Balances on Nonreactive Processes

Balances on Nonreactive Processes

Energy Balances on Nonreactive Processes
Objectives: Calculate DH (and DU ) from heat capacity equations, graphs, charts, tables associated with the change in T, P, phase and mixing process (given the initial and final states of materials). Become familiar with steam tables and psychometric charts Understand the concept of reference state for enthalpy values in the data source, identify the reference states. Convert the expression for heat capacity from one unit to another.

Enthalpy and Enthalpy Change
The evolved H2 pushes back the atmosphere; work is done at constant pressure. Enthalpy is the sum of the internal energy and the pressure-volume product of a system: H = U + PV Mg + 2 HCl  MgCl2 + H2 For a process carried out at constant pressure, Q = DU + PDV so Q = DH Most reactions occur at constant pressure, so for most reactions, the heat evolved equals the enthalpy change.

Properties of Enthalpy
Enthalpy is an extensive property. It depends on how much of the substance is present. Two logs on a fire give off twice as much heat as does one log. Since U, P, and V are all state functions, enthalpy H must be a state function also. Enthalpy changes have unique values. DH = Q Enthalpy change depends only on the initial and final states.

Reference State Enthalpy (H) and internal energy (U) are not absolute values ….. But values relative to a value at the reference state (T&P) Energy balance of a control volume between 2 states requires only the energy difference (DU or DH) …. So Uref or Href can be at any reference state

Reading Assignment

Example 1 Ĥ=-2751kJ/kg and specific internal energy =-2489 kJ/kg
What reference state was used to generate the specific internal energies and enthalpies of steam? Suppose water vapor at 300oC and 5 bar is chosen as a reference state at which Ĥ is defined to be zero. Relative to this state, what is the specific enthalpy of liquid at 75oC and 1 bar. What is the specific internal energy of liquid at 75oC and 1 bar? (Use Table B.7) Ĥ=-2751kJ/kg and specific internal energy =-2489 kJ/kg

Hypothetical Process Path
We will learn how to calculate DH and DU changes associated with certain process specifically: Changes in P at constant T and state of aggregation Changes in T at constant P and state of aggregation Phase changes at constant T and P – melting, solidifying, vaporizing, condensing, and sublimation Mixing of two liquids or dissolving of a gas or a solid in a liquid at constant T and P. Chemical reaction at constant T and P Calculate the overall DH using the summation of each DHi steps in system.

Suppose that we wish to calculate DĤ for a process in which solid ice at -5oC and 1 atm is converted to water vapor at 300oC and 5 atm. Since Ĥ is a state property, we could simply subtract DĤ at the initial state from DĤ at the final state DĤ = Ĥ (vapor, 300oC, 5 atm) - Ĥ (solid, -5oC, 1 atm) Suppose that we do not have such a table, our task is then to construct a hypothetical process path from the solid ice at -5oC and 1 atm to the water vapor at 300oC and 5 atm

Ice -5oC, 1 atm True Path Vapor 300oC, 5 atm Hypothetical Path Ice 0oC, 1 atm Vapor 300oC, 1 atm Liquid 0oC, 1 atm Liquid 100oC, 1 atm Vapor 100oC, 1 atm

Reading Assignment

Example 2 Calculate the specific enthalpy requirement to transform water liquid at 0oC and 0.01 bar to water vapor at 300oC and 5 bar DĤ = Ĥ (vapor, 300oC, 5 bar) - Ĥ (liquid, 0oC, 0.01bar) Table B.7, Ĥ (300oC, 5 bar) = 3065 kJ/kg Table B.5, Ĥ (0oC, 0.01 bar) ~ 0 kJ/kg Hence, DĤ ~ 3065 kJ/kg Suppose that we do not have such a table, consruct a hypothetical process from water liquid at 0oC and 0.01 bar to water vapor at 300oC and 5 atm

Example 2 - Hypothetical Process Path of enthalpy requirement to generate steam at 300oC, 5 bar

Working session 1 Cyclohexane vapor at 180oC and 5 atm is cooled and condensed to liquid cyclohexane at 25oC and 5 atm. The change enthalpy for the condensation of cyclohexane at 80.7oC and 1 atm is known. Construct a hypothetical process path of the above process O2 at 170oC and 1 atm, and CH4 at 25oC and 1 atam are mixed and react completely to form CO2 and H2O at 300oC and 1 atm. The enthalpy change for the reaction occuring at 25oC and 1 atm is known.

