Mathematics and Chemistry Calculations PDF

Mathematics Selected Safety Calculations and their General Considerations Please click and select from the following subjects: 1. Chemical Hazards 2. Linear Kinematics 3. Dynamic of Particles 4. Pressure 5. Archimedes Principle 6. Inferential Statistics 7. Set Theory, probability, and Statis...

Mathematics Selected Safety Calculations and their General Considerations Please click and select from the following subjects: 1. Chemical Hazards 2. Linear Kinematics 3. Dynamic of Particles 4. Pressure 5. Archimedes Principle 6. Inferential Statistics 7. Set Theory, probability, and Statistics 8. Structural Hazards 9. References  Chemistry Please click to select from the following subjects: 1. Density and Specific Gravity 2. Temperature Conversion 3. Atoms and Isotopes 4. Molecular Weight and Mole 5. Balanced Chemical Equations 6. Valence and Chemical Bonding 7. Oxidation-Reaction Reductions 8. Terminologies and Calculations related to “Solutions” 9. Avogadro’s Number 10. Vapor Pressure and Heat 11. Gas Laws Cl2  Density and specific gravity Density is defined as the mass of a substance contained in one unit of its volume. mass of object Density = mass per unit volume = volume of object Specific gravity of a liquid or solid is defined as the ratio of the mass of the liquid or solid to the mass of an equal volume of water. The ratio of density of a gas or vapor to density of air is called vapor density. mass of solid or liquid specific gravity = mass of an equal volume of water at 4 C Example A cylindrical object with a height of 2 feet and a diameter of 1 foot has a mass of 3 lbs. What is the density of this object? What is the value of its specific gravity? Density of water = 62.4 lbm/ft3. Solution First we calculate the volume of the cylinder. volume  (3.14 R 2 )  H Where R is the radius and H is the height of cylinder volume  (3.14  0.5 2 )  2 volume  1.57 ft 3 mass density  volume 3 lbs density  1.57ft 3 density  1.91 lbs / ft 3 density of the object specific gravity  density of water 1.91 specific gravity   3.06  10 2 62.4  Temperature conversion To convert between Celsius and Fahrenheit temperature scales use:  F  32 = C 1.8 where C is degrees Celsius and F is degrees Fahrenheit. To convert to absolute temperature scales use: o R = oF + 460 o K = oC + 273 where oR and oK denote the Rankine and Kelvin temperature scales respectively. Example Convert 20 C to the Rankine temperature scale. Solution We know R  F  460 First we convert degrees Celsius to degrees Fahrenheit.   F  32  20 1.8 or  F  32  36  68  R  68  460  528  Atoms All matter is made up of atoms. An atom contains a central nucleus made up of positively charged particles, the protons, and electrically neutral particles, called neutrons. Almost all of the mass of an atom is contained in the nucleus which is surrounded by negatively charged particles, the electrons, rotating around it in orbits, as shown in Figure 1. The number of electrons and protons in any atom are equal, so that the atom, in its normal state, is neutral. The number of protons in any atom is called the atomic number, and the number of protons and neutrons the atomic mass. ns tr o  ec el  protons neutrons nucleus Figure 1. - Atomic Structure.  Isotopes Isotopes are atoms of a given matter with the same atomic number, but different atomic mass, i.e., the number of protons in all atoms are equal but the number of neutrons are different.  Molecular weight The sum of the atomic weight of different atoms making a compound is called the molecular weight of that compound.  Mole The average chemical experiment involves the reaction of an enormous number of molecules. In order to simplify calculations, one mole of a substance is defined as the mass of that substance which contains exactly 6.023  1023 molecules (Avogadro’s number). In order to convert the mass of a given substance to moles, we divide the mass by the molecular weight of that substance, thus: mass Mole = molecular weight mass expressed in grams Gram mole = molecular weight mass expressed in pounds Pound mole = molecular weight mass expressed in kilograms Kilogram mole = molecular weight Example A drum contains 90 lbs of water (H2O). Calculate the number of pound moles, gram moles, and kilogram moles of water in the drum. The atomic weight of hydrogen is 1 and the atomic weight of oxygen is 16. Solution First we calculate the molecular weight of water. H2O: molecular weight = (2  1) + 16 = 18 90 Pound moles of water  5 18 In order to calculate the gram moles of water, we have to convert the weight of water in the drum to grams. 454 grams 90 lbs   40860 grams of water 1 lb 40860 gram moles of water   2270 18 In order to calculate the kilogram moles of water in the drum, we have to convert grams of water into kilograms of water. 1 kilo gram 2270 grams   2.27 1000 grams (we are using two significant figures) Example If we have 2 lb moles of sulfuric acid (H2 SO4), how many kilograms of sulfuric acid do we have? Atomic weights: H = 1, S = 32, O = 16 Solution First, we calculate the molecular weight of sulfuric acid: H2SO4: molecular weight = (2  1) + (32) + (4  16) molecular weight = 98. We know that: lbs H 2 SO 4 lb mole H 2 SO 4  molecular weight or lbs H 2 SO 4  (lb mole H 2 SO 4 )  (molecular weight) lbs H 2 SO 4  2  98  196 Now, we have to convert pounds of sulfuric acid into kilograms of sulfuric acid. We know that there are 2.2 lbs in one kilogram. Therefore: 196 kilo grams of suluric acid   89 2.2 Balanced Chemical Equations, Weight, and Mole Relations A chemical equation must be balanced before any weight or mole calculations can be performed on the reactants or products. In order to balance a chemical equation, one must ensure that the number of atoms of different substances comprising ALL molecules of reactants is equal to the number of atoms of substances comprising molecules of the products. For example, consider the reaction between ethane and oxygen to form carbon dioxide and water. C2 H6 + O2 CO2 + H2 O The above chemical equation represents the reaction between ethane and oxygen; however, it is not a balanced chemical equation. In order to balance the above equation, note that we have 2 atoms of carbon present in the reactant molecule C2H6; therefore we need 2 molecules of CO2 to make the number of carbon atoms in the reactant and product molecules equal. In the same manner, since we have 6 atoms of hydrogen in the reactant molecules, we must have three molecules of water to make the number of hydrogen atoms equal. Since we have 7 atoms of oxygen in the product molecules, we need 7/2 O2 or 3.5 O2 in the reactant molecules. The balanced equation is: 7 C2 H6 + O2 2 CO2 + 3 H2 O 2 Once a chemical equation is balanced, we can carry out either weight or mole calculations for different reactant and product molecules. The numbers in front of each reactant and product molecule are called stoichiometric coefficients. The above balanced equation can be interpreted in a number of ways: 1. One molecule of ethane reacts with 7/2 molecules of oxygen to form 2 molecules of CO2 and 3 molecules of water. One gram mole of ethane reacts with 7/2 gram moles of oxygen to form 2 gram moles of carbon dioxide and 3 gram moles of water.  2. 30 grams of ethane (M.W. = 30) reacts with 3.5  32 grams of oxygen (M.W. = 32) to form 2  44 grams of carbon dioxide (M.W. = 44) and 3  18 grams of water (M.W. = 18). It is important to note that in a balanced chemical equation, the number of moles of reactants and products are not necessarily equal. In other words MOLES ARE NOT NECESSARILY CONSERVED in a chemical reaction. For example, in the reaction between ethane and oxygen (above chemical equation), the total number of moles of reactants equals 4.5 while the total number of moles of products equals 5. Obviously 4.5  5. It is equally important to note that in a balanced chemical equation, THE MASS OF REACTANTS AND PRODUCTS MUST BE EQUAL. In the above reaction, the mass of reactants is 30 grams of ethane plus 3.5  32 grams of oxygen (total = 142 grams of reactant), forming 88 grams of CO2 and 54 grams of water (total = 142 grams of product). Therefore, we can make the following statement that is essential in solving problems dealing with any chemical reaction: IN ANY CHEMICAL REACTION, MOLES ARE NOT NECESSARILY CONSERVED, HOWEVER, MASS IS ALWAYS CONSERVED. CO2 Valence and Chemical Bonding The electrons in the outer orbit of an atom are called the valence electrons, which are responsible for chemical reactions. Chemical reactions occur either as a result of complete transfer of one or more valence electrons from one atom to another or as a result of sharing valence electrons among different atoms.  Ionic bonds When one or more valence electrons are transferred from one atom to another, the atom that has lost the electrons becomes positively charged (positive ion) and the atom that has gained the electrons becomes negatively charged (negative ion). A strong electrical force will be developed between the positive and negative ions which form a new compound. This electrical force is known as the ionic bond. Since the number of electrons lost by one atom exactly equals the number of electrons gained by the other atom, the compound, as a whole, is electrically neutral. For example, sodium and chlorine can react to form sodium chloride (table salt), by transfer of one electron from the valence orbit of sodium to the valence orbit of chlorine and forming an ionic bond between the sodium and chlorine ions as shown in Figure 2. Although in a molecule of sodium chloride the sodium ion is positively charged and the chlorine ion is negatively charged, sodium chloride as a compound is electrically neutral because the number of electrons lost by the sodium atom equals the number of electrons gained by the chlorine atom. Ionic Bond valance valance shell shell chlorine ion sodium ion Figure 2. Ionic bond.  