CHBE 244: Chemicaland Biological Engineering Thermodynamics I Tutorial 3: Second law of thermodynamics Feb 2024 Instructions: Have with you a copy of the STEAM TABLES PROBLEM 1 Consider a piston-cylinder setup that contains 1.5 moles of isopentane (ideal gas with temperature independent properties) at 1 MPa and 100°C. Consider expansion to 0.1 MPa and 50°C. Calculate: (i). The total entropy changes (ii). Analyze this process as a combination of an isothermal (first) and isobaric (second) step to calculate the expansion-contraction work, WEc and the thermal energy needed Q in each step (iii). Calculate the totaAU PROBLEM 2 A cylinder fitted with a movable piston contains water at 3 MPa, 50% quality at which point the volume is 20L. The water now expands to 1.2 MPa as a result of receiving 600kJ of heat from a large source at 300lt is claimed that the water does 124kJ of work during the process. (i). Calculate the mass volume, the mass of water, and intensive internal energy , entropy at the initial state (1). (ii.) Calculate the temperature and the intensive internal energy , entropy at the final state (2). (iii.). Is the process feasible (perform an entropy analysis). PROBLEM 3. Consider a piston-cylinder assembly that initially contains 0.5kg of steam & and 10MPa. For the isothermal expansion to a final pressure of 0.1MPa determine (i.). What is the maximum possible work in kJ and what is the entropy change of the environment in kJ/K? (ii) Repeat (i) using the ideal gas model. Compare the results 1
PROBLE 1 (solution) IT T=373K P=1 MPa T2=3;23 K Pz = p.1 M Pa ... n=1.5 moles of isopentane ( ideal gas) From the Table (back cover of textbook Cp= 14.28 R for this cnp (i) ?S = Glu Tz/T1 - RMu Bz = 14.28 R M /-Rm Br P AS = Q.2475R = 2.0585 J/mol AS = NAS > DS = 3,088 J/K R= 8.31447 J/mol.K (ii) Isothermal AU = Q+WEL Q =- WEC AU=0 WEL =- pdV 1nt, WEC= RT lu P/p1, P1=P2 WEG =- 858.86 R or WEC =7,141 J/mol WE = nWEC = WEC =- 10,711.5 J Q=+10,711.5J Iso baric, The energy balance reduces to dH=Q Q= Cp (T2-T1) => Q=14.28Rx(-50)=Q=5,936.5 J/mol Q=nQ = =- 8,904.8 J Note T11 = T1 For isobaric WEC =- P2 (V2-V1)= - P2 (RTZ _RTI) P=P2 P2 WEC =- R (T2-T1) = +50R = 415.7 J/mol WEL=nWEL7 WEC= 623.6 J (iii) DU = Q +WEL => DU = (10,711.5-8,904.8)+ AU = - 8,280.8 J (-10,71.5+623.6) or largely due to temperature change AV=nCv(T2-T1) 2
PROBLEM 2 (solution) QH P = 3MPa 9, =0.5 Y = 20L W == - 124 KJ P2 = 1.2 MPa Q = 600KJ TH=300? (i) Sat. steam table at P1=3 MPa, V= 0.001217 m/kg uh=1004.69 KJ/Kg, U=2603.16 KJ/ V = 0.0667 m'/kg sh= 2.6456 KJ/Kgk, 5"=6.1856 KJ/Kg.K. Kg V1= (1-9) v+qVv => V1=0.034 m3/kg v=(1-9)0+quv =>V1=1803.93 KJ/Kg S = (1-9) 5+ qs = == 4.4156 KJ/kg 3 Mass of water m =