Entropy Generated During an Adiabatic Chemical Reaction

1. Entropy Generated During an Adiabatic Chemical Reaction

If a chemical reaction is irreversible under adiabatic conditions, then, according to Second law, δσ > 0. The question is “why?”. Combustion is usually a process of oxidation of

fossil fuels to CO 2 and H 2 O. Fuel and oxygen are called reactants in this case, while CO 2 an

H 2 O are called products. The chemical bonds between the atoms in fuel are broken dring com- bustion, energy is converted into thermal energy which is eventually stored as te, ve, re (ran- dom energy) in a large number of quantum states. Hence, the entropy of the products is higher than that of the reactants. When combustion occurs in an adiabatic vessel, the energy is con- served. In such a process, the entropy difference between the products and reactants is due to the entropy generated during the process.

k. Example 11 One kmole of CO, 0.5 kmole of O 2 , and 1.88 kmole of N 2 enter an adiabatic reactor and produce CO 2 due to the exothermic reaction of the monoxide and oxygen. Deter- mine the temperature and entropy of the products assuming that CO reacts in varying amounts, i.e., 0.1, 0.2, …, 0.9, 1 kmole.

Solution For a steady adiabatic process that involves no work,

dE cv /dt = ˙ Q cv – W ˙ cv + Σ k,i N ˙ k h ,k – Σ k,e N ˙ k h k e .

(A)

Denoting h R = Σ k,i N ˙ k h ,k,

¯h P = Σ k,e N ˙ k h k , Eq. (A) may be expressed in the form

h P – h R = 0. (B) For every kmole of CO consumed, a half kmole of oxygen is consumed and a kmole

of carbon dioxide is produced. For instance, in the case 0.2 moles of CO react,

h R =1 × h CO (298 K) + 0.5 × h O 2 (298 K) + 1.88 h N 2 (298 K) =1 ×(–110530) + 0 + 0×(–393520)+1.88×0 = –110530 kJ, and

(C)

h P = 0.8 × h CO (T) +0.4 × h O 2 (T) +0.2 × h CO 2 (T) + 1.88 × h N 2 (T) =0.8 ×(–110530+(h T –h 298 )) + 0.4 ×(0+(h T –h 298 )) + 0.200 ×(–393520+(h T –h 298 )) + 1.88 × (0+(h T –h 298 )).

(D) Equating Eqs. (C) and (D) one can iteratively solve for T, which in this case is 846.6

K. Figure 5 presents the variation of T and entropy generated versus the moles of CO that are burned. The corresponding entropy balance equation is

(E) which for steady state and adiabatic conditions assumes the form

dS cv /dt = ˙ Q cv /T b + Σ k,i N ˙ k s k – Σ k,e N ˙ k s k + σ ˙

0=˙ Q cv /T b + Σ k,i N ˙ k s k – Σ k,e N ˙ k s k + σ ˙ , or σ ˙ =+ Σ k,e N ˙ k s k – Σ k,i N ˙ k s k (F) Denoting S R =( ΣN k s k ) i ,S P =( ΣN k s k ) e , and ˆs k (T,P,X k )= s k (T,p k ) = s o k – R ln

(p k /1), we obtain σ =S P –S R .

(G) Since S R is specified and σ > 0, S P must increase with the burned fraction. For in-

stance, if 0.2 moles of CO react, the product stream contains 0.8 moles of CO, 0.4 moles of oxygen, 0.2 moles of the dioxide, and 1.88 moles of nitrogen which is inert. In that case,

p CO =N CO P/N = 0.8 ×1/(0.8+0.4+0.2+1.88) = 0.244 bar. Similarly, p O 2 = 0.122, p CO 2 = 0.0.061, and p N 2 = 0.573. Hence,

S P = 0.8 × s CO (T, p co ) +0.40 × s O 2 (T, p O 2 )+0.20 × s CO 2 (T, p CO 2 )+

1.88 × s N 2 (T, p N 2 ) = 0.8 ×( s o CO (846.6 K) – 8.314 × ln 0.57/1) +

T adiab

,K 1500 T adiab

kJ/kmole CO

CO,burnt Fr

Figure 5: Variation of the adiabatic temperature and entropy with the CO burned fraction.

×( 0.4 s o (846.6 K) – 8.314 s O o 2 ×ln 0.122/1) + 0.2×( CO 2 (846.6 K) – 8.314 × ln 0.061/1)+1.88 ×( s o (846.6 K)–8.314 ×ln 0.573/1) = 779.3 kJ K –1 N 2 . Figure 5 plots the variation in S p with respect to the CO burned fraction.

l. Example 12 Determine the entropy generated during the oxidation of glucose within the cells of the human body assuming that the reaction takes place at 310 K under steady state conditions using 400 % excess air that is at 300 K and 1 bar. The entropy of glucose at 310 K is 288.96 kJ kmole –1 K. Determine the entropy generation per kmole of glu- cose burned assuming that no work is done during the purely biochemical process. The typical consumption rate of glucose is 0.31 g min –1 for a 65 kg human. If the ac- cumulated entropy generation over the lifetime of any biological species is limited to 10000 kJ kg –1 K –1 , obtain the life expectancy for a human. The value of “s” for glu- cose at 298 K, 1 bar is 212 kJ kmole –1 K –1 .