Covalent bonds Some chemical reactions take place when atoms “share” their valance electrons to complete the electron requirement in the valance shell of each atom. This type of bonding among different atoms is known as “covalent bonds.” Covalent bonds among atoms are much weaker than the ionic bonds. An example of formation of covalent bonds is the reaction between hydrogen and oxygen to form water. Oxygen has 6 electrons in its valence shell that requires 8 electrons and hydrogen has 1 electron in its valence shell which requires 2 electrons. Therefore, two atoms of hydrogen can share their one electron with the oxygen atom to complete the number of electrons in all atoms involved as shown in Figure 3.     H O H     Figure 3. Formation of a molecule of water by covalent bonds.  Atomic absorption Atomic absorption is an analytical technique which takes advantage of the characteristic absorption by metals of certain wavelength of light.  Electronegativity Electronegativity is a property associated with certain atoms which causes unequal sharing of electrons during bonding. This results in shared electrons spending more time with one atom and less time with the other. Large differences in electronegativity result in complete transfer of electrons from one atom to another and formation of an ionic bond.  Heat of reaction There is a certain amount of energy associated with each chemical bond. In a chemical reaction, the sum of the energy of the chemical bonds in the products is usually different from the bond energies in the reactants. Depending on the magnitude of the bond energies involved, a chemical reaction can produce or consume energy. This energy is called the heat of reaction. Chemical reactions that consume energy are called endothermic, and chemical reactions that produce energy are called, exothermic.  Isomers Substances that have the same numbers of atoms of each element per molecule but differ in spatial arrangement of the atoms are known as isomers of each other.  Crystals A crystalline substance is an ORDERED array of its component atoms, ions, or molecules.  Oxidation-reduction reactions Oxidation refers to the loss of electron(s) by an atom involved in a chemical reaction. Reduction is the gain of electron(s) by an atom involved in a chemical reaction. Oxidation is always accompanied by reduction. These reactions are more commonly known as Redox reactions. Fe  Fe+2 + 2 electrons (oxidation) Cl2 + 2 electrons  2 Cl- (reduction) O2  Terminology and Calculations Related to “Solutions”  Solute Dissolved substance.  Solvent The substance in which the solute is dissolved.  Concentration The concentration of solutions can be expressed either in physical or chemical units. Physical units of concentration The concentration of a solution in physical units can be expressed in the following ways:  By weight of solute per unit volume of solution. For example, 10 grams of NaCl (table salt) per liter of solution.  By weight percentage. For example, 10% NaCl solution contains 10 grams of NaCl per 100 grams of solution.  By the weight of solute per weight of solvent. For example, 10 grams of sugar in 100 grams of water. 1. Some of the more common chemical units of concentration  Molarity is the number of gram moles of the solute contained in one liter of solution.  Mole fraction is the number of moles of solute divided by the sum of number of moles of solutes and solvents. Example 1000 liters of brine solution contains 292.5 pounds of salt (NaCl). What is the molarity of the solution. Atomic weights Na=23, Cl = 35.5 Solution We have to find the number of gram moles of NaCl in one liter of the solution.  molecular weight (NaCl)  23  35.5  58.5 292.5 pound moles (NaCl)  5 58.5 454 gram moles gram moles (NaCl)  5pound moles   2270 1 pound mole 2270 gram moles in one liter of solution  molarity   2.27 1000 Example A brine solution has been prepared by adding 234 grams of NaCl to 1800 grams of water (H2O). What is the mole fraction of NaCl in the solution?  Solution First, we have to calculate the number of gram moles of both water and salt (NaCl). molecular weight of water  18 molecular weight of NaCl  58.5 (see above example) 1800 gm moles of water   100 18 234 gm moles of NaCl  4 58.5 mole fraction of NaCl  number of moles of NaCl  number of moles of NaCl  Number of moles of water 4 4   3.84  10  2 100  4 104  Dissolved solute Amount of dissolved solute = volume  concentration in volume percent/100 Amount of dissolved solute = weight  concentration in weight percent/100 Example How many pounds of sulfuric acid exists in 100 ft3 of a 10 percent by volume sulfuric acid solution? The density of sulfuric acid is 70 lbs/ft3. Solution First we calculate the volume of sulfuric acid. Volume of sulfuric acid = 100  0.1 = 10 ft3. Next, since we know that each cubic foot of sulfuric acid weighs 70 lbs (density = 70 lbs/ft3), we can calculate the weight of sulfuric acid: 3 lbs weight of sulfuric acid  10 ft  70 3  700 lbs ft Example How many pound moles of salt exist in 2000 lbs of a 10 percent by weight brine solution. The molecular weight of salt is 58.5. Solution First we calculate the weight of salt: 2000  01.  200 lbs We know that pound moles are the weight-expressed in lbs divided by the molecular weight. 200 lb moles of salt   3.42 58.5 Avogadro’s Number Avogadro’s number is the number of atoms of an element in one gram mole of that element. The value of Avogadro’s number is 6.024  1023 (this value will be provided by the BCSP at the time of the examination and there is no need to memorize this number). This means that if we count the number of atoms in one gram mole of any element, the result is 6.024  1023 (indeed a very large number). For example, the atomic weight of oxygen is 16, and based on our definition of gram mole, one gram mole of oxygen weights 16 grams. Avogadro’s number, therefore, indicates that if we count the number of atoms of oxygen in 16 grams of oxygen, (one gram mole), the result is 6.024  1023 atoms of oxygen. Example The atomic weight of sulfur is 32. Calculate the number of atoms of sulfur contained in 64 kilogram of sulfur. Solution Based on Avogadro’s number, we know the number of atoms of sulfur in one gram mole of sulfur (6.024  1023). In order to solve this problem, we have to calculate how many gram moles of sulfur are contained in 64 kilogram of sulfur. weight in grams gram mole  atomic or molecular weight 1000 grams 64 kilo gram   64,000 grams 1 kilo gram 64,000 grams gram moles of sulfur   2,000 32  Now, each gram mole of sulfur contains 6.024  1023 atoms of sulfur. number of atoms of sulfur  2000  6.024  10 23  12.048  10 26 atoms Example The atomic weight of sodium is 23. Calculate the weight in grams of one atom of sodium. Solution Based on Avogadro’s number, we know that one gram mole of sodium which is equivalent to 23 grams of sodium contains 6.024  1023 atoms. In other words, we know that 6.024  1023 atoms of sodium has a weight equal to 23 grams. Therefore: 23 the weight of one sodium atom  23  3.82  10 23 grams 6.024  10 (indeed a very very small number) Example One drop of seawater contains 300 billion atoms of sodium (atomic weight 23) in the form of salt. How many drops of seawater do we need to have 23 kilogram of sodium contained in salt? Solution First we have to calculate the number of atoms of sodium in 23 kilogram of sodium. 23 kilo gram of sodium  23,000 grams of sodium  23,000  1000 gram moles of sodium. 23 We know that each gram mole of sodium contains 6.024  1023 atoms of sodium. Therefore: The number of atoms of sodium  1000  6.024  10 23  6.024  10 26 Now each drop of seawater contains 300 billions (300  109) atoms of sodium in the form of salt. Therefore, the number of drops of seawater needed is: 6.024  1026 11  2.008  1015 3  10 2NO  Vapor pressure Vapor pressure of a liquid is defined as the pressure exerted by the vapor when the liquid and vapor phases have reached equilibrium. Vapor pressure always increases with an increase in temperature. Heat  Heat Heat is energy in transfer.  Specific heat The amount of heat expressed in calories which is required to raise the temperature of 1 gram of substance by one degree Celsius or the amount of heat expressed in BTU (British Thermal Units) which is required to raise the temperature of one pound of substance by one degree Fahrenheit. Example The specific heat of a substance is 2 Btu/lb. How much heat in Btu is needed to raise the temperature of 5 lbs of this substance by 10 degrees Fahrenheit? Solution From the definition of specific heat (above) we can say the heat needed to raise the temperature of 5 lbs by one degree is: Btu 5 lbs  2  10 Btu lb amount of heat needed to raise the temperature by 10 degrees is: 10 Btu  10  100 Btu  Heat of fusion The amount of heat required to melt 1 unit mass of the solid at its melting point without changing its temperature.  Heat of vaporization The amount of heat required to vaporize 1 unit mass of liquid at its boiling point without changing its temperature.  Heat of sublimation The amount of heat required to convert 1 unit mass of a solid to vapor without changing its temperature. Heat of sublimation = heat of fusion + heat of vaporization Gas Laws: Basic Definitions and Relationships  Density Density is mass per unit volume: m = v where: is density m is mass v is volume Example A spherical tank is half-filled with a liquid that has a density of 70 lbs/ft3. If the mass of liquid in the tank is 2100 lbs, calculate the volume of the tank. Solution mass density  volume 2100 70  volume 2100 volume   30ft 3 70 However, the above value is only half of the total volume of the tank. Total volume of tank = 30  2 = 60 ft3.  