112student_CHAPTER 4 (1)

CHAPTER 4-ENERGY BALANCE

FOR NONREACTIVE SYSTEM

CHE243-MATERIAL AND ENERGY BALANCE AND SIMULATION

Forms of EnergyThree component of total energy of a system

Kinetic energy (Ek) energy due to the translational motion of the system as a whole

relative to some frame of reference (usually the earth’s surface) or to rotation of the system about some axis.

Potential energy (Ep) energy due to the position of the system in a potential field

(such as a gravitational or electromagnetic field).

Internal energy (U) all energy possessed by a system other than kinetic and

potential energy; or

Energy due to translation, rotation, vibration & electromagnetic interactions of the molecules, atom and subatomic particle within the system.

Transfer of Energy In closed system (i.e. no mass is transferred across the

system boundaries while the process is taking place), energy may be transferred between such a system and its surroundings in two ways as heat or work.

Heat Energy that flows as a result of temperature difference between

a system and its surroundings. The direction of flow is always from a higher temperature to a

low one. Heat is defined as positive when its transferred to the system

from the surroundings.

Work energy that flows in response to any driving force other a

temperature difference, such as a force, a torque or a voltage Work is defined as positive when it is done by the system on

the surroundings.

First Law of Thermodynamics

Law of conservation of energy, which state that energy can neither be created nor destroyed.

General form of first law of thermodynamics

Inlet Energy + Heat - Outlet Energy – Work = Accumulation

Inlet energy and outlet energy is summation/total of all energy such as potential, kinetic and internal energy

Kinetic Energy Equation (Ek)

Kinetic energy, Ek (J) of an object of mass m (kg) moving with velocity u (m/s) relative to the surface of the earth is

If the fluid enters a system with a mass flow rate (kg/s) and uniform velocity u (m/s), the rate at which kinetic energy (J/s) is transported into the system is

2

2

1muEk

2

2

1umEk

m

Potential Energy Equation (Ep) Gravitational potential energy, Ep

if the fluid enters a system with a mass flow rate (kg/s) and an elevation z relative to the potential energy reference plane.

Normally we are interested in the change of potential energy during energy balance calculation

mgzEp

gzmEp

)( 1212zzgmEEE ppp

m

Energy Balances on Closed System Closed system

no mass is transferred across the system boundaries while the process is taking place

Energy balance

Final System Energy – Initial System Energy

= Net Energy Transferred to the System

Initial energy system = Ui + Eki + Epi

Final energy system = Uf + Ekf + Epf

Net energy transfer = Q-W

(Uf-Ui) + (Ekf-Eki) + (Epf-Epi) = Q-W

WQEEU pk

Energy Balances on Closed System When applying energy balance equation to a given process, the following

point must be aware;

1. The internal energy of a system depends almost entirely on the chemical composition, state of aggregation (solid, liquid, or gas), and temperature of the system materials. If no temperature changes, phase changes, or chemical reactions occur in a closed system and if pressure changes are less than a few atmospheres, then ∆U ≈ 0.

2. If a system is not accelerating, then ∆Ek = 0. If a system is not rising or falling, then ∆Ep = 0.

3. If a system and its surroundings are at the same temperature or the system is perfectly insulated, then Q = 0. The process is then termed adiabatic.

4. Work done on or by a closed system is accomplished by movement of the system boundary against a resisting force or the passage of an electrical current or radiation across the system boundary. If there no moving parts or electrical current at the system boundary, then W = 0.

Energy Balances on Open System In open system, mass is transferred across the system

boundaries while the process is taking place. Therefore work must be done on open system to push mass in

and work is done on the surrounding by mass that emerges from the systems.

Both work terms must be include in the energy balance for open system

The net work done by an open system

- shaft work - rate of work done by the process fluid on a moving part within the system such as a pump rotor.- flow work- rate of work done by the fluid at the system outlet

minus rate of work done by the fluid at the system inlet.

sW

flW

fls WWW

Energy Balances on Open System ^ symbol is used to denote the specific property ( property

divided by mass or by mole) such as specific internal energy (Û kJ/kg), specific volume (m3/kg) and so on.

One important property for energy balance on open system is specific enthalpy (Ĥ kJ/kg).

Sometimes, universal gas law constant can be used as a conversion factor to evaluate specific enthalpy.

VPUH ˆˆˆ

Energy Balances on Open System

The first law of thermodynamics for an open system at steady state

input=output

Upon derivation, the equation reduced to,

If a process has a single input and output stream, the expression for simplifies to

WEEQ joutput

jinput

spk WQEEH

H

HmHHmH inoutˆ)ˆˆ(

A common practice is to arbitrarily designate a reference state for asubstance at which U or H is declared to equal zero, and then tabulateU and/or H for the substance relative to the reference state. *Note: ReferFelder pp. 339 and 359

In Chapter 7 (Felder), U and H are state properties of a species; theirvalues depend only on the state of the species – primarily on itstemperature and state of aggregation (solid, liquid or gas) and, to alesser extent, on its pressure (and for mixtures of some species, on itsmole fraction in the mixture).