Working session 1 Solution
a. Lower P isothermally to 1 atm, cool at 1 atm to 80.7°C, condense at 80.7°C and 1 atm, cool liquid at 1 atm to 25°C, raise pressure to 5 atm. b. Keeping pressure constant at 1 atm. cool O2 to 25°C, mix O2 and CH4 at 25°C, carry out reaction at 25°C, raise product gas to 300°C.

Procedure for Energy Balance Calculations
Perform all required material balance calculations Write the appropriate form of energy balance (closed or open system) and delete any of the terms that are either zero or negligible for the given process systems Choose a reference state – phase, temperature, and pressure – for each species involved in the process For water look at the steam tables Choose the reference state used to generate the table Choose the inlet or outlet states as the reference state of the species (so that at least one or may be set to zero)

For a closed constant-volume system, construct a table with columns for initial and final amounts of each species (mi or ni) and specific internal energies relative to the chosen reference states For an open system, construct a table with columns References : Ac ( l,20oC, 5 atm), N2 (g, 25oC, 1 atm) Substance nin (mol/s) Nout (mol/s) Ac (v) 66.9 3.35 Ac (l) - 63.55 N2 33.1

Calculate all required values of internal energy and enthalpies and insert in the appropriate places in the table. Calculate

Calculate any work, kinetic energy, or potential energy terms that you have not dropped from the energy balance. Solve the energy balance for whichever variable is unknown from the equation given below (often Q )

Changes in Pressure at Constant Temperature
Internal energy is nearly independent of pressure on solid and liquid at fixed temperature, as is specific volume. If the pressure of liquid and solid change at constant temperature, you may write Both and are independent of pressure for ideal gas. Consequently, you may generally assume and for a gas undergoing isothermal pressure changes unless gas at temperature well below 0oC or well above 1 atm are involved

Sensible Heat and Heat Capacities
Sensible heat signifies heat that must be transferred to raise or lower the temperature of a substance or mixture of substances. The quantity of heat required to produce a temperature change in a system is given by the first law of thermodynamics, neglecting the changes in kinetic and potential energies and as well as work.

Heat Capacity @ Constant Volume
DU for constant V, and changes in T DT T1 T1 + DT T (oC) slope of the curve, Cv(T). Only “linear” in a very narrow range

Heat Capacity @ Constant Volume
DU for variable T and V Solid, liquid & ideal gas - Isothermal process Ideal gas: exact Solid or liquid - good approximation Real gas: valid only if V is constant

Example 3 : Evaluation of an Internal Energy Change from Tabulated Heat Capacity
Calculate the heat required to raise 200 kg of nitrous oxide from 20oC to 150oC in a constant-volume vessel. The constant-volume heat capacity of N2O in this temperature range is given by the equation where T is in oC.

Example 3 : Evaluation of an Internal Energy Change from Tabulated Heat Capacity
Calculate the heat required to raise 200 kg of nitrous oxide from 20oC to 150oC in a constant-volume vessel. The constant-volume heat capacity of N2O in this temperature range is given by the equation where T is in oC. The energy balance for this closed system

Heat Capacity @ Constant Pressure
DH for constant P and variable T slope of the curve, Cp (T). Only “linear” in a very narrow range DT T1 T1 + DT T (oC)

DH for variable T and P For a process A(T1, P1)  A(T2, P2) we may construct two-step process path For the first step change (DĤ1) in pressure at constant temperature

For the second step change (DĤ2) in temperature at constant pressure Hence, DĤ = DĤ1 + DĤ2 Solid, liquid & ideal gas Ideal gas: Exact Nonideal gas : Exact only if P constant Solid or Liquid

