fde 211 material & energy balances
TRANSCRIPT
FDE 211Material & Energy Balances
Instructor: Dr. Ilgin Paker YikiciFall 2015
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
Material& Energy Balances
Moles and Molecular Weight
Mass Fraction and Mole Fraction
Concentration
Pressure Measurement
Types of Pressures
Standard Temperature and Pressure
Process and Process Variables
Process Flow Sheet
Process Units
Process Streams
Density, Mass, and Volume
Mass and Volumetric Flow Rates
Agenda
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.
Process and Process Variables
Schematic of a process unit with single input and single output stream.
Process Diagram
Process UnitInput output
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.
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.
Definitions
Density (ρ) is defined as mass (m) per unit volume (V) of a substance. The proper units reflect mass/length3 (e.g., kg/m3, g/cm3, and lbm/ft3):
ρ = m/V
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.
Density
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 (ρref) of a reference substance at a specific condition:
SG=ρ/ ρref density of substance/ ref density of a reference substance
Specific volume & gravity
The reference commonly used for solids and liquids is water at 4°C
For gas, air is the reference
whose density ρref = ρH20,4C=1.0 g/cm3 = 1000 kg/m3 = 62.43 lbm /ft3
For example, the specific gravity of a substance at 20 oC:
SG= ρsub at 20 oC/ ρref at 4 oC
Specific gravity
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.
Mass and Volumetric Flow Rates
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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
Mass and Volumetric Flow Rates
Material transports through a pipe
Problem:
Nitrogen from a tank flows at a rate of 6 lbm/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 N2 is 28 g/mol or 28 lbm/lbmol. Assume the ideal gas law.
(Gas constant, R = 10.73 ft3 psia/lbmol.R)
Example Volumetric Flow Rate
PV= nRT
the equation of state of a hypothetical ideal gas
Reminder: Ideal Gas Law
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
Example Solution
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Solution Flow Chart
Heater
Tank
filled
with
Nitrogen
6 lbm/min at −350°F Nitrogen gas
At 150°F with P=600 psia
Molar flow rate
n= 6 lbm/min x lbmol/28 lbm = 0.214 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.
Example Solution
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Absolute gas exit temperature T = 150 + 459.7 = 609.7 R
Gas constant, R = 10.73 ft3 psia/lbmol.R
Gas volumetric flow rate= (0.214 lbmol/min x 10.73 ft3
psia/lbmol.R x 609.7 R)/600 psia = 2.33 ft3/min
Example Solution
The specific volume is obtained by dividing the volumetric flow rate by the mass flow rate:
ν = V/m = 2.33 ft3/min / 6 lbm/min = 0.39 ft3/lbm
Solution
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The specific molar volume is obtained by dividing the volumetric flow rate by the molar flow rate:
Vm = V/n = 2.33 ft3/min / 0.214 lbmol/min = 10.9 ft3/lbmol
Solution
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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 6.023 ×1023 (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.
Moles and Molecular Weight
Molecular Weight
The average molecular weight based on mole fraction is:
MW = yAMWA + yBMWB
The average molecular weight based on mass fraction is:
1/MW = xA/MWA + xB/MWB
Where,
MW= average molecular weight of the molecule
yi =mole fraction of component i in the mixture
xi =mass fraction of component i in the mixture
Molecular weight
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.
Problem
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:
Solution
MW = yAMWA + yBMWB = 0.3 × 78 + 0.7 × 92 = 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
Solution
Molar flow rate of benzene = 0.3 x 1.14 mol/min = 0.342 mol/min
Molar flow rate of toluene = 0.7 x 1.14 mol/min = 0.80 mol/min
Solution
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: xA = mass of A/total mass
Mole fraction: yA = moles of A/total mole content
Remember mass and mole fractions are unitless.
Composition of Streams
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 0.80. If the inside diameter of the pipe is 0.10 m what is the flow rate of the mixture in kg/s, kmol/s?
Problem
Known quantities: Mixer velocity, specific gravity, and inner pipe diameter
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 = ρsubstrate / ρreferance
ρreferance = 1000 kg/m3 so ρsubstrate = 0.80 x 1000 kg/m3 = 800 kg/m3
m= ρsubstrateνA = 800 kg/m3 x 3 m/s x (π x (0.10 m)2)/4)= 18.85 kg/s
Solution
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Since we know the total mass flow rate, we can compute the molar flow rate of methanol and ethanol:
nmethanol= 0.3 x 18.85 kg/s / 32.04 kg/kmol = 0.176 kmol/s
Ethanol molar flow rate:
nethanol= 0.7 x 18.85 kg/s / 46.07 kg/kmol = 0.28 kmol/s
The total molar flow rate then is:
ntotal = nmethanol + nethanol = 0.176 + 0.282 = 0.458 kmol/s
Solution
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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/cm3 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?
Mass Fractions- Problem
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:
mi = yi * ntotal * MWi
mi : the component mass flow rate
yi : the component mole fraction
ntotal : the total molar flow rate
MWi : the component molecular weight
Solution
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Mass flow rate of benzene:
mb = 0.3 * 10 mol/s * 78 g/mol = 234 g/s
Mass flow rate of toluene:
mt = 0.5* 10 mol/s * 92 g/mol = 460 g/s
Mass flow rate of xylene:
mx = 0.2* 10 mol/s * 106 g/mol = 212 g/s
The total mass flow rate is the sum of these three:
mtotal = mb + mt + mx = 234 + 460 + 212 = 906 g/s
Solution
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The mass fractions are the individual mass flow rates divided by the total mass flow rate;
Mass fraction of benzene: xB = 234 g/s /906 g/s = 0.258
Mass fraction of toluene: xt = 460 g/s /906 g/s = 0.508
Mass fraction of xylene: xx = 212 g/s /906 g/s = 0.234
Solution
The volumetric flow rate obtained by dividing the mass flow rate by the density is;
V= mtotal/ ρ = 0.906 g/s / 0.87 g/cm3 = 1041 cm3/s
Solution
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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
Concentrations
If 100 g of air consists of 77% by weight of nitrogen (N2, molecular weight = 28), and 23% by weight of oxygen (O2, molecular weight = 32),
Calculate;
a. the mean molecular weight of air
b. mole fraction of oxygen
c. concentration of oxygen in mol/m3 and kg/m3, if the total pressure is 1.5 atm and the temperature is 25°C.
