Loading in 5 sec....

Chapter 8 Balances on Nonreactive ProcessesPowerPoint Presentation

Chapter 8 Balances on Nonreactive Processes

- By
**arva** - Follow User

- 1098 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Chapter 8 Balances on Nonreactive Processes' - arva

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Chapter 8 Balances on Nonreactive Processes

8.1 Elements of Energy Balances Calculations

8.1a Reference States – A Review

We can never know the absolute values of and for a species at a given state. Fortunately, we never need to know the absolute values of and at specified

states; we only need to know and for specified changes of state, and we

can determine these quantities experimentally.

We may therefore arbitrarily choose a reference state for a species and determine

for the transition from the reference state to a series of other states.

If we set equal to zero, then for a specified state is the specific internal energy at that state relative to the reference state. The specific enthalpies at each state can then be calculated from the definition, , provided that the

specific volume ( ) of the species at the given temperature and pressure is known.

The values of and in the steam tables were generated using this procedure. The reference state was chosen to be liquid water at the triple point [H2O(l,

0.01C, 0.00611 bar)], at which point was defined to be zero.

This does not mean that the absolute value of for water vapor at 400C and 10.0 bar is 2958 kJ/kg. It means that of water vapor at 400C and 10.0 bar is 2958 kJ/kg relative to

water at the reference state, or

The specific enthalpy of water vapor at 400C and 10.0 bar is

10 bar

0.307 m3

103 L

8.31410-3 kJ/(molK)

kg

1 m3

0.08314 Lbar/(molK)

Test Yourself p. 359

8.1b Hypothetical Process Paths vapor at 400

and are state properties of a species; that is, their values depend only on the state of the species – primarily on its temperature, state of aggregation (solid, liquid or gas), and, to a lesser extent, on its pressure (and for mixture

of some species, on its mole fraction in the mixture).

A state property does not depend on how the species reached its state. Consequently, when a species passes from one state to another, both and

for the process are independent of the path taken from the first state to

the second one.

In most of this chapter and in chapter 9, we will learn how to calculate internal energy

and enthalpy changes associated with certain processes; specifically,

1.Changes in P at constant T and state of aggregation (Section 8.2).

2.Changes in T at constant P and state of aggregation (Section 8.3).

3.Phase changes at constant T and P – melting, solidifying, vaporizing, condensing,

sublimating (Section 8.4).

4.Mixing of two liquids or dissolving of a gas or a solid in a liquid at constant T and P

(Section 8.5).

5.Chemical reaction at constant T and P (Chapter 9).

Once we know how to calculate and for these five step of processes, we can calculate these quantities for any process by taking advantage of the fact that and are state properties. The procedure is to construct a hypothetical process path from the initial state to

the final state consisting of a series of steps of the given five steps.

Having done this, we calculate for each of the steps, and then add the for the steps

to calculate for the total process. Since is a state property, calculated for the hypothetical process path – which we constructed for convenience – is the same as for the

path actually followed by the process.

For example, we wish to calculate

for a process in which solid phenol at 25C and 1 atm is converted to

phenol vapor at 300C and 3 atm.

However, we do not have such a table.

Step 1,3, 5 Type 2 (change in T at constant P)

Step 2,4 Type 3 (change in phase at constant T and P)

Step 6 Type 1 (change in P at constant T)

Test Yourself p. 361

8.1c Procedure for Energy Balance Calculations five step of processes, we can calculate these quantities for any process by taking advantage of the fact that and are state properties. The procedure is to construct a

The procedure to follow for the energy balance calculation.

1.Perform all required material balance calculations.

2.Write the appropriate form of the energy balance(closed or open system) and delete any

of the terms that are either zero or negligible for the given process system.

3.Choose a reference state – phase, temperature, and pressure – for each species involved

in the process.

4.For a closed 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 state ( ).

For an open system, construct a table with columns for inlet and outlet stream

component flow rates ( or ) and specific enthalpies relative to the chosen reference

states.

5.Calculate all required values of ( or ) and insert the values in the appropriate

places in the table.

6.Calculate

7.Calculate any work, kinetic energy, or potential energy terms that you have not dropped

from the energy balance.

8.Solve the energy balance for whichever variable is unknown (often Q or ).

Closed System

Open System

Closed System

Open System

Example 8.1-1 the values in the appropriate

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 flowchart shown below.

The process operates at steady state. Calculate the required cooling rate.

Solution

1.Perform required material balance calculations. None are required in this example.

2.Write and simplify the energy balance.

3.Choose reference states for acetone and nitrogen.