When a species passes from one state to another, both U and H forthe process are independent of the path taken from the first state tosecond one – Hypothetical Process Path *Note: Refer Felder pp. 360

ENERGY balance for non reactive system

Reference State

HYPOTHETICAL PROCESS PATHS

CHANGES IN P AT

CONSTANT T

CHANGES IN T AT

CONSTANT P

PHASE CHANGE

OPERATIONS

Internal energy (U) is nearly independent of pressure for solids andliquids at a fixed temperature, as is specific volume (V) . *Note: ReferFelder pp. 365 – 366

If pressure of a solid and liquid changes at constant temperature

U = 0

H = [U + (PV)] = [U + PV + V P] = [V P]

Both U and H independent of pressure for ideal gases – may assumeU = 0 and H = 0 for a gas undergoing an isothermal pressure changeunless gas temperature below 0 0C or well above 1 atm are involved.

CHANGES IN P AT CONSTANT T

CHANGES IN TEMPERATURE

SENSIBLE HEAT AND

HEAT CAPACITIES

HEAT CAPACITY

FORMULAS

ESTIMATION OF HEAT

CAPACITIES

The term sensible heat signifies that heat must be transferred to raiseor lower the temperature of a substance or mixture of substances.*Note: Refer Felder pp. 366

The quantity of heat required to produce a temperature change:

Q = U (closed system)

Q = H (open system)

Heat capacity at constant volume – Cv. At constant volume:

SENSIBLE HEAT AND HEAT CAPACITIES

dTTCU

T

T

v

^

2

1

Suppose both temperature and the volume of a substance change. Tocalculate U – break the process into 2 steps ( a change in V atconstant T followed by a changes in T and constant V):


21

22211121

^^^

UU

UUU

V,TAV,TAV,TA^^

For ideal gas and (to a good approximation) liquid and solids, Udepends only on T. In step 1, T is constant, U1 = 0.

Step 2 – V is constant:


dTTCU

T

T

v

^

2

1

Ideal gas: ExactSolid or liquid: good approximationNon ideal gas: valid only if V is constant

• Heat capacity at constant pressure – Cp. At constant pressure:


dTTCH

T

T

p

^

2

1

For first step – refer section 8.2 (Felder), as T is constant, H1 = 0 (forideal gas), H1 = V P (for solid or liquid).

Step 2 – P is constant:

dTTCH

T

T

p

^

2

1

Ideal gas: ExactNon ideal gas: valid only if P is constant

dTTCPVH

T

T

p

^^

2

1

Solid or liquid

Example

Calculate the heat required to raise 200 kg nitrous oxide from 200C to1500C in a constant – volume vessel. The constant – volume heatcapacity in this temperature range is given by the equation:

Cin T where

T109.42 0.855C kJ/kgC0

40v

Answer

kJ 312 24 UΔ^

Heat capacities are functions of temperature and frequently expressedin polynomial form (Cp = a + bT + cT2 + dT3). *Note: Refer Felder pp. 369

constant Gas : R

R C C : Gases Ideal

C C : Solids and Liquid

v p

v p

HEAT CAPACITY FORMULAS

Example (Felder pp369)

Assuming ideal gas behavior, calculate the heat must be transferred ineach of the following cases.

1. A stream of nitrogen flowing at a rate of 100 mol/min is heatedfrom 200C to 1000C

2. Nitrogen contained in 5 – liter flask at an initial pressure of 3 baris cooled from 900C to 300C

Neglect the changes in kinetic energy

min

kJ 233.3HΔQ nH

Answer

1.

2. kJ 621.0UnΔUQ^

Kopp’s rule – simple empirical method for estimating the heatcapacity of a solid or liquid near 200C. *Note: Refer Felder pp. 372

Use Data in Table B.10 for Cp of atom compound

Example: heat capacity of solid Ca(OH)2

Cmol

J89.5 is value True

Cmol

J 79

Cmol

J 9.6217226C

C2C2CC

0

00OHCap

HpaOpaCapaOHCap

2

2

ESTIMATION OF HEAT CAPACITIES

For heat capacities of certain mixture – may use these rules:

Rules 1 : For a mixture of gases or liquids, calculate the total enthalpychange as the sum of the enthalpy changes for the pure mixturecomponent

Rules 2 : For a highly dilute solutions of solids or gases in liquids,neglect the enthalpy change of solute.


For heat capacities of certainmixture: (Cp)mix (T)


component i of capacityheat C

component i of fraction moleor mass y

mixture of capacityheat C

where

TCy TC

pi

i

mix p

componentsmixture

all piimix p

For enthalpycalculation:

2

1

T

T

^

dTH TCmix p


Calculate the heat required to bring 150 mol/h of a stream containing60% C2H6 and 40% C3H8 by volume from 00C to 4000C.

