CHAPTER 4HEAT EFFECT

Consider the process of manufacturing ETHYLENE GLYCOL (an antifreeze agent) from ethylene :

-Vaporization

-Heating

Ethylene (liquid)

(-30oC ?)

Ethylene + O2 ethylene oxide +H2O(vapor) (gas) (gaseous)

ETHYLENE GLYCOL

-Reaction-Cooling

-Reaction-Mixing (dissolution)-Condensation-Cooling/Heating

The heat effects involved are :1. Sensible heat (heating, cooling)2. Heat of reaction (ethylene + O2 eth-oxide, etc.)3. Latent heat (vaporization / condensation)4. Heat of mixing (dissolution, etc.)

4.1. SENSIBLE HEAT EFFECT Occurs due to TEMPERATURE CHANGE.

(but NO phase change; no chemical reaction; NO CHANGE in composition).

H = H(T,P)Whence : dH = We know that :

The term : can be NEGLECTED if :a. The process is at a CONSTANT PRESSURE, orb. H ≠ H (P), regardless the process.

H can be independent of P [H ≠ H (P)] for :- Ideal gases (exactly true)- Low pressure gases- Solids- Liquids outside the critical region

In those case : dH = Cp. dTor, ∆H =

Approximate true

For NON-FLOW process, mechanically reversible, const. pressure :

Q = ∆H

For STEADY-FLOW process, where ∆Ep & ∆Ek ≈ 0, and Ws = 0 :

Q = ∆H

Similarly, we can express :U = U (T,V)

Whence : dU =

dU = Cv dT +

The term : can be neglected if :a. The process is at CONSTANT VOLUME, orb. U ≠ U (V), regardless the process. This is

TRUE for :◦ Ideal gases◦ Incompressible fluids

In those case : dU = Cv dTor, ∆U =

Note : the common direct engineering application is to STEADY FLOW heat transfer ; i.e. Q = ∆H =

In the evaluation of thermodynamics properties (e.g. H & U), Cp used is Cp for IDEAL GASES.

Cp = Cp (T)

For an ideal gas ; Cp = Cp (T) can be expressed in various term, e.g. :

; or

; or

Table 4.1 (S&V, 4th ed.), provides the data of parameters A, B, C, & D of various substances.

Cvig can be calculated from the relationship : Cp = Cv + R

Mean heat capacityDefined as : Cpmh =

If : Cp = ,Then, Cpmh/R = A + BT + CT+2 + DT-2

Where, Tam = = arithmetic mean

temperature

In this case, ∆H is given by : ∆H = Cpmh (T2-T1)

or, T2 =

Example 4-1:

What is the final temperature when 4x105 Btu are added to 25 lbmole of ammonia, initially at 500oF in a steady process at approximately 1 atm?

Solution :

If ∆H is the enthalpy change per lbmole, then :

Q = n.H

∆H = Q/n = 4x105 Btu / 25 lbmole

= 16000 Btu/lbmole

T2 =

conversion factor from (oR)to (oK)

For ammonia,

; T [=] oK

[See table 4.1 s&v)

Hence:

With : R = 1,986 Btu/lbmoleoR

T1 =

Tam = ½ (T1+T2) = ½ (533,15 + T2)

By trial T2 (T2 ≥ T1), we obtain T2 = 1250,1 K (or 1790,5 oF)

Example 4.2The molar heat capacity for methane in the ideal gas state is : Where T [=] oK. Develop for temperature in oC.

Solution : ToK = toC + 273,15

Therefore : Or :

The heat capacity of the gas MIXTURE in the ideal gas state is given by:

Cpig

mixture = yA Cpig

A + yB Cpig

B +yC Cpig

C + …

= ∑yi Cpig

i

Where :Cp

igA, Cp

igB, Cp

igi = heat capacity of gas A, B,

and i in ideal gas state, respectively.

yA, yB, yi = mole fraction of gas A, B, and I, redpectively.

4.2. STANDART HEAT OF REACTIONHeat of reaction occurs due to changes in molecular structure. Differences in molecular structure differences in energy content.

STANDARD HEAT :Consider reaction : aA + bB l L + mM

Standard heat of reaction : enthalpy change when a mole of A and b mole of B in their standard states at temperature T reacts to form l moles of L and m moles of M in their standard states, also at temperature T.

- Pressure : 1 bar (=105 Pa 1 std atm)

- Temperature : usually 25oC (298 K)Standard state

- States :1. gases : pure

substance in ideal gas state

2. liquid or solid : actual pure liquid or solid

Example :Standard heat of reaction in ammonia

synthesis : N2 + H2 NH3 ; ∆Ho

298 = -46110 J/mole NH3

If expressed in per mole of N2 reacted :

N2 + 3H2 2NH3 ; ∆Ho298 = -92220 J/mole

N2 rxn

Standard heat of formationHeat of formation : the heat effect resulted from a formation reaction, based on 1 mole of the compound formed.Formation reaction : a reaction which forms a single compound from its constituent elements.

