Course 2 Mathematical Tools and Unit Conversion Used in Thermodynamic Problem Solving


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1 Course Mathematical Tools and Unit Conversion Used in Thermodynamic Problem Solving 1 Basic Algebra Computations 1st degree equations  =0 Collect numerical values on one side and unknown to the otherside to find the root nd degree equations  =0 Use discriminant equations, = ± 4 Complex equations containing higher degree logarithmic, exponential terms log = 0 Use graphical method I. Separate the equation into two parts y1 = log y II. Prepare a table of y1 versus y with values of x within the domain III. Plot both y1 and y versus x on the same graph 1
2 x y1 y y1 y Numerical method for complex equations Successive iteration steps: Seperate the equation into two in such a way to have one side of the equation with terms that are easy to solve = log y 1 y Start the iteration by supplying an initial value for x Replace the numerical value of x into y and obtain a numerical value for y = Solve y 1 for x y 1 Compare the new x with the previous x If the two values are different go to the second step using the new x value Repeat the process until the two successive x values are equal 4
3 Example Calculate the mole fraction of a gas x, in a mixture from the equation: = 0 y 1 y = x y1 y x 0,1000 0,014 0,014 0,90 0,90 0,10 0,10 0,814 0,814 0,164 0,164 0,99 0,99 0,1715 0,1715 0,96 0,96 0,174 0,174 0,97 0,97 0,1750 0,1750 0,975 5 Thermodynamics is a science dealing with energy changes, state changes and property changes = = = Engineering requires accurate calculations, taking into account even the infinitely small changes = = = = Differential equations and their integrals are tools for engineers to make detailed analysis of problems involving changes big or small Continuous paths of reversible processes also necessitate the use of differential equations 6
4 Work done by compression of a gas in a cylinder by a piston 1 atm P ext atm >>, = Irreversible process gets less work out of the system than possible,5,5 1,5 1 0,5 0 1 P ext P ext More work done by Irreversible process In steps 1 atm atm reversible 1 atm atm atm, = Reversible process Supplies maximum work,5,5 1,5 1 0, , 1,6 1,9,,5,8 7 Total differential equations z = f(x, y) = Useful relationships can be developed from total differential equation by taking z constant 0 = dividing by dy 0 = = 1 = dividing by Transformation formula 8 4
5 Line integrals Integrating a simple function such as = between the limits gives the area under the curve Thermodynamic functions often contain more than one independent variable Solutions to thermodynamic problems are often done using paths or curves Total change in z can be obtained between the limits by integrating dz along a given curve = = = 8, 10 =,, =,,, y f x f 9 Steps: I. Use the curve equation to calculate variables in terms of each other Main function: = Path: y = I. Replace them in the integral in such a way to have only one variable in each part of the integral = II. Evaluate the integral between the limits, =, = = 4 Example Calculate the change in z between (0,0) and (1,1) for = along the following paths a) y=x b) y=x c) (0,0)(1,0)(1,1) 10 5
6 Exact differentials The value of Δz becomes independant of the path when certain conditions are met This situation is encountered when dz is an exact differential =,, For a differential to be exact, it has to meet the Euller s criteria:, =, Example Calculate the change in z between (0,0) and (1,1) for the function = ( ) ( ) along the following paths: a) y=x b) y=x c) (0,0)(100,0)(100,1)(1,1) 11 Units and conversion factors Mass  absolute quantity of matter m mass, kg Velocity  distance per unit time where: d m v velocity in m/sec v t sec d distance in meters t time in sec Acceleration  the rate of change of velocity with respect to time dv a dt m sec Force  the mass multiplied by the acceleration kg  m F ma or Newton sec ma F kn Newton = 1 kgm/sec Newton  the force required to accelerate 1 kg mass at the rate of 1 m/sec 1 6
7 Weight  the force due to gravity W mg W mg 1000 kn Specific weight  the weight per unit volume. N Where: g gravitational acceleration, m/sec At standard sea level condition g = 9.81 m/sec Density  the mass per unit volume Where; m kg  density in kg/m OR g/cm V m m mass in kg V volume in m Specific volume  the volume per unit mass or the reciprocal of its density V 1 m Where: m kg  specific volume in m /kg W mg g kn Where: V 1000V 1000 m  specific weight in KN/m L For Liquids : SL Specific gravity or relative density w For liquids it is the ratio of density to that of water at standard G temperature and pressure For Gases : SG For gases it is the ratio of density to that of either air or hydrogen AH at some specified temperature and pressure 1 Pressure force per unit area P F A ma kg A sec m or Newton per meter squared or Pascal Atmospheric Pressure  the absolute pressure exerted by the atmosphere At Standard Condition 1 atm = kpa = 1.0 kg/cm = MPa = Bar = 760 mm Hg = 76 cm Hg = 14.7 lb/in = 10. m of H O = 9.91 in of Hg =.88 ft of H O kg m W Fl mal PV or sec Newton meter or liter atm or Joule Work force acting through a distance 14 7
8 Energy the capacity to perform work Kinetic energy Force applied to a body to move a distance dw madl dv dl dw m dl m dt dt dw mvdv W m mv mv1 W 1 EK mv Newton meter or Joule Potential energy Upward force exerted on a body to raise to an elevation E Newton meter or Joule Power Energy output rate of one Joule per second v v1 F ma mg mgh dv v v 1 vdv m W F h h ) mg( h h ) P ( 1 1 Newton meter / second or Watt 15 Temperature the degree of intensity of heat Measured with liquidinglass thermometers where the liquid expands when heated Temperature scale of the SI system is based on the ideal gas as thermometric fluid RT v ave M Where; R gas constant = joules/degree mole M Molar mass kg/mole Ideal gas is made up of particles or molecules Each particle in a gas has kinetic energy All collisions are perfectly elastic Volume of the particles is insignificant There are no interactions between particles The average kinetic energy of the particles is a function of only absolute temperature The volume of the gas is zero at absolute zero The temperature relates to the average velocity of the whole sample, as there is one temperature for the sample 16 8
9 Temperature Conversion 17 Gas constant R Ideal gas obeys Boyle s law, Charle s law and Avogadro s law 1 Ideal Gas Equation of State: = = = = Volume per grammole of ideal gas at 0 C and 1 atm is.414 liters according to Avogadro s law Thus = = =
10 Heat  the form of energy transferred from hot to cold objects It is energy in transit, not stored in the system as heat but as kinetic and potential energy of the atoms The rate of heat transfer from one body to another is proportional to the difference in temperature 1 calorie = the quantity of heat required to raise the temperature of 1 gram of water by 1 C = Joules Specific Heat (C) Specific Heat or Heat Capacity is the amount of heat required to raise the temperature of a 1 kg mass by 1C Q=m*C*ΔT 19 Example 1 What quantity of heat is required to change the temperature of 00 g of lead from 0 to C? C Pb = 10 J/kg.K Example  An oven applies 400 kj of heat to a 4 kg of a substance causing its temperature to increase by 80 C 0. What is the specific heat capacity? Example  How many grams of iron at 0 0 C must be heated to C in order to be able to release 1800 cal of heat as it returns to its original temperature? C Pb = 11 cal/kg.k 0 10
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