CHAPTER 4 SINGLE-PHASE SYSTEMS ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga Sem 1, 2015/2016

CHAPTER 4SINGLE-PHASE SYSTEMS

ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

Sem 1, 2015/2016

Introduction

Before carrying out a complete material balance, we usually need to determine various physical properties of materials in order to derive additional relationship among the system variables.

As an example we need the density to relate the volumetric flow rate to mass flow rate or vice versa.

3 ways to obtain the values of physical properties (such as density, vapor pressure, solubility, heat capacity, etc)

1. Handbook or database

- Perry’s Chemical Handbook, CRC Handbook of Chemistry & Physics, TRC Database in Chemistry & Engineering, etc

2. Estimation using empirical correlations

3. Experimental work

When you heat a liquid or a solid it normally expands (i.e., its density decreases).

In most process applications, however, it can be assumed with little error that solid and liquid densities are independent of temperature.

Similarly, changes in pressure do not cause significant changes in liquid or solid densities; these substances are therefore termed incompressible.

To determine the density of a mixture of liquids or a solution of a solid in a liquid is from experimental data.

Perry's Chemical Engineers' Handbook provides data for mixtures and solutions of a number of substances on pp. 2-99 through 2-118 and lists additional sources of data on p. 2-99

Liquid and Solid Densities

In the absence of data, the density of a mixture of n liquids (AI, A2,…. An) can be estimated from the component mass fractions [xi] and pure-component densities [σ1] in two ways.

First, we might assume volume additivity-that is, if 2 mL of liquid A and 3 mL of liquid B are mixed, the resulting volume would be exactly 5 mL. Making this assumption and recognizing that component masses are always additive leads to the formula

Second, we might simply average the pure-component densities, weighting each one by the mass fraction of the component:

n

i i

ix

1

1

n

iiix

1

Ideal Gases

Equation of state Relates the molar quantity and volume of a gas to temperature

and pressure. Ideal gas equation of state

Simplest and most widely used Used for gas at low pressure and high temperature Derived from the kinetic theory of gases by assuming gas

molecules:

1. have a negligible volume;

2. Exert no forces on one another;

3. Collide elastically with the wall of container

or The use of this equation does not require to know the gas species:

1 mol of an ideal gas at 0˚C and 1 atm occupies 22.415 liters, whether the gas is argon, nitrogen, mixture of propane and air, or any other single species or mixture of gases

nRTPV RTnVP

Ideal Gas Equation of State

P = absolute pressure

V = volume of the gas

n = number of moles of gas

R = gas constant which the unit depend on unit of P, V, n, T

T = absolute temperature

Ideal gas equation of state can also be written as

Which ; specific molar volume of gas. Unit for gas constant, R

or

nRTPV

RTVP ˆnVV /ˆ

etemperaturmole

volumepressureRfor Unit

etemperaturmole

energyRfor Unit

Ideal Gas Equation of State

The ideal gas equation of state is an approximation.

It works well under some conditions = at temperatures above about 0oC and pressures below about 1 atm-but at other conditions its use may lead to substantial errors.

Here is a useful rule of thumb for when it is reasonable to assume ideal gas behavior.

Rule of thumb for when it is reasonable to assume ideal gas behavior.

Let Xideal be a quantity calculated using ideal gas equation of state (X can be P (absolute), T (absolute), n or V

Error is estimated value is ε

Let’s say quantity to be calculate is ideal specific molar volume,

If error calculated satisfies this criterion, the ideal gas equation of state should yield an error less than 1%

%100

true

trueideal

X

XX

gasesother mole)-lb/ ft (320 L/mol 20

gasesdiatomicmole)lb/ ft (80L/mol 5ˆ%13

3

idealVif

Example: The Ideal Gas Equation of State

One hundred grams of nitrogen is stored in a container at 23.0oC and 3.00 psig.Assuming ideal gas behavior, calculate the container volume in liters. The ideal gas equation of state relates absolute temperature,

absolute pressure, and the quantity of a gas in moles. We therefore first calculate

and (assuming Patm = 14.7 psia) P = 17.7 psia. Then from the ideal gas equation of state

Unfortunately, the table of gas constants at the back of this book does not list the value of R with this particular set of units. In its absence, we use an available value and carry out the necessary additional unit conversions.

