First Law

1. First Law

Consider a control volume (illustrated in Figure 3 ) whose boundary is penetrated by k material streams that bring in various species at the rates N ˙ 1 , N ˙ 2 , N ˙ 3 … into the volume. The

chemical processes within the CV are governed by the First law of thermodynamics, i.e.,

(19) The methalpy e Tk , = ( h + ke + pe ) k includes the enthalpy of formation, thermal enthalpy, as

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

well as the kinetic and potential energies wherever appropriate. Methalpy, e Tk , = ( h + ke

k V k /2000, pe k =M k g k Z k /1000 (in SI units), and M k V k /(2 g c J), pe k =M k g k Z k /(g c J) (in English units). If the latter two energies in the methalpy are neglected (as they often can be), then

+ pe ) k The values of ke k =M 2

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

Let N ˙ F denote the molal flow rate of fuel through one of the inlet streams that flow into a combustion chamber. Dividing Eq. (19) by N ˙ F and simplifying the resultant expression we obtain the relation (1/ N ˙ F ) dE cv /dt = ( ˙ Q cv / N ˙ F )– ( W ˙ cv / N ˙ F )+ Σ k,I ( N ˙ k / N ˙ F ) ¯e T,k – Σ k,e ( N ˙ k / N ˙ F ) ¯e T,k For SSSF processes d( )/dt = 0, and

(˙ Q cv / N ˙ F )– ( W ˙ cv / N ˙ F )+ Σ k,i ( N ˙ k / N ˙ F ) ¯e T,k – Σ k,e ( N ˙ k / N ˙ F ) ¯e T,k = 0. Replacing the subscript “i” by R and “e” by P to indicate the reactant and product streams,

respectively, ¯e T,R = Σ k,i ( N ˙ k / N ˙ F ) ¯e T,k and ¯e T,P = Σ k,e ( N ˙ k / N ˙ F ) ¯e T,k

and we obtain the following more convenient form

(20) Neglecting the potential and kinetic energies, we obtain the relation

q – w = e T,P – e T,R .

where h R denotes the enthalpy of the reactant stream Σ k,i ( N ˙ k / N ˙ F ) h ,k , and h P the enthalpy of

the product stream Σ k,e ( N ˙ k / N ˙ F ) h k .

f. Example 6 Glucose, i.e., C 6 H 12 O 6 , is oxidized in the human body in its cells. Air is inhaled at 298 K, and a mixture of air and metabolism products (CO 2 ,H 2 O, O 2 and N 2 ) is exhaled at 310 K. Assume that glucose is supplied steadily to the cells at 25ºC and that 400% excess air is utilized during its consumption. If the breathing rate of humans is 360 L/hr, determine the amount of heat loss from the human body, 6 h

f,glucose = –1.26 ×10 kJ kmole -1 , and the higher heating value of glucose is 15628, kJ kg -1 . Solution

The reaction equation with 400 % of excess air (or 5 × theoretical air) is

C 6 H 12 O 6 +5 × 6 (O 2 + 3.76 N 2 ) → 6CO 2 +6H 2 O + 24 O 2 + 112.8 N 2 . We now conduct an energy balance, i.e.,

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

Ignoring the breathing (PdV) and other forms of work 0= ˙ Q cv +1 ×(–1.26×10 6 )–6 ×(–393520 + 9807 – 9364) –6 ×(–241820 + (10302 – 9904)) – 24 × (0 + 9030 – 8682) –112.8 ×(09014 – 8669).

Hence, = –2.5 ×10 6 kJ kmole q -1 cv of glucose.

M F = 6 ×12.01 + 12×1.01 + 6×16 = 180.2 kg kmole -1 . Therefore, q -1

cv = 13872 kJ kg of glucose. The air consumption is 360 L hr -1 or 0.360 m 3 hr -1 ×(24.45 m 3 kmole -1 ) = 0.0147

kmole hr -1 so that the glucose consumption is 0.0147 ÷142.8 = 0.0001031 kmole hr -1 , or

0.0001031 kmole × 180.2 kg kmole -1 × 1000 g kg -1 hr -1 ÷ 60 min hr -1 = 0.31 g min -1 . The heat loss rate Q ˙ cv = m ˙ F q cv = 0.31 g min -1 × 13872 J g -1 ÷ 60 s min -1 = 72 W. We can interpret this result that the hypothetical human body considered in this ex-

ample has the same energy consumption as a 70 W light bulb. In reality, the body burns a mixture of glucose (which is a carbohydrate with a H/C ratio of two) and fats (e.g., palmitic acid). Palmitic acid has a higher heating value HHV = 39125.5 kJ kg -1 and participates in the energy consumption through the stoichiometric reaction

C 15 H 31 COOH + 23 O 2 → 16CO 2 + 16 H 2 O.

Our typical food consists of carbohydrates (HHV = 18000 kJ dry kg -1 ), fatty acids (HHV = 40000 kJ dry kg -1 ), and proteins (HHV = 22000 kJ dry kg -1 ) of which almost

96 %, 98 % and 78 % are metabolized, respectively. In general, the larger the amount of moisture in food, the lower is its heating value. Fats have a lower moisture content and, therefore, have a larger heating value while the carbohydrates having relatively more moisture have a smaller value.

g. Example 7 Determine the amount of carbon dioxide (which is a greenhouse gas) emitted if coal is used as a fuel. Assume that the chemical formula for coal is CH

O , its gross heating value is 30,000 kJ kg –1

, and the power plant efficiency is 30%. Compare your answer with the result for natural gas (CH 4 ) that has a gross heating value of 50,000 kJ kg –1 . Solution The chemical reaction governing the burning of coal is

CH 0.8 O 0.3 + 1.05 O 2 → CO 2 + 0.4 H 2 O. The amount of CO 2 produced per kg of coal consumed is

44.01/(12.01 + 0.8 ×1.01 + 16×0.3) = 2.498 kg per kg of coal. Note that 1 kWh = 1 kJ s –1 × 3600 s hr –1 × 1 hr –1 = 3600 kJ.