Air Compositions- Problem
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.
Solution
The mass of N2 and O2 that 100 g of air contains:
Mass of N2 = 0.77 × 100 g = 77 g
Mass of O2 = 0.23 × 100 g = 23 g
Number of moles of N2 = 77 g/ 28 g/mol = 2.75 mol
Number of moles of O2 = 23 g/32 g/mol = 0.72 mol
Total number of moles = 2.75 + 0.72 = 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
Solution- Part a
Mole fraction of oxygen (O2) = 0.72/3.47 = 0.21, and this is also the volume fraction.
xi = ni/n = Vi/V = Pi/P = yi
Thus, mole fraction (xi) and volume fraction (yi) for an ideal gas mixture are equivalent.
Solution- Part b
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 0.082056 atmL/mol K
PV = nRT and molar concentration = n/V
T= 25 oC+ 273 = 298 K
First calculate the total molar concentration:
n/V = P/RT= 1.5 atm/(0.082056 atmL/mol K)(298 K) = 0.060134 mol/L
Convert mol/L into mol/m3 so 0.060134 mol/K x 1000L/1 m3 = 61.343 mol/m3
Solution- Part c
We determined the total molar concentration. If we multiply the total molar concentration by the mole fraction of O2, we will find O2 molar concentration:
61.343 mol/m3 x 0.21 = 12.88 mol/m3
The O2 mass concentration is the molar concentration multiplied by the molecular weight:
no2 x MWo2= 12.88 mol/m3 x 32g/mol x 1kg/1000g
= 0.41 kg/m3
Solution- Part c
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/m3. Calculate the concentration of salt in this solution as a
a. weight/weight fraction
b. weight/volume in kg/L
c. mole fraction
Concentration of Salt in Water-Problem
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.
Solution
Weight fraction of salt = 20 kg/(100 kg+20kg) = 0.167
Solution- Part a
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/m3 means that 1 m3 of solution weighs 1323 kg.
Volume of mixture = (100 kg + 20 kg)/1323 kg/m3= 0.091 m3
Weight of salt/volume = 20 kg/0.091 m3= 220 kg/m3
Weight/volume in kg/L = 220kg/m3 x m3/1000L= 0.22 kg/L
Solution- Part b
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 = 5.56 + 0.34 = 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
Solution- Part c
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/m2 or Pascal (Pa), dyn/cm2, 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= Po + ρgh
where P0 is the pressure exerted on the top of the column, g is the acceleration of gravity
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, Patm.
The absolute pressure, Pabs, 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: Pabs = Patm + Pgauge
Pressure measurement
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/m2 (or Pa)
CGS unit: dyn/cm2
AES unit: lbf/ in2 (or psi)
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).
Fluid Pressure
Force on a side hole in an
open tank
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.
Hydrostatic Pressure
Pressure at the bottom of an open
tank
1. Atmospheric pressure, Patm, is the pressure caused by the weight of the earth’s atmosphere. Often atmospheric pressure is called barometric pressure.
2. Absolute pressure, Pabs, 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, Pgauge, is pressure relative to atmospheric pressure.
4. Vacuum pressure, Pvac, is a gauge pressure that is below atmospheric pressure. It is used so that a positive number can be reported.
Types of 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.
Pressure
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 = 1.013x105 N/m2 (or Pa) = 101.3 kPa = 1.013 kPa = 14.696 psi (or lbf/in2) = 29.92 Hg = 33.91 ft H20
Pressure Measurement
Standard temperature and pressure are used widely as a standard reference point for expression of the properties and processes of ideal gases:
Standard temperature TS = 0 °C ⇒ 273 K
Standard pressure Ps = 1 atm = 760 mmHg = 101.3 kPa
Standard molar volume ṽS = 22.4 m3 (STP)/kmol = 22.4 L (STP)/mol = 359 ft3 (STP)/lbmol
(STP) = The standard temperature and pressure
One molecule of an ideal gas at 0°C and 1 atmoccupies 22.4 L.
Standard Temperature and Pressure
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
Pressure-Sensing Devices
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
Pressure Equations
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:
Pabs = Pgauge + Patm where
Pabs is absolute pressure, Pgauge is gauge pressure, and Patm is atmospheric pressure
Pressure Summary
A pressure gauge on a tank reads 25 psi. What is the absolute pressure, considering the atmospheric, Patm =14.7 psi?
Problem
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
Pabs=25 psig + 14.7 psi = 39.7 psia = 40 psia
Solution
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?
Problem
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,
Pvac=Patm − Pabs. Thus, the absolute pressure is
Pabs = Patm - Pvac = 760 − 200= 560 mmHg
Solution
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.
Problem
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/m3.
ρ=ρS x ρH2O = 0.8 x 1000 kg/m3 = 800 kg/m3
P = 100 kPa+(800 kg/m3)(9.81 m/s2)(0.50m)x(1 N/1 kgm/s2)x(1 kPa/1000 N/m2) = 103.924 kPa
Solution