N2 (g, 25C, 1 atm)

Ac (l, 20C, 5 atm)

4.Construct an inlet-outlet enthalpy table. the values in the appropriate

5.Calculate all unknown specific enthalpies.

6.Calculate the values in the appropriate

7.Calculate nonzero work, kinetic energy, and potential energy terms.

Nothing to do in this step.

8.Solve the energy balance for

Heat must be transferred from the condenser at a rate of 2320 kW to achieve the required cooling and condensation.

8.2 Changes in Pressure at Constant Temperature the values in the appropriate

It has been observed experimentally that internal energy is nearly independent of pressure for solids and liquids at a fixed temperature, as is specific volume.

If the pressure of a solid or liquid changes at constant temperature

For a gas undergoing an isothermal pressure change unless gases at temperature well below 0C or well

above 1 atm are involved.

ideal gas

If tables of or are available for the gas, there is of course no need to make this assumption.

If gases are far from ideal or if they undergo large pressure changes, you must either use tables of thermodynamic properties (such as the steam tables for water)

or thermodynamic correlations to determine or .

Test Yourself p. 366

8.3 Changes in Temperature the values in the appropriate

8.3a Sensible Heat and Heat Capacities

Sensible Heat : 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 specified temperature change in a system

can be determined by the appropriate form of the first law of thermodynamics:

We have neglected kinetic and potential energy

changes and work.

To determine the sensible heat requirement for a heating or cooling process, you must therefore be able to determine U or for the specified temperature change.

The specific internal energy of a substance depends strongly on temperature. If the temperature is raised or lowered in such a way that the system volume remains constant, the specific internal energy might

vary as shown in the right plot:

heat capacity at constant volume

Suppose both temperature and volume of a substance change. To calculate , you may break the process into two steps – a change in at constant T followed

by a change in T at constant .

is a state property

ideal gas: exact

solid or liquid: a good approximation

nonideal gas: valid only if V is constant.

Example 8.3-1 To calculate , you may break the process into two steps – a change in at constant T followed

Calculate the heat required to raise 200 kg of nitrous oxide from 20C to 150C 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 C.

Solution

The energy balance for this closed system is

heat capacity at constant pressure To calculate , you may break the process into two steps – a change in at constant T followed

Suppose both temperature and pressure of a substance change. To calculate , you may break the process into two steps – a change in P at constant T followed

by a change in T at constant P.

is a state property

ideal gas: exact

nonideal gas: valid only if P is constant.

solid or liquid

tabulated enthalpy

thermodynamic relation for variations of with P

Test Yourself p. 368

8.3b Heat Capacity Formulas To calculate , you may break the process into two steps – a change in at constant T followed

Heat capacities are functions of temperature and are frequently expressed in

polynomial form

Values of the coefficients a, b, c, and d are given in Table B.2 of Appendix B for a number of species at 1 atm, and listings for additional substances are

given on pp.2-161 to 2-186 of Perry’s Chemical Engineers’ Handbook.

Simple relationships exist between Cp and Cv in two cases:

The relationship between Cp and Cv for nonideal gases is complex.

Example 8.3-2 To calculate , you may break the process into two steps – a change in at constant T followed

Assuming ideal gas behavior, calculate the heat that must be transferred in each of the

following cases.

1.A stream of nitrogen flowing at a rate of 100 mol/min is heated from 20C to 100C.

2.Nitrogen contained in a 5-liter flask at an initial pressure of 3 bar is cooled from

90C to 30C.

Solution

From table B.2 the heat capacity of N2 at a constant pressure of 1 atm is

1.

2. To calculate , you may break the process into two steps – a change in at constant T followed

Example 8.3-3 To calculate , you may break the process into two steps – a change in at constant T followed

Fifteen kmol/min of air is cooled from 430C to 100C. Calculate the required heat removal rate using (1) heat capacity formulas from Table B.2 (2) specific enthalpies Table B.8.

Solution

(1) The hard way

Table B.8

(2) The easy way To calculate , you may break the process into two steps – a change in at constant T followed

Table B.8

Table B.8

15.0 kmol

1 min

1 kW

103 mol

-9.98 kJ

1 kmol

s

min

1 kJ/s

mol

Test Yourself p. 371

8.3c Estimation of Heat Capacities To calculate , you may break the process into two steps – a change in at constant T followed

The polynomial expressions for Cp in Table B.2 are based on experimental data for the listed compounds and provide a basis for accurate calculations of enthalpy changes. Several approximate methods for estimating heat capacities in the absence of tabulated

formulas are presented.

