FDE 211 Material & Energy Balances. Instructor: Dr. Ilgin Paker Yikici Fall 2015

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1 FDE 211 Material & Energy Balances Instructor: Dr. Ilgin Paker Yikici Fall 2015

2 Material& Energy Balances first step in understanding a process leads to a better understanding of a process forces the engineer to be 100% complete locates imbalances serves as a starting point for analysis/design serves as a model for change

3 Agenda Process and Process Variables Process Flow Sheet Process Units Moles and Molecular Weight Mass Fraction and Mole Fraction Concentration Process Streams Density, Mass, and Volume Pressure Measurement Types of Pressures Mass and Volumetric Flow Rates Standard Temperature and Pressure

4 Process and Process Variables A process is any operation or series of operations that causes physical or chemical changes in a substance or a mixture of substances. Process Flow Sheet: A process flow sheet (PFD) is a sequence of process units connected by process streams. It shows the flow of materials and energy through the process units.

5 Process Diagram Input Process Unit output Schematic of a process unit with single input and single output stream.

6 Definitions Process Unit: A process unit is an apparatus/equipment in which one of the operations that constitutes a process is carried out. Each process unit has associated with it a set of input and output process streams, which consists of materials that enter and leave the unit.

Typically, these streams are labeled with information

7 Definitions A process stream is a line that represents the movement of material to or from process units. Typically, these streams are labeled with information regarding the amount, composition, temperature, and pressure of the components.

8 Density Density (ρ) is defined as mass (m) per unit volume (V) of a substance. The proper units reflect mass/length 3 (e.g., kg/m 3, g/cm 3, and lb m /ft 3 ): ρ = m/v

9 Density Density can be used as a conversion factor to relate the mass and volume of a substance. Densities of pure solids and liquids are essentially independent of pressure and vary slightly with temperature. For most compounds, density decreases with temperature (volume expansion). Solids and liquids are incompressible, which means that density is constant with change in pressure. Gases (vapors) are compressible, which means that density changes as pressure changes.

10 Specific volume & gravity Specific volume (1/ρ) is defined as volume occupied by a unit mass of a substance. The proper units reflect length3/mass (reciprocal of density). Specific gravity (SG) is the ratio of the density (ρ) of a substance to the density (ρ ref ) of a reference substance at a specific condition: SG=ρ/ ρ ref density of substance/ ref density of a reference substance

11 Specific gravity The reference commonly used for solids and liquids is water at 4 C For gas, air is the reference whose density ρ ref = ρ H20,4C =1.0 g/cm 3 = 1000 kg/m 3 = lb m /ft 3 For example, the specific gravity of a substance at 20 o C: SG= ρ sub at 20 o C/ ρ ref at 4 o C

12 Mass and Volumetric Flow Rates A flow rate is the rate (on a time basis) at which a material is transported through a process line: Mass flow rate (ṁ) = mass/time Volume flow rate (V) = volume/time where the dot above m and V refers to a flow rate, which is relative to time.

13 Mass and Volumetric Flow Rates 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 particular stream or vice versa Material transports through a pipe

14 Example Volumetric Flow Rate Problem: Nitrogen from a tank flows at a rate of 6 lb m /min at 350 F, then enters a heater where it is heated, and leaves the heater as a gas at 150 F and 600 psia. Calculate the volumetric flow rate and the specific volume of the gas leaving the heater. Molecular weight of N 2 is 28 g/mol or 28 lb m /lbmol. Assume the ideal gas law. (Gas constant, R = ft 3 psia/lbmol.r)

15 Reminder: Ideal Gas Law PV= nrt the equation of state of a hypothetical ideal gas

16 Example Solution Known quantities: Nitrogen mass flow rate, inlet temperature, exit temperature, and pressure Find: The volumetric flow rate of gas leaving the heater, specific volume. Analysis: Use the ideal gas law, PV= nrt Volumetric flow rate (V) of gas leaving the heater V =nrt/p

17 Solution Flow Chart Tank filled with Nitrogen 6 lb m /min at 350 F Nitrogen Heater gas At 150 F with P=600 psia

18 Example Solution Molar flow rate n= 6 lb m /min x lbmol/28 lb m = lbmol/min Molar flow rate is a measure of the amount of substance (the number of molecules) that passes through a given area perpendicular to the flow in a given time.

