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Energy Balance Introduction

Functional form: How to evaluate ],,[ 222 XPThmix ?

mixHXPT HHHH ˆˆ*40.0ˆ*60.0ˆ2 Δ++= (1)

where 2H is molar enthalpy (kJ/gmol) of mixture at 40ºC and 1 atm.

PTH is molar enthalpy (kJ/gmol) of pure n-pentane at 40ºC and 1 atm.

HXH is molar enthalpy (kJ/gmol) of pure n-hexane at 40ºC and 1 atm.

mixHΔ is molar enthalpy of mixing that is assumed zero for ideal solution.

],,,1,40[ˆˆˆ oPT

oPT

oPTPT

oPTPT PhPTliqatmCHHH ⇐Δ+= o (2)

zero └───────┘ n-PT reference state

],,,1,40[ˆˆˆ oHX

oHX

oHXHX

oHXHX PhPTliqatmCHHH ⇐Δ+= o (3)

zero └────────┘ n-HX reference state

where oPTH is molar enthalpy of pure n-pentane at arbitrarily-selected reference state. oHXH is molar enthalpy of pure n-hexane at arbitrarily-selected reference state.

PTHΔ is molar enthalpy change to go from n-PT reference state to 40ºC and 1 atm.

HXHΔ is molar enthalpy change to go from n-HX reference state to 40ºC and 1 atm.

Enthalpy Change of a Pure Substance is represented by an exact differential:

∫∫∫ ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−+=i

oj

i

oj

j

oj

P

P P

T

T

PhjP

H

H

dPTVTVdTcHdˆˆˆˆ

,

ˆ

ˆ

(4)

where PhjPc ,ˆ is molar heat capacity of comp. j for a liquid, gas, or solid in Table B.2 of F&R, 3rd Ed.

Thus the molar liquid enthalpy changes for pure n-pentane and pure n-hexane are:

Units of kJ/gmol Excel APEx Add-In Function

(∫ −− ×+×=ΔC

TPT

oPT

dTHo40

53 1068.43104.155ˆ ) enthalpy("n-pentane",Tpt,T2,"C","l")

( )∫ −×=ΔC

THX

oHX

dTHo40

3103.216ˆ enthalpy("n-hexane",Tpt,T2,"C","l")

T2 = 40ºC, Tpt = , Thx = oPTT o

HXT Ppt = 1 atm, Phx = 1 atm

v07.08.04 © 2007, Michael E. Hanyak, Jr., All Rights Reserved Page 7-19

/* Quest P4.W1M.A - Energy Balance Introduction Example Problem */

CHEG 200, Michael Hanyak, October 27, 2005

Mathematical Model/* total */ n1 + n2 - n3 = 0

/* pt = pentane */ 0.40*n1 + 0.60*n2 - n3pt = 0/* hx = hexane */ 0.60*n1 + 0.40*n2 - n3hx = 0

/* pt mol frac */ n3pt = n3*x3pt/* hx mol frac */ n3hx = n3*x3hx

/* energy bal'n */ n1*h1 + n2*h2 - n3*h3 = 0

/* molar enthalpies of mixtures 1, 2, and 3 in kJ/gmol */

h1 = 0.40*( Hpt + dHhat("n-Pentane_l",Tpt,T1) ) + 0.60*( Hhx + dHhat("n-Hexane_l",Thx,T1) )h2 = 0.60*( Hpt + dHhat("n-Pentane_l",Tpt,T2) ) + 0.40*( Hhx + dHhat("n-Hexane_l",Thx,T2) )h3 = x3pt*( Hpt + dHhat("n-Pentane_l",Tpt,T3) ) + x3hx*( Hhx + dHhat("n-Hexane_l",Thx,T3) )

/* Givens: */ n1 = 45 // gmol/hn2 = 55 // gmol/h T1 = 20 // C, inlet Stream 1T2 = 40 // C, inlet Stream 2

/* Ref. Temp: */ Tpt = 20 // reference state for pure pentane: Tpt, C & Ppt = 1 atm/* Ref. Temp: */ Thx = 20 // reference state for pure hexane: Thx, C & Phx = 1 atm

/* Ref. enthalpy: */ Hpt = 0 // molar enthalpy for pure pentane at Tpt, C & Ppt = 1 atm/* Ref. enthalpy: */ Hhx = 0 // molar enthalpy for pure hexane at Thx, C & Phx = 1 atm

E-Z Solve Numerical SolutionT1 20 CT2 40 CT3 30.8092 C

n2 55 gmol/hn3 100 gmol/h

n3hx 49 gmol/hn3pt 51 gmol/h

x3hx 0.49 mol fracx3pt 0.51 mol frac

h1 0 kJ/gmolh2 3.7524 kJ/gmolh3 2.06382 kJ/gmol

v07.08.04 © 2007, Michael E. Hanyak, Jr., All Rights Reserved Page 7-20

Energy Balance Introduction

Why Arbitrary Component Reference States? ( no reactions )

Begin with the energy balance to mix two process streams to form a third one. These three streams labeled 1, 2, and 3 are mixtures of two chemical components—n-pentane (PT) and n-hexane (HX). When we determine the molar enthalpy of a mixture ( ,ˆ

iH kJ/kgmol), we can arbitrarily pick a reference state for each component (

jj PT & ) and assign zero to the molar enthalpy of the pure component at that

reference state ( ˆ 0jH = ). Why can we do this? Let’s look at the following derivation for the answer. Eqn # Derivation Comments

1 0ˆˆˆ

332211 =−+ HnHnHn start with the energy balance

2 mixHXHXPTPT HHxHxH ˆˆˆˆ

,1,11 ∆++= [ ]1111 ,,ˆ XPThmixH =

Multiplying Eq. 2 with ,1n gives:

3 HXHXPTPT HnHnHn ˆˆˆ,1,111 += ideal solution [ ]snPThmixHn j ',,ˆ

,11111 =

Molar enthalpy of pure n-pentane in Eq. 3 is calculated by:

4 [ ] [ ]PTPTPTPTPTPTPTPTPT PhPTPhPTHPhPTHH ,,,,ˆ,,ˆˆ

111 ⇐∆+= units of kJ/kgmol

term is set arbitrarily term is the enthalpy change to go from to zero for pure n-PT pure n-PT reference state to mixture state

Simplifying the notation in Eq. 4 gives:

5 [ ]11,ˆˆˆ PTHHH PTPTPT ∆+= molar enthalpy for pure n-PT

Similarly, we can find the molar enthalpy of pure n-hexane by:

6 [ ]11,ˆˆˆ PTHHH HXHXHX ∆+= molar enthalpy for pure n-HX

Substituting Eqs. 5 and 6 into Eq. 3 gives:

7 [ ]{ } [ ]{ }11,111,111 ,ˆˆ,ˆˆˆ PTHHnPTHHnHn HXHXHXPTPTPT ∆++∆+= [ ] hkJsnPThmixHn j /',,ˆ,11111 =

Applying the same procedure used in Eqs. 2 thru 7, the energy contents of Streams 2 and 3 are:

8 [ ]{ } [ ]{ }22,222,222 ,ˆˆ,ˆˆˆ PTHHnPTHHnHn HXHXHXPTPTPT ∆++∆+= [ ] hkJsnPThmixHn j /',,ˆ,22222 =

9 [ ]{ } [ ]{ }33,333,333 ,ˆˆ,ˆˆˆ PTHHnPTHHnHn HXHXHXPTPTPT ∆++∆+= [ ] hkJsnPThmixHn j /',,ˆ,33333 =

where PTH in Eqs. 7, 8, and 9 is at the reference state of

PTT and PTP for n-pentane.

where HXH in Eqs. 7, 8, and 9 is at the reference state of

HXT and HXP for n-hexane.

Substituting Eqs. 7, 8, and 9 into the energy balance of Eq. 1 and factoring the reference enthalpy terms gives:

10 { } { } +⋅−++⋅−+ HXHXHXHXPTPTPTPT HnnnHnnn ˆˆ,3,2,1,3,2,1 component reference states

[ ] [ ] +∆⋅+∆⋅ 11,111,1 ,ˆ,ˆ PTHnPTHn HXHXPTPT

[ ] [ ] −∆⋅+∆⋅ 22,222,2 ,ˆ,ˆ PTHnPTHn HXHXPTPT

[ ] [ ] 0,ˆ,ˆ33,333,3 =∆⋅−∆⋅ PTHnPTHn HXHXPTPT energy balance equation

v07.08.04 © 2007, Michael E. Hanyak, Jr., All Rights Reserved Page 7-21

Energy Balance Introduction

Arbitrary Component Reference States, Cont’d ( no reactions ) Eqn # Derivation Comments

Remember that the two component material balances are:

11 0,3,2,1 =−+ PTPTPT nnn for n-pentane

12 0,3,2,1 =−+ HXHXHX nnn for n-hexane

Substituting Eqs. 11 and 12 into the energy balance of Eq. 10 gives:

13 { } { } +⋅+⋅ HXPT HzeroHzero ˆˆ Each

jH is multiplied by 0

[ ] [ ] +∆⋅+∆⋅ 11,111,1 ,ˆ,ˆ PTHnPTHn HXHXPTPT

[ ] [ ] −∆⋅+∆⋅ 22,222,2 ,ˆ,ˆ PTHnPTHn HXHXPTPT

[ ] [ ] 0,ˆ,ˆ33,333,3 =∆⋅−∆⋅ PTHnPTHn HXHXPTPT energy balance equation

Conclusions

In the energy balance of Eq. 13, each component reference enthalpy of jH is multiplied by its

component balance. Thus an jH is always multiplied by zero. Because of this fact, we can arbitrarily

pick a reference temperature ( jT ) and pressure (

jP ) for a component and assign any value to its

reference enthalpy ( jH ). For convenience, we usually assign zero to a component reference enthalpy,

ˆ 0jH = . The reference temperature (

jT ) and pressure ( jP ) for each component does not have to be the same;

they all could be different. However, once you pick a reference state for a component, you must use that component reference state in any enthalpy mixture (hmix) calculations.

In the energy balance, the enthalpy changes ( sH j 'ˆ∆ ) are calculated from the reference state of the components to the state of the mixtures. In Eq. 13, they are expressed fully as follows:

Form 1.PT [ ]

PTPTPTPT PhPTPhPTH ,,,,ˆ111 ⇐∆ units of kJ/kgmol

1.HX [ ]HXHXHXHX PhPTPhPTH ,,,,ˆ

111 ⇐∆ units of kJ/kgmol

2.PT [ ]PTPTPTPT PhPTPhPTH ,,,,ˆ

222 ⇐∆ units of kJ/kgmol

2.HX [ ]HXHXHXHX PhPTPhPTH ,,,,ˆ

222 ⇐∆ units of kJ/kgmol

3.PT [ ]PTPTPTPT PhPTPhPTH ,,,,ˆ

333 ⇐∆ units of kJ/kgmol

3.HX [ ]HXHXHXHX PhPTPhPTH ,,,,ˆ

333 ⇐∆ units of kJ/kgmol

Each of these functional forms for a pure component can be represented by either a graph, a table, a set of equations, or a computer program.

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