Compressible fluid The volume changes significantly with changes in pressure or temperature. For example, by pressurizing a gas, we can reduce its volume substantially. Therefore, gases are considered to be compressible.  Incompressible fluid The volume does not change significantly with changes in pressure or temperature. For example, if we increase the pressure on a liquid, its volume does not change substantially. Therefore, liquids are considered to be incompressible.  Mole Mass divided by the molecular weight. Example What is the number of pound moles of octane (C8 H18) in 114,000 grams of octane. The atomic weight of carbon is 12 and that of hydrogen is 1. Solution Step 1: Calculate the molecular weight of octane (8  12)  (18  1)  114 Step 2: Since we are interested in pound moles of octane, we have to convent the weight of octane from grams to pounds. 1 lb 114,000 grams   251 lbs 454 grams Step 3: weight in lbs pound moles  molecular weight 251 lb moles   2.2 114  Mixture density 1 x1 x2 =  1 2 where: 1 is the density of component 1, 2 is density of component 2, etc. x1 is the mass fraction of component 1, x2 is the mass fraction of component 2, etc. is the mixture density. Example A uniform solution of sulfuric acid and water contains 20 percent by weight sulfuric acid. If the density of sulfuric acid is 70 lbs/ft3 , and the density of water is 62 lbs/ft3; what is the volume in ft3 that 100 lbs of this mixture occupies? Solution First we calculate the mixture density 1 x1 x2   1 2 1  70 2  62 x1  0.2 x 2  0.8 1 0.2 0.8   70 62 lbs  63.45 ft 3 A mixture density of 63.45 lbs per cubic foot indicates 63.45 pounds of this mixture occupies a volume equal to one cubic foot. Therefore, the volume occupied by 100 pounds of this mixture is equal to: 100/ 63.45 = 1.58 cubic feet  Ideal gas law PV = nRT where p is the absolute pressure (gauge pressure plus atmospheric pressure) v is the volume occupied by the gas n is number of moles of gas R is the universal gas constant whose value depends on the units selected for P, V, n, and T T is the absolute temperature (degrees Fahrenheit plus 460 or degrees centigrade plus 273).  Absolute pressure and temperature scales Absolute pressure = gauge pressure + atmospheric pressure There are two absolute temperature scales used in solving gas law problems: degrees Kelvin = oC + 273 degrees Rankine = oF + 460  The gas constant The value of the gas constant in ideal gas law depends on the units chosen for pressure, temperature, number of moles, and volume. The following table lists the values of gas constant that are more commonly used in solving gas law problems. PV Value of Gas Constant, R  nT pressure  volume R has units of moles  temperature Absolute Pressure Volume Temp. moles Atm psia mm Hg in Hg ft H2O g 0.00290 0.0426 2.20 0.0867 0.0982 K lb 1.31 19.31 999 39.3 44.6 ft3 g 0.00161 0.02366 1.22 0.0482 0.0546 R lb 0.730 10.73 555 21.8 24.8 g 0.08205 1.206 62.4 2.45 2.78 K lb 37.2 547 28,300 1113 1262 liters g 0.0456 0.670 34.6 1.36 1.55 R lb 20.7 304 15,700 619 701 Source: BCSP Candidate Handbook Example What value of gas constant we should use if pressure is expressed in psia., volume in ft3, moles in gram moles, and temperature in degrees Rankine? Solution From the table above, we choose the column under ft3 for volume, next column, we choose R for temperature, we choose in the third column from the left g moles, then we move to the fifth column from the left for pressure in psia and read the value of R. psia  ft 3 R  0.02366 g moles   R Example What is the value of R if pressure is expressed in mm Hg (millimeter of mercury), volume in liters, moles in lb moles, and temperature in degrees Kelvin? Solution Using the same procedure as the above example mm Hg  liters R  28,300 lb moles  K Example Calculate the volume occupied at 25 C and 2 psig (pounds per square inch gauge) by the gas evolved from 1 ft3 of solid carbon dioxide lbs (density  95 ). The atomic weight of carbon is 12, and that of oxygen is ft 3 16. Solution V ? T  25 C P  2 psig lbs  95 ft 3 Let’s write the ideal gas law: PV  nRT We have to calculate n (number of moles of carbon dioxide). Since we have the volume and density of solid CO2 we can calculate its mass. mass  density  volume lbs mass  95 3  1ft 3  95 lbs ft In order to calculate the number of moles (n), we need to calculate the molecular weight of CO2. Molecular weight of CO2 = 12 + 2  16 = 44 Now we can calculate the number of moles of CO2. mass of CO 2 in lbs lb moles CO 2  molecular weight 95 lb moles CO 2   2.16 44 Before using the gas law, we have to convert the pressure and temperature to absolute scales.  P  2 psig  14.7  16.7 psia T  25 C  273  298  K Now, we have to select a value for R (gas constant). Our selected value must have the same units for pressure, temperature, volume, and moles. Since our pressure is in psia, temperature is in degrees Kelvin, and moles are in lb moles. From the table above we select the following value of R. ft 3  psia R  19.31 lb mole   K PV  nRT nRT V P (2.16)  19.31  298 V 16.7 V  744ft 3 Since the selected value of R has volume in ft3, the volume calculated also has units of ft3. Example A 100 ft3 container of hydrogen is under 1000 psig pressure at a temperature of 80 F. Calculate the mass of hydrogen in pounds. The atomic weight of hydrogen is 1. Solution P  1000 psig  14.7  1014.7 psia T  80  F  460  540  R V  100ft 3 mass of hydrogen in lbs  ? PV  nRT In order to calculate the mass of hydrogen in pounds, we need to calculate the number of pound moles of hydrogen and multiply that by the molecular weight of hydrogen. PV n RT In order to solve for n, we must select a value of R which has psia for unit of pressure, ft3 for volume, lb moles for number of moles, and degrees Rankine for temperature. From the table for values of gas constant, we have: psia  ft 3 R  10.73 lb moles  R PV n RT (1014.7)  (100) n 10.73  540 n  17.51 lb moles  Now, we know that molecular weight of hydrogen (H2) is 2. In order to obtain pounds of hydrogen, we multiply the number of lb moles by the molecular weight. lbs of H 2  17.51  2  35.02  Standard conditions Reference temperatures of 0 °C and pressure of 1 Atm are commonly referred to as standard temperature and pressure (STP). Example Calculate the volume in liters occupied by 1 gram mole of a gas under STP condition. Solution P  1 atm T  0 C  273  K n  1 kg mole V ? PV  nRT Since the problem asks for volume in liters, it is more convenient to use a value atm  liters of R that has units of. From the summary table for the gm mole   K values of gas constant, we have:  atm  liters R  0.08205 gm mole   K nRT V P 1  0.08205  273 V  22.4 liters 1 Note 1: When a gas is under atmospheric pressure, the absolute pressure is the atmospheric pressure. In other words, we do not add atmospheric pressure to atmospheric pressure. Note 2: In the above example, we did not specify the type of the gas. It could be hydrogen, nitrogen, or for that matter, any other gas. Therefore, we can make the following statement: One gram mole of any ideal gas under standard temperature and pressure (STP) conditions occupies a volume equal to 22.4 liters. Example Calculate the volume in ft3 occupied by 1 pound mole of a gas under STP conditions. Solution P  1 atm V ? n  1 lb mole T  0 C  32  F  492  R We choose the following value of R (see summary table for R) which is consistent with the units given for other parameters.  atm  ft 3 R  0.730 lb mole   R PV  nRT nRT V P 1  0.730  492 V  359ft 3 1 Note: Once again, since we did not specify the type of gas in the above example, we can make the following statement: One pound mole of any ideal gas occupies a volume equal to 359 cubic feet under STP conditions. Using the procedure outlined above, we can prove that one kilogram mole of any gas occupies a volume equal to 0.02415 cubic meters under STP conditions. We can summarize the results in the following table: The PVT values for standard conditions in different system of units are tabulated below: Standard conditions for gases System Ts ps Vs ns SI 273oK 1 atm 0.02415 m3 1 kg mole CGS 273oK 1 atm 22.415 liters 1 gram mole American 492oR 1 atm 359.05 ft3 1 lb-mole Engineering  Combined gas law When the state of an ideal gas is changed from conditions 1 to 2, the relationship between pressure, temperature, and volume is given by: P1V1T2 = P2V2T1 or PV 1 1 PV  2 2 T1 T2 Where P1, V1, T1 are absolute pressure, volume, and absolute temperature under conditions 1, and P2, V2, T2 are absolute pressure, volume and absolute temperature under condition 2. Example An ideal gas is contained in a 100 ft3 expandable container which is under 50 psig pressure at a temperature of 70 F. If we double the volume of this container and raise its temperature to 100 F, what is the value of new pressure? Solution V1  100ft 3 P1  50 psig or 50  14.7  64.7 psia T1  70  F  460  530  R V 2  200ft 3 (we doubled the initial volume ) P2  ? T2  100  F  460  560  R P1V1 P2V2  T1 T2 Substitute the known values: 64.7  100 P2  200  530 560 or 64.7  100  560  P2  200  530 64.7  100  560 P2  200  530 P2  34.18 psia Note 1: When we are dealing with changes in the state of an ideal gas, we eliminate the gas constant (R) from the equation.  Note 2: Remember that in solving any gas law problem, you must always use the ABSOLUTE PRESSURE (gauge pressure plus atmospheric pressure) and the ABSOLUTE TEMPERATURE (degrees Kelvin or Rankine).  