19 Example Solution Absolute gas exit temperature T = = R Gas constant, R = ft 3 psia/lbmol.r Gas volumetric flow rate= (0.214 lbmol/min x ft 3 psia/lbmol.r x R)/600 psia = 2.33 ft 3 /min

20 Solution The specific volume is obtained by dividing the volumetric flow rate by the mass flow rate: ν = V/m = 2.33 ft 3 /min / 6 lb m /min = 0.39 ft 3 /lb m

21 Solution The specific molar volume is obtained by dividing the volumetric flow rate by the molar flow rate: V m = V/n = 2.33 ft 3 /min / lbmol/min = 10.9 ft 3 /lbmol

22 Moles and Molecular Weight Atomic weight is the mass of an atom of an element. Mole is the amount of the species whose mass in grams is numerically equal to its molecular weight. One mole of any species contains approximately (Avogadro s number) molecules of that species. Molecular weight is the sum of the atomic weights of the atoms that constitute a molecule of the compound (same as molar mass); units are of the form kg/kmol, g/mol, or lb/lbmol. Molecular weight is the conversion factor that relates the mass and the number of moles of a quantity of a substance.

23 Molecular Weight

24 Molecular weight The average molecular weight based on mole fraction is: MW = y A MW A + y B MW B The average molecular weight based on mass fraction is: Where, 1/MW = x A /MW A + x B /MW B MW= average molecular weight of the molecule y i =mole fraction of component i in the mixture x i =mass fraction of component i in the mixture

25 Problem A liquid stream flowing at 100 g/min contains 0.3 mole fraction benzene (Mw = 78) and the balance 0.7 mole fraction toluene (Mw = 92). Calculate the molar flow rate of each component in the stream.

26 Solution Known quantities: Stream mass flow rate and component mole fraction. Find: The molar flow rate of benzene and toluene. Analysis: Use the average molecular weight (Mw) based on mole fraction to convert the mass flow rate to mole flow rate:

27 Solution MW = y A MW A + y B MW B = = 87.8 g Convert the stream total mass flow rate to mole flow since the stream composition is in mole fraction: Total molar flow rate of the stream= mass/mw= 100g/min x mol/87.8 g = 1.14 mol/min

28 Solution Molar flow rate of benzene = 0.3 x 1.14 mol/min = mol/min Molar flow rate of toluene = 0.7 x 1.14 mol/min = 0.80 mol/min

29 Composition of Streams Process streams occasionally contain one substance; more often they consist of mixtures of liquids or gases, or solutions of one or more solutes in a liquid solvent. The following terms may be used to define the composition of a mixture of substances, including a species A: Mass fraction: x A = mass of A/total mass Mole fraction: y A = moles of A/total mole content Remember mass and mole fractions are unitless.

30 Problem A mixture of methanol (CH3OH, Mw = 32.04) and ethanol (C2H5OH, Mw = 46.07) is flowing through a circular pipe at a flow rate of 3.0 m/s. The mixture contains 30.0 wt% methanol and 70.0 wt% ethanol. The specific gravity of the mixture is If the inside diameter of the pipe is 0.10 m what is the flow rate of the mixture in kg/s, kmol/s?

31 Solution Known quantities: Mixer velocity, specific gravity, and inner pipe diamet Find: The molar flow rate of the mixture. Analysis: The mass flow rate is obtained by multiplying its density by the volumetric flow rate. Specific gravity (SG) = 0.80 = ρ substrate / ρ referance ρ referance = 1000 kg/m 3 so ρ substrate = 0.80 x 1000 kg/m 3 = 800 kg/m 3 m= ρ substrate νa = 800 kg/m 3 x 3 m/s x (π x (0.10 m) 2 )/4)= kg/s

32 Solution Since we know the total mass flow rate, we can compute the molar flow rate of methanol and ethanol: n methanol = 0.3 x kg/s / kg/kmol = kmol/s Ethanol molar flow rate: n ethanol = 0.7 x kg/s / kg/kmol = 0.28 kmol/s The total molar flow rate then is: n total = n methanol + n ethanol = = kmol/s

33 Mass Fractions- Problem A liquid stream containing only benzene, toluene, and xylene flows through a conduit that has a square cross-section. The total flow rate of the stream is 10 mol/s. If the mole fraction of benzene (Mw = 78) is 0.3, that of toluene (Mw = 92) is 0.5, and that of xylene (Mw = 106) is 0.2, what is the mass fraction of each of the three components? If the density of the liquid is 0.87 g/cm 3 and the conduit is 0.10 m on a side, what is the mass fraction of each component and the volumetric flow rate of the liquid inside the conduit?

