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# ME235_Exam2_F03.pdf

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Department
Mechanical Engineering
Course
MECHENG 235
Professor
Donald Siegel
Semester
Winter

Description
ME 235 EXAM 2, NOVEMBER 20, 2003 A. ATREYA and C. BORGNAKKE, ME Problem 1 (25 Points) A 5 kg granite rock was placed over a fire and warms to 200 C. A pot (neglect mass) o with cold water at 15 C is now placed on top of the rock piece and it should be brought to the boiling T. Assume the fire is put out and we have no external heat transfer. a) What is the mass of water I can have in the pot? b) Find the entropy generation in the overall process. Problem 2 (35 Points) A piston/cylinder device has 0.01 kg of saturated liquid ammonia at 0 C. It is now expanded in a reversible constant pressure process to a final state of quality 50%. The heat transfer is from a Carnot cycle device with the other side connected to the ambient at 25 C. a) Find the total heat transfer to the ammonia. b) Find the work the ammonia gives out. c) Find the work to/from the Carnot cycle device d) Find the total entropy generation in the process. Problem 3 (40 Points) A piston cylinder contains 0.01 kg air at 290 K, 96 kPa. A reversible ad iabatic compression brings it to 1/10 of the initial volume. After this process combustion adds 1000 kJ/kg of energy (assume heat transfer) in a constant P process to state 3. a) plot the two process in a P-v and T-s diagram. b) Find P and T for state 2. c) Find T for state 3. d) Find the work in the two processes. ME 235 EXAM 2, SOLUTION NOVEMBER 20, 2003 A. ATREYA and C. BORGNAKKE, ME Problem 1 Energy Eq.: m ( u - u ) + m ( u - u ) = 0 - 0 rock 2 1 H2O 2 1 m rock rock T2- T1)rock+ m H2O ( uf2 uf1 H2O = 0 Entropy Eq.: m ( s - s ) + m ( s - s ) = 0 + S rock 2 1 H2O 2 1 1 2 gen From the energy eqution we get: T - T m = m C 1 2 = 5 × 0.89 × 200 - 100 = 1.25 kg H2O rock rockuf2 uf1 418.91 - 62.98 T 2 373.15 s ( 2 - 1 rock= Crockln T = 0.89 × ln473.15= -0.2113 kJ/kg K 1 s ( 2 - 1 H2O = sf2 s f11.3068 – 0.2245 = 1.0823 kJ/kg K 1 2 gen= 5 × (-0.2113) + 1.25 × 1.0823 = 0.296 kJ/K Problem 2 CV Ammonia (CM) Energy Eq.: m( u2- u1) =1 2- W1 2 2 dQ Entropy Eq.: m( s 2 s 1 =  + 1 2 gen ⌡1 T Process: Reversible => S = 0 1 2 gen
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