Using first law of thermodynamics for a constant pressure system: Using first law of thermodynamics for a constant volume system:

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1 TUTORIAL-10 Solution (19/04/2017) Thermodynamics for Aerospace Engineers (AS1300) Properties of ideal gas, their mixtures and adiabatic flame temperature For air take c v = kj/kg K and c p = kj/kg K and γ =1.4 Q1. 12 kg of certain ideal gas (molecular weight 25 kg/kmol) at the same temperature and pressure is contained separately in a frictionless piston cylinder assembly (piston mass unknown) and a rigid container. It is desired to raise the temperature of both systems by 15 C. What is the difference the amount of heat that has to be supplied to the systems? Solution Given, mass of a certain ideal gas M = 12 kg is raised to a temperature by 15 C. The molecular weight of the gas is 25 kg/kmol (MW gas ) Two different systems was considered. First system is a piston cylinder assembly container. Here pressure of the system remain constant throughout the process in which temperature of the system is raised by 15 C. Using first law of thermodynamics for a constant pressure system: δq = U + δw = C v T + Pdv = C v T + R gas T = C P T = H (units J/kg) [1.1] gas So, in a constant pressure process the heat given to the system can be calculated from the change in enthalpy across the process. The second system is a rigid cylinder. The volume of the container remains constant. So, it is a constant volume process. There is no work interaction in this process as the volume doesn t change across a process. Using first law of thermodynamics for a constant volume system: δq = U + δw = C v T + Pdv = C v T + 0 = C v T = U (units J/kg) [1.2] gas So, the difference of the amount of heat that has to be supplied to the systems can be calculated using eqn [1.1] and [1.2]. δq diff = H U = M gas (C P gas C vgas ) T = M gasr gas T [1.3] R δq diff = M gas R gas T = M universal gas T = = kj Ans MW gas 25

2 Q2. The molar analysis of a gas mixture at 350 K, 100 kpa is 70 % N2, 20 % CO2 and 10 % O2. Determine (a) the analysis in terms of mass fractions, (b) partial pressure of each component, and (c) the volume occupied by 25 kg of mixture. Solution Given P = 100 kpa, T = 350 K, X N2 = 0.7, X CO2 = 0.2, X O2 = 0.1. (a) Mass fraction Y i = (X i W i ) ( j X j W j ), where W i is the molecular weight of species i and the molecular weight of the mixture is W mix = j X j W j. Using the above formula, calculate W mix = 31.6 g/mol. Also, Y N2 = , Y CO2 = and Y O2 = (b) Partial pressure of gas i, P i = X i P. Therefore, P N2 = 70 kpa, P CO2 = 20 kpa and P O2 = 10 kpa. (c) Given m = 25 kg. We have calculated the molecular weight to be W mix = 31.6 g/mol. Using the perfect gas law, V = mr mix T P = mr u T P W mix, where R u is the universal gas constant. Substitute the values to obtain V = m 3.

3 Q3. 10 kg of a mixture having an analysis on a mass basis of 50 % N2, 30 % CO2 and 20 % O2 is compressed adiabatically from 100 kpa, 280 K to 500 kpa, 387 K. Determine the work interaction. The specific heats at constant volume of the three components are respectively 747, 750 and 696 J/kg K. Solution Given mass of the mixture m = 10 kg. The mass based fraction of the components of the mixture is also given. So the mass fraction (Y i ) of the components of the mixture can be obtained as follows: Y N2 = 0.5 Y CO2 = 0.3 Y O2 = 0.2 The corresponding mass of the components in the mixture can be obtained by multiplying mass fraction (Y i ) with the total mixture mass (m). The calculation is as follows: m N2 = m Y N2 = = 5 kg m CO2 = m Y N2 = = 3 kg m O2 = m Y N2 = = 2 kg The specific heats at constant volume for the components of the mixture are as follows: J C v = 747 N 2 kg K J C v = 750 CO 2 kg K J C v = 696 O 2 kg K The gas is compressed adiabatically. The initial and final temperature and pressure are given below. P initial = 100 kpa & T initial = 280 K P final = 500 kpa & T final = 387 K Using first law of thermodynamics for an adiabatic process where heat transfer to the system is zero, the expression becomes: δq = U + δw = C v T + Pdv = 0 (unit J/kg) [3.1]

