Chemical Engineering Thermodynamics 141 - Midterm I Solutions, Exams of Engineering Chemistry

Download Chemical Engineering Thermodynamics 141 - Midterm I Solutions and more Exams Engineering Chemistry in PDF only on Docsity! Name: SID: Discussion Session: Chemical Engineering Thermodynamics 141 -- Fall 2007 Tuesday, October 9, 2007 Midterm I - 70 minutes 110 Points Total Closed Book and Notes (20 points) 1. Consider 1 mole of an ideal gas with constant heat capacities. For each of the following processes, deduce which of the listed properties or quantities remain constant and check those boxes. ∆T ∆P ∆V Q W ∆U ∆H ∆S Reversible adiabatic expansion Reversible isothermal expansion Constant-pressure heating Constant-volume cooling Joule-Thomson process Solution ∆T ∆P ∆V Q W ∆U ∆H ∆S Reversible adiabatic expansion 0 0 Reversible isothermal expansion 0 0 0 Constant-pressure heating 0 Constant-volume cooling 0 0 Joule-Thomson process 0 0 0 0 0 (20 points) 2. In state 1, a piston-cylinder assembly contains 3 kg of saturated steam of quality 0.1 at 45.47 kPa. Heat is added at constant pressure while the piston moves outward. This continues until it reaches state 2 where the quality is 0.3, at which point the piston becomes stuck and cannot move any further. Heat continues to be added at constant volume until the pressure reaches 133.90 kPa (state 3). i) Sketch qualitatively the path on a PV diagram. Label each state 1, 2, and 3. ii) Using the attached data for saturated steam: a. Calculate the quality of the steam at final state (state 3). b. Calculate the net heat added to the system. Solution The first step from quality 0.l to 0.3 (state 1 to 2) is at constant pressure, so the heat is equal to the change in enthalpy. Referring to the saturated steam table: Q = m*(H2 - H1) = (3kg) [0.3(2642.1) + 0.7(330.7) – 0.1(2642.1) – 0.9(330.7) kJ/kg] = 1386.84 kJ = Q(1->2) For state 2 to state 3, the piston is stuck so the volume is constant. We can calculate the specific volume at states 2 and 3 which is the same in both states but we calculate from state 2 since we know the quality at state 2 but not at state 3: V2 = 0.3(3541.3) + 0.7(1.029) cm 3 /g = 1063.11 cm 3 /g = V3 We can now calculate the quality, x, at state 3: V3 = 1063.11 cm 3 /g = x(1288.9) + (1-x)1.050 cm 3 /g => x = 0.825 Since the process from state 2 to 3 is constant volume, the change in internal energy is equal to the heat. So again, using the steam table: Q = m*(U3 – U2) = (3kg) [0.825(2515.7) + 0.175(452.7) – 0.3(2481.1) – 0.7(330.7)] = 3536.565 kJ = Q(2->3) Total Q = Q(1->2) + Q(2->3) = 1386.84 + 3536.565 = 4923.41 kJ 45.47kPa 133.9kPa 1 2 3 V P (40 points) 4. Consider a cycle that consists of the following reversible processes: (a) Isobaric cooling from T1 to T2 (b) Adiabatic compression from T2 to T3 (c) Isobaric heating from T3 to T4 (d) Adiabatic expansion from T4 to T1 Assuming that the working fluid is an ideal gas with constant heat capacities: i) Sketch qualitatively the paths on a PV diagram and a TS diagram. Label each state with 1, 2, 3, and 4 according to their respective temperatures T1, T2, T3, and T4. ii) Calculate the heat and work exchanged (using the sign convention U = Q - W), and the change in the entropy of the system for parts (a) and (b), in terms of the given temperatures. iii) Show that: T2 T1 = T3 T4 Solution i) The Brayton cycle consists of the above processes. P V S T A B C D A B C D 1 2 3 4 1 2 3 4 Gas turbine power plants use the Brayton cycle, often in conjunction with a bottoming Rankine cycle to make use of the energy in the high-temperature exit stream. ii) a. Isobaric Cooling QA = CP (T2 - T1) < 0 W A = PdV Path A ∫ = P R P dT = R(T2 − T1)∫ ∆S = Cp ln(T2/T1) b. Reversible Adiabatic Compression QB = 0 dU = -PdV = CVdT WB = CV(T2 - T3) ∆S = 0 since reversible and adiabatic iii) 4 3 1 2 1 2 4 3 1 4 2 3 1 4 2 3 C A 4 3 4 3 2 1 2 1 4 4 3 3C 2 2 1 1A T T = T T :Therefore V V V V and V V V V :constant a is Since V V V V P P :have we steps reversible adiabatic the From V V T T and V V T T :therefore V T V T R P and V T V T R P :have we steps isobaric the From == γ       =      = == ==== γγ You could also solve this problem using an entropy balance knowing that the overall entropy change for the cycle is zero since entropy is a state function then set delta S for the two isobaric processes equal to find the temperature relation directly => Cp ln(T2/T1) = Cp ln(T3/T4) => T2/T1 = T3/T4 SSI8K LESLS SSIS ROBLES seraa V [=] em? gt U[=] kd kot H [=] kd kort Table F.1, Saturated Steam, $1 Units (Continued) SPECIFIC VOLUME V INTERNAL ENERGY U ENTHALPY H ‘sat. ‘Sat. sat. ‘sat. 427.4 2081.4 2506.8 4275 2251.6 2679.1 : : f a8 ABD8E OE : : : é 2888 i au age $338 Bae SF] kJ kot K+ aR 2 £ #5 #5 80982 Be abd g § $8 2232 Se be > & aes 8288 g Rs