Processes and Process Variables

Processes and Process Variables average molecular weight = M y i M i Component Mol Fraction, y i M i N 2 0.579 28 O 2 0.263 32 CO 2 0.158 44 Total 1.0 M av = 31.6 Average molecular weight = 31.6 ECB3013: Material and Energy Balances Ir. Abdul Aziz Omar 4

Density The density of a substance is its mass per unit volume. The symbol most often used for density is ρ. Mathematically, density is defined as mass divided by volume: ρ = m v where ρ is the density, m is the mass, and v is the volume. ρ = mass in kg or g volume in m 3 or cm 3

Examples 1: 1) Calculate the density of an object with a mass of 7.2 g and a volume of 24 cm 3. 2) Calculate the density of an object with a mass of 45.4 lb and a volume of 0.125 cm 3. WE KNOW HOW TO SOLVE IT 1.0 lb = 454 g

Examples: 1) Calculate the density of an object with a mass of 0.072 kg and a volume of 24 cm 3. 2) Calculate the density (in kg/m 3 ) of an object with a mass of 2.5 lb and a volume of 0.125 m 3. 1.0 lb = 454 g

Example 3: Determine the volume of wood has a density of 0.6 g/cm 3 and a mass of 1.2 g. ρ = m v v = m ρ = 1.2 g 0.6 g/cm 3 = 2 cm 3

Class Work 3.1 1) Calculate the density of an object with a mass of 0.1 lb and a volume of 24 cm 3. 2) Determine the volume of wood has a density of 0.6 g/cm 3 and a mass of 120g.

Specific volume specific volume (ῡ) = v m = 1 ρ Example 1: Calculate the density and specific volume of an object with a mass of 60 g and a volume of 2.5 cm 3. ρ = m v = 60 g 2.5 cm3 = 24 g/cm3 ῡ = v m = 1 ρ = 1 24 =0.042 cm3 /g

Specific volume specific volume (ῡ) = v m = 1 ρ Example 2: Calculate the volume and specific volume for 20 g of n-propyl alcohol. The density of n-propyl alcohol is 0.8 g/cm 3. v = m ρ = 20 g 0.8 g/cm3 = 25 g/cm3 ῡ = v m = 1 ρ = 1 0.8 = 1.25 cm3 /g

Specific gravity Example 1: If the density of iron is 7.85 g/cm 3, calculate the specific gravity of iron. (Water density (@ 4 C) is 1000 kg/m 3 )

Example 2: If dibromopentane (DBP) has a specific gravity of 1.57, what is the density in: (a) g/cm 3 (b) lb m /ft 3 and (c) kg/m 3

density Volumetric flow rate MW

Moles & Concentrations Mole is a unit of measurement used in chemistry to express amounts of a chemical substance, defined as the amount of any substance that contains as many elementary entities, as there are atoms in 12 grams of pure carbon-12 ( 12 C) the isotope of carbon with relative atomic mass 12. A mole is simply a unit of measurement. Units are invented when existing units are inadequate. Chemical reactions often take place at levels where using grams wouldn't make sense, yet using absolute numbers of atoms/molecules/ions would be confusing, too.

Mole and Mass Fraction mole fraction = (moles of a species)/(total moles) y i n n i T mass fraction = (mass of a species)/(total mass) = weight fraction w i m m i T ECB3013: Material and Energy Balances Ir. Abdul Aziz Omar 24

EXAMPLE 1 Component No. Mole Mol Fraction, y i N 2 55 0.579 O 2 25 0.263 CO 2 15 0.158 Total 95 1.0 25

EXAMPLE 2 A mixture of gases has the following composition by mass: O 2 16% CO 4.0% (x O2 =0.16 g O 2 /g total) CO 2 17% N 2 63% What is the molar composition? Solution Basis: 100 g of the mixture

Mass and Volumetric Flow Rate The rate at which a material is transported through a process line is the flow rate of that material. The flow rate of a process stream may be expressed as a mass flow rate (mass/time) or as a volumetric flow rate (volume/time). Suppose a fluid (gas or liquid) flows in the cylindrical pipe shown below, where the shaded area represents a section perpendicular to the direction of flow. If the mass flow rate of the fluid is ṁ (kg/s), then every second m kilograms of the fluid pass through the cross section. If the volumetric flow rate of the fluid at the given cross section is V (m 3 / s), then every second V cubic meters of the fluid pass through the cross section. m V m V The density of a fluid can be used to convert a known volumetric flow rate of a process stream to the mass flow rate of that stream or vice versa.

Flow Rate Measurement A flowmeter is a device mounted in a process line that provides a continuous reading of the flow rate in the line. Two commonly used flowmeters the rotameter (a) and orifice meter (b) are shown schematically in the Figure below. The rotameter is a tapered vertical tube containing a float; the larger the flow rate, the higher the float rises in the tube. The orifice meter is an obstruction in flow channel with a narrow opening through which the fluid passes. The fluid pressure drops (decreases) from the upstream side of the orifice to the downstream side; the pressure drop.