34 Solution Known quantities: Mixture mole fraction of benzene, toluene, and xylene, density of mixture and conduit diameter, and molar flow rate are known. Find: The mass fraction of each component. Analysis: The mass flow rate for the three components is equal to their respective molar flow rate times their molecular weight: m i = y i * n total * MW i m i : the component mass flow rate y i : the component mole fraction n total : the total molar flow rate MW i : the component molecular weight

35 Solution Mass flow rate of benzene: m b = 0.3 * 10 mol/s * 78 g/mol = 234 g/s Mass flow rate of toluene: m t = 0.5* 10 mol/s * 92 g/mol = 460 g/s Mass flow rate of xylene: m x = 0.2* 10 mol/s * 106 g/mol = 212 g/s The total mass flow rate is the sum of these three: m total = m b + m t + m x = = 906 g/s

36 Solution The mass fractions are the individual mass flow rates divided by the total mass flow rate; Mass fraction of benzene: x B = 234 g/s /906 g/s = Mass fraction of toluene: x t = 460 g/s /906 g/s = Mass fraction of xylene: x x = 212 g/s /906 g/s = 0.234

37 Solution The volumetric flow rate obtained by dividing the mass flow rate by the density is; V= m total / ρ = g/s / 0.87 g/cm 3 = 1041 cm 3 /s

38 Concentrations Concentrations can be expressed in many ways: weight/weight fraction (w/w), weight/volume fraction (w/v), molar concentration (M), and mole fraction. The weight/weight concentration is the weight of the solute divided by the total weight of the solution, and this is the fractional form of the percentage composition by weight. The weight/volume concentration is the weight of solute divided by the total volume of the solution. The molar concentration is the number of moles of the solute, expressed in moles, divided by the volume of the solution. The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all species present in the solution

39 Air Compositions- Problem If 100 g of air consists of 77% by weight of nitrogen (N 2, molecular weight = 28), and 23% by weight of oxygen (O 2, molecular weight = 32), Calculate; a. the mean molecular weight of air b. mole fraction of oxygen c. concentration of oxygen in mol/m 3 and kg/m 3, if the total pressure is 1.5 atm and the temperature is 25 C.

40 Solution Known quantities: Mass of air and its composition in weight fraction Find: The air composition. Analysis: The number of moles is obtained by dividing the mass by molecular weight.

41 Solution- Part a The mass of N 2 and O 2 that 100 g of air contains: Mass of N 2 = g = 77 g Mass of O 2 = g = 23 g Number of moles of N 2 = 77 g/ 28 g/mol = 2.75 mol Number of moles of O 2 = 23 g/32 g/mol = 0.72 mol Total number of moles = = 3.47 mol The mean molecular weight of air is the total mass/total moles. Therefore, the mean molecular weight of air: 100 g/3.47mol = 28.8 g/mol

42 Solution- Part b Mole fraction of oxygen (O 2 ) = 0.72/3.47 = 0.21, and this is also the volume fraction. x i = n i /n = V i /V = P i /P = y i Thus, mole fraction (x i ) and volume fraction (y i ) for an ideal gas mixture are equivalent.

43 Solution- Part c In the ideal gas equation, n is the number of moles present, P is the pressure (in atm, in this instance), and T is the temperature (in K, in this case). The corresponding value of R is atml/mol K PV = nrt and molar concentration = n/v T= 25 o C+ 273 = 298 K First calculate the total molar concentration: n/v = P/RT= 1.5 atm/( atml/mol K)(298 K) = mol/l Convert mol/l into mol/m 3 so mol/k x 1000L/1 m 3 = mol/m 3

44 Solution- Part c We determined the total molar concentration. If we multiply the total molar concentration by the mole fraction of O 2, we will find O 2 molar concentration: mol/m 3 x 0.21 = mol/m 3 The O 2 mass concentration is the molar concentration multiplied by the molecular weight: n o2 x MW o2 = mol/m 3 x 32g/mol x 1kg/1000g = 0.41 kg/m 3