Kopp’s rule is a simple empirical method for estimating the heat capacity of a solid or liquid at or near 20C. According to the rule, Cp for a molecular compound is the sum of contributions (given in Table B.10) for each

atomic element in the compound.

true value is 89.5 J/(molC)

Suppose we wish to calculate the enthalpy change associated with a change in temperature undergone by a mixture of substances. Enthalpies and heat capacities of certain mixtures are tabulated in standard references. Lacking such data, we may use the following

approximation:

Rule 1 : For a mixture of gases or liquids, calculate the total enthalpy change as the sum of

the enthalpy changes for the pure mixture components. The enthalpy changes

associated with the mixing of the components are neglected.

Rule 2 : For 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 may often be simplified by calculating a heat capacity for the mixture in the following manner:

(Cp)mix = heat capacity of the mixture

yi = mass or mole fraction of the ith component

Cpi = heat capacity of the ith component

If Cpi and (Cp)mix are expressed in molar units, then yi must be the mole fraction of the ith component, and if the heat capacities are expressed in

mass units, then yi must be the mass fraction of the ith component.

(Cp)mix is known

Valid to the extent that enthalpies of mixing may be neglected.

Example 8.3-4 cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

Calculate the heat required to bring 150mol/h of a stream containing 60% C2H6 and 40% C3H8 by volume from 0C to 400C. Determine a heat capacity for the mixture as part of the problem.

Solution

Table B.2

150 mol

34.89 kJ

mol

h

Test Yourself p. 373

8.3d Energy Balances on Single-Phase Systems cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

We are now in position to perform energy balances on any processes that do not involve phase changes, mixing steps for which enthalpy changes cannot be neglected,

or chemical reactions.

If a process involves heating or cooling a single species from T1 to T2, the procedure

is straightforward:

1. Evaluate

2. For a closed system

For a open system

3.Substitue for U, H, or in the approximate energy balance equation to

determine the required heat transfer, Q, or heat transfer rate, .

correcting for pressure changes if necessary.

Closed System

Open System

Example 8.3-5 cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

A stream containing 10% CH4 and 90% air by volume is to be heated from 20C to 300C. Calculate the required rate of heat input in kilowatts if the flow rate of the gas is 2.00103 liters (STP)/min.

Solution

Basis: Given Flow Rate

2000 L (STP)

1 mol

22.4 L (STP)

min

neglecting effect of pressure on enthalpy cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

neglecting heats of mixing of gases

Table B.8

776 kJ

1 min

1 kW

60 s

min

1 kJ/s

Example 8.3-6 cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

A gas stream containing 8.0 mole% CO and 92.0 mol% CO2 at 500C is fed to a waste heat boiler, a large metal shell containing a bundle of small-diameter tubes. The hot gas flows over the outside of the tubes. Liquid water at 25C is fed to the boiler in a ratio 0.200 mol feedwater/mol hot gas and flowed inside the tubes. Heat is transferred from the hot gas through the tube walls to the water, causing the gas to cool and the water to heat to its boiling point and evaporate to form saturated steam at 5.0 bar. The steam may be used for heating or power generation in the plant or as the feed to another process unit. The gas leaving the boiler is flared (burned) and discharged to the atmosphere. The boiler operates adiabatically – all the heat transferred from the gas goes into the water, as opposed to some of it leaking through the outside boiler wall. The flowchart for an assumed basis of 1.00 mol

feed gas is shown below.

What is the temperature of the exiting gas?

Solution cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

adiabatic

8.3e Numerical Integration of Tabulated Heat Capacities cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

If a function relation for Cp(T) is available, such as one of polynomial of Table B.2, the integration can be carried out analytically; and if tabulated specific enthalpies are available for the substance being heated or cooled, a simple subtraction

replaces the integration.

The only information you have about Cp is its value at a series of temperatures that span the range from T1 to T2. The question is how to estimate the value of the integral from

these data.

A better solution is to use one of the many existing quadrature formulas – algebraic expressions that provide estimates of the integrals of tabulated data. Several such

formulas are presented and illustrated in Appendix A.3.

8.4 Phase Change Operations cooling of a mixture of known composition may often be simplified by calculating a heat capacity for the mixture in the following manner:

Phase changes such as fusion and vaporization are usually accompanied by large changes in internal energy and enthalpy. Heat transfer requirements in phase change operations consequently tend to be substantial, since (closed

constant-volume system) or (open system).

