non-reactive energy balance 1

7/28/2019 Non-Reactive Energy Balance 1

http://slidepdf.com/reader/full/non-reactive-energy-balance-1 1/31

! A common practice is to arbitrarily designate a reference state for a

substance 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 a

lesser 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 ΔU and ΔH for

the process are independent of the path taken from the first state tosecond one – Hypothetical Process Path *Note: Refer Felder pp. 360

CHAPTER 4- ENERGY balance for non

reactive system

Reference State



CHANGES IN P AT CONSTANT T


reactive system

CHANGES IN T AT CONSTANT P

PHASE CHANGE OPERATIONS




reactive system

! 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

ΔU = 0

ΔH = [ΔU + Δ(PV)] = [ΔU + PΔV + V ΔP] = [V ΔP]

! Both U and H independent of pressure for ideal gases – may assumeΔU = 0 and ΔH = 0 for a gas undergoing an isothermal pressure change

unless gas temperature below 0 0C or well above 1 atm are involved.

CHANGES IN P AT CONSTANT T




reactive system

SENSIBLE HEAT AND HEAT CAPACITIES

HEAT CAPACITY FORMULAS

ESTIMATION OF HEAT CAPACITIES




reactive system

! 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 = ΔU (closed system)

Q = ΔH (open system)

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


( )dT T C U

T

T

v

^

∫ =Δ

2

1




reactive system

! Suppose both temperature and the volume of a substance change. Tocalculate ΔU – break the process into 2 steps ( a change in V at

constant T followed by a changes in T and constant V):


( ) ( ) ( )

21

222111

21

^ ^ ^

U U

U U U

V ,T AV ,T AV ,T A^ ^

Δ+Δ=Δ

⎯ ⎯→ ⎯ ⎯ ⎯→ ⎯ ΔΔ




reactive system

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

Step 2 – V is constant:


( )dT T C U

T

T

v

^

∫ =Δ2

1

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




reactive system

• Heat capacity at constant pressure – C p. At constant pressure:


( )dT T C H T

T

p

^ ∫ =Δ2

1

! For first step – refer section 8.2 (Felder), as T is constant, ΔH 1 = 0 (for

ideal gas),Δ

H 1 = V Δ

P (for solid or liquid). Step 2 – P is constant:

( )dT T C H

T

T

p

^

∫ =Δ2

1

Ideal gas: ExactNon ideal gas: valid only if P

is constant

( )dT T C P V H

T

T

p

^ ^

∫ +Δ=Δ2

1

Solid or liquid




reactive system

! 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:

CinT whereT 109.420.855CkJ/kgC

0

40v

−

×+=⋅

! Solution

For vessel – closed system. Q = ΔU

For constant volume: ( )dT T C U

T

T

v

^

∫ =Δ2

1




reactive system

! Solution

( )

( ) ( )[ ]

( )

kJ 31224kg200kgkJ 121.5591U Δ

kg

kJ 10.4091111.15U Δ

20150104.7120-1500.855T 2109.42 0.855T U Δ

dT T 109.420.855U Δ

^

^

224-

150

20

24-^

150

20

4-^

=×=

+=

−×+=⎥⎦

⎤⎢⎣

⎡ ×+=

×+= ∫




reactive system

! Heat capacities are functions of temperature and frequently expressedin polynomial form (C p = a + bT + cT 2 + dT 3). *Note: Refer Felder pp. 369

constantGas:R

RCC: GasesIdeal

CC:SolidsandLiquid

v p

v p

+=

=

HEAT CAPACITY FORMULAS




reactive system

! 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




reactive system

! Solution

Case 1

System: open system → Q = ΔH

From Table B.2, Appendix B, the heat capacity of N2 at constantpressure of 1 atm:

3-122-8-50 p T 102.871T 100.5723T 100.21990.02900CmolkJ/C ×−×+×+=⋅




reactive system

! Solution

( )

( ) ( ) ( ) ( )[ ]

minkJ 233.3

minmol100

molkJ 2.333H ΔQ

Finally,

kJ/mol2.333H Δ

0.000070.0020.0112.32H Δ

20100100.717820100100.190820100100.110020-1000.02900H Δ

T 4

102.871T

3

100.5723 T

2

100.2199 0.02900T H Δ

dT T 102.871T 100.5723T 100.21990.02900H Δ

^

^

^

4412-338-225-^

100

20

412-

38-

25-^

100

20

312-28-5-^

=×==

=

−++=

−×−−×+−×+=

⎥⎦

⎤⎢⎣

⎡ ×−

×+

×+=

×−×+×+= ∫

Case 2 and Example 8.3 – 3 (pp 371): DIY




reactive system

! 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 C p of atom compound

! Example: heat capacity of solid Ca(OH)2

( )

( )( )

( ) ( )

Cmol

J

89.5isvalueTrue

Cmol

J 79

Cmol

J 9.6217226C

C2C2CC

0

00OH Ca p

H paO paCa paOH Ca p

2

2

⋅

⋅

=

⋅

++=

++=





reactive system

! 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.





reactive system

! For heat capacities of certainmixture: (C p)mix (T)


( ) ( )

( )

componentiof capacityheatC

componentiof fractionmoleormass y

mixtureof capacityheatCwhere

T C yT C

pi

i

mix p

componentsmixture

all piimix p

=

=

=

= ∑

! For enthalpycalculation:

( ) ( ) ∫ =Δ2

1

T

T

^

dT H T Cmix p




reactive system


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

Solution

System: open system → Q = ΔH

From Table B.2, Appendix B, the heat capacity of component:

C pi yi C pi (C p)mix

kJ/mol.0CC2H6 C3H8 C2H6 C3H8

a 0.04937 0.06803 0.02962 0.0272 0.05682

b 13.92 x 10-5 22.59 x 10-5 8.352 x 10-5 9.036 x 10-5 17.388 x 10-5

c -5.618 x 10-8 -13.11 x 10-8 -3.3708 x 10-8 -5.244 x 10-8 -8.6148 x 10-8

d 7.280 x 10-12

31.71 x 10-12

4.368 x 10-12

12.684 x 10-12

17.052 x 10-12




reactive system

!Solution

( )

( ) ( ) ( )

( )

h

kJ 5237

h

mol150

mol

kJ 34.91H Δ

mol

kJ 34.910.10911.837813.910422.728H Δ

mol

kJ

0-400

4

1017.052

0-4003

108.6148 0-400

2

1017.388 0-4000.05682

H Δ

T 4

1017.052

T 3

108.6148

T 2

1017.388

0.05682T H Δ

dT T 1017.052T 108.6148T 1017.3880.05682H Δ

^

^

4412-

338-

225-

^

400

0

412-

38-

25-^

400

0

312-28-5-^

=×=

=+−+=

⎥

⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢

⎣

⎡

×+

×−

×+

=

⎥⎦

⎤

⎢⎣

⎡ ×+

×

−

×+=

×+×−×+= ∫




reactive system

! 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) ΔH m (T,P) – specific enthalpy difference between the solid and liquid forms of aspecies at T and P

2. Heat of vaporization ΔH v (T,P) – specific enthalpy

difference between the liquid and vapor forms of a species at T andP


1) Latent Heats

C G f




reactive system

! Trouton’s rule – a simple formula for estimating a standard heat of

vaporization (ΔH v at normal boiling point); provide an estimate of ΔH v accurate to within 30%. *Note: Refer Felder pp. 381


2) Estimation and Correlation of Latent Heats

liquidtheof pointboilingnormal:T

alcoholsweightmolecularlowwater, (K)0.109T

liquidsnonpolar(K)0.088T (kJ/mol)

b

b

b

→≈

→≈Δ

^

v H

CHAPTER 4 ENERGY b l f




reactive system

! Chen’s equation – provides roughly 2% accuracy. *Note: Refer Felder

pp. 381



( )( )

(atm) pressurecritical

(K)etemperaturcritical

(K) pointboilingnormal

T T 1.07

Plog0.02970.0327T T 0.0331T (kJ/mol)

cb

c10cbb

: P

:T

:T

H

c

c

b

^

v

−

+−=Δ





reactive system

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



solidtheof pointmeltingnormal:T compoundsorganic (K)0.050T

compoundsinorganic (K)0.0025T

elementsmetallic (K)0.0092T (kJ/mol)

m

m

m

m

→≈

→≈

→≈Δ

^

m H





reactive system

! Watson’s colleration – a useful approximation for estimating ΔH v at

T 2 for known value at T 1. *Note: Refer Felder pp. 382



etemperaturcritical

T T

T T )(T )(T

1c

2c12

:T

H H

c

.^

v

^

v

380

⎟⎟

⎠

⎞⎜⎜

⎝

⎛

−

−Δ=Δ





reactive system


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 data

provided in Table B.1 to solve the problem.

HO W ? ? ? Analyze the information….

! P1 , T 1 = 7 bar, 250C → Liquid Phase

! P2 , T 2 = 7 bar, 3000C → Gas Phase

! T b at 7 bar = 1460C





reactive system


! There is phase change operation, however the ΔH v at 1460C and7 bar is not provided in the question.

! From Table B.1, T b = 68.740C, ΔH v = 28.85 kJ/mol at P = 1 atm or1.013 bar

! As stated in question, the effect of pressure on

enthalpy is neglected





reactive system


! Path method

C6H14 (l)

250

C, 7 bar

C6H14 (l)

1460

C, 7 bar

C6H14 (v)

1460

C, 7 bar

1

DH1=òCpdt (liq) C6H14 (v)

3000

C, 7 bar

2

DH2=DHv

3

DH3=òCpdt (vap)

C6H14 (l)

250C, 1 atm

C6H14 (l)

68.740C, 1 atm

C6H14 (v)

68.740C, 1 atm

C6H14 (v)

3000C, 1 atm

4

DH4= VDP (liq)

5

DH5=òCpdt (liq)

6

DH6=DH v

7

DH7=òCpdt (vap)

8

DH8= 0 (vap)

! True process path – 1 → 2 → 3 : ΔH v at 1460C and 7 bar is notprovided

! Alternative process path – 4→ 5→ 6→ 7→ 8





reactive system


! Calculate the enthalpy for each of stream number.

! Stream 4

! Since the effect of pressure on enthalpy is neglected, ΔH = 0

04

4

=Δ

Δ=Δ

H

P V H





reactive system

! Stream 5

! Find C p for liquid at Table B.2

( )

mol

kJ 9.461ΔH

2568.7410216.3dT 10216.3ΔH

5

3

68.74

25

3

5

=

−×=×=

−−

∫

! Stream 6

mol

kJ 28.85ΔH ΔH v6 ==

CHAPTER 4 ENERGY balance for non




reactive system

! Stream 7

! Find C p for gas at Table B.2

molkJ 47.191ΔH

dT T 1057.66T 1023.92T 1040.8510137.44ΔH

7

300

68.74

3122853

7

=

×+×−×+×= ∫ −−−−

! Stream 8

0=8

ΔH

CHAPTER 4 ENERGY balance for non




reactive system

! Total enthalpy

mol

kJ 85.502ΔH

047.19128.859.4610ΔH

ΔH ΔH ΔH ΔH ΔH ΔH 87654

=

++++=

++++=

! For overall process

kW 2.375Qs3600

h1

mol

kJ

85.502h

mol

100H nQ=

××=Δ=

non-reactive energy balance 1

Documents