45 Concentration of Salt in Water- Problem Salty water is prepared by mixing salt (NaCl, MW = 58.5 g/mol) in water (Mw = 18 g/mol). A solution is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kg/m 3. Calculate the concentration of salt in this solution as a a. weight/weight fraction b. weight/volume in kg/l c. mole fraction

46 Solution Known quantities: Mass of salt and mass of water, mixture density. Find: (a) Weight/weight fraction, (b) weight/volume in kg/l, and (c) mole fraction. Analysis: Weight fraction is obtained by dividing the mass of each component by the total mass.

47 Solution- Part a Weight fraction of salt = 20 kg/(100 kg+20kg) = 0.167

48 Solution- Part b Salt concentration is the weight of salt/total volume. The total volume is the mass of the mixture divided by mixture density. A density of 1323 kg/m 3 means that 1 m 3 of solution weighs 1323 kg. Volume of mixture = (100 kg + 20 kg)/1323 kg/m 3 = m 3 Weight of salt/volume = 20 kg/0.091 m 3 = 220 kg/m 3 Weight/volume in kg/l = 220kg/m 3 x m 3 /1000L= 0.22 kg/l

49 Solution- Part c Moles of water and salts are obtained by dividing the mass of 100kg/each component by its respective molecular weight: Moles of water = 100 kg/18 kg/kmol = 5.56 kmol Moles of salt = 20 kg/58.5kg/kmol =0.34 kmol Total number of moles = = 5.9 kmol Mole fraction is obtained by dividing moles of each component by the total number of moles: Mole fraction of water = 5.56 kmol/5.9 kmol = 0.94 Mole fraction of salt = 5.56 kmol/5.9 kmol = 0.06

50 Pressure Measurement Pressure is the ratio of force perpendicular to the area on which the force acts. Pressure units are force units divided by area units (e.g., N/m 2 or Pascal (Pa), dyn/cm 2, and lbf/in. 2 or psi). The pressure at the base of a vertical column of fluid (nonmoving) of density ρ and height h, called the hydrostatic pressure, is given by P= P o + ρgh where P 0 is the pressure exerted on the top of the column, g is the acceleration of gravity

51 Pressure measurement The earth atmosphere can be considered as a column of fluid with zero pressure at the top. The fluid pressure at the base of this column (e.g., at sea level) is the atmospheric pressure, P atm. The absolute pressure, P abs, of a fluid is the pressure relative to a perfect vacuum (P = 0). Many pressure measurement devices report the gauge pressure of the fluid, which is the pressure of the fluid relative to atmospheric pressure (reference pressure). A gauge pressure of 0 indicates that the absolute pressure of the fluid is equal to the atmospheric pressure. The relationship for converting between absolute and gauge pressure is given by: P abs = P atm + P gauge

52 Pressure Pressure is defined as force per unit area. Pressure can be expressed as a relative pressure or absolute pressure. The units of pressure are; SI unit: N/m 2 (or Pa) CGS unit: dyn/cm 2 AES unit: lb f / in 2 (or psi)

53 Fluid Pressure Food engineers are most often interested in pressures that are caused by a fluid. If a fluid is flowing through a horizontal pipe and a leak develops, a force must be applied over the area of the hole that causes the leak to stop the fluid from leaking. This pressure is called the fluid pressure (the force applied must be divided by the area of the hole). Force on a side hole in an open tank

54 Hydrostatic Pressure If a vertical container contains a fluid, the mass of the fluid will exert a force on the base of the container. This pressure is called the hydrostatic pressure. Hydrostatic pressure is the pressure caused by the mass of a fluid. Pressure at the bottom of an open tank

55 Types of pressure 1. Atmospheric pressure, P atm, is the pressure caused by the weight of the earth s atmosphere. Often atmospheric pressure is called barometric pressure. 2. Absolute pressure, P abs, is the total pressure. An absolute pressure of 0.0 is a perfect vacuum. Absolute pressure must be used in all calculations, unless a pressure difference is used. 3. Gauge pressure, P gauge, is pressure relative to atmospheric pressure. 4. Vacuum pressure, P vac, is a gauge pressure that is below atmospheric pressure. It is used so that a positive number can be reported.