Flow Rate Mass flow rate Mass/time Volumetric flow rate Volume/time m V m V

Example 1 A solution contains 15% A by mass (x A = 0.15) and 20 mole% B (y B = 0.20): 1.Calculate the mass of A in 175 kg of the solution. 175 kg solution 0.15 kg A kg solution = 26 kg A 2.Calculate the mass flow rate of A in a stream of solution flowing at a rate of 53 lb m /h. 53 lb m 0.15 lb m A h lb m = 8.0 lb m A/h If a mass or molar unit such as lb m in 53 lb m /h is not followed by the name of a species, the unit should be understood to refer to the total mixture or solution rather than to a specific component. 3.Calculate the molar flow rate of B in a stream flowing at a rate of 1000 mol/min. 1000 mol 0.20 mol B min mol = 200 mol B/min

4.Calculate the total solution flow rate that corresponds to a molar flow rate of 28 kmol B/s. 28 kmol B 1 kmol solution s 0.20 kmol B = 140 kmol solution/s 5.Calculate the mass of the solution that contains 300 lb m of A. 300 lb m A 1 lb m solution 0.15 lb m A = 2000 lb m solution 28 kmol B 0.2 kmol B X 1.0 kmol solution

Concentrations

Concentration There are quantities describe concentrations: 1) Mass concentration: C = m V (kg/m3 ) 2) Molar concentration: C = n V (mol/m3 ) 3) Number concentration: C = Ni V 1 {Ni = number of entities of constituent, always = 1} m3

Concentration 4) Part Per Million (ppm): Usually describes the concentration of very dilute substances like body fluid (blood):

Example 1: A solution contains Cu 2+ ions at a concentration of 3 10-4 M. What is the Cu 2+ concentration in ppm. Solution: Concentration of Cu 2+ ions = 3 10-4 M = 3 10-4 mol/l Mwt of Cu 2+ = 63.55 g/mol Concentration Mwt = 3 10-4 mol/l 63.55 g/mol = 1.9 10-2 g/l 1 ppm = mg/l

Example 2: Calculate the moles of CuSO 4 in 250 ml of 0.02M copper sulfate solution. Solution: n = M V = 0.02 250 1000 = 0.005 mol

Example 3: Calculate the volume of 0.08 M KBr solution containing 1.6 moles of potassium bromide. Solution: V = n M = 1.6 0.8 = 2.0 L

Example 4 A 0.50-molar aqueous solution of sulfuric acid flows into a process unit at a rate of 1.25 m 3 /min. The specific gravity of the solution is 1.03. Calculate (1) the mass concentration of H 2 SO 4 in kg/m 3, (2) the mass flow rate of H 2 SO 4 in kg/s, and (3) the mass fraction of H 2 SO 4. Solution 1. C H SO kgh 2 3 2 4 m SO 4 0.50 mol H 2 SO 4 98 g L mol 1 kg 10 3 g 10 3 L 1 m 3 = 49 kg H 2 SO 4 /m 3 2. kgh 2SO s 4 m H SO 2 4 1.25 m 3 min 49 kg H 2 SO 4 m 3 1 min 60 s = 1.0 kg H 2 SO 4 /s 1000kg kg 3. solution (1.03) 1030 3 3 m m m solution kg s 1.25 m 3 solution 1030 kg 1 min min m 3 solution 60 s = 21.46 kg/s x m m 1.0kgH2SO4 / s kgh2so4 0. 048 21.46kgsolution / s kgsolution H2SO4 H2SO4 solution

Temperature conversion F (Fahrenheit) = 1.8 C + 32 K (Kelvin) = C + 273.15 C (Celsius) = ( F 32)/1.8 R (Rankin) = 1.8 K Re (Reaumur) = 0.8 C Complete the table below with the proper equivalent temperature: ⁰C ⁰F K ⁰R Re 40 40 25 77 298 437 20 698 69.8 233 420 32 425 797 1257 340 235 390 38.4 188

HW 3.2: Complete the table below with the proper equivalent temperature. ⁰C ⁰F K ⁰R Re 160 530 294 50 1010-198

Example 2 What is the pressure 30.0 m below the surface of a lake? Atmospheric pressure (the pressure at the surface) is 10.4 mh 2 O, and the density of water is 1000 kg/m 3. Assume that g is 9.807 m/s 2. Solution P P0 gh P 10.4 m H 2 O 1.013 10 5 N/m 2 10.33 m H 2 O + 1000.0 kg/m 3 9.807 m s 2 30.0 m 1 N 1 kg.m/s 2 = 3.96 10 5 N/m 2 (Pa) = 396 kpa P h ( mh 2 O) 10.4mH 2 O 30.0mH 2 O 40. 4mH 2 O CLASS WORK Verify that the two calculated pressures are equivalent.