Isothermal changes in an ideal gas When an ideal gas undergoes a change at constant temperature, T1 and T2 in our combined gas law equation are equal P1V1 P2V2  T1 T2 but, T1  T2 (constant temperature). Therefore: P1V1 = P2V2 where: P1 and V1 are pressure and volume under conditions 1 P2 and V2 are pressure and volume under conditions 2. Example A vessel at 70 F contains an ideal gas under 2000 psig. If we transfer this gas to another vessel with a volume 4 times the original vessel. What is the value of the new pressure of the gas? Assume isothermal conditions. Solution T1  T2  70 F (isothermal) P1  2000 psig  14.7  2014.7 psia P2  ? V2  4V1 (volume of second vessel, V2 , is 4 times the volume of original vessel, V1 ) 1 1  PV PV 2 2 P1 V2  4 P2 V1 2014.7 4 P2 P2  503.67 psia Isobaric changes in an ideal gas When an ideal gas undergoes changes under constant pressure, the values of P1 and P2 in the combined gas law are equal P1V1 P2V2  T1 T2 P1  P2 (isobaric) V1 V2  T1 T2 or V1T2 = V2T1 Example The volume of a sample of gas is 10 liters at 300 C. What is its volume at 150 C? Assume isobaric conditions. Solution V1  10 liters T1  300 C  273  573  K V2  ? T2  150 C  273  423  K P1  P2 (const ant pressure) V1T2  V 2T1 10  423  V2  573 V 2  7.38 liters Isometric changes in an ideal gas When an ideal gas undergoes changes under constant volume conditions, the values of V1 and V2 are equal in the combined gas law. P1V1 P2V2  T1 T2 or P1 P2  T1 T2 or P1T2  P2T1 Example The gauge on a hydrogen cylinder reads 2000 psig at a temperature of 70 F. What is the pressure in the cylinder at 110 F? Assume isometric conditions. Solution P1  2000 psig  14.7  2014.7 psia T1  70  F  460  530  R T2  110  F  460  570  R P2  ? P1T2  P2T1 2014.7  570  P2  530 P2  2166.7 psia Note: In using combined gas law equation, we have to use absolute pressure and temperature. It does not matter what absolute scales we choose as long as we maintain consistency in the selected units. In order to demonstrate this concept, let’s rework the above example using Kelvin temperature scale instead of Rankine. We should still end up with the same result.  P1  2014.7 psia T1  70  F  21.1 C  294.1  K T2  110  F  43.3 C  316.3  K P2  ? P1T2  P2T1 2014.7  316.3  P2  294.1 P2  2166.7 Which is the same result we obtained using Rankine temperature scale.  Mole fraction The number of moles of a component in a mixture divided by the total number of moles nA yA = nt where: yA is the mole fraction of component A in a mixture nA is the number of moles of component A nt is the total number of moles  Example 10 grams of hydrogen is mixed with 420 grams of nitrogen. What is the mole fraction of hydrogen in the gas mixture? The molecular weights of hydrogen and nitrogen are 2 and 28 respectively. Solution First we have to calculate the number of moles of hydrogen and nitrogen in the gas mixture. 10 gram moles of hydrogen  5 2 420 gram moles of nitrogen   15 28 total number of moles  5  15  20 5 mole fraction of hydrogen   0.25 20 Note: The sum of mole fractions of all components is a mixture must equal to 1. To demonstrate this concept, let’s calculate the mole fraction of nitrogen in the above example. 15 mole fraction of nitroen   0.75 20 mole fraction of hydrogen  mole fraction of nitrogen  0.25  0.75  1.0  Partial pressure The pressure exerted by a component of a mixture ALONE at the same volume and temperature as that of the mixture. The partial pressure of a component in an ideal gas mixture is equal to its mole fraction multiplied by the total pressure  P A = yA P t where PA is partial pressure of component A in the gas mixture yA is mole fraction of A in the mixture Pt is total pressure Example 10,000 lbs of a gas mixture of methane (CH4), ethane (C2H6), and hydrogen (H2) is under 2,000 psig pressure. The mixture contains 10 percent by weight methane, 5 percent hydrogen, and the balance ethane. Calculate the partial pressure of each component of this gas mixture. The atomic weight of Carbon (C) is 12, and the atomic weight of hydrogen (H) is 1. Solution First we have to calculate, the weight of each component in the gas mixture: weight of CH 4  10,000  0.1  1000 lbs weight of H 2  10,000  0.05  500 lbs weight of C 2 H 6  10,000  1000  500  8500 lbs Next, we calculate the molecular weight of each component: molecular weight of CH 4  12  (4  1)  16 molecular weight of H 2  2  1  2 molecular weight of C 2 H 6  (2  12)  (6  1)  30 We can now calculate the number of moles and the mole fraction of each component of the mixture. 1000 moles of CH 4   62.50 lb moles 16 500 moles of H 2   250.00 lb moles 2 8500 moles of C 2 H 6   283.33 lb moles 30 total number of moles  595.83 62.50 moles fraction of CH 4   0.105 595.83 250 moles fraction of H 2   0.419 595.83 283.33 moles fraction of C 2 H 6   0.475 595.83 Notice that the sum of mole fractions is equal to 1. partial pressure of CH 4  0.105  2000  210 psig partial pressure of H 2  0.419  2000  838 psig partial pressure of C 2 H 6  0.475  2000  950 psig  Partial volume The volume that would be occupied by a component of a mixture ALONE at the pressure and temperature of the mixture. The partial volume of a component of an ideal gas mixture equals its mole fraction in the mixture multiplied by the total mixture volume: V A = yA V t where: VA is the partial volume of component A yA is mole fraction of component A in the mixture Vt is the total mixture volume Example The mole fraction of oxygen in air is approximately 0.21. Assuming that air contains only oxygen and nitrogen, calculate the partial volume of nitrogen in 1000 ft3 of air. Solution Since the sum of mole fraction of oxygen and nitrogen must equal to 1, we can write: mole fraction of nitrogen = 1.00  0.21 = 0.79 partial volume of nitrogen = 0.79  1000 = 790 ft3 Note 1: The sum of partial pressures of all components in a gas mixture must equal the total mixture pressure. Note 2: The sum of partial volumes of all components in a gas mixture must equal the total volume of the mixture. Linear Kinematics  Linear kinematics The analysis of linear motion without regard to its origin and effects is called linear kinematics.  Average speed The average speed of an object is equal to the distance it covers in a certain time period divided by that amount of time. sfinal - sinitial V = t final - t initial where: V = average speed s = distance t = time Example A car travels a distance of 950 miles in 18 hours. What is the average speed of this car?  Solution 0 950 miles 18 hours In this case: s initial  0 s final  950 miles t final  18 hours t initial  0 950  0 V 18  0 miles Average Speed  52.78 hr  Constant acceleration When acceleration is constant the equations relating distance and speed to time take the following forms: 2 V 2  V0  2as V  V0  at 1 s  V0t  at 2 2 Where: V = speed Vo = initial speed t = time a = acceleration s = distance Example An object has an initial speed of 5 ft/sec and an acceleration of 2 ft/sec2. How far does this object travel after 10 seconds? Solution Step 1: In solving problems that require use of equations, it is best to write down all of our known and unknown values. This would greatly help in determining which equation we have to use to solve the problem. In this case: V0  5 ft / sec a  2 ft / sec 2 t  10 sec s? Step 2: We compare our known and unknown values to the terms in the above equations. In this case, we notice that in the equation: at 2 s  V0t  2 All terms are known except s (distance). Therefore, we have one equation with one unknown that we can easily solve. Step 3: Substitute for known values and calculate the unknown value.  at 2 s  V0t  2 2  100 s  5  10  2 s  50  100 s  150 ft Note 1: Once we have an equation with a number of terms, the question is that how many different types of problems can be based on that equation? In order to solve one equation, we can only have one unknown. Therefore, generally speaking, the number of problems, which can be based on a given equation, is equal to the number of terms in the equation. For example in the equation: at 2 s  V0t  2 we have 4 terms (s, V0, t, a). Therefore, we can design four different problems based on the above equations. Problem Number 1: s (distance), V0 (initial speed), t (time) are known, calculate the acceleration. Problem Number 2: s, V0, a are given, calculate the time. Problem Number 3: V0, t, a are given, calculate the distance. Problem Number 4: s, t, a are given, calculate initial speed V0. Note 2: In solving any equation, care must be exercised to maintain consistency in units. For example, in the equation: at 2 s  V0t  2 the unit for distance appears in three different terms. These terms are s (which has units of distance, ft., meters, miles, etc.), V0 which has units of distance divided by time, ft/sec, miles/hr, etc.), and acceleration (which has units of distance divided by time squared, ft/sec2, miles/hr2, etc). ONCE WE HAVE SELECTED A UNIT FOR DISTANCE (for example feet), WE HAVE TO USE THAT UNIT IN ANY TERM IN THE EQUATION WHERE DISTANCE APPEARS. This means that in the equation: at 2 s  V0t  2 We can not use feet for distance (s), meters per second for initial speed V0, or centimeters per second squared for acceleration. Once we have selected a unit for a given dimension (such as distance), we have to convert all other terms where that dimension appears to our selected unit. Example An object starting from rest with an acceleration of 7200 ft/min2 after 10 seconds has a speed equal to: a) 36 ft/sec b) 72,000 ft/sec c) 72 ft/sec d) 20 ft/sec Solution Select (d). Step1: Let’s write our known and unknown values. V0  0 a  7200 ft / min 2 t  10 seconds V ? Step 2: Compare the terms with the terms in equations for constant acceleration. We notice that if we use the equation: V  V0  at All terms are known except the final speed (V). Therefore we have one equation with one unknown which we can easily solve. Step 3: We notice that the unit of time t is given in seconds (t = 10 seconds), while the unit of time in acceleration is expressed in minutes. Since the answer to the problem uses seconds as unit of time, we have to convert the unit of time in acceleration from minutes to seconds: ft (1 min) 2 ft a  7200   2 min 2 (60 sec) 2 sec 2 Step 4: Substitute the known values in the equation and calculate the unknown value: V  V0  at V  0  2  10 ft V  20 sec Example The final speed of an object starting from rest with an acceleration of 2 ft/sec2 after 100 ft is: a) 200 ft/sec b) 50 ft/sec c) 20 ft/sec d) 400 ft/sec Solution Select (c). Step 1: V ? a  2 ft / sec 2 s  100 ft V0  0 Step 2: In the equation: V 2  V02  2as All terms are known except V (final speed). Step 3: Substitute for known values and calculate the unknown value (V).  