Specific Heat Capacity Formulas
Cp = a + bT + cT2 + dT3 values for a, b, c and d are obtain from Table B.2 Perry’s Handbook (section 3) When reading the coefficients from Table B.2 do not mistake their orders of magnitude 72.4 is read from column labeled b means that b = 72.4 x 10-5 Simple relationship exist between Cp and Cv for two cases: For liquids and solids: Cp ~ Cv Ideal gases: Cp = Cv + R (R = gas constant)

Specific Heat Capacity Formula
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is (1) The above formula is sometimes simplified as (2)

Example 4: Evaluation of Cp & Cv
The constant-pressure heat capacity of carbon monoxide (CO) is given by the expression Write an expression for the heat capacity at constant volume for CO, assuming ideal gas behaviour Derive an expression for Cp(BTU/Ib-mole.oF) as a function of T(oF). (Remember that the temperature unit in the denominator refers to a temperature interval)

Example 4: Evaluation of Cp & Cv
Write an expression for the heat capacity at constant volume for CO, assuming ideal gas behaviour Derive an expression for Cp(BTU/Ib-mole.oF) as a function of T(oF). (Remember that the temperature unit in the denominator refers to a temperature interval) Gas constant R=1.987 cal/mol-oC

Working session 2 Estimate the specific enthalpy of steam at 350oC and 100 bar relative to steam at 100oC and 1 atm using The steam tables Table B.2 and assuming ideal gas behaviour Table B.8 What is the physical significance of the difference between the values of Ĥ calculated by the two methods?

Example 5 : Evaluation of DH using Heat Capacity
Fifteen kmol/min of air is cooled from 430oC to 100oC. Calculate the required heat removal rate using Heat Capacity from Table B.2 Specific Enthalpies from Table B.8 Air (g, 430oC)  Air (g, 100oC) With DEk, DEp, and Ws deleted, the energy balance is Assuming ideal gas behavior, so that the pressure changes do not effect specific enthalpy

Solution : The Hard Way (using Heat Capacity from Table B.2)
Integrate the heat capacity formula in Table B.2

Solution : The Easy Way (using Specific Enthalpies from Table B.8)
Use Tabulated enthalpies form Table B.8 ( …. What is reference temp.?) Table B.8 and B.9 apply strictly for heating and cooling at a constant pressure of 1 atm and may also be used for nonisobaric heating and cooling of ideal gas

Example 6 : Evaluation of DU using Heat Capacity relationship
Fifteen kmol of air is cooled form 430oC to 100oC in a constant-volume vessel. Calculate the required heat removal rate using Cv = Cp - R & Heat Capacity from Table B.2 Air (g, 430oC)  Air (g, 100oC) With DEk, DEp, and Ws deleted, the energy balance is Assuming ideal gas behavior

Solution : The Hard Way (using CV & Cp relationship and Heat Capacity from Table B.2)
substitute Cv by Cp - R & apply the heat capacity formula in Table B.2 R (kJ/mol-C)

Solution : The Easy Way (using CV & Cp relationship and Heat Capacities from Table B.8)
Use H = U + PV correlation & Table B.8

Working session 3 (similar Problem 8.3)
The constant-volume heat capacity of hydrogen sulfide (H2S) at low pressures is given by the expression where T is in oC. A quantity of H2S is kept in a piston-fitted cylinder with initial temperature, pressure and volume equal to 25oC, 2 atm and 3 liters, respectively. Calculate the heat (kJ) required to raise the gas temperature from 25oC to 1000oC if the heating takes place at constant volume. Calculate the heat (kJ) required to raise the gas temperature from 25oC to 1000oC if the heating takes place at constant pressure. What would piston do during the heating process in part (b). What is the physical significance of the difference between the Values of Q calculated in part (a) and part (b)? Home work

Working session 4 - Problem 8.9
Chlorine gas is to be heated from 100oC and 1 atm to 200oC Calculate the heat input (kW) required to heat a stream of the gas flowing at 5 kmol/s at constant pressure Calculate the heat input (kJ) required to raise the temperature of 5 kmol of chlorine in a closed rigid vessel from 100oC and 1 atm to 200oC What is the physical significance of the numerical difference between tha values calculated in part (a) and part (b)? Home work

