1. Submitted by: 130190105045 Abhimanyu. C. Menon 130190105044 130190105043

2. In practice, the mixture of gases are often heated or cooled. The sensible heat change of an ideal gaseous mixture can be calculated by calculating the mixture properties from the properties of pure gases present in the mixture. For an ideal gas mixture, the molar heat capacity of a gas mixture at a constant pressure is given by C° mp mix = ∑y i C ⁰ mpi

3. Where y i is the moles fraction of the i th component and C mpi is the molar heat capacity of the i th component. In this equation C mp mix can be expressed as a function of temperature. For a real gas mixture, Lee at al 12 have proposed an additive rule for estimation of critical properties which can be utilized for evaluation of residual heat capacity. This rule is different from Kay’s additive rule for estimation of pseudo-critical properties For a limited range of pressure and temperature, enthalpy of a real gas mixture can be represented by a model, incorporating C mp mix and pressur parameter p.

5. Example: Pyrites fines are roasted in a chamber plant for making sulphuric acid. The gases leaving the roaster are 502 ⁰ C and have molar composition SO 2 7.09%,SO 3 0.45% and N2 81.91%. Calculate the heat content of 1kmol gas mixture over 25 ⁰ C, using the heat capacity data. Solution: Basis 1kmol of gas mixture (a) In the first instance, take the value of C mp =f(t) Heat change Q = C mp mix dt

6. Sulphur dioxide ySO 2 C mpSO 2 = 0.0709(24.7706 + 62.9481*10-3T – 44.2582*10-6T 2 + 11.122*10-9T 3 ) = 1.7562 + 4.4630*10-3T – 3.1379*10- 6T 2 +0.7885*10-9T 3 Oxygen yO 2 C mpO 2 = 2.7457 + 1.2402*10-3T – 0.24871*10- 6T 2 – 0.0593*10-9 T 3 Sulphur trioxide ySO 3 C mpSO 3 = 0.0992 + 0.5473*10-3T – 0.4134*10- 6T 2 + 0.1097*10-9T 3

7. Nitrogen yN 2 C mPN 2 = 24.2379 – 4.2110*10-3T +10.7891*10-6T 2 – 4.0693*10-9T 3 Summing up, C mp mix = y i C mpi = 28.839 + 2.0395*10-3T + 6.9907*10-6T 2 – 3.2304*10-9 T 3 T 1 = 298.15K and T 2 = 775.15K Using eqn. Q = 15016.7kJ/kmol

8. Heat Capacity of Liquid Mixtures For an immiscible liquid mixture, pure component value are additive and a weighted average is used for calculations. For miscible systems additivity is not generally true although for similar substances and for system with small heats of mixing, additivity gives fairly good results. A number of plots are available for aqueous solutions in literature. The heat capacity of mixtures of liquid metals or fused salts can be predicted within about 10% accuracy by the additivity rule.

9. For petroleum oils, the American Petroleum Institute has compiled extensive data. For a defined liquid hydrocarbon mixture, addivity rule is recommended by API. Thus, for immiscible liquid mixtures and also for liquid hydrocarbon mixtures, C 1 = ∑w i.C 1i Where w i = mass fraction of i th component, and C u = heat capacity of the i th component.

11. Example : A mixture of aniline and water, containing 11.8 aniline, is sub cooled in the overhead condenser of the distillation column from 100 to 40 ⁰ C with the help of cooling water at the rate 8000 kg/h. Find the heat removal rate of the sub cooling zone of the confessor. Solution : Basis 8000 kg/h mixture is to be cooled Aniline and water are practically immiscible in each other and hence the additivity rule can be used for calculations. Aniline in the mixture, q n1 = 0.118*8000 = 944kg/h ≡ 10.137 kmol/h

12. Water in the mixture, qn2 = 8000 – 944 = 7056 kg/h ≡ 391.67 kmol/h T1 = 373.15 K and T2 = 313.15 K Heat extraction rate, Φ = 10.137(12 376.2 – 4354.7 + 3996.9) + 391.67(3050.7 + 4387.1 – 4472.3 + 1584.8) = 10.137*12 018.4 + 391.67*4550.3 = 121 830.5 + 1782 216 = 190 4046.5 kJ/h ≡ 528.9 kW Heat capacity of aniline and water at 25C are 2.0515 and 4.1868 kJ/(kg*K) respectively. Id it is assumed to be constant over the temperature range of 313.15 K to 373.15 K,

13. Φ = (944*2.0515 + 7056*4.1868)(373.15 – 313.15) ≡ 524.65 Error = (528.9 – 524.65)100/528.9 = 0.8%

14. Latent Heats When matter changes from one phase to another, the latent heat is either absorbed or rejected. For eg, ice melts at 0C by supplying heat to produce water. Similarly, water at 101.325kPascal (760 torr) and 100 ⁰ C produces vapor if heat is consciously supplied to it. The former is called the Latent heat of fusion, while the latter is called latent heat of vaporation. In some cases, such as Iodine crystals, camphor dry ice, etc., vapor is produced from the solids, process is called the latent heat of sublimation.

15. The heat supplied to melt the solid to liquid or removed to convert the liquid into solid per kilogram is called the latent heat of fusion, represented by the symbol λ f. The heat supplied to convert liquid to vapor at constant pressure is called the latent heat of vaporization, represented by the symbol λ v In the case of liquids/vapors, the boiling/saturation point changes with changes in pressure, and hence the latent heat of vaporization also varies with pressure from the foregoing discussion, it is clear that the units of latent heat are kJ/kg. As in the case of heat capacity, molar latent heat are expressed in kJ/mol.

16. The variation of latent heat of vaporization with pressure/temperature is of considerable interest to the process industry. For this reason, a reliable equation for the correlation of pressure and the corresponding saturation temperature must be available. An equation proposed by Antoine is as follows: Log 10 Pv = A – B (T+C) Where P v is the vapor pressure in bar, T is the temperature in K and A,B and C are species specific constants.

18. Example : For O – xylene, calculate (a) latent heat of vaporizaton at Tb using Riedel equation, and (b) latent heat of vapourization at 25C using Watson equation. Solution: (a) For o – xylene, p c = 3732 kPa and T c = 630.3 K T b = 417.6 K T Br = 417.6/630.3 =0.6625 λ v = 36 945kJ/kmol (b) T 1 = 298.15 K λ v1 = 42 928 kJ/kmol