V 2  (0) 2  2  2  100 V 2  400 V  20 ft / sec  Motion due to constant gravitational acceleration The equations for speed, time, and distance covered for falling objects due to constant acceleration of gravity are: V  V0  gt 1 s  V0t  gt 2 2 2 V 2  V0  2 gs Where: g = acceleration of gravity 32.2 ft/sec2 V0 = initial speed V = instantaneous speed s = distance t = time Example An object is dropped from the top of a tower. It takes 2 seconds for this object to reach the ground. What is the speed of this object just as it reaches the ground? What is the height of the tower? Neglect air resistance. Solution First we have to recognize that since the object is dropped, it is moving under acceleration of gravity. Therefore, we need to focus on the equations for constant acceleration of gravity. Step 1: Write all known and unknown values. V0  0 (object dropped from rest) V ? t  2 seconds s? Step 2: Find an equation, which has the speed as the only unknown value. We notice that in the equation: V  V0  gt all terms are known except V (final speed). Therefore, we can solve this equation for V. ft V  0  32.2  2  64.4 sec Step 3: Now we have to calculate the height of the tower, which is the distance traveled by this object. Again we have to find an equation that the only unknown term in it is the distance (s). We notice that in the equation: 1 s  V0 t  gt 2 2 all terms are known except distance (s). We also notice that in the equation: 2 V 2  V0  2 gs all terms are known except for distance (s). Therefore, we can use either of the above equations to calculate the distance traveled by the object (height of the tower). Let’s use both of the above equations and see if we obtain the same results. Step 4: 1 s  V0 t  gt 2 2 1 s  0  2  (32.2)  4 2 s  64.4 ft (height of the tower) Step 5: Let’s use the equation: 2 V 2  V0  2 gs to calculate s. (64.2) 2  (0) 2  2  32.2  s (64.4) 2 s 2  32.2 s  64.4 ft. which is the same result obtained in step 4. Note: The above example clearly shows that many times we have more than one choice of equations to calculate a given unknown value. Example A worker drops a 3 pound wrench while working on a scaffold 50 feet above the ground. How long does it take for the wrench to reach the ground? What is the speed of this wrench just when it reaches the ground? Neglect air resistance. Solution We notice that once the wrench is dropped, it is moving under the acceleration of the force of gravity. Step 1: Write all known and unknown values. s  50 ft t ? V ? V0  0 (Wrench dropped from rest, i.e. no initial volocity) Step 2: In the equations for motion due to constant acceleration of gravity, we have to find an equation which has all terms known except the unknown term of our interest (in this case t or V). We notice that in the equation: 2 V 2  V0  2 gs all terms are known except V. Step 3: Substitute the known values in the above equation and calculate the value of final speed (V). V 2  (0) 2  2  32.2  50 V 2  3220 ft V  56.7 sec Step 4: In order to calculate the time, we can use the equation: V  V0  gt or 1 s  V0 t  gt 2 2 In both of the above equations, the only unknown term is time t. Let’s use both of the above equations to calculate t; we should end up with the same result. Step 5: Substitute for known values in the equation. V  V0  gt 56.7  0  32.2  t 56.7 t  1.76 seconds 32.2  Step 6: Substitute for known values in the equation: 1 2 s  V0 t  gt 2 1 50  (0)  t   32.2 t 2 2 1 50   32.2 t 2 2 50  2 t2  32.2 100 t2  32.2 t  1.76 seconds which is the same result obtained in Step 5. Note: Sometimes a problem gives information that is not necessary to solve the problem. In this case, the mass of the wrench (3 pounds) had no role in solving this problem. Dynamics of Particles  First law of motion In the absence of any force, an object at rest will remain at rest and an object in motion will continue in motion in a straight line at constant speed.  Second law of motion When a non-zero force is acting on an object, the object will accelerate in the direction of the force with acceleration directly proportional to the magnitude of the force and inversely proportional to its mass. ma F gc Where: F = force m = mass a = acceleration gc = constant of proportionality It is important to remember that the value of gc depends on the system of units selected. In the metric system (CGS and MKS), the units of force and mass are defined in such a way to make the value of gc equal to 1. In the American Engineering System and British Engineering, however, the value of constant gc must be calculated in such a way that the numerical values of one pound force (lbf) and one pound mass (lbm) are equal on the surface of the earth, where acceleration of gravity is 32.2 ft/sec2. As a result, the value of gc in the metric system is equal to 1. In the American Engineering System and British Engineering, however, the value of gc = 32.2. Important Note: When using the second law of motion in the American Engineering System or British Engineering, THE VALUE OF gc IS EQUAL TO 32.2 In SI and Metric Absolute system of units: THE VALUE OF gc IS EQUAL TO 1. Note: In the British Engineering system of units, the unit of mass is called “slug” which is mass expressed in pounds divided by 32.2 The following table summarizes different systems of units. Force Mass Length Time British Engineering Pound Slug Foot Second (32.2 lb) (ft) (sec) American Pound Pound Mass Foot Second Engineering Force (lbf) (lbm) (ft) (sec) Metric Absolute Dyne Gram Centimeter Second (CGS) (gm) (cm) (sec) SI Units Newton Kilogram Meter Second (kg) (m) (sec) Let’s work out a couple of example problems to demonstrate the above concept. Example An object with a mass of 2 grams accelerates at the rate of 1 cm/sec2. Calculate the amount of force exerted on this object. Solution Step 1: Write all known and unknown values. F ? m  2 grams a  1 cm / sec 2 Step 2: Substitute in the second law equation. ma F gc 2 1 F  2 dyne 1 Note 1: Since we are in Metric Absolute System (see above table), the unit of force is expressed in dynes. Note 2: Since we are NOT working in the American Engineering or British Engineering Systems, the value of gc = 1. Example Calculate the amount of force (in lbf) exerted on an object having a mass of 4 lbm and accelerating at the rate of 64.4 ft./sec2. Solution F ? m  4 lbm a  64.4 ft / sec 2 Clearly, we are working in the American Engineering System of Units (see above table). Therefore, the value of gc is equal to 32.2 ma F gc 4  64.4 F 32.2 F  8 lbf Note 1: The BCSP will provide the formulae, constants, tables, and conversions at the time of the examination. However, the second law of motion in the BCSP handout is expressed as F  ma You have to remember one of two things: a) When working in the American Engineering or British Engineering System of units, divide mass by 32.2, or b) Write the above formula as ma F gc and realize that the value of gc is 32.2 when working in the American Engineering or British Engineering System of units, and is equal to 1 in SI or Metric Absolute system of units. Note 2: Let’s see what answer we obtain for the above example if we don’t realize that we are working in the American Engineering System of units and that the unit of mass is (lbm / 32.2) or the value of gc = 32.2. F  ma F  4  64.4 F  257.6 lbf Which is clearly the WRONG ANSWER.  Friction The frictional force exerted by one surface on another surface depends on the normal force pressing the two surfaces together, and the nature of the surfaces: F= N Where: F = frictional force = coefficient of friction N = normal force Example A rectangular box with a weight of 100 lbs. is on the floor in a warehouse. Assuming that the coefficient of friction between the box and the floor is 0.3, what is the value of minimum horizontal force that can put this box in motion? Solution F ?  0.3 N  100 lbs F N F  0.3  100 F  30 lbs Example An object is resting on a floor with the coefficient of friction equal to 0.4. If the value of the minimum horizontal force required to put this object in motion is 50 lbs, what is the weight of this object? Solution F  50 lbs  0.4 N ? F N 50  0.4 N 50 N 0.4 N  125 lbs  Example The frictional force for an object sitting on a horizontal surface is 50 lbs. If the weight of the object is 100 lbs, what is the coefficient of friction? Solution F  50 lbs N  100 lbs ? F N 50   100  0.5  Work When a force F acts upon a body through a distance S, the following formula for calculation of work is used: W = FS Where W is the amount of work done on the body. Work has units of force times distance, for example ft.  lbf, newton.  meter (joule). Example If we exert a constant force of 10 newtons to move an object for 1000 meters, what is the amount of work that we have performed on this object? Solution W  FS W  10  1000 W  10,000 joules Example A worker lifts an object, which has a mass of 20 lbm to an elevation of 30 ft. above the ground. Calculate the amount of work performed by the worker. Solution m  20 lbm h  30 ft W ? W  mgh Important note: Remember that since we are working in the American Engineering system of units, lbm must be divided by 32.2. 