T=273+23=296K

14x2=28

Pabs=Patm+Pgauge = 14.7 + 3 =17.7 psig

nRTPV

Standard Temperature and Pressure (STP)

A way to avoid the use of gas constant, R when using ideal gas equation For ideal gas at arbitrary temperature, T and pressure, P

For the same ideal gas at standard reference temperature, Ts and standard reference pressure, Ps (refer to STP).

Divide eq. 1 to eq. 2

Value of standard conditions (Ps, Ts, Vs) are known, above equation can be used to determine V for a given n or vice versa

Standard cubic meters (SCM) : m3 (STP) Standard cubic feet (SCF) : ft3 (STP) Let say 18 SCMH mean 18 m3 (STP)/h

nRTPV

sss RTVP ˆ

sssT

Tn

VP

PV

ˆ

Standard Conditions for Gases

System Ts Ps Vs ns Vs (in STP)

SI 273K 1atm 0.022415 m3 1 mol 22.4 m3/kmol

CGS 273K 1atm 22.415 L 1 mol 22.4 L/mol

American 492˚R 1atm 359.05ft3 1 lb-mole 359.05 ft3 /lb-mole

Engineering

Example: Conversion from Standard

ConditionsButane (C4HIO ) at 360°C and 3.00 atm absolute flows into a reactor at a rate of 1100 kg/h. Calculate the volumetric flow rate of this stream using conversion from standard conditions.

Solution:As always, molar quantities and absolute temperature and pressure must be used. C4H10 = 12(4)+1(10)

=58 kg/kmol

T=360+273= 633K

sssT

Tn

VP

PV

ˆ

Example: Effect of T and P on Volumetric Flow

RatesTen cubic feet of air at 70°F and 1.00 atm is heated to 610°F and compressed to 2.50 atm. What volume does the gas occupy in its final state?

Solution:Let 1denote the initial state of the gas and 2 the final state. Note that n1 = n2 (the number of moles of the gas does not change). Assume ideal gas behavior.P1V1=nRT1

P2V2=nRT2

Example: Standard and True Volumetric Flow

RatesThe flow rate of a methane stream at 285°F and 1.30 atm is measured with an orifice meter. The calibration chart for the meter indicates that the flow rate is 3.95 X 105SCFH. Calculate the molar flow rate and the true volumetric flow rate of the stream.

SOLUTIONRecall that SCFH means ft3(STP)/h.Standard cubic feet (SCF)

Note that to calculate the molar flow rate from a standard volumetric flow rate, you don't need to know the actual gas temperature and pressure.

The true volumetric flow rate of the methane is calculated using this method (T1 = 492°R, P1 = 1.0 atm, V1 = 3.95 X lO5 ft3/h) to actual conditions (T2 = 745°R, P2 = 1.30 atm, V2= ?). We therefore obtain

285oF=285+459.67=745oR

At ideal gas condition, T=0oC and P=1atmSoT=0oC=273.15K=1.8x273.15=492oR

Ideal Gas Mixture

Suppose nA moles of species A, nB moles of species B, nc moles of species C and so on, contained in a volume, V at temperature, T and pressure, P Partial pressure, pA

The pressure that would be exerted by nA moles of species A alone in the same total volume, V at the same temperature, T of the mixture.

Pure component volume, vA

The volume would be occupied by nA moles of A alone at the same total pressure, P and temperature, T of the mixture.

Ideal gas mixture Each of the individual species component and the mixture as whole

behave in an ideal manner

Pyp AA

Vyv AA

Ideal Gas Mixture

Dalton’s Law The summation of partial pressure of the

component of an ideal gas mixture is equal to total pressure

Amagat’s Law

Volume fraction = vA/V; percentage by volume (%v/v)= (vA/V )x 100%

For an ideal gas mixture, the volume fraction is equal to the mole fraction of the substance:

70% v/v C2H6 = 70 mole% C2H6

PPyyyppp CBACBA ....)(.....

VVyyyvvv CBACBA ....)(.....

Example: Material Balance on an

Evaporator CompressorLiquid acetone (C3H60) is fed at a rate of 400 L/min into a heated chamber, where it evaporates into a nitrogen stream. The gas leaving the heater is diluted by another nitrogen stream flowing at a measured rate of 419 m3(STP)/min. The combined gases are then compressed to a total pressure P = 6.3 atm gauge at a temperature of 325°C. The partial pressure of acetone in this stream is Pa = 501 mm Hg. Atmospheric pressure is 763 mm Hg.1. What is the molar composition of the stream leaving the compressor? 2. What is the volumetric flow rate of the nitrogen entering the

evaporator if the temperature and pressure of this stream are 27°C and 475 mm Hg gauge?