A stream of water vapor flowing at a rate of 250 mol/h is brought from 600oC and 10 bar to 100oC and 1 atm Estimate the cooling rate (kW) using (i) steam tables, (ii) Table B.2 and (iii) Table B.8 Which of the answers is part (a) is most accurate? Why? What is the physical significance of the numerical difference between the values calculated with methods (i) and (ii)? Home work

Estimation of Heat Capacities
You may use the following approximation to find heat capacity if you don’t have enough data: For a mixture of gases and liquids, calculate the total enthalpy change as the sum of the enthalpy changes for the pure mixture components. For a highly dilute solutions of solids or gases in liquids, neglect the enthalpy change of the solute. The more dilute the solution, the better this approximation. The calculation of enthalpy changes for the heating or cooling of a mixture of known composition can be simplified using this formula:

Estimation of Heat Capacities
Kopp’s rule is an empirical method for estimating heat capacity of a solid or liquid at or near 20oC. Cp of a molecular compound is the sum of contributions of each element in the compound. The values for the coefficients are given in Table B.10 Heat capacity of solid calcium hydroxide, Ca(OH)2 is estimated as

Example 7- Heat Capacity of a Mixture
Calculate the heat required to bring 150 mol/h of a stream containing 60% C2H6 and 40% C3H8 by mole from 0oC to 400oC. Determine the heat capacity for the mixture as part of the problem solution. The polynomial heat capacity formula for ethane and propane given in Table B.2 are substituted:

Alternatively …. calculate DH of each component, multiply by its mole fraction, respectively & then sum them up, i.e If the potential and kinetic energy changes and shaft work are neglected, the energy balance becomes As usual, we assumed that the gases are sufficiently close to ideal for the formula for Cp at 1 atm is valid.

Working session 6 - Problem 8.19 & 8.21 (Modified)
A gas mixture containing one-third methane by volume and the balance oxygen at 350oC and 3 bar. Calculate the specific enthalpy of this stream in kJ/kg relative to pure component at 25oC and 1 atm. State clearly all assumptions. Propane is to be burned with 15% excess air. Before entering the furnace the air and propane are uniformly mixed and preheated from 0oC and 300oC. At what rate (BTU/h) must heat be transferred to the propane-air mixture if the feed rate of propane is 3820 SCMH [m3(STP)/h]? mixer 3820 SCMH [m3(STP)/h Propane Air 15% in excess 300oC 0oC Q (BTU/h) preheater

Working session 6 - 8.21 (Modified)
Molar flow rate of propane , Stoichiometric combustion of propane, C3H O2  3CO2 + 4H2O Molar flow rate of 15% excess air, Simplified energy bal. around preheater What are reference states for propane & air? How do you calculate the value of Ĥ?

Energy Balances on a Single-Phase Systems
If the process only involves heating or cooling a single species from T1 to T2, the procedure is straight forward. Evaluate the For a closed system at constant volume For a closed system at constant pressure For a open system, calculate Correct pressure Changes if necessary

Air (g, 430oC) Cooler Air (g, 100oC)
Example 8 - Recalling example 5 to demonstrate the use of reference states Fifteen kmol/min of air is cooled from 430oC to 100oC. Calculate the required heat removal rate (KW) Air (g, 430oC) Cooler Air (g, 100oC) With DEk, DEp, and Ws neglected and assuming ideal gas behavior, so that the pressure changes do not effect specific enthalpy the energy balance is Q

Recalling example 5 to demonstrate the use of reference states
Reference : Air (g, 25oC, 1 atm) Substance nin (kmol/min) Ĥin nout (kmol/min) Ĥout Air 15 Ĥ1 Ĥ2

Recalling example 5 to demonstrate the use of reference states
Hypothetical Process Path Air (g, 430oC) Air (g, 100oC) Reference state Air, 25oC State two possible reference states that can be used to solve the problem and compare your previously calculated answer

