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Unformatted text preview: 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. i. Starting at 227oC, 1000 kPa with a volume of 0.1 m3 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 Pv1.4 = constant, ending with a pressure of 200 kPa which is state 3. a. b. c. d. What is the mass of air and what is the pressure at state 2. Find the work and heat transfer needed in the process from state 1 to state 2. What is the final volume and the temperature at state 3. 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 Cp0 = 1.004 & Cv0 = 0.717. 2. [35%] A piston cylinder has the water volume separated into VA = 0.2 m3 and VB = 0.3 m3 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 contains R-134a at 15oC, x = 0.6 and a volume of 0.02 m3. It is heated up to 60oC at which point the specific volume is 0.03002 m3/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 Problem 2: Figure for Problem 3: P o P o cb mp A:H2O B:H2O g R-134a SOLUTIONS Problem I C.V. air Energy: control mass. Air assumed as ideal gas. E2 - E1 = m(u2 - u1) = 1Q2 - 1W2 ....[3pts]
1W2 Process equation: V = C; State 1: m = = P dV = 0 ....[3pts] PV 1000 0.10 = = 0.697 kg ....[3pts] RT 0.287 500 State 2: V2 = V1; m2 = m1 P2 = P1T2 / T1 = 10001500/500 = 3000 kPa [3pts] Work done during the constant volume process from state 1 to state 2 is zero. Heat transfer from the energy equation becomes:
1Q2 = m(u2 - u1) = mCv(T2 - T1) = 0.6970.717(1500 500) = 499.75 kJ [5pts] The polytropic expansion process from state 2 to state 3: P2V21.4 = P3V31.4 V3 = V2 (P2/P3)1/1.4 = 0.1(3000/200)0.7143 = 0.692 m3 ....[4pts] P3V3 = mRT3 T3 = P3V3/mR T3 = (2000.692)/(0.6970.287) = 691.87 K ....[4pts] Now the work term becomes:
2 W3 = 1 (P3V3 - P2V2 ) = mR (T3 - T2 ) (1 - n) (1 - n) ....[5pts]
0.697 0.287 (691.87 - 1500) = 404.14kJ = (1 - 1.4) Heat transfer from the energy equation becomes
2Q3 = m(u3 - u2) + 2W3 = mCv (T3 - T2) + 2W3 = 0.6970.717(691.87 1500) + 404.14 = 0.27 kJ 0 (adiabatic) ....[5pts] Problem II Take the water in A and B as CV. Continuity: m2 - m1A - m1B = 0 Energy: Process: m2u2 - m1Au1A - m1Bu1B = 1Q2 - 1W2 P2 = Peq = constant = P1A as piston floats and mp, Po do not change State 1A: Two phase. Table B.1.2 v1A = 0.001127 + 0.75 0.19332 = 0.146117 m3/kg, u1A = 761.67 + 0.75 1821.97 = 2128.15 kJ/kg State 1B: Table B.1.3 v1B = 0.14184 m3/kg, u1B = 2692.26 kJ/kg => m1A = V1A/v1A = 1.3688 kg, m1B = V1B/v1B = 2.115 kg State 2: 1000 kPa, 200oC sup. vapor => v2 = 0.20596 m3/kg, u2 = 2621.9 kJ/kg m2 = m1A + m1B = 3.4838 kg => V2 = m2v2 = 3.4838 0.20596 = 0.7175 m3 So now
1W2 = 1Q2 = P dV = Peq (V2 - V1) = 1000 (0.7175 - 0.5) = 217.5 kJ m2u2 - m1Au1A - m1Bu1B + 1W2 = 3.4838 2621.9 - 1.3688 2128.15 - 2.115 2692.26 + 217.5 = 744 kJ Problem III Take CV as the R-134a. m2 = m1 = m ; m(u2 - u1) = 1Q2 - 1W2 State 1: T1, x1 => Two phase so Tbl B.5.1: P1 = Psat = 489.5 kPa v1 = vf + x1 vfg = 0.000805 + 0.6 0.04133 = 0.0256 m3/kg u1 = uf + x1 ufg = 220.1 + 0.6 166.35 = 319.91 kJ/kg State 2: (T, v) Superheated vapor, Tbl B.5.2. P2 = 800 kPa m3/kg, u2 = 421.2 kJ/kg V2 = mv2 = 0.78125 0.03002 = 0.02345 m3 Work is done while piston moves at linearly varying pressure, so we get
1W2 = P dV = area = Pavg (V2 - V1) = 0.5(P2 + P1)( V2 - V1) = 0.5 (489.5 + 800)(0.02345 - 0.02) = 2.22 kJ Heat transfer is found from the energy equation
1Q2 = m(u2 - u1) + 1W2 = 0.78125 (421.2 319.91) + 2.22 = 81.36 kJ P 2 P2 R-134a P 1 1
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This note was uploaded on 05/05/2008 for the course ME 235 taught by Professor Borgnakke during the Winter '07 term at University of Michigan.
- Winter '07