4MA017 Mechanical Engineering principles Thermodynamics and Fluid mechanics Mixture of gases Session outcome • At the end of the session, you will gain the understanding of: οMixture of gases οDalton’s law of partial pressure οExamples Mixture of gases and vapour • The mixture of ideal gases behaves as a unique ideal gas with its equivalent molecular mass and gas constant. • Note that the mixture of ideal gas also obey the ideal gas equation of state. • Hence, ππ ππ = ππ π π ππ • Where ππ and ππ are the mixtures of pressure and temperature respectively. Dalton’s law of partial pressure • States that, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases. • Mathematically, it is written as; π ππ = ΰ· ππ = π1 + π2 + π3 + β― + ππ π=1 Where ππ = • Therefore, ππ π π ππ ππ ππ π π ππ π1 π 1 ππ π2 π 2 ππ π3 π 3 ππ ππ = = + + +β― ππ ππ ππ ππ • Since ππ and ππ are equal. Dalton’s law of partial pressure • Therefore, ππ π π = π1 π 1 + π2 π 2 + π3 π 3 + β― • Also, partial pressure ratio to the total pressure is expressed as π1 ππ = π1 π 1 ππ π π and π2 ππ = π2 π 2 ππ π π • And ππππ£πππ ππ πππ ππππ π‘πππ‘, Θ ππππππππ πππ ππππ π‘πππ‘, π = πππππ πππ π , π Example 1. A mixture of gases has the following properties by mass: 40% π2 25% π»2 35% πΆπ2 π = 100πππ π = 17 β Calculate a. Partial pressure of each of the gas. b. The specific volume of the mixture. Solution: a. Partial pressure of each gas Specific gas constant, R, of each gas constituent is; π ΰ· 8.314 π π2 = = = 0.2969ππ½/πππΎ ππ2 28 Solution π ΰ· 8.314 π π»2 = = = 4.157ππ½/πππΎ ππ»2 2 π ΰ· 8.314 π πΆπ2 = = = 0.1889ππ½/πππΎ ππΆπ2 12 + (16 × 2) Total specific gas constant, π π , for the mixture; π π = 0.4 × 0.2969 + 0.25 × 4.157 + 0.35 × 0.1889 = 1.2241ππ½/πππΎ Pressure ratio of gas constituent ππ2 ππ2 π π2 0.2969 = = 0.4 × = 0.097 ππ ππ π π 1.2241 ππ»2 ππ»2 π π»2 4.157 = = 0.25 × = 0.84899 ππ ππ π π 1.2241 Solution ππΆπ2 ππΆπ2 π πΆπ2 0.1889 = = 0.35 × = 0.054 ππ ππ π π 1.2241 Partial pressure of each gas; ππ2 = 0.097 × 100πππ = 9.7πππ ππ»2 = 0.84899 × 100πππ = 84.899πππ ππΆπ2 = 0.054 × 100πππ = 5.4πππ b. Specific volume of the mixture in m3/kg From the equation of state ππ = ππ π But specific volume, π£ = π π π π π = π π Solution Therefore, specific volume of the mixture π π π 1.2241 × 290 π£π = = π 100 = 3.5499π3/ππ Example • A mixture of gas contains the following by mass: 40% π2 , 35% πΆπ2 and 25% π2 . • Calculate the partial pressure of each of the gas in kPa and the specific volume of the mixture in m3/kg, if the total pressure is 150kPa and at temperature of 20β. Take Θ = 8.314kJ/kmol.K. Solution: specific gas constant of each gas π ΰ· 8.314 π π2 = = = 0.2969ππ½/πππΎ ππ2 28 π ΰ· 8.314 π πΆπ2 = = = 0.1889ππ½/πππΎ ππΆπ2 44 π ΰ· 8.314 π π2 = = = 0.2598ππ½/πππΎ ππ2 16 ∗ 2 Solution Total specific gas constant of the mixture π π = 0.4 × 0.2969 + 0.35 × 0.1889 + 0.25 × 0.2598 = 0.2498ππ½/πππΎ Partial pressure of each of the gas in kPa ππ2 π π2 