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Mechanical Engineering
Donald Siegel

ME 235 EXAM 1, OCT 12, 2004 A. Atreya and C. Borgnakke Exam Rules: Open Book, Not Open Notes. Name: __________________________________ I have observed the honor code and have neither given nor received aid on this exam. Signature:____________________________________________ 1. [35%] Air sucked into an internal combustion engine piston cylinder arrangement essentially goes through two different processes. 3 i. Starting at 227 C, 1000 kPa with a volume of 0.1 m the air is heated up to 1500K in a constant volume process by combustion ending at state 2. ii. Now the air is expanded in a polytropic process with Pv 1.= constant, ending with a pressure of 200 kPa which is state 3. a. What is the mass of air and what is the pressure at state 2. b. Find the work and heat transfer needed in the process from state 1 to state 2. c. What is the final volume and the temperature at state 3. d. Find the work and heat transfer in the process from state 2 to state 3. You may use constant specific heats to solve this problem C p01.004 & C = v0717. 3 2. [35%] A piston cylinder has the water volume separated into VA = 0.2 m and VB = 0.3 m by a stiff membrane. The initial state in A is 1000 kPa quality x = 0.75 and in B it is 1600 kPa and 250 C. Now the membrane ruptures and the water comes to a uniform state with 200 C. a) Find the final pressure. b) Find the final volume c) Find the work in the process. d) Find the heat transfer in the process 3. [30%] A piston cylinder arrangement with a linear spring acting on the piston o 3 o contains R-134a at 15 C, x = 0.6 and a volume of 0.02 m . It is heated up to 60 C at which point the specific volume is 0.03002 m 3/kg. a) Find the final pressure. b) Find the final volume c) Find the work in the process. d) Find the heat transfer in the process Figure for Figure for Po P Problem 2: o Problem 3: cb m p A:H2O g B:H2O R-134a SOLUTIONS Problem I C.V. air Æ control mass. Air assumed as ideal gas. Energy: E2- E1= m(u -2u )1= Q1-2W 1.[2pts] Process equation: V = C; 1W 2 ∫P dV = 0 ….[3pts] PV 1000×0.10 State 1:m = = = 0.697 kg ….[3pts] RT 0.287×500 State 2: V2= V 1 m =2m ⇒ P1= P ×T2/ T1= 1200×1500/500 = 3000 kPa [3pts] Work done during the constant volume process from state 1 to state 2 iszero. Heat transfer from the energy equation becomes: Q = m(u - u ) = mC (T - T ) = 0.697×0.717×(1500 – 500) = 499.75 kJ [5pts] 1 2 2 1 v 2 1 The polytropic expa
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