Working session 7 - Problem 8.31 (Modified)
In the manufacture of nitric acid (NO3), ammonia (NH3) and preheated air are mixed to form a feed gas containing 10 mole% NH3 at 600oC. NH3 enters the mixer at 25oC and 520 kg/h and the preheated air enters the mixer at 700oC. Calculate the rate of heat loss (kW) from the mixer to its surrounding. NH3 25oC 520 kg/h n1 mol/s Air 700oC n2 mol/s NH3 –Air mixture 600oC n3 mol/s 0.1 mol NH3/mol 0.9 mol Air/mol

Solution to Working session 7 - Problem 8.31 (Modified)
Basis : 520 kg/hr of NH3 (30.6 kmol/hr or 8.5 mol/s) Summary of mass balance calculations: Assumptions : steady state, DKE~DPE = 0) ……. Q = DH = Σ(noutĤout)- Σ(ninĤin) NH3 25oC 8.5 mol/s NH3 –Air mixture 600oC 85 mol/s 0.1 mol NH3/mol 0.9 mol Air/mol Air 700oC 76.5 mol/s

Reference : (Air & Substance nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol) NH3 8.5 Ĥ1 Ĥ3 Air 76.5 Ĥ2 Ĥ4

Reference : 700oC & 25oC) Substance nin (mol/s) Ĥin (kJ/mol) nout (mol/s) Ĥout (kJ/mol) NH3 8.5 Ĥ1 Ĥ3 Air 76.5 Ĥ2 Ĥ4

Example 9 - Problem 8.25 Saturated steam at 300oC is used to heat a counter-currently flowing stream of methanol from 65oC to 260oC in an adiabatic heat exchanger. The flow of the methanol is 5500 L(STP)/min. and the steam condenses and leaves the heat exchanger as water liquid at 90oC. Calculate the rate of heat transfer from the water to the methanol (kW). What is the reference state for methanol? Calculate the required flow rate of the entering steam in m3/min. What is the reference state for water? Saturated steam, 300oC Liquid water, 90oC 5500 L(STP)/min methanol, 65oC methanol, 260oC

Processes Involving Phase Changes
Processes of phase change Liquid to vapor – vaporization Vapor to liquid – condensation Solid to liquid - fusion (melting) solid to vapor -sublimation The phase boundary between liquid and gas does not continue indefinitely. Instead, it terminates at a point on the phase diagram called the critical point. This reflects the fact that, at extremely high temperatures and pressures, the liquid and gaseous phases become indistinguishable, in what is known as a supercritical fluid. In water, the critical point occurs at around Tc = K ( °C), pc = MPa (217.75 atm) and ρc = 356 kg/m³

Latent heat Heat required to transform a substance from one phase to another phase at constant temperature or pressure Latent heat of 1. vaporization (liquid  vapor), DĤv 2. fusion or melting (solid liquid), DĤm 3. sublimation (solid  vapor), DĤs Table B.1 has some values at normal conditions (1 atm) & Perry’s Chemical Handbook

Enthalpy changes of a pure substance

If a phase change takes place in an open system, the heat of vaporization (DĤvap) of a substance may be determined from Table B.1 (for normal boilng point) or estimated using Trouton’s rule, Chen’s equation, Clausius-Clapeyron equation or Watson’s correlation If a phase change takes place in a closed system, DÛ will be used to substitute into the energy balance equation assuming ideal gas behaviour For phase change such as fusion involving only solids or liquids, changes in is usually negligible, so DÛ ≈ DĤ

Example 10 - Problem 8.36 Calculate the specific enthalpy (kJ/mol) of n-hexane vapor at 200oC and 2 atm relative to n-hexane liquid at 20oC and 1 atm, assuming ideal gas behaviour. Show clearly the process path you used the ideal gas assumption. Liquid 68.74oC, 1 atm Vapor 68.74 oC, 1 atm 200oC, 2 atm 200oC, 1 atm True Path liquid 20oC, 1 atm Hypothetical Path

Example 10 - Problem 8.36

Benzene vapor at 580oC is cooled and converted to a liquid at 25oC in a continuous condenser. The condensate is drained into 1.75-m3 drums, each of which takes 2 minutes to fill. Calculate the rate (kW) at which heat is transferred from the benzene in the condenser. Process path :- (vapor, 580oC) (vapor, Tbp, 1 atm=80.1oC) (liquid, Tbp, 1 atm=80.1oC) (liquid, 25oC)

Solution to Working session 8 - Problem 8.38

Estimation & correlation of latent heats
There are some estimation methods on pp Trouton’s rule – estimate a standard latent heat of vaporization at the normal boiling point Chen’s equation Accurate to within 30% where Tb is the normal boiling point of the liquid provides roughly 2% accuracy where Tb and Tc are the normal boiling point and critical temperature in kelvin and Pc is the critical pressure in atmospheres.