Solution The chemical reaction can be represented by the equation

C 6 H 12 O 6 +5 ×6(O 2 + 3.76 N 2 ) → 6 CO 2 +6H 2 O + 24 O 2 + 112.8 N 2 . We will apply the entropy balance equation

dS cv /dt = ˙ Q cv /T b + Σ k,i N ˙ k s k – Σ k,e N ˙ k s k + σ ˙

(A)

N 2 = 0.79 bar, and since s k (300 K) has values of 205.03 and 191.5 kJ kg –1 K –1 for O 2 and N 2 , respectively, s k (T, p k ) for these two species is, respectively, 218.01 and 193.46 kJ kg –1 K –1 . Therefore, the inlet entropy is given as

at steady state. At the inlet p O 2 = 0.21 bar, and p

S i = Σ k,i N ˙ k s k = 1 ×212+30×218.01+30×3.76×193.4662 = 28663 kJ K –1 per kmole per second of glucose consumed.

(B) At the exit (at 310K in units of kJ kg –1 K –1 ), (B) At the exit (at 310K in units of kJ kg –1 K –1 ),

148.8 kmole. Therefore,

X O 2 = 24 ÷148.8 = 0.161, Similarly X N 2 = 0.758, X CO 2 = 0.0403, and Y HO 2 = 0.0403.

s –1 O 2 (310 K, p O 2 ) = 205.066 – 8.314 × ln (1× 0.16) = 220.2 kJ K per kmole of O 2 .

The corresponding values of the other entropies at the exit (at 310K in units of kJ kmole –1 K –1 ) are

s N 2 =193.84, s CO 2 = 240.42, and s HO 2 = 215.49. S e = Σ k,e N ˙ k s k = 6 ×24.42 + 6×215.49 + 24×220.2 + 112.8×193.64 = 29900 kJ K –1 per kmole per second of glucose consumed.

Applying this value in the context of Eqs. (A) and (B) and assuming the heat loss from an average human body to be 2.5 ×10 6 kJ per kmole of glucose consumed (ex- ample 6) ,

0 = –2.5 ×10 6 ÷310 + 28633– 29900 + σ ˙ , i.e., σ –1 ˙ = 9300 kJ K per kmole per second of glucose consumed, or σ ˙ = 9300 ÷180.2 = 52 kW K –1 per kg per second of glucose consumed.

The typical consumption rate of glucose is 1.031 ×10 –4 kmole per hr (i.e., 0.31 g min –1 ) for a person weighing 65 kg. Therefore, the entropy generated per second due to the irreversible metabolic activity is 52 ×0.31÷(1000×60) = 2.67×10 –4

kW K –1 . The entropy generated per unit mass of that person is 2.67 ×10 –4 ÷65 = 4.1×10 –6 kW kg –1

K –1 . Since the total entropy that can be generated σ m = 10,000 kJ kg –1 K –1 , the life- time of that person is

10000 kJ kg –1 K ÷(4.1×10 kJ s kg K ×3600 s hr ×24 hr day × 365 days year –1 ) = 77 years.

Remarks The entropy generated per unit mass of a human is 4.1 ×10 –6 kW kg –1 K –1 . The human

body has more than 60 trillion cells, which are constantly repaired and restored to an “original” state, but with some imperfections (e.g., aging). The cells require energy from metabolism to do so. The imperfections accumulate over time affecting the cell

6 performance. If the size of each is of the order of 1 mm, there are 10 –3 cells m . As- sume that the density of humans is roughly 1000 kg m –3 and σ ˙ cell = 4.1x10 –9 kW K –1

per cell. Other organisms may have to burn more energy per cell (see Chapter 2) and hence generate more entropy. Consequently, they may have a shorter life span. If T i = 37ºC, the only irreversibility is due to the chemical reaction. Typically, the entropy advection terms are small. Therefore,

σ ˙ =-˙ Q cv /T b = ˙ m Fb , HV /T b

is a function of body temperature and generally in- creases with an increase in body temperature (e.g., thorough fever, jogging, etc.).

where the fuel burn rate ˙ m Fb ,

Under normal conditions, the term ˙ m Fb , HV is the basal metabolic rate (BMR). The BMR seems to decrease due to aging. However it cannot decrease below a critical

BMR (1 W kg –1 ) which supports life functions. The corresponding critical entropy BMR (1 W kg –1 ) which supports life functions. The corresponding critical entropy

≈ σ m T b / ˙q cv , the higher the specific metabolic rate, the lower is the life expectancy. It can be shown from scaling laws that ˙q cv , ≈Cm –n B (n = 1/3 from the theory in Chap- ter 2 and from experiments n= 0.26 and C = 0.003552 for analyses SI units). There

fore, the life span, t life (years) ≈ (3.17x10 -8 σ

m m B T b /C) from theory. From experi- ments t life (years) = 11.8 m k B where k =0.20. Assuming n ≈ k, T b = 310 K, C= 0.003552, σ m = 4265 kJ kg –1 K –1 rather than 10,000 as we had assumed. The earlier empirical law seems to predict a lower life span. Similarly, the warmer the body

temperature, the longer the life span assuming that biological/metabolical reactions are unaffected (this is not generally true). Experiments on water fleas have revealed an increase in their life spans as the ambient temperature is increased from 5 to 15ºC, but this reduces thereafter, since their metabolism is affected. If the human body is unhealthy, it requires a higher energy or metabolic rate to cure itself, and, consequently, the entropy generated is larger for a sick person, which shortens the anticipated life span.