibrium': 'Equilibrium' means that there are no spatial differences in the variables that describe the condition of the system, also called the 'state' of a system, such as its pressure, temperature, volume, and mass (P, T, V, m), and that any changes which occur do so infinitesimally slowly. The laws of thermodynamics are applicable only to equilibrium states which means that the state does not really change significantly with time, differences in variables between the state of a system and its surroundings are of infinitesimal magnitude and that within the system itself there are no spatial variations of the variables that determine its state. Using thermodynamics, we can predict the amount of energy needed to change a system from an equilibrium state to another. For example it will take about 75 kJ to change 1 kg of air at 25oC and 1 atm to 100oC and 1 atm. It will take much more energy, about 2257 kJ, to change 1 kg of water at 100oC and 1 atm to water vapor (steam) at the same temperature and pressure. 75 kJ requiredAir, 1 atm 25 C 1 kg o Air, 1 atm 100 C 1 kg o 2257 kJ required Water 1 atm 100 C 1 kg o Steam 1 atm 100 C 1 kg o State 1 State 2 State 1 State 2 Figure 1.1 Energy required changing air or water from state 1 to state 2. 1.2 Property A property is a macroscopic characteristic of a system such as pressure, temperature, volume, and mass. At a given state each property has a definite value independent of how the system arrived at that state. The properties of air in state 1 shown in Figure 1.1 are: pressure at 1 atm, temperature at 25oC, and mass of 1 kg. A property can be classified as extensive or intensive. An extensive property depends on the size of the system while an intensive property is independent on the size of the system. Consider systems (1) and (2) shown in Figure 1.2 both at 100oC and 1 atm containing 2 and 5 kg of steam, respectively. 1-5 Consider a hole in the wall of a tank or a pipe as shown in Figure 1.4-2. The fluid pressure p may be defined as the ratio F/A, where F is the minimum force that would have to exerted on a frictionless plug in the hole to keep the fluid from emerging1 P(N/m )2 P(N/m )2 A(m )2 F(N) F(N) A(m )2 Fluid Fluid flowing through a pipe Figure 1.4-2 Fluid pressure in a tank and a pipe. The pressure at a given position measured relative to absolute zero pressure or absolute vacuum is called the absolute pressure. Most pressure-measuring devices are calibrated to read zero in the atmosphere as shown in Figure 1.4-3. These pressure gages indicate the difference between the absolute pressure and the local atmospheric pressure. Pressures below atmospheric pressure are called vacuum pressures and are measured by vacuum gages that indicate the difference between the atmospheric pressure and the absolute pressure. Absolute, gage, and vacuum pressures are all positive quantities and are related to each other by Pgage = Pabs − Patm Pvac = Patm − Pabs 01 2 3 4 5 6 7 8 9 10 kPa Pabs = 0 Pabs = 0 Patm Patm Pabs Pgage Pabs Pvac Patm Patm Figure 1.4-3 Absolute, gage, and vacuum pressures. Two common pressure units are the bar and standard atmosphere: 1 bar = 105 Pa = 0.1 Mpa = 100 kPa 1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bar = 14.696 psi 1 R. M. Felder and R. W. Rousseau, Elementary Principles of Chemical Processes, Wiley, 2000, p.54. 1-6 1.5 Variation of pressure with elevation Fluid at rest cannot support shearing stress. It can only support normal stress or pressure that can result from gravity or various other forces acting on the fluid. Pressure is an isotropic stress since the force acts uniformly in all directions normal to any local surface at a given point in the fluid. An isotropic stress is then a scalar since it has magnitude only and no direction. By convention, pressure is considered a negative stress because it is compressive, whereas tensile stresses are positive. The direction of pressure force is always pointing inward the control volume. We now investigate how the pressure in a stationary fluid varies with elevation z as shown in Figure 1.5-1. P|z P|z+ z∆ ∆z z A Figure 1.5-1 Forces acting on control volume A∆z Applying a momentum or force balance in the z-direction on the control volume A∆z we obtain ΣFz = maz = 0 AP|z − AP|z+∆z − ρg A∆z = 0 Dividing the equation by the control volume A∆z and letting ∆z approach zero we obtain 0 lim →∆z z PP zzz ∆ −∆+ || = dz dP = − ρg (1.51) Equation (1.5-1) is the basic equation of fluid statics. It can be integrated if the density and the acceleration of gravity are known functions of elevation. We will assume g a constant since the change in elevation is usually not significant enough for g to vary. The integration will depend on the variation of density. If the density is not a constant, a relation between ρ and z or P must be obtained. For constant density fluids, equation (1.5-1) can be easily integrated
2 1 P P dP = − ρg
2 1 z z dz  P2 − P1 = − ρg(z2 − z1) This equation can also be written as 1-7 P2 + ρgz2 = P1 + ρgz1 = Φ = constant The sum of the local pressure P and static head ρgz is called the potential Φ that is constant at all points within a given incompressible fluid. Example 1.5-1. 2 ---------------------------------------------------------------------------------- The manometer system shown in Figure 1.5-2 contains oil and water, between which there is a long trapped air bubble. For the indicated heights of the liquids, find the specific gravity of the oil. The two sides of the U-tube are open to the atmosphere. h = 2.5 ft1 h = 0.5 ft2 h = 1.0 ft3 h = 3.0 ft4 Oil Air Water P1 P2 Figure 1.5-2 A manometer system with oil, air, and water Solution ------------------------------------------------------------------------------------------ The pressure P1 at the water air interface on the left side of the U-tube has the same value as the pressure P2 in the water at the same elevation on the right side of the U-tube. P1 = Patm + ρogh1 + ρagh2 P2 = Patm + ρwg (h4 − h3) If the pressure due to the weight of air is neglected compared to that of oil and water, we can determine the specific weight so of oil as follow ρogh1 = ρwg (h4 − h3)  so = w o ρ ρ = 1 34 h hh − = 5.2 0.10.3 − = 0.80 2 Wilkes, J., Fluid Mechanics for Chemical Engineers, Prentice Hall, 1999, p. 28