Clausius-Clapeyron equation
3. Watson’s correlation

Example 11 - Problem 8.44 Estimate the heat of vaporization (kJ/mol) of benzene at a pressure of 100 mm Hg, using The normal boiling point given in Table B.1, the boiling point at 100 mm Hg as determined by the Antoine equation and Watson’s correlation

Example 11 - Problem 8.44 Estimate the heat of vaporization (kJ/mol) of benzene at a pressure of 100 mm Hg, using The Clausius-Clayperon equation and the boiling points at 50 mm Hg and 150 mm Hg as determined from the Antoine equation

Example 11 - Problem 8.44 True Path
Estimate the heat of vaporization (kJ/mol) of benzene at a pressure of 100 mm Hg, using The normal boiling point given in Table B.1, the boiling point at 100 mm Hg and the heat capacity data given in Table B.2 liquid 26.1oC, 100 mm Hg True Path Vapor 26.1oC, 100 mm Hg liquid 26.1oC, 760 mm Hg Vapor 26.1oC, 760 mm Hg liquid 80.1oC, 760 mm Hg Vapor 80.1 oC, 760 mm Hg

Example 11 - Problem 8.44

Energy Balances on Process involving Phase Changes
An energy balance on a process in which a component exists in two phases Choose a reference state for that component by specifying both a phase and a temperature Calculate the specific enthalpy of the component in all process streams relative to a reference state If the substance is a liquid at its reference state a vapor in a process stream, Ĥ may be calculated using process path calculations

Example 12 - Problem 8.50 (modified)
A mixture of n-hexane vapor and air leaves a solvent recovery unit and flows through a 70 cm-diameter duct at a velocity of 3 m/s. At sampling point in the duct the temperature is 40oC, the pressure is 850 mm Hg, and the dew point of the gas sample is 25oC. The gas is fed to a condenser in which it is cooled at constant pressure, condensing 60% of the hexane in the feed. Calculate the required condenser outlet temperature and cooling rate (KW). 70 cm-ID duct V=3 m/s T= 40oC P= 850 mm Hg Tdp= 25oC nf = ? mol/s Yh,f=? mol hexane/mol Yair,f=? mol air/mol P= 850 mm Hg Tc= ? oC nc = ? mol/s Yh,c=? mol hexane/mol Yair, c=? Mol air/mol condenser 60% of Hexane feed nhex = ? mol/s

Problem 8.50 (modified) Assuming an ideal gas air-hexane mixture entering the condenser, nf = PV/RT = mol/s Tdp= 25oC , thus the partial pressure of condensable species (i.e. hexane), using Antoine equation, p*h = mm Hg hence, p*h = Yh,f P …… Yh,f= mol hexane/mol and Yair,f= mol air/mol then, nhex = (0.6)(0.178)(50.3) = mol/s hexane condensed ! and nc = – = mol/s Yair, c= (0.822)(50.3)/ = mol air/mol Yh,c= 0.08 mol hexane/mol p*h = Yh,f P …… Yh,f= (0.08)(850) = 68 mm Hg from Antoine equation, Tc(p*h = 68 mm Hg ) = 7.83 oC

Problem 8.50 (modified) Energy balance for open system, neglecting changes in kinetic and potential energy and enthalpy is independent of pressure change, References : Hexane ( liquid,7.8oC), Air (gas, 25oC) Substance nin (mol/s) Ĥin(kJ/mol) nout (mol/s) Ĥout(kJ/mol) Hexane (v) 8.95 3.58 Hexane (l) - 5.37 Air 41.35

