Reading Problems , 13.19, 13.22, 13.30, 13.39, , 13.78, 13.81, 13.84, ,

Transcription

1 Non-Reacting Gas Mixtures Reading Problems , 13.19, 13.22, 13.30, 13.39, , 13.78, 13.81, 13.84, , Introduction homogeneous gas mixtures are frequently treated as a single compound the individual properties of inert gases tend to behave as a single pure substance Definitions m - total mass of a mixture m i - mass of the i th component of the mixture n - total number of moles of a mixture n i - number of moles of the i th component of the mixture X i - mass fraction of the i th component of the mixture Y i - mole fraction of the i th component of the mixture M - molecular weight of the mixture (molar mass) M i - molecular weight of the i th component of the mixture R - universal gas constant kj/kmole K R - gas constant for a particular gas R i - gas constant of the i th component of the mixture P - mixture pressure P i - partial pressure of the i th component of the mixture V - mixture volume V i - partial volume of the i th component of the mixture u - specific internal energy of the mixture u i - specific internal energy of the i th component of the mixture u - specific internal energy of the mixture per mole of the mixture (same as for h, s, c p and c v ) 1

2 Formulations the total mass of a mixture, m is the sum of the masses of its components m = m 1 + m m j = m i the relative amounts of the components present in the mixture can be specified in terms of mass fractions X i = m i m X i = 1 the total number of moles in a mixture, n is the sum of the number of moles of each of the components n = n 1 + n n j = n i the relative amounts of the components present in the mixture can be specified in terms of mole fractions Y i = n i n Y i = 1 m 1 and n i are related by the molecular weight M i m i = n i M i Therefore the total mass is m = n i M i the mixture molecular weight can be calculated as a mole fraction average of the component molecular weights M = m n = n i M i n = Y i M i 2

3 X i and Y i are also related by the molecular weights X i Y i = (m i/m) (n i /n) = ( ) ( mi n ) n i m = ( ) ( ) 1 M i M Therefore X i = M i Y i M X M i i = Y i Y i M i P-V-T Relationships for Ideal Gas Mixtures Amagat Model (law of additive volumes) The volume of the gas mixture is equal to the sum of the volumes each gas would occupy if it existed at the mixture temperature and pressure. V = V i Dalton Model (law of additive pressures) The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at T and V. By combining the results of the Amagat and Dalton models i.e. (1) and (2), we obtain for ideal gas mixtures P i P = V i V = n i n Therefore, Amagat s law and Dalton s law are equivalent to each other if the gases and the mixture are ideal gases. 3

4 Ideal Gas Law for a Mixture The gas constant can be expressed as R = X i R i The mixture molecular weight can be written as M = and Y i M i X i Y i = M i M X i R i = Y i M i M R i = Y i R M = Y ir The relative mass fractions and mole fractions can be related in terms of the gas constant as R i Y i = X i X i R i Mixture Properties Extensive properties such as U, H, c p, c v and S can be found by adding the contribution of each component at the condition at which the component exists in the mixture. U = U i = m i u i = m X i u i = mu = n i u i = n Y i u i = nu where u is the specific internal energy of the mixture per mole of the mixture. 4

5 u = X i u i h = X i h i c v = X i c vi c p = X i c pi s = X i s i Changes in internal energy and enthalpy of mixtures u 2 u 1 = X i (u 2 u 1 ) i = h 2 h 1 = X i (h 2 h 1 ) i = T2 T 1 c v dt = c v (T 2 T 1 ) T2 T 1 c p dt = c p (T 2 T 1 ) s 2 s 1 = X i (s 2 s 1 ) i = c p ln T 2 T 1 R ln P 2 P 1 These relationships can also be expressed on a per mole basis. Entropy Change Due to Mixing of Ideal Gases when ideal gases are mixed, a change in entropy occurs as a result of the increase in disorder in the systems if the initial temperature of all constituents are the same and the mixing process is adiabatic temperature does not change but entropy does S = ( m A R A ln P A P + m BR B ln P ) B P + = m i R i ln P i P = R n i ln Y i 5