0.2969 ππ2 = × ππ = 0.4 × × 150πππ = 71.313πππ ππ π π 0.2498 ππΆπ2 π πΆπ2 0.1889 ππΆπ2 = × ππ = 0.4 × × 150πππ = 39.701πππ ππ π π 0.2498 ππ2 π π2 0.2598 ππ2 = × ππ = 0.4 × × 150πππ = 39.001πππ ππ π π 0.2498 b. Specific volume of the mixture π π π 0.2498 × 293 π£π = = ππ 150 = 0.4879π3/ππ Example • A vessel of volume 0.4m3 contains 0.5kg of carbon dioxide and 1kg of air at 27 0C. a. Calculate the partial pressure of each constituent gas in kPa b. Calculate the total pressure in the vessel in kPa The gravimetric analysis of air is to be taken as 22% oxygen and 78% Nitrogen. Assume the following molar masses: Carbon dioxide CO = 28kg/kmol Oxygen O2 = 32kg/kmol Nitrogen N2 = 28kg/kmol Universal gas constant Θ = 8.314kJ/kmol.K Solution V=0.4m3, mass of CO =0.5kg, mass of air =1kg, T=27 0C=300K a. Partial pressure of each constituent in kPa Mass of oxgen in air = 22% x 1kg = 0.22kg Mass of nitrogen = 78% x 1kg = 0.78 From the EOS, pV=mRT But specific gas constant, R = Therefore, ππ = ΰ· ππ π π π ΰ· π ΰ· ππ π π= π. π Solution Partial pressure of each gas ΰ· ππΆπ π π 0.5 × 8.314 × 300 πππ = = = 111.348πππ πππ π (12 + 16) × 0.4 ΰ· ππ2 π π 0.22 × 8.314 × 300 ππ2 = = = 42.869πππ ππ2 π (16 × 2) × 0.4 ΰ· ππ2 π π 0.78 × 8.314 × 300 ππ2 = = = 173.703πππ ππ2 π (14 × 2) × 0.4 b. Total pressure in the vessel in kPa π = ππΆπ + ππ2 + ππ2 = 111.348 + 42.869 + 173.703 = 327.92πππ. Test your understanding • The intake of a jet engine acts as a diffuser. Air enters at a pressure of 60kPa and temperature of -200C and a velocity of 230 m/s. the air leaves the diffuser with a velocity of 100m/s. calculate; i. The pressure of the air leaving the diffuser in kPa. ii. The temperature of the air leaving the diffuser in K. • Take ππ = 1.005ππ½/πππΎ π12 − π22 = 2ππ (π2 − π1 ) πΎ π2 π2 πΎ−1 =( ) π1 π1 πΎ = 1.4 Solution ππ = 1.005ππ½/πππΎ, πΎ = 1.4 i. Temperature of air leaving the diffuser, π2 From π22 − π12 = 2ππ (π2 − π1 ) 2302 − 1002 = 2 × 1.005 × π2 − 253 42900 = 2.01 × 103 × π2 − 253 42900 = 2.01 × 103 π2 − 508530 42900 + 508530 π2 = = 274.34πΎ 3 2.01 × 10 Solution ii. Pressure of air leaving the diffuser in kPa πΎ π2 π2 πΎ−1 =( ) π1 π1 π2 274.34 1.4 =( )1.4−1 400 253 π2 = 400 × 1.328 = 79.68πππ. Question 2: • Oil enters a 18mm diameter pipe of length 20m with an inlet pressure π1 of 400kPa. The volumetric flow rate of the oil is Q = 3.4 X 10-5 m3/s. a. Calculate the flow velocity of the oil in m/s. b. Calculate the exit pressure π2 of the oil in kPa. For oil, assume dynamic viscosity π = 0.045 ππ /π2 . Solution: d=18mm, L=20m, π1 = 400πππ, παΆ = 3.4 × 10−5 π3 /π a. Flow velocity of oil; Since volumetric flow, παΆ = π΄π Therefore, π = παΆ π΄ where π΄ = ππ 2 4 Solution παΆ 4παΆ 4 × 3.4 × 10−5 π= = = = 0.134π/π 2 2 2 ππ ππ π × (0.018) 4 b. Exit pressure, π2 , of the oil The pressure difference in pipe, βπ = π1 − π2 = Therefore; 60 − π2 = π2 = 400-11.911 32×0.045×20×0.134 0.0182 400 − π2 = 11911π/π2 π2 = 388πΎππ 32ππΏπ π2