Problem 8.50 (modified) Inlet stream Ĥhexane (v) =
= – = kJ/mol Ĥair = = kJ/mol (Table B.2) or alternatively Table B.8 Outlet stream Ĥair = = kJ/mol Ĥhexane (v) = = – = kJ/mol Ĥhexane (l) = 0 kJ/mol

Problem 8.50 (modified) References : Hexane ( liquid,7.8oC), Air (gas, 25oC) Substance nin (mol/s) Ĥin(kJ/mol) nout (mol/s) Ĥout(kJ/mol) Hexane (v) 8.95 37.463 3.58 32.728 Hexane (l) - 5.37 Air 41.35 0.436 -0.499 rate of heat removed from the hexane-air mixture which is transferred to cooling medium

Problem 8.50 (modified) – using different reference states
Recalculate the required cooling rate using different reference state, e.g. 40oC and gas phase for both substances References : Hexane (gas, 40oC), Air (gas, 40oC) Substance nin (mol/s) Ĥin(kJ/mol) nout (mol/s) Ĥout(kJ/mol) Hexane (v) 8.95 ? 3.58 Hexane (l) - 5.37 Air 41.35

Problem 8.50 (modified) Inlet stream
Ĥhexane (v) = 0 kJ/mol Ĥair = 0 kJ/mol Outlet stream Ĥair = = kJ/mol Ĥhexane (l) = = – = kJ/mol Ĥhexane (v) = = kJ/mol

Problem 8.50 (modified) – using different reference states
Recalculate the required cooling rate using different reference state, e.g. 40oC and gas phase for both substances References : Hexane (gas, 40oC), Air (gas, 40oC) Substance nin (mol/s) Ĥin(kJ/mol) nout (mol/s) Ĥout(kJ/mol) Hexane (v) 8.95 3.58 -4.735 Hexane (l) - 5.37 Air 41.35 -0.935

Working session 9 - Problem 8.52 (modified)
A liquid containing 50 mole% benzene (Bz) and the balance toluene (Tl) at 25oC is fed to a continuous single-stage evaporator at a rate of 1320 mol/s. The liquid and vapor streams which are in equilibrium with each other leaving the evaporator at 95oC. The liquid product stream contains 42.5 mole% benzene. Using Raoult’s law, calculate the vapor product compositions, the system pressure (atm). Estimate the heating requirement for this process in kW. Evaporator F = 1320 mol/s T= 25oC 50 mole% Bz 50 mole% Tl V mol/s T= 95oC YBz mol Bz/mol (1-YBz ) mol Bz/mol L mol/s T= 95oC 0.425 mol Bz/mol 0.575 mol Tl/mol

Raoult’s law, Yi P = Xi p*i (T) Using Antoine eq. to estimate the partial pressure of ; p*Bz(95oC)= mm Hg and p*Tl(95oC)= mm Hg Benzene ; YBz P = (1176.6) …….. (1) Toluene; YTl P = (476.9) …………(2) YBz + YTl = 1 …………………………(3) Solve eqs. (1)/(2) & (3) , YBz = YTl = 0.354 Hence, P = 774 mm Hg (1.018 atm), V = 448 mol/s , L=872 mol/s

References : Benzene ( liquid,25oC), Toluene (liquid, 25oC) Substance nin (mol/s) Ĥin(kJ/mol) nout (mol/s) Ĥout(kJ/mol) Benzene (l) 660 370.6 ? Toluene (l) 501.4 Benzene (v) 289.4 Toluene (v) 158.6

Outlet liquid stream ĤBenzene (l) = = kJ/mol ĤToluene (l) = = kJ/mol Outlet vapor stream ĤBenzene (v) = = = kJ/mol ĤToluene (v) = = – = kJ/mol

References : Benzene ( liquid,25oC), Toluene (liquid, 25oC) Substance nin (mol/s) Ĥin(kJ/mol) nout (mol/s) Ĥout(kJ/mol) Benzene (l) 660 370.6 9.838 Toluene (l) 501.4 11.777 Benzene (v) 289.4 39.913 Toluene (v) 158.6 46.054