20 W (32.2) 30 32.2 W  600 ft  lbf Example What is the amount of work performed on an object with a mass of 10 kilogram raised to an elevation of 20 meters above the grounds?  Solution m  10 kg h  20 meters g  9.8 meters / sec 2 W  mgh W  (10)  (9.8)  (20) W  1960 joules Note: Since we are not working in the American Engineering or British Engineering system of units, we did not divide mass by 32.2.  Energy The kinetic energy of an object moving with a velocity V and having a mass m is: mV 2 K.E.  2 The potential energy of an object of mass m located at a height h above the surface of the earth is: P.E. = mgh Example A car with a mass of 4000 kilograms is traveling with a constant speed of 60 miles per hour. Calculate the kinetic energy of this car in units of joule. Solution m  4,000 kg V  60 miles / hr K.E.  ? Since joule is the unit of energy in the SI System of units, we have to convert the speed of the car from miles per hour to meters per second. miles 5230 ft 30.48 1m 1 hr 60     hr 1 mile 1 ft 100 cm 3600 sec 60  5280  30.48 (we have rounded off to two   26.82 m / sec significant figures) 100  3600 mV 2 K.E.  2 4000  26.82 2 K.E.  2 K.E.  1438624.8 joules Example An object with a mass of 10,000 lbm is traveling with a speed of 20 ft/sec. What is the kinetic energy of this object in ft. lbf? Solution m  10,000 lbm V  20 ft / sec K.E.  ? mV 2 K.E.  2 K.E.  62111.8 ft  lbf Note: Since we are working in the American Engineering System of units, we had to divide lbm by 32.2 to come up with the correct answer. Example What is the potential energy of an object with a mass of 20 kilogram at a point 500 meters above the ground level? Solution m  20 kg h  500 meters P.E.  ? P.E.  mgh P.E.  20  9.8  500 P.E.  98,000 joules Note: The value of acceleration of gravity in the SI System is 9.8 meters per second squared (m/sec2).  Example What is the potential energy of an object 100 feet above ground level? The mass of the object is 500 lbm. Solution m  500 lbm h  100 ft P.E.  ? P.E.  mgh 500  32.2  100 P.E.  32.2 P.E.  50,000 ft  lbf Note: Once again, since we are working in the American Engineering System, we had to divide pound mass (lbm) by 32.2. Also note that the value of acceleration of gravity in this system of units is 32.2 ft/sec2.  Moment Moment is defined as force times distance. When the amount of force or distance changes, the following relationship applies: F2 F1 F1  d 1 = F 2  d 2 (Assuming that moment generated by d2 F1 and F2 are equal) d1 Where F is force and d is distance. Example A 20 ft. long beam is fixed at one end inside a wall. A force of 100 lbs is applied to the beam at a location 3 ft. away from the wall. What force should be applied to the end of the beam to generate the same moment as the 100 lbs force? Solution F1  100 lbs d1  3 ft F2  ? d 2  20 ft F1d1  F2 d 2 100  3  F1  20 300 F2  20 F2  15 lbs  Rotational motion The linear velocity of an object in rotational motion can be calculated from: V = C  RPM Where V is linear velocity C is the circumference of the rotating object’s path RPM is revolutions per minute. Example A grinding wheel with a diameter of 2 ft. has a rotational velocity of 500 rpm. What is the tangential velocity at any point on the circumference of the wheel in feet per second? Solution d (diameter )  2 ft RPM  500 V ? V  C  RPM C  3.14 d C  3.14  2  6.28 ft V  6.28  500 V  3140 ft / min Note: Since we have used “minutes” for the unit of time in the above equation (RPM; revolutions per minute), when we calculate velocity (V), it has units of feet per minute. Since the problem asks for V in feet per second, we divide our answer by 60 (60 seconds in one minute). 3140 V  52.3 ft / sec 60 Pressure Pressure is the ratio of force to the area on which the force acts. Pressure has units of force per area such as newton/m2 (pascal) or lbf/in2 (psi).  Hydrostatic pressure Consider a column of fluid as shown in Figure 4. It can be shown that the hydrostatic pressure at the base of the column is: P = Po + gh where P = pressure at the base of column Po = pressure on the upper surface of the column = fluid density g = acceleration of gravity h = the height of the fluid column, and P0 h Figure 4. Pressure at the base of a fluid column. Example A tank 20 feet in diameter and 50 feet high contains water that has a density of 62.4 lbm/ft3. Assuming that the tank is open to the atmosphere and that the barometric pressure is 14.7 psi, calculate the total pressure at the bottom of the tank in pounds per square foot. Solution h  50 ft  62.4 lbm / ft3 P0  14.7 lbs / in2 P? P  P0  gh In order to maintain consistency of units, we have to convert 14.7 pounds per square inch to pounds per square foot. lbs 144 in2 P0  14.7 2  in 1 ft 2 P0  2116.8 lbs / ft 2 Important Note: Since density has units of ”mass” per “volume”, and because we are working in the American Engineering System, we have to divide its value by 32.2.  62.4 P  2116.8   32.2  50 32.2 P  5236.8 pounds per ft 2 ( psf ) or convert ing back to psi lbs 1 ft 2 P  5236.8 2  1 ft 144 in 2 P  36.36 lbs / in 2 ( psi)  Fluid pressure measurement Bourdon gauge: a hollow tube closed at one end and bent into a C configuration. The open end is exposed to the fluid whose pressure is to be measured. Manometer: a U-shaped tube partially filled with a liquid of known density. When the ends of tube are exposed to different pressures, the fluid level drops in the high-pressure arm and rises in the low pressure arm. The difference between the pressures can be calculated from the measured difference between the liquid levels in each arm. Archimedes’ Principle When an object is submerged in a fluid (liquid or gas), the buoyant force on the submerged object is equal to the weight of fluid displaced by the object. Example A cylindrical object having a diameter of 1 foot and a length of 2 feet is completely submerged in water, which has a density of 62.4 lbm/ft3. What is the buoyant force exerted on this object? 1 ft. 2 ft. Water Solution According to the Archimedes’ Principle, the buoyant force is equal to the weight of the water that is displaced by the cylinder. The amount of water displaced, however, is equal to the volume of cylinder. Therefore, we need to calculate the weight of water that has a volume equal to the volume of the cylinder. Step 1: Calculate the volume of the cylinder. D2 V H 4 Where V is the volume, D is the diameter, and H is the height of the cylinder. (3.14)  (1) 2 V 2 4 V  1.57 ft 3 Step 2: Calculate the weight of the water that has a volume equal to 1.57 ft3. We know that weight (force) is equal to mass multiplied by the acceleration of gravity. W  mg where W is the weight of an object, m is its mass, and g is the acceleration of gravity. We also know that mass is equal to density multiplied by volume. m V where m is the mass, is the density, and V is the volume. Therefore, we can write: W  Vg In this case:   62.4 lbm / ft3 V  1.57 ft3 g  32.2 ft / sec2 Before we use the above equations, we have to remember that since we are working in the American Engineering System, mass ( V) must be divided by 32.2. 62.4  1.57  32.2 W 32.2 W  97.97 lbs or approximately 100 lbs  Apparent Weight The apparent weight of an object, which is submerged in a fluid, is equal to its real weight minus the buoyant force exerted on the object by the displaced fluid. Example In the above example if the real weight of the object is 250 lbs, what is its apparent weight when completely submerged in water? Solution apparent w eight  real weig ht  weight of displaced fluid apparent w eight  250  100  150 This means that the cylinder, which has a weight of 250 lbs, weighs only 150 lbs when completely submerged in water. Example An object that weighs 400 lbs is completely submerged in a liquid. The apparent weight of the object is 300 lbs. What is the specific gravity of the object? Solution Specific gravity of an object is defined as the weight of a given volume of the object divided by the weight of the SAME volume of water. In this case, we know that the object weighs 400 lbs. Therefore, the question is what is the weight of water with the same volume as the object? Using Archimedes’ Principle, we know that the weight of water displaced is the difference between the real weight of the object and its apparent weight. In other words: weight of water  400  300  100 lbs 400 specific gravity  4 100 Inferential Statistics The Normal Probability Distribution A large number of random variables observed in everyday life and nature form a frequency distribution which is approximated by a bell-shape curve called the normal probability distribution.    The equation representing a normal probability distribution is: ( x  ) 2 1 2 2 f ( x)  e 2 Where x is the value of a random variable in a population. For example if the height of people in a given city is normally distributed, then x in the above equation represents a given height and f(x) represents the proportion of people with that height. Let’s assume that the height of people in a small city of 10,000 people is normally distributed, and let’s further assume that 500 people are 6 feet tall. This means that for a value of x equal to 6, f(x) is equal to 500/10,000.  