The overall efficiency = Work output/Gross heat value released by combustion, i.e., per kWh of power produced

0.3 = 3600 kJ ÷ (gross heat value released by combustion). Therefore, the

gross heat value released by combustion per kWh = 3600 ÷0.3 = 12000 kJ. Consequently, the gross heat value released by combustion per kWh = 3600 ÷0.3 = 12000 kJ. Consequently, the

amount of CO 2 released = 2.498

× 0.4 = 0.999 kg per kWh of power. In the case of methane, the

Reac

CO 2 produced per kg of fuel

Prod

consumed = 44.01 ÷16.05 = 2.74, and the fuel consumed per kWh = 12000 ÷50000 = 0.24 kg. Therefore, the

CO 2 produced = 0.24 × 2.74 = 0.658 kg kWh –1 or 0.183 kg MJ –1 .

Figure 3: Illustration of the First Law Remarks

for a chemically reacting system. The CO 2 emitted when natural gas is used

as a fuel is reduced due to the higher heating value of methane. The Boie equation is an empirical relation that can be used to determine the HHV of many C-H-N-O fuels including solid and liquid fuels. The relation is

HHV, kJ kg –1 = 35,160 ×C + 116,225×H – 11,090×O +6,280×N +10,465×S where C, H, O, N, and S denote the mass fractions of carbon, hydrogen, oxygen, ni-

trogen, and sulfur in the fuel. Using this relation, the formula for the mass of CO 2 emitted per MJ of heat input is

kg of CO 2 per MJ of heat input = C ×44.01×1000÷ (35,160 ×C + 116,225×H – 11,090×O +6,280×N +10,465×S), or kg of CO 2 per MJ of heat input = 1000 ÷ (798.9 + 2640.9 (H/C) –252 (O/C) + 142.7 (N/C) + 237.8 (S/C)).

Therefore, the higher the H/C ratio and the lower the O/C ratio, the lower the CO 2 emission to the atmosphere.

h. Example 8 Consider the metabolism of glucose (i.e., C 6 H 12 O 6 ) in the human body that breathes in air at 298 and exhales a mixture of CO 2 ,H 2 O, O 2 , and N 2 at 37ºC. Assume that there is a steady supply of glucose to the human body at 25ºC, that 400% excess air is in- haled, and h f,glucose = –1.26 ×10 6 kJ kmole –1 . If the inhalation occurs at a rate of 360 l

hr –1 , determine the amount of heat loss from the human body. Solution 400% excess air is equivalent to 5 times theoretical air, i.e.,

C 6 H 12 O 6 +5 ×6(O 2 + 3.76 N 2 ) → 6CO 2 + 6H 2 O + 24O 2 + 112.8N 2 . The energy balance is

(19) Neglecting the potential and kinetic energies,

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

0=˙ Q cv – W ˙ cv + Σ k,i N ˙ k h ,k – Σ k,e N ˙ k h k

Ignoring the breathing (PdV) and other forms of work,

0 =˙ 6 Q cv +1 × (–1.26×10 )+0–6 ×(–393520+443)– ×(–241820+398)–24×(0+348)–112.8×345, i.e., 6

¯q 6 c.v. =(˙ Q/ N ˙ F) = –2.5x10 kJ per kmole of glucose. Since the molecular weight of glucose is

6 ×12.01 + 12×1.01 + 6×16 = 180.2 kg kmole –1 , q cv = - 13872 kJ per kg of glucose.

The air consumption is 360 l hr –1 or 0.360 m 3 hr –1 ÷ (24.45 m 3 kmole –1 ) = 0.0147 kmole hr –1 .

Therefore, the glucose consumption per hr = 0.0147 ÷142.8 = 0.0001031 kmole hr –1 , or

0.0001031 kmole ×180.2 kg kmole –1 × 1000 g kg –1 hr –1

= 18.6 g hr –1 or 0.31 g min –1 .

The heat loss rate =˙ Q cv = m ˙ F q cv = 0.31 g min –1 × 13872 J g –1 ÷ (60 s min –1 ) = 72 W.

Remarks The human body is a complex system. For a detailed analysis of this problem we must determine the work required for a body to function and a knowledge of the chemical enthalpies of the constituents of the metabolic process. Hence, due to change in meta- bolic activity, increased amount of heat may be liberated which will be exhibited through composition of nasal gases. If glucose is not transported to cells or is not metabolized, its concentration will increase in the bloodstream, which is known as the diabetic condition.

The body burns 0.7 kJ min –1 kg –1 of energy during aerobic dancing, 0.8 kJ min –1 kg –1 while running a mile in 9 minutes, 1.2 kJ min –1 kg –1 for a 6 minute mile, and 0.3 kJ min –1 kg –1 while walking. Due to a change in metabolic activity, an in- creased amount of heat may sometimes be liberated (e.g., fever), which will also be manifested through the composition of the human exhalation gases.