Answer

h

kJ5230HnHΔQ

Latent heat – the specific enthalpy change associated with thetransition of a substance from one phase to another at constanttemperature and pressure. *Note: Refer Felder pp. 378

Latent heats for the two most commonly encountered phase changesare defined as follows:

1. Heat of fusion (or heating of melting) Hm (T,P) – specificenthalpy difference between the solid and liquid forms of aspecies at T and P

2. Heat of vaporization Hv (T,P) – specific enthalpy differencebetween the liquid and vapor forms of a species at T and P

PHASE CHANGE OPERATIONS

1) Latent Heats

Trouton’s rule – a simple formula for estimating a standard heat ofvaporization (Hv at normal boiling point); provide an estimate of Hv

accurate to within 30%. *Note: Refer Felder pp. 381


2) Estimation and Correlation of Latent Heats

liquid the ofpoint boiling normal :T

alcoholsweight molecular low water, (K) 0.109T

liquidsnonpolar (K) 0.088T (kJ/mol)

b

b

b

^

vH

Chen’s equation – provides roughly 2% accuracy. *Note: Refer Felderpp. 381



(atm) pressure critical

(K) etemperatur critical

(K)point boiling normal

TT1.07

Plog 0.02970.0327TT 0.0331T (kJ/mol)

cb

c10cbb

:P

:T

:T

H

c

c

b

^

v

A formula for approximating a standard heat of fusion. *Note: ReferFelder pp. 381



solid the ofpoint melting normal :T

compounds organic (K) 0.050T

compounds inorganic (K) 0.0025T

elements metallic (K) 0.0092T (kJ/mol)

m

m

m

m

^

mH

Watson’s colleration – a useful approximation for estimating Hv at T2

for known value at T1. *Note: Refer Felder pp. 382



etemperatur critical

TT

TT)(T )(T

1c

2c12

:T

HH

c

.^

v

^

v

380


100 mole/h of liquid n-hexane (C6H14) at 250C and 7 bar is vaporizedand heated to 3000C at constant pressure. Neglecting the effect ofpressure on enthalpy, estimate the rate at which heat must be supplied.Given the boiling temperature of n-hexane at 7 bar is 1460C. Use dataprovided in Table B.1 to solve the problem.

kWQ 38.2

ENERGY balance for non reactive system

PROCEDURE CALCULATION

SIMPLE ENERGY

BALANCE CALCULATION

MIXING AND SOLUTION

Perform all required material balance calculation. *Note: Refer Felderpp. 361

Write the appropiate form of energy balance (close or open system)and delete any terms that are either zero or negligible for givenprocess system.

Choose a reference state – phase, temperature and pressure – for eachspecies involved in the process

Construct inlet – outlet enthalpy table (open system) or inlet – outletinternal energy table (close system)


Calculate all required values (U or H) and insert the values intoappropriate table

Calculate Q for open or close system

Calculate any work, kinetic energy or potential energy – terms notdropped from energy balance

Solve the energy balance for unknown variables


ini

outi

initiali

finali

ini

outi

initiali

finali

mmor nnH :system Open

mmor nnU :system Close

iiii

iiii

HHHH

UUUU

Example – Energy balance on a gas preheater (Felder pp374)

A stream containing 10% methane (CH4) and 90% air by volume is to beheated from 200C to 3000C. Calculate the required rate of heat input inkW if the flow rate of the gas is 2.00 x 103 liters (STP)/min

kW 12.93 @ s

kJ12.93Q

SIMPLE ENERGY BALANCE CALCULATION

Example – Partial vaporization of mixture (Felder pp383)

An equimolar mixture of benzene (B) and toluene (T) at 100C is fedcontinuously into an evaporator which the mixture is heated to 500C.The liquid product is 40 mole% B, and the vapor product is 68.4 mole%B. Given the vapor pressure of product is 34.8 mm Hg. Neglecting theeffect of pressure on enthalpy, estimate the heat must be transferred tothe mixture per g-mole of feed.

kJ 17.7Q

SIMPLE ENERGY BALANCE CALCULATION

Heat of mixing and solution can be analyzed when 2

different liquids are mixed or when a gas or solid is

dissolved in a liquid.

Heat of solution, - enthalpy change for a process

in which 1 mole of a solute (gas/solid) is dissolved in r

moles of a liquid solvent at constant T

As r become large, approach a limiting value known

as heat of solution at infinite dilution

Heat of mixing has the same meaning as the heat of

solution when the process involves mixing two fluids

MIXING AND SOLUTION

),(ˆ rTHs

sH

MIXING AND SOLUTION

Example – production of hydrochloric acid (Felder pp397)

Hydrochloric acid is produced by absorbing gaseous HCl (hydrogenchloride) in water. Calculate the heat that must be transferred to orfrom an absorption unit if HCl(g) at 100°C and H2O(l) at 25°C are fed toproduce 1000kg/h of 20wt% HCl(aq) at 40°C

kJ/h1035.3Q 5

Date post:	12-Dec-2015
Category:	Documents
Upload:	sharifah-zulaikha-benyahya
View:	8 times
Download:	0 times

112student_CHAPTER 4 (1)

Documents