In the equation representing the normal distribution,  is the population mean and  is the population standard deviation. Although in solving the CSP examination problems dealing with normal distribution we do not use the above equation directly (as will be discussed later), we do need to understand how to calculate the mean () and the standard deviation () for a given population. These can be calculated from the following relationships: n  xi i 1  n where n is the sample size and xi represents the values of the random variable. n 2  (x i  ) i 1  n1  is called the standard deviation and 2 is called the variance of the sample distribution. Example 1 Calculate the mean and the standard deviation for the following sample of x’s: xi xi (xi )2  3 1.33 1.77 5 0.67 0.45 2 -2.33 5.43 7 2.67 7.13 6 1.67 2.79 3 -1.33 1.77 = 4.33 sum=19.34 our sample size is 6, we can write: 19.34 standard deviation    61 or  = 1.966. Characteristics of the Normal Probability Distribution 1. The area under the normal probability distribution curve between two values x1 and x2 represents the probability that a randomly selected value would fall between x1 and x2. For example, the shaded area under the curve in the  following diagram is the probability that a randomly selected variable would fall between x1 and x2. x1 x2 2. The area under the normal probability curve represents probability. The maximum value that a probability function can assume is 1 (which means the event will certainly happen). Therefore, the total area under the normal probability distribution curve is equal to 1. Area=1 3. The normal probability distribution curve is symmetrical around its mean. This indicates that half of the total area under the curve (0.5) is to the right of the mean and the other half is to the left of the mean. 0.5 0.5  (mean) 4. The shape of the normal distribution is determined by the value of its standard deviation. Large values of standard deviation reduce the height of the curve and increase its spread while small values of standard deviation increase the height of the curve and decrease its spread. Small value of Standard Deviation Large value of Standard Deviation 5. Approximately 68% of the area under the normal distribution curve lies within  one standard deviation of the mean. About 95% of the area under the curve falls within  two standard deviations of the mean and almost all within  three standard deviations of the mean. 68% of total area    95% of total area     >99% of total area    Types of Problems Dealing With the Normal Probability Distribution  One type of problem dealing with normal probability distribution is to find the probability that a randomly selected variable from the population (with known values of mean and standard deviation) assumes a value between x1 and x2. In order to solve this type of problem we proceed as follows: Step 1: Calculate the value of Z1 from x1   Z1   Step 2: Calculate the value of Z2 from x2   Z2   Step 3: Table 3 (below): which is the table of areas for the normal probability distribution provides the areas (probabilities) under the curve between Z = 0 and a given value of Z1. Important Note: The areas under the normal distribution curve listed in Table 3 are the areas under the curve between Z = 0 and a given value of Z1. Z=0 Z1 Obtain the area between Z = 0, Z1 and Z2 from Table 3. This table will be provided by the BCSP at the time of the examination. Step 4: Since for the problem at hand we are interested in the area under the curve between Z1 and Z2, subtract the area obtained for Z1 from that obtained for Z2 if both Z1 and Z2 are positive or they are both negative. This value represents the area under the curve between Z1 and Z2 or the probability that a randomly selected variable would assume a value between x1 and x 2 (remember Z1 and Z2 were calculated form x 1 and x 2). If Z1 and Z2 have opposite signs (i.e.: one is positive and the other is negative) add the areas obtained for Z1 and Z2. Due to symmetry, the area for a negative value of Z1 is the same as the positive value of Z1 except that a negative Z1 has its area to the left of Z = 0. Example 2 A population is normally distributed with a mean of 50 and a standard deviation of 10. What is the probability that a randomly selected variable from this population falls between 35 and 40? Step 1: 35  50 Z1 =  15. 10 Step 2: 40  50 Z2 =  10. 10 Step 3: From Table 3, the area between Z = 0 and Z1 = 1.5 is 0.4332 and the area between Z = 0 and Z2 = 1.0 is 0.3413. Therefore the area under the curve between above values of Z1 and Z2 is: 0.4332  0.3413 = 0.0919 or the probability that a randomly selected variable from this population falls between 35 and 40 is 0.0919 or 9.19 per cent. This problem can be demonstrated graphically as follows.  35 40 =50  Another type of problem dealing with normal probability distribution is to find the probability that a randomly selected variable would assume a value larger or smaller than a given value x1. In order to solve this type of problem, we proceed as follows: Step 1: Calculate the value of Z1 from x1   Z1 =.  Step 2: If Z1 has a positive value two situations may arise: 1. The problem asks for probability of a randomly selected variable being greater than Z1. In this case subtract the area obtained from 0.5 2. Probability of a randomly selected variable being smaller than Z1. Add 0.5 to the area obtained from the table. If Z1 has a negative value: 1. For probability of the random variable being greater than Z1, add 0.5 to the area obtained from the table. 2. For probability of random variable being smaller than Z1, subtract the area obtained form 0.5. Example 3 For a population with a mean of 200 and a standard deviation of 30, what is the probability that a randomly selected variable assumes a value less than 250? Solution 250  200 Z1 =  166. 30 From the table of normal distribution (Table 3) the area between Z0 and Z1 = 1.66 is 0.4515. Therefore, the probability that a randomly selected variable from this population assumes a value less than250 is 0.5 plus 0.4515 or 95.15 per cent. Example 4 A population has a mean of 500 and a standard deviation of 50. What is the probability that a randomly selected variable from this population has a value larger than 600? Solution 600  500 Z1 =  2.0 50 From the table of normal distribution the area between Z0 and Z1 = 2.0 is 0.4772. However, for this problem we need the area to the right of Z1 which is: 0.5 – 0.4772 = 0.0228 which means that the probability that a randomly selected variable has a value larger than 600 is 0.0228 or 2.28 percent.  How to use Table 3: The left column lists values of Z with one decimal point. The second decimal for Z is selected from the top row. For example, the area between Z = 0 and Z1 =1.23 is equal to 0.3907. z 0 1 2 3 4 5 6 7 8 9 0.0.0000.0040.0080.0120.0160.0199.0239.0279.0319.0359 0.1.0398.0438.0478.0517.0557.0596.0636.0675.0714.0754 0.2.0793.0832.0871.0910.0948.0987.1026.1064.1103.1141 0.3.1179.1217.1255.1293.1331.1368.1406.1443.1480.1517 0.4.1554.1591.1628.1664.1700.1736.1772.1808.1844.1879 0.5.1915.1950.1985.2019.2054.2088.2123.2157.2190.2224 0.6.2258.2291.2324.2357.2389.2422.2454.2486.2518.2549 0.7.2580.2612.2652.2673.2704.2734.2764.2794.2823.2852 0.8.2881.2910.2939.2967.2996.3023.3051.3078.3106.3133 0.9.3159.3186.3212.3238.3264.3289.3315.3340.3365.3389 1.0.3413.3438.3461.3485.3508.3531.3554.3577.3599.3621 1.1.3643.3665.3686.3708.3729.3749.3770.3790.3810.3830 1.2.3849.3869.3888.3907.3925.3944.3962.3980.3997.4015 1.3.4032.4049.4066.4082.4099.4115.4131.4147.4162.4177 1.4.4192.4207.4222.4236.4251.4265.4279.4292.4306.4319 1.5.4332.4345.4357.4370.4382.4394.4406.4418.4429.4441 1.6.4452.4463.4474.4484.4495.4505.4515.4525.4535.4545 1.7.4554.4564.4573.4582.4591.4599.4608.4616.4625.4633 1.8.4641.4649.4656.4664.4671.4678.4686.4693.4699.4706 1.9.4713.4719.4726.4732.4738.4744.4750.4756.4761.4767 2.0.4772.4778.4783.4788.4793.4798.4803.4808.4812.4817 2.1.4821.4826.4830.4834.4838.4842.4846.4850.4854.4857 2.2.4861.4864.4868.4871.4875.4878.4881.4884.4887.4890 2.3.4893.4896.4898.4901.4904.4906.4909.4911.4913.4916 2.4.4918.4920.4922.4925.4927.4929.4931.4932.4934.4936 2.5.4938.4940.4941.4943.4945.4946.4948.4949.4951.4952 2.6.4953.4955.4956.4957.4959.4960.4961.4962.4963.4964 2.7.4965.4966.4967.4968.4969.4970.4971.4972.4973.4974 2.8.4974.4975.4976.4977.4977.4978.4979.4979.4980.4981 2.9.4981.4982.4982.4983.4984.4984.4985.4985.4986.4986 3.0.4987.4987.4987.4988.4988.4989.4989.4989.4990.4990 Table 3. Table of The Normal Distribution; source: BCSP Candidate Handbook Set Theory, Probability and Statistics Definitions and Concepts  Sample space A set of ALL possible outcomes of an experiment. For example the sample space for the experiment tossing a coin is the set of all possible outcomes shown as {head, tail}. This sample space has 2 elements. By the same token, the sample space for the experiment rolling a die is the set of all possible outcomes of the experiment, which is {1, 2, 3, 4, 5, 6}. This sample space has 6 elements. Example What is the sample space for the experiment drawing a card from a deck of 52 cards? Solution Once again a sample space is the set of all possible outcomes of the experiment. In this case, there are 52 possibilities, and therefore, the sample space has 52 elements. Example Suppose a government agency must decide where to locate three new nuclear research laboratories, and that (for a certain purpose) it is of interest only how many of these facilities will be located in Colorado. Solution The set of all possible outcomes is {0, 1, 2, 3} which means that none, one, two or three of the research facilities may be located in Colorado. This sample space has four elements and can be shown graphically as follows: 0 1 2 3 Sample Space  Selecting elements If sets M1, M2,..., Mk contain, respectively N1, N2,...,Nk elements, there are N1  N2... Nk ways of selecting first an element from M1, then an element from M2,..., and finally an element from Mk. In other words, the sample space for  selecting first an element from M1, then an element form M2,…, and finally an element from Mk has N1  N2  …  Nk elements. Example In an oil company the list of candidates for the President and Vice President has been narrowed down to 15. In how many different ways a President and a Vice President can be elected from the set of these 15 candidates? Solution The set for electing a president has 15 elements. However, once a President has been elected, there are only 14 candidates remaining for Vice President. Therefore, the set for the Vice President has 14 elements. Therefore, the number of different ways this election can be carried out is: 15  14  210 In other words, the sample space for this election has 210 possible outcomes.  