Example :(a) C + ½ O2 + 2H2 CH3OH (formation reaction)

(b) H2O + SO3 H2SO4 (NOT a formation reaction)

Same examples of heat of formation :CO2(g) : C(s) + O2(g) CO2(g) ∆Hf

o298 = -393509 J

H2 (g) : since H2 is an element ∆Hfo298 = 0

CO(g) : C(s) + ½ O2(g) CO(g) ∆Hfo298 = -110525 J

H2O(g) : H2(g) +1/2 O2(g) H2O(g) ∆Hfo298 = -241818 J

In many cases, heat of reaction can be calculated from heat of formation, e.g.

CO2(g) + H2(g) CO(g) + H2O(g) ∆Hfo298 = ?

CO2(g) C(s) + O2(g) ∆Hfo298 = 393509 J

C(s) + ½ O2(g) CO(g) ∆Hfo298 = -

110525 JH2(g) + ½ O2(g) H2O(g) ∆Hf

o298 = -241818 J

+CO2(g) + H2(g) CO(g) + H2O(g) ∆Hf

o298 =

41166 J

Standard heat of combustionA combustion reaction is defined as a reaction between an element or compound and oxygen to form specified combustion products.

In some cases, heat of formation can be calculated from heat of combustion (which is measurable), e.g.

4C(s) + 5H2(g) C4H10(g)

4C(s) + 4O2(g) 4CO2(g) ∆Hfo298 = 4(-393509)

5H2(g) + 2½ O2(g) 5H2O(l) ∆Hfo298 = 5(-

241818)4CO2(g) + 5H2O(l) C4H10(g) + 6½ O2(g) ∆Hf

o298 =

2877396+

4C(s) + 5H2(g) C4H10(g) ∆Hfo298 = -125790 J

TEMPERATURE DEPENDENCE OF ∆Ho

the general chemical reaction may be written as:lv1lA1 + lv2lA2 + … lv3lA3 + lv4lA4 + …

Where : lv1l = stoichiometric coefficient of Ai

lv1l = + for products

- for reactants

Example : N2 + 3H2 2 NH3

vN2 = -1 ; vH2 = -3 ; vNH3 = +2

The std. rate of rxn :∆H ≡ ∑ vi . Hi

o

Hio = [enthalpy of species –i in its standard state]

For std. reactions, products and reactants are always at 1 bar (std. pressure); std. state enthalpy is only a function of temperature,

dHio = Cpi

o dT

For total products & reactants,∑d(viHi

o) = ∑viCpio dT

d(∆Ho) ∆Cpo

Hence, d ∆Ho = ∆Cpo dTthe integration gives : ∆Ho

T = ∆Hoo +

Note : ∆Cpo = ∑ viCpio

= [∑(vCp)products - ∑(vCp)reactants]

4.3. LATENT HEAT The latent heat of vaporization is a function of temperature only :

∆H = T. ∆V clayperon equation*)

Where : ∆H = latent heat of pure substance ∆V = volume change

accompanying the phase changePsat = vapor pressure = slope of vapor-pressure vs

temperature curve at temperature of interest.∆V ≈ (Vsat vapor – Vsat liquid)

Useful method for prediction of (∆H)vaporization at normal boiling point :

(Riedel eqn.)

Where : Tn = normal boiling point ; ∆Hn = molar latent heat of

vaporization at Tn ;Pc = critical pressure, bar ;Trn = reduced temperature at Tn

Estimation of (∆H)vap. of PURE LIQUID at any temperature from the KNOWN value at a single temperature, proposed by Watson (1943) :

Another eqn. for predicting latent heat of vaporization : (kistiawkowsky)

Where : Tn = normal boiling point, oK∆H [=] cal/gmol

*)CLAUSIUS-CLAYPERON Eqn. for LOW-PRESSUREAt low pressure :

∆V = Vvapor – Vliq ≈ Vvapor

hence : At low pressure : vapor IDEAL GAS

Vvapor

Or : ∆ H =

Example IV-1 :What is the maximum temperature that can be reached by the combustion of methane with 20% excess air? Both methane & air enter the burner at 25oC.

The reaction is : CH4 + 2O2 CO2 + 2H2O(g)

∆Hfo298 = -802625 J

Example IV-2 :One method for the manufacture of “synthesis gas” (primarily a mixture of CO and H2) is the catalytic reforming of CH4 with steam at high temperature and atmospheric pressure :

CH4(g) + H2O(g) CO(g) + 3H2(g)

The only other reaction which occurs to an appreciable extent is the water-gas-shift reaction :

CO(g) + H2O(g) CO2(g) + H2(g)

If the reactants are supplied in the ratio : 2 mole steam to 1 mole CH4, and if heat is supplied to the reactor so that the products reach a temperature of 1300 K, the CH4 is completely converted and the product stream contains 17,4 % mole CO. assuming the reactants to be preheated to 600 K, calculate the heat requirement for the reactor.

Data:CH4(g) + H2O(g) CO(g) + 3H2(g) ∆Hf

o298 = 205813 J

CO(g) + H2O(g) CO2(g) + H2(g) ∆Hfo298 = -41166 J