6 Psychrometrics studies involving mixtures of dry air and water vapour for T 50 C (P sat 13 kp a) h h(t ) water vapour can be treated as an ideal gas Definitions Moist Air a mixture of dry air and water vapour where dry air is treated as if it were a pure component the overall mixture is given as P = mrt V Total Pressure P = P a + P w P a = m ar a T V P w = m wr w T V where P a is the partial pressure of air and P w is the partial pressure of water vapour. Typically m w << m a. Relative Humidity - φ φ = P w(t ) P sat (T ) = vapour pressure at the prevailing T saturation pressure at the prevailing T 6

7 If P w = P sat (T ) the mixture is said to be saturated. Specific Humidity (Humidity ratio) - ω ω = m w m a = mass of water vapour mass or air = M w n w M a n a = M w (P w V /RT ) M a (P a V /RT ) = M w M a = ( ) Pw P a ( ) Pw P a In addition ω can be written as ω = ( ) Pw P a = ( ) Pw P P w = ( φpsat P φp sat ) which can be rearranged in terms of relative humidity φ = P sat P ω ω + M w M a = P ω P sat (ω ) Dry Bulb Temperature - the temperature measured by a thermometer placed in a mixture of air and water vapour Wet Bulb Temperature thermometer surrounded by a saturated wick if air/water vapour mixture is not saturated, some water in the wick evaporates and diffuses into the air cooling the water in the wick as the temperature of the water drops, heat is transferred to the water from both the air and the thermometer 7

8 the steady state temperature is the wet-bulb temperature Sling Psychrometer - a rotating set of thermometers one of which measures wet bulb temperature and the other dry bulb temperature. T DB and T W B are sufficient to fix the state of the mixture. The Psychrometric Chart where the dry bulb temperature is the temperature measured by a thermometer place in the mixture and the wet bulb temperature is the adiabatic saturation temperature. 8

9 An Adiabatic Saturator How can we measure humidity? Conservation of Mass ṁ a,1 = ṁ a,3 air (1) ṁ w,1 + ṁ w,2 = ṁ w,3 water (2) Conservation of Energy (ṁh) a,1 + (ṁh) w,1 + (ṁh) w,2 = (ṁh) a,3 + (ṁh) w,3 (3) By definition ω 1 = ω 3 = ) (ṁw ṁ a 1 ) (ṁw ṁ a 3 (4) (5) From (2) and (1) ) (ṁw,2 ṁ a,1 = ṁ w,3 ṁ a,1 }{{} ṁ a,3 ) (ṁw,1 ṁ a,1 = ω 3 ω 1 9

10 Dividing (3) by ṁ a,1 and noting m a1 = m a3 and m w 2 m a1 = ω 3 ω 1 h a,1 + ω 1 h w,1 + (ω 3 ω 1 ) h w,2 = h a,3 + ω 3 h w,3 (6) ω 1 = (h a,3 h a,1 ) + ω 3 (h w,3 h w,2 ) (h w,1 h w,2 ) If we assume: i) air is an ideal gas and (h a,3 h a,1 ) = c pa (T 3 T 1 ) ii) (h w,3 h w,2 ) = h g h f = h fg (T 3 ) iii) h w,1 h g (T 1 ) iv) h w,2 = h f (T 2 ) = h f (T 3 ) then we can write ω 1 as a function of T 1 and T 3 only ω 1 = c p a (T 3 T 1 ) + ω 3 h fg (T 3 ) h g (T 1 ) h f (T 3 ) 10

11 PROBLEM STATEMENT: Outdoor air at T 1 = 10 C and φ = 30% is first to be heated to T 2 = 22 C and then humidified to T 3 = 25 C and φ 3 = 60% at atmospheric pressure. The volumetric flow rate is V = 45 m 3 /min. i) find the heating requirement, Q ii) find the mass flow rate of steam, ṁ steam required to complete the process. iii) what would you expect if liquid water was sprayed instead of steam? PROBLEM STATEMENT: A mixture of 5 kg carbon dioxide and 10 kg nitrogen is at 27 C and kp a. Determine: a) the volumetric analysis [% by volume] of this mixture b) the specific heat of the mixture, c v, c p [kj/kg K] and c v, c p [kj/kmol K] c) the heat transfer [kj] to reduce the temperature to 0 C, if the mixture is confined in a rigid tank d) the increase in availability [kj] of the mixture during this cooling process 11