Events Probabilities are always associated with occurrence or nonoccurrence of events; such as getting one head in four flips of a coin. Events can be considered as a subset of a sample space. For example, if we are interested in the event of getting a 6 in rolling one die, our sample space has 6 elements {1, 2, 3, 4, 5, 6}, and the event of interest (getting a 6) is a subset of this sample space with one element {6}.  Example If we are interested in the event drawing an ace of hearts from a deck of 52 cards, how many elements our event set has? How about our sample space? Solution Our sample space has 52 elements (the set of all possible outcomes), and our event set has only one element {ace of hearts}.  Mutually exclusive events When there are no common elements among events (see Figure 2). sample space A B Figure 2. Sample space and mutually exclusive events. In the above figure, there are no common elements between events A and B, and therefore, they are mutually exclusive events. In other words two events are considered to be mutually exclusive if WHEN ONE EVENT OCCURS, THE OTHER CAN NOT OCCUR, AND VICE VERSA. Two mutually exclusive events can not occur simultaneously. Example Let A and B represent two events in rolling a die A: {to get an odd number} i.e. {1, 3, 5} B: {to get an even number} i.e. {2, 4, 6}  The two events A and B described above are said to be mutually exclusive because in one roll of a die we can only get either an even or an odd number but not both. If event A happens, B cannot happen and vice versa Simple Events An event that can not be decomposed is called a simple event. Example In order to more clearly understand the above definition, let’s go back to our experiment of rolling a die. Let’s further consider the following events: A1: {getting an odd number}, i.e., {1, 3, 5} A2: {getting an even number}, i.e., {2, 4, 6} A3: observing a 1, i.e., {1} A4: {2} A5: {3} A6: {4} A7: {5} A8: {6} In this example we notice that there is a difference between events A1, A2 and events A3 through A8. Event A1 (getting an odd number) occurs if any of the events A3 (getting a 1), A5 (getting a 3), or A7 (getting a 5) occur. Therefore, we can decompose event A1 into simple events A3, A5, and A7. Similarly, event A2 (getting an even number) can be decomposed into simple events A4, A6, and A8.  Independent Events Two or more events are said to be independent of each other if the occurrence (or non-occurrence) of one has no effect on the occurrence or non-occurrence of the others. Example Consider tossing a coin and rolling a die. The events getting a head on the coin and number 3 on the die are independent of each other because the occurrence (or non-occurrence) of one event has no effect on the occurrence (or non- occurrence) of the other event.  Probability Now that we are familiar with fundamentals of set theory, and have defined events and various types of events in a given sample space, we can talk about the probability of one or more events. The probability associated with an event is a measure of belief that the event will occur on the next repetition of the experiment. For example when we say that the probability of getting head in one toss of a coin is ½ (0.5 or 50 percent), what we are really saying is that there is a 50 percent chance that the next toss of a coin will be a head. Given a finite sample space S and an event A in that sample space we can say: 1. The probability of event A is a number between zero and 1. Zero means that event A can not happen and a probability of 1 means that event A will certainly happen. 2. The probability of sample space (set of ALL possible outcomes of an experiment) is 1. If we show probability of event A with P(A), and probability of the sample space with P(S), we can write: 0  P A  1 PS   1  Calculation of Probabilities of Events in a Sample Space The probability of an event A in a finite sample space is equal to the number of simple events in A (see definition of simple events discussed earlier) divided by the total number of simple events in the sample space. Note that we are making the assumption that all simple events in the sample space have the same probability of occurrence. Example What is the probability of getting a head in one flip of a coin? Solution First we have to find out how many simple events our sample space has. Remember that sample space, by definition, is the set of all possible outcomes of an experiment. In this case, our sample space S has only 2 elements: S: {head, tail} Next we have to find out how many simple events there are in our event of interest. In this case our event of interest (head) has only one element: A: {head} Probability Number of simple events in event A = of event A Number of simple events in sample space In this case Probability of getting a head =  = 0.5 or 50 percent Example What is the probability of drawing a 4 out of a 52 card deck? Solution Once again we have to find out how many simple events our sample space and the event of our interest have. Our sample space, in this case, is comprised of 52 simple events, and our event of interest is comprised of 4 simple events (there are four 4’s in a deck of cards). Therefore, we can say: Probability of 4   0.077 or 7.7 percent drawing a 4 52 Example What is the probability of getting an even number in one roll of a die? Solution Our sample space S is comprised of 6 simple events (because there are only 6 numbers on a die). The event of getting an even number is comprised of 3 simple events (2, 4, 6). Therefore, we can say: Probability of 3 1 getting an    0.5 or 50 percent even number 6 2 Example What is the probability of drawing a king OR an ace from a deck of cards?  Solution There are 4 kings and 4 aces in a deck of cards. Therefore, the event of interest to us (drawing an ace or a king) is made of 8 simple events. Our sample space, on the other had, is made of 52 simple events. The probability in this case is: 8  0.154 or 15.4 percent 52  Calculation of Probabilities of Mutually Exclusive Events Once again, remember that two events are said to be mutually exclusive if, when one event occurs, the other can not, and vice versa. For example the events getting an even number and the event getting an odd number in one roll of a die are mutually exclusive. Axiom If A and B are two mutually exclusive events, the probability of either A or B occurring is the SUM of the probabilities of A and B. Example If the proportion of voters favoring legislation is 0.38, and the proportion of voters who are undecided is 0.22, what is the proportion of voters who are either in favor of the legislation or undecided? Solution In this case, the two events are mutually exclusive. Because a voter can not be in favor of the legislation and, at the same time, be undecided. The probabilities are additive. 0.38 + 0.22 = 0.60  Example What is the probability of getting 1 or 6 in one roll of a die? Solution These two events are mutually exclusive because if we get 1 we can not get 6 and vice versa. The probabilities, in this case, are additive. The probability of getting 1 is  (sample space has 6 elements, and the event has only one element). Similarly, the probability of getting 6 is . Therefore, the probability of getting 1 or 6 is: 1 1 2 1     0.33 or 33 percent 6 6 6 3 Example What is the probability of drawing an ace of hearts or a king of spades or a 4 of diamonds from a 52 deck of cards? Solution These events are obviously mutually exclusive because if one event occurs, the other events can not occur. The probabilities are, therefore, additive. In this case the probability is: 1 1 1 3     0.05 or 5 percent 52 52 52 52  Calculation of Probabilities of Independent Events Definition Two events are said to be independent if the occurrence or non-occurrence of one event has no effect on the outcome of the other event. Example If we roll a pair of dice and the events of our interest are getting 1 on one die and 6 on the other die, we have two independent events; because getting (or not getting) 1 on the first die has no effect on getting (or not getting) 6 on the second die. Example If we draw two cards from two separate decks of cards and our event of interest is getting two aces, we have two independent events; because getting an ace from the first deck of cards has no effect on getting an ace from the second deck of cards. Axiom If A and B are two independent events, the probability of both events occurring is the product of probabilities of events A and B. if we show probability of A with P(A) and probability of B with P(B), the probability of A and B occurring is: P(A) P(B) Example What is the probability that two cards drawn from two separate decks of cards are both aces? Solution The events getting an ace from the first deck and an ace from the second deck are independent. Let’s focus on the first deck of cards. Our sample space has 52 elements and our event has 4 elements (there are 4 aces in a deck of cards). The probability of drawing an ace from the first deck is: 4 52  Similarly, the probability of drawing an ace from the second deck is: 4 52 Since these two events are independent, the probability that both cards are aces is: 4 4   0.0059 or 0.6 percent 52 52 A probability of 0.5 percent