Working session 10 Solid soaked with liquid hexane are dried by being contacted with nitrogen at an elevated temperature. The gas stream leaving the dryer is at 80oC, 1 atm absolute and 50% relative saturation. One of the several possibilities for recovering the hexane is to send the stream to a cooling condenser. The gas stream leaving the condenser would contain 15 mole % hexane, and hexane condensate would be recovered at a rate of 1.5 kmol/min. the condenser would be operated at a pressure of 1 atm absolute. Calculate the required condenser cooling rate (kW). Hexane-Nitrogen mixture 80oC, 1 atm RS=50% Yhex,f mol hexane/mol Ynitrogen,f mol nitrogen/mol nf kmol/min Condenser Hexane-Nitrogen mixture ? oC, 1 atm 0.05 mol Hexane/mol 0.95 mol Nitrogen/mol nv kmol/min Hexane condensate 1.5 kmol/min

Working session 10 Relative saturation, RS= pi/pi*(T) Inlet stream
The only condensable species is hexane, thus RS = 0.5 = phex/phex*(80oC) Using Antoine equation, p*hex = mm Hg, and hence, phex = mm Hg Yhex,f= mol hexane/mol and Ynitrogen,f= mol nitrogen/mol Outlet stream Assuming the condenser is in thermal equilibrium, p*hex = Yhex P …… Yh,f= (0.15)(760) = 114 mm Hg from Antoine equation, Tc(p*h = 114 mm Hg ) = oC

Working session 10 then, nf = 2.31 kmol/min and nv = kmol/min References : Hexane (liquid, 18.64oC), Nitrogen (gas, 25oC) Substance nin (kmol/min) Ĥin (kJ/mol) nout (kmol/min) Ĥout Nitrogen (v) 1.624 1.602 0.6885 -0.185 Hexane (v) 0.686 41.562 0.1215 Hexane (l) 1.5

Working session 10 Inlet stream Ĥhexane(v) =
= = kJ/mol Ĥnitrogen = = kJ/mol (Table B.2) or alternatively Table B.8 Outlet stream Ĥnitrogen = = kJ/mol Ĥhexane (v) = = – = kJ/mol Ĥhexane (l) = 0 kJ/mol

Additional working session - Rework Example 8
Additional working session - Rework Example using different reference states Acetone (denoted as Ac) is partially condensed out of a gas stream containing 66.9 mole% acetone vapor and the balance nitrogen. Process specifications and material balance calculations lead to the following flowchart. The process is steady state. Calculate the required cooling rate (kW). Use the inlet conditions as reference states. Compare your answer with that calculated from example 8.1-1

Additional working session - Rework Example using different reference states References : Ac (l, 20oC, 5 atm), Nitrogen (g, 25oC, 1 atm) Substance nin (mol/s) Ĥin (kJ/mol) nout Ĥout Ac (v) 66.9 Ĥ1 3.35 Ĥ3 Ac (l) - 63.55 Ĥ4 N2 33.1 Ĥ2 Ĥ5

Additional working session - Rework Example using different reference states

Additional working session - Rework Example using different reference states References : Ac l, 20oC, 5 atm), Nitrogen (25oC, 1 atm) Substance nin (mol/s) Ĥin (kJ/mol) nout Ĥout Ac (v) 66.9 35.64 3.35 32.02 Ac (l) - 63.55 N2 33.1 1.165 -0.145

Additional working session - Rework Example using different reference states References : Ac (g, 65oC, 1 atm), Nitrogen (g, 65oC, 1 atm) Substance nin (mol/s) Ĥin (kJ/mol) nout Ĥout Ac (v) 66.9 Ĥ1 3.35 Ĥ3 Ac (l) - 63.55 Ĥ4 N2 33.1 Ĥ2 Ĥ5

Additional working session - Rework Example using different reference states

Additional working session - Rework Example using different reference states References : Ac (g, 65oC, 1 atm), Nitrogen (g, 65oC, 1 atm) Substance nin (mol/s) Ĥin (kJ/mol) nout Ĥout Ac (v) 66.9 3.35 -3.611 Ac (l) - 63.55 N2 33.1 -1.310

Balances on Nonreactive Processes

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