4 δw = U = mc v T (unit J) [3.2] There are two method to calculate the total work interaction. The first method is to find the specific heat at constant volume for the mixture (C v mix ). The expression for the specific heat at constant volume for mixture is as follows: C v mix = Y ic v i i = = J kg K Note: If the specific heats for the components are given per mass basis then multiply with mass fraction (Y i ) similar to the above method to get the mixture specific heats. But if the specific heats are specified per Kmol basis then multiply by the corresponding mole fraction (X i ) to get the mixture specific heats. So, using the method 1 the work interaction of the system is as follows: δw = U = mc v T = ( ) = J mix Ans The second method is to calculate the work interaction for each component of the mixture. The mass of the components of the mixture has been already calculated in the above steps. The corresponding specific heats can also be used to calculate the answer. δw = U = (m N2 C v N 2 T + m N2 C v N 2 T + m N2 C v N 2 T) = ( ) ( ) = J Ans

5 Q4. Calculate specific molar enthalpy of equimolar mixture of N2 and H2O (vapour) at 100 C. The heat of formation of H2O is kj/kmol. The Cp,H2O = 40 kj/kmol K and Cp,N2 = 33 kj/kmol K Solution Given T = 373 K, h f,h2 O 0 = kj/kmol, C P,H2 O = 40 kj/kmol K and C P,N2 = 33 kj/kmol K. h N2 = h 0 f,n2 + C P,N2 (T 298) = C P,N2 (T 298) = kj/mol Similarly, h H2 O = h f,h2 O 0 + C P,H2 O(T 298) = 239 kj/mol Also, it is given that X N2 = 0.5 and X H2 O = 0.5. Therefore, h mix = X N2 h N2 + X H2 Oh H2 O = kj/mol. Thus, specific molar enthalpy of the mixture = kj/mol.

6 Q5. Two gases A and B are separated by a wall in a rigid container. The initial temperature and pressure of the gases in both the container are equal. There are na and nb moles of gases are present in the container A and B respectively. The wall separating the gases is now removed and the gases are allowed to mix. The gases in the containers are non-reacting gases. Calculate change in entropy for the above mixing process. Solution: There are two methods in which this question can be solved. First method is to derive the expression for entropy from first and third law of thermodynamics. Using first law of thermodynamics: δq = U + δw = C v T + Pdv (unit J/kg) [5.1] As it is mentioned in the question that the gases are non-reacting gases which suggests that the temperature of the system will not change during mixing. The final temperature will be same as the initial temperature. Using this information in eqn [5.1] the expression can be simplified as follows: δq = U + δw = 0 + m Pdv = PdV (unit J) [5.2] where v is specific volume (m 3 /kg) and V is volume (m 3 ) From third law of thermodynamics following expression can be obtained: S = δq reversible T [5.3] δq reversible = T S [5.4] So, using eqn [5.2] and [5.4] the expression for entropy change can be written as follows: ds = PdV = n R universal V dv [5.5] The expression in the eqn [5.5] was obtained by using ideal gas law of PV=nRuniversalT. Let us take the initial volume of gas A and B in the container were VA & VB respectively. Using ideal gas law VA & VB can be expressed in terms of pressure (P), temperature (T) and no. of moles of gas A and B (na and nb). PV A = n A R universal T [5.6]