Liquid acetone (C3H60) is fed at a rate of 400 L/min into a heated chamber, where it evaporates into a nitrogen stream. The gas leaving the heater is diluted by another nitrogen stream flowing at a measured rate of 419 m3(STP)/min. The combined gases are then compressed to a total pressure P = 6.3 atm gauge at a temperature of 325°C. The partial pressure of acetone in this stream is Pa = 501 mm Hg. Atmospheric pressure is 763 mm Hg.

1. What is the molar composition of the stream leaving the compressor?

SOLUTION

Basis: Given Feed Rates Assume ideal gas behavior. Let n1, n2, ... (mol/min) be the molar flow rates of each stream

Atmospheric pressure is 763 mm Hg.

You should be able to examine the flowchart and see exactly how the solution will proceed.1.Calculate n2 (from the given volumetric flow rate and a tabulated density of liquid acetone), n3 (from the ideal gas equation of state), and Y4 (= Pa/P).

2. Calculate n4 (overall acetone balance), n1 (overall mass balance), and V1 (ideal gas equation of state).



Calculate Molar Flow Rate of AcetoneCalculate n2 (from the given volumetric flow rate and a tabulated density of liquid acetone),From Table B.1 in Appendix B, SG(acetone)=0.791 ..so

So, density of liquid acetone is 0.791 g/cm3 (791 g/L), so that


SG = substance/ref ref@H2O(l) (4˚C) = 1.000 g/cm3

MW acetone=58.08 g/mol


Appendix B (Page 628)

Determine Mole Fractions from Partial PressureCalculate Y4 from (= Pa/P).In the stream leaving the compressor,



Calculate n3 from STP InformationCalculate n3 (from the ideal gas equation of state)using STP


System Ts Ps Vs ns Vs (in STP)

SI 273K 1atm 0.022415 m3 1 mol 22.4 m3/kmol

CGS 273K 1atm 22.415 L 1 mol 22.4 L/mol

American 492˚R 1atm 359.05ft3 1 lb-mole 359.05 ft3 /lb-mole

Engineering

Calculate n4 from overall acetone mole balance


Overall Mole BalanceCalculate n1 (from overall mole balance n1 + n2 + n3 = n4


2. What is the volumetric flow rate of the nitrogen entering the evaporator if the temperature and pressure of this stream are 27°C and 475 mm Hg gauge?

Calculate V1 (from ideal gas equation of state)

Equation of State for Nonideal Gases

Critical temperature (Tc)- the highest temperature at which a species can exist in two phases (liquid and vapor), and the corresponding pressure is critical pressure (Pc)

Other definition: highest temperature at which isothermal compression of the species vapor results in the formation of a separate liquid phase.

Critical state- a substance at their critical temperature and critical pressure.

Species below Pc: Species above Tc- gas Species below Tc- vapor

Species above Pc and above Tc- supercritical fluids

Virial Equation of State Expresses the quantity PV/RT as a power series in the inverse of specific

volume.

Virial equation of state

B,C,D- second, third, fourth virial coefficient respectively (function of TEMPERATURE)

Truncated virial equation at the second term of yields

Tr=T/Tc ω – acentric factor from Table 5.3-1

Tc,Pc from Table B.1

....ˆˆˆ

1ˆ

32

V

D

V

C

V

B

RT

VP

V

B

RT

VPˆ

1ˆ

6.102.4110

422.0083.0;

172.0139.0);(

rrc

c

TB

TBBB

P

RTB

Example: The Truncated Virial Equation

Two gram-moles of nitrogen is placed in a three-liter tank at -150.8°C.

Estimate the tank pressure using the ideal gas equation of state and then using the virial equation of state truncated after the second term.

Taking the second estimate to be correct, calculate the percentage error that results from the use of the ideal gas equation at the system conditions.

SOLUTION:

IDEAL GAS EQUATION STATE

Two gram-moles of nitrogen is placed in a three-liter tank at -150.8°C.

Estimate the tank pressure using the ideal gas equation of state and then using the virial equation of state truncated after the second term.