8.4a Latent Heats

- The specific enthalpy change associated with the transition of a substance from one phase to another at constant temperature and pressure is known as the latent heat of
- the phase change.
- Latent heats for the two most commonly encountered phase changes are defined as
- follows:
- 1.Heat of fusion (or heat of melting). is the specific enthalpy difference
- between the solid and liquid forms of a species at T and P.
- 2.Heat of vaporization. is the specific enthalpy difference between the
- liquid and vapor forms of a species at T and P.

Tabulated values of these two latent heats, such as those in Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

referred to as standard heats of fusion and vaporization.

The latent heat of a phase change may vary considerably with the temperature at which the change occurs but hardly varies at all with the pressure at the transition point.

For example:

When using a tabulated latent heat, you must therefore be sure that the phase change in question takes place at the temperature for which the tabulated value is reported,

but you may ignore moderate variations in pressure.

Example 8.4-1 Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

At what rate in kilowatts must heat be transferred to a liquid stream of methanol at its normal boiling point to generate 1500 g/min of saturated methanol vapor?

Solution

Table B.1

1500 g CH3OH

1 min

1 kW

1 mol

35.3 kJ

32.0 g CH3OH

60 s

min

1 kJ/s

mol

Phases changes often occur at temperatures other than the temperature for which the latent heat is tabulated. When faced with this situation, you must select a hypothetical process path that permits the available data to be used.

Example 8.4-2 Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

One hundred g-moles per hour of liquid n-hexane at 25C and 7 bar is vaporized and heated to 300C at constant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must be applied.

Solution

The temperature at which the vapor pressure of n-hexane is 7 bar (104 psia) is a approximately 295F (146C).

Figure 6.1-4

Table B.1

1 L Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

0.008314 kJ/(molK)

(1.013-7.0) bar

86.17 kg

0.08314 Lbar/(molK)

0.659 kg

1000 mol

0.2163 kJ

(69-725)C

molC

100 mol Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

1 kW

85.5 kJ

1 h

mol

h

1 kJ/s

3600 s

If a phase change takes place in a closed system, you must evaluate for the

phase changes to substitute into the energy balance equation.

For phase changes such as fusion, which involve only liquids and solids, changes in are

generally negligible compared to changes in , so that

For vaporization, for the vapor (which equals RT if ideal gas behavior may be assumed) is normally orders of magnitude greater than for the liquid, so that , and

Test Yourself p. 381

8.4b Estimation and Correlation of Latent Heats Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

A simple formula for estimating a standard heat of vaporization ( at the normal boiling point) is Trouton’s rule : (30% accuracy)

Tb : normal boiling point

Another formula that provides roughly 2% accuracy is Chen’s equation:

Tb : normal boiling point (K)

Tc : critical temperature (K)

Pc : critical pressure (atm)

Latent heat of vaporization may be estimated from vapor pressure data by using the Clausius-Clapeyron equation,

Clapeyron equation

A useful approximation for estimating at T Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are 2 from a known value at T1 is Watson’s correlation:

Tc : critical temperature

Another formula that provides roughly 2% accuracy is Chen’s equation:

Example 8.4-3 Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

The normal boiling point of methanol is 337.9 K, and the critical temperature of this substance is 513.2 K. Estimate the heat of vaporization of methanol at 200C.

Solution

Trouton’s rule

(The measured value is 35.3 kJ/mol, Chen’s equation yields 37.2 kJ/mol, so in this unusual case Trouton’s rule provides the better estimate.)

Watson’s correlation

The measured value is 19.8 kJ/mol.

Test Yourself p. 382

8.4c Energy Balances on Processes Involving Phase Changes Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

When writing an energy balance on a process in which a component exists in two phases, you must choose a reference states for that component by specifying both a phase and a temperature and calculate the specific enthalpy of the component in all process streams

relative to this state.

If the substance is a liquid at its reference state and a vapor in a process stream, may be calculated as outlined in Section 8.4a: that is, bring the liquid from the reference temperature to a point at which is known, vaporize the liquid, bring the vapor to the process stream temperature, and sum the individual enthalpy changes for the three steps.

Example 8.4-4 Table B.1 and on pp. 2-151 through 2-160 of Perry’s Chemical Engineers’ Handbook, usually apply to a substance at its normal melting or boiling point – that is, at a pressure of 1 atm. The quantities are

An equimolar liquid mixture of benzene (B) and toluene (T) at 10C is fed continuously to a vessel in which the mixture is heated to 50C. The liquid products is 40 mole% B, and the vapor product is 68.4 mole% B. How much heat must be transferred to the

mixture per mole of feed?

Solution

Basis: 1 mol Feed

Degree-of-Freedom Analysis

3 unknown variables (nV, nL, Q)

-2 material balances

-1 energy balance

0 degree of freedom

Download Presentation

Connecting to Server..