7 PV B = R universal T [5.7] From eqn [5.6] and [5.7] the following expression can be obtained. V A V B = n A [5.8] V A +V B V B = n A+ [5.9] V A +V B V A = n A+ n A [5.10] The eqn [5.5] can be used to calculate the total entropy change in the mixing process. Individual contribution from each gas can be summed to obtain this. The gas A will go from the initial volume of VA to the final volume of VA+VB. Similarly, the gas B will go from the initial volume of VB to the final volume of VA+VB. So, the final answer is as follows: ds Total = ds A + ds B = V A +V B V A n A R universal V dv + V A +V B V B R universal V dv S Total = S A + S B = n A R universal ln ( V A + V B ) + n V B R universal ln ( V A + V B ) A V B Using the relationship given in the eqn [5.9] and [5.10] the final answer becomes as follows: S Total = S A + S B = n A R universal ln ( n A+ ) + n R universal ln ( n A+ ) Ans A The second method is to use the formula for entropy change in a reversible process derived in the class room. There are two forms of the formula. In the expression where volumes are need to be put use the values mentioned in the above method. For the pressure part, take the partial pressure of the gas A and B to calculate the individual contributions from the gases. S A = n A C v ln ( T ) + n T A R universal ln ( V A+V B ) = n A C P ln ( T ) n T A R universal n A ln ( P A ) = n P A R universal ln ( ) = n n A +n A R universal ln ( n A+ B V A n A ) In the above expression P A P = n A n Total = n A n A +

8 Similarly, for gas B S B = C v ln ( T ) + n T B R universal ln ( V A+V B ) = C P ln ( T ) n T B R universal ln ( P B ) = n P B R universal ln ( ) = n n A + R universal ln ( n A+ B V B ) In the above expression P A P = n A n Total = n A n A + So, the total entropy change is as follows: S Total = S A + S B = n A R universal ln ( n A+ ) + n R universal ln ( n A+ ) Ans A

9 Q6. Consider the combustion of a) hydrogen and air (consider only oxygen and nitrogen) and b) hydrogen and oxygen (no nitrogen), with initial reactants at temperature of 298K and total pressure of 1 atmosphere. Assuming constant pressure combustion and that there is no dissociation of products. Compute the adiabatic flame temperatures for the following reaction/s φh (O N 2) φh 2 O + ( 1 2 φ 2 ) O N 2 for φ 1 (1) φh (O N 21 2) H 2 O + (φ 1)H N 42 2 for φ > 1 (2), where is the equivalence ratio Take heat of formation, H kj / kmol and constant specific heat, f H 2 O C 40 kj / kmolk, C 30 kj / kmolk, C 35 kj / kmolk and C 33 kj / kmolk P( H2O) P( H2 ) P( O2 ) P( N2 ) Solution The process is adiabatic and the pressure remains constant. Therefore, enthalpy is a constant. h reactants = h products (3) Enthalpy of a species i is given by h i = h f,i 0 + C P,i (T 298) (4) where h i is the enthalpy per mole of the species. We know that h f,o2 0 = h f,n2 0 = h f,h2 0 = 0 and the initial temperature T i = 298 K. Therefore h reactants = 0 in this problem. Our aim is to find the final temperature T f, which is the adiabatic flame temperature. (a) Now, for φ 1, use equation (4) in equation (3) to obtain 0 = φ{ h f,h2 O 0 + C P,H2 O(T f 298)} + ( 1 2 φ 2 ) { h f,o C P,O2 (T f 298)} { h f,n C P,N2 (T f 298)} Rearrange the above equation and obtain the following expression for T f φ h f,h2 O 0 T f = [φc P,H2 O + ( 1 2 φ 2 ) C P,O C P,N 2 ] Substitute the value of φ and evaluate T f. For φ > 1, proceed as above using equation (2) to obtain the expression for T f as given below.

10 T f = h f,h2 O 0 [C P,H2 O + (φ 1)C P,H C P,N 2 ] (b) When nitrogen is absent, one can use the above expressions for T f without the term C P,N 2 in the denominator. Both formulas satisfy for φ = 1 : T f = K in air and T f = 6344 K in pure oxygen.