SOLUTION:

VIRIAL EQUATION OF STATEThe virial equation solution procedure is as follows:

1. Table B.1 ====> (Tc)N2 = 126.2 K, (PC)N2 = 33.5 atm

2. Table 5.3-1 ====> ωN2 = 0.040

3. Tr = T/Tc = 122.4K/126.2 K = 0.970

Appendix B (Table B.1)

6.102.4110

422.0083.0;

172.0139.0);(

rrc

c

TB

TBBB

P

RTB

V

B

RT

VPˆ

1ˆ


Pideal= 6.73 atmP=6.19 atm

Cubic Equations of State Refer as cubic equation because when the

equation is expanded, it becomes third order equation for the specific volume

To evaluate volume for a given temperature and pressure using cubic equation of state, we need to do trial and error procedure.

Two famous cubic equation of statea) Van der Waals equation of stateb) Soave-Redlich-Kwong (SRK) equation of state

Van der Waals Equation of State

In the van der Waals derivation, the term a/V2 accounts for attractive forces between molecules.

b is a correction accounting for the volume occupied by the molecules themselves

2ˆˆ V

a

bV

RTP

c

c

c

c

P

RTb

P

TRa

864

27 22

Soave-Redlich-Kwong (SRK) equation of state

)ˆ(ˆˆ bVV

a

bV

RTP

2

2

2

1561.055171.148508.0

)]1(1[/

08664.0)(

42747.0

m

TmTTT

P

RTb

P

RTa

rcr

c

c

c

c

where the parameters a, b, and α are empirical functions of the critical temperature and pressure (Tc and Pc from Table B.1),

the Pitzer acentric factor (ω from Table 53-1) and the systemtemperature.

The following correlations are used to estimate these 3 parameters

Example: The SRK Equation of State

A gas cylinder with a volume of 2.50 m3 contains 1.00 kmol of carbon dioxide at T = 300 K. Use the SRK equation of state to estimate the gas pressure in atm.

SOLUTION:

The specific molar volume is calculated as

2

2

2

1561.055171.148508.0

)]1(1[/

08664.0)(

42747.0

m

TmTTT

P

RTb

P

RTa

rcr

c

c

c

c

From Table B.1, Tc = 304.2 K and Pc = 72.9 atm, and from Table 5.3-1, w = 0.225. The parameters in the SRK equation of state are evaluated using Equations 5.3-7 through 5.3-12:

)ˆ(ˆˆ bVV

a

bV

RTP

Compressibility Factor Equation of State

or

If z=1, equation become ideal gas equation of state

Value of z is given in Perry’s Chemical Engineering Handbook pg. 2.140- 2.150.

Alternatively; can use generalized compressibility chart Figure 5.4-1 – generalized compressibility chart Fig. 5.4-2 to Fig. 5.4-4 – expansion on various region

in Fig. 5.4-1

zRTVP ˆRT

VPz

ˆ

Step to Read Compressibility Factor

1. Find Tc and Pc2. If gas is either Hydrogen or Helium, determine adjusted

critical temperature and pressure from Newton’s correction equation

3. Calculate reduce pressure and reduce temperature of the two known variables

4. Read off the compressibility factor from the chart

c

cidealr RT

VPV

Pc

P

Tc

TTr

ˆ;Pr;

atmPPKTT cacc

ac 88

Nonideal Gas Mixtures

Kay’s Rule: estimation of pseudocritical properties of mixture as simple average of pure a component critical constants

Pseudocritical temperature (Tc’)

Tc’= yATcA + yBTcB +……

Pseudocritical pressure (Pc’)

Pc’= yAPcA + yBPcB +……

Pseudocritical reduced temperature (Tr’)

Tr’= T/Tc’

Pseudocritical reduce pressure (Pr’)

Pr’= P/Pc’

Compressibility factor for gas mixture, Zm P

RTzV mˆ

Example: Kay’s Rule

A mixture of 75% H2 and 25% N2 (molar basis) is contained in a tank at 800 atm and -70°C.Estimate the specific volume of the mixture in L/moI using Kay's rule.

atmPPKTT cacc

ac 88

Pseudocritical temperature (Tc’)Tc’= yATcA + yBTcB +……

Pseudocritical pressure (Pc’)Pc’= yAPcA + yBPcB +……

A mixture of 75% H2 and 25% N2 (molar basis) is contained in a tank at 800 atm and -70°C.

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CHAPTER 4 SINGLE-PHASE SYSTEMS ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga Sem 1, 2015/2016