CHBE 244: Chemical and Biological Engineering Thermodynamics I Assignment 2 Solution: First law of thermodynamics I Problems: (4 Questions in total) Q1 Internal energy and enthalpy changes of steam (40 marks) Steam undergoes a state change from 600°C and 5 MPa to 350? and 0.4 MPa. Determine AH and AU using the following methods: (a.) Steam table data (b.) Ideal gas assumptions (Be sure to use the ideal gas heat capacity for water.) 1. Solution (40 marks) (a.) From steam table, · At state 1, it is a superheated steam. From superheated steam table, at 600°? and 5 MPa, H1 = 3666.8 KJ/kg, U1 = 3273.3 KJ/kg · At state 2, it is also a superheated steam. From superheated steam table, at 350°C and 0.4 MPa, H2 = 3170 KJ/kg, U2 = 2884.4 KJ/kg AU = U2 - U1 = (2884.4 - 3273.3) = - 388.9 KJ/kg (1) AH = H2 - H1 = (3170 - 3666.8) = - 496.8 KJ/kg (2) 1
(b.) Ideal gas assumptions (Be sure to use the ideal gas heat capacity for water.) Temperature 2 Intermediate 3 (Final) 1 (Initial) Pressure Figure 1: Superheated steam cooling pathway . This can be divided into 2 processes: Isothermal contraction from state 1 to state 2 to an intermediate at state 2 then followed by isobaric cooling from state 2 to state 3 (Figure 1) · For step 1 (1 -> 2): Since it is a constant temperature process, the changes in enthalpy AHI = 0, same as for the changes in internal energy AUI = 0 · Determine the specific volume at intermediate state 2 (V2): cm3 = 18148 V2 = P2 RT2 0 cm3 . MPa (600 9- 273.19 = 0.01815 m3 mol mol (3) = 8.314 K . mol . For step 2 (2 -> 3): It is a constant pressure process, assuming the water as ideal gas (remember this is a gas phase water), by using the heat capacity polynomial equations, we can determine the changes in enthalpy (AHII ) · From the thermochemical data table in Lira's textbook appendix E.I., the capacity constant of water in gas phase are: A = 32.24, B = 0.001924, C = 1.055E-05 and D =- 3.596E-09 AHII = H18 - Hig Z_T3 T2 CpdT = Z T3 T2 (A+ BT + CT2 + DT3)dT (4) 2 = AHII = A(T3 - T2) + 3 B (T2 - T?) + (T3 - T?) + (T} - T24) AHI= (32.24)(623.15 - 873.15) + 0.001924 2 (623.152 - 873.152)+ 1.055E - 05 (623.153 - 873.153) + -3.596E - 09 4 (5) (623.154 - 873.154) (6) 3 =- 9522.91 gmol J 2
· Determine the specific volume at state 3 (V3): V3 = P 0 = 8.314 K . mol cm3 . MPa (350(9-424593)K) m3 = 0.01295 = 12952 cm3 mol mol (7) · For changes in internal energy (AUn ) in step 2, we can use eqn. (8.) to determine: 0 I AUT= AFRI -[(PV)3- (PV)2] = m9522.91 Pa ?