56 Pressure The units psi and atm often carry a suffix a or g to indicate absolute or gauge (relative) pressure, respectively. Thus, by psig, gauge pressure in psi is meant, and by psia, absolute pressure in psi is meant.

57 Pressure Measurement Standard atmosphere is defined as the pressure equivalent to 760 mmhg or 1 atm at sea level and at 0 C. Other units used for the standard atmosphere (atm) are; 1 atm= 760 mmhg= 76 cmhg = 1.013x10 5 N/m 2 (or Pa) = kpa = kpa = psi (or lb f /in 2 ) = Hg = ft H 2 0

58 Standard Temperature and Pressure Standard temperature and pressure are used widely as a standard reference point for expression of the properties and processes of ideal gases: Standard temperature T S = 0 C 273 K Standard pressure P s = 1 atm = 760 mmhg = kpa Standard molar volume ṽ S = 22.4 m 3 (STP)/kmol = 22.4 L (STP)/mol = 359 ft 3 (STP)/lbmol (STP) = The standard temperature and pressure One molecule of an ideal gas at 0 C and 1 atm occupies 22.4 L.

59 Pressure-Sensing Devices The types of pressure-sensing devices include the Bourdon gauge, diaphragm capsule, and capacitance sensor, column of fluid, manometer, barometer, silicon diaphragm, and semiconductor strain gauges. The latter are used for better response and increased sensitivity. A manometer is a U-shaped device that uses a fluid having greater density than other fluids in the process unit. Manometer operation is based on the fact that hydrostatic pressure at the same level in the same fluid must be the same in each leg. To understand how a manometer works, we must understand how to determine the hydrostatic pressure caused by a mass of a column of fluid

60 Pressure Equations P =F/A = mg/a = ρvg/a = ρahg/a = ρgh So whenever we need to determine the hydrostatic pressure caused by a mass of fluid, it is simply given by P=ρgh

61 Pressure Summary Gauge pressure is the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it. Atmospheric pressure does add to the pressure in any fluid not enclosed in a rigid container. The total pressure, or absolute pressure, is thus the sum of gauge pressure and atmospheric pressure: P abs = P gauge + P atm where P abs is absolute pressure, P gauge is gauge pressure, and P atm is atmospheric pressure

62 Problem A pressure gauge on a tank reads 25 psi. What is the absolute pressure, considering the atmospheric, Patm =14.7 psi?

63 Solution Known quantities: Pressure gauge. Find: The gauge pressure. Analysis: Absolute pressure is obtained by adding the gauge pressure and local atmospheric pressure. The gauge reads the gauge pressure directly. Therefore, the absolute pressure P abs =25 psig psi = 39.7 psia = 40 psia

64 Problem The pressure gauge on a tank reads 20 cmhg vacuum. The atmospheric pressure in that tank area is 760 mmhg. What is the absolute pressure in the tank?

65 Solution Known quantities: Vacuum pressure. Find: The absolute pressure. Analysis: The gauge reads a vacuum gauge pressure directly. The pressure is measured below atmospheric pressure relative to the latter and in the reverse direction toward zero absolute pressure, that is, P vac =P atm P abs. Thus, the absolute pressure is P abs = P atm - P vac = = 560 mmhg

66 Problem The fluid used in a manometer has a specific gravity of 0.80, and the column manometer height is 0.50 m. The manometer is used to measure the pressure in a pipe. The local atmospheric pressure is 100 kpa. Determine the pressure in the column.

67 Solution Known quantities: Fluid specific gravity and column height are known. Find: The pressure in the manometer column. Analysis: The density of the fluid is obtained by multiplying its specific gravity by the density of water, which is 1000 kg/m 3. ρ=ρ S x ρ H2O = 0.8 x 1000 kg/m 3 = 800 kg/m 3 P = 100 kpa+(800 kg/m 3 )(9.81 m/s 2 )(0.50m)x(1 N/1 kgm/s 2 )x(1 kpa/1000 N/m 2 ) = kpa

KNOWN: Mass, pressure, temperature, and specific volume of water vapor.

KNOWN: Mass, pressure, temperature, and specific volume of water vapor. .0 The specific volume of 5 kg of water vapor at.5 MPa, 440 o C is 0.60 m /kg. Determine (a) the volume, in m, occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c)