MECH 310 Exam 1, Fall 2009

MECH 310 Thermodynamics I November 25, 2009 American University of Beirut, Fall 2009 Handout # Quiz 1 Quiz 1 • This is an open book 90 minutes exam. • It is recommended that you read the whole exam before you start solving. • Make sure that the units are consistent. • Write your name and section number on both the question and answer sheets. • Clearly identify your control mass / control volume. • State any assumptions you need and provide a convincing justification. Problem 1 (40 points) A rigid cylinder of volume 1 m3 contains 1 kg of a saturated liquid+vapor water mixture at 100 C. The water undergoes a process with a final state of saturated vapor. The cylin- der is fitted with a piston which is either free to move or held stationary, depending on the process. For each of the processes (a), (b) and (c): (1) Find the initial and final states, work produced by and heat transferred into water, (2) Sketch the process on a p − v and T − v diagrams, (3) Explain whether the piston is free to move or held stationary. (a) An isobaric process. (b) An isothermal process. (c) An isometric process. 1 kg water V1 = 1 m3 pa depending on process, piston is either free to move or held stationary Final State: saturated vapor Figure 1: Schematic for problem 1. 1 Problem 2 (30 points) A well-insulated rigid tank contains 5 kg of saturated liquid+vapor mixture of water at 100 kPa at an initial unknown quality, x1. An electric resistor, placed in the tank, is connected to a 110-V source and a current of 8 A flows through the resistor when the switch is turned on. It took 10 minutes to completely vaporize the liquid. Determine the following: (a) The volume of the tank in m3. (b) The initial quality x1 (c) Sketch the process on a p-v diagram Note that when electric current I flows through a resistor, it dissipates heat at a rate of I V (Watts), where V is the potential difference across the resistor. Note: This problem may require an iterative solution. Do not carry out more than two iterations. 5 kg water p1 = 100 kPa initial quality x1 8 A 110 V Figure 2: Schematic for problem 2. 2 Problem 3 (30 points) The rigid container shown in Figure 1 consists of two air compartments separated by a piston of area A = 250 cm2. The piston conducts heat but experiences no friction with the container walls. Compartment B is thermally insulated from the surrounding. The piston is supported by linear spring with a spring constant k = 100 kN/m. Initially the temperature in both compartments is 300 K and the volumes of the two compartments are VA1 = 4 liters and VB1 = 6 liters. The mass of air in compartment A is 20 grams and that in compartment B is 10 grams. Air is assumed to behave as an ideal gas with cv = 0.72 kJ/kg.K and R = 0.287 kJ/kg.K. (a) Find the initial air pressure in each compartment. (b) Is the spring in the initial state under tension or compression? Find l1 − l0 where l1 is length of spring at the initial state and l0 is its length when it is under no stress. Heat is slowly added to compartment A (while compartment B is kept insulated) until the final volume of air in compartment B is VB2 = 5.5 liters. (c) Find final temperature and pressure in each compartment, TA2, pA2, TB2, pB2. (d) Find the work produced by air in both compartments W → A+B and the heat added Q← A . A B Figure 3: Schematic for problem 3. 3 Problem 1: Solution Initial state: T1 = 100C, v1 = V1/m = 1 m3/kg. From tables p1 = 1 atm, vf = 0.0010435 m3/kg, vg = 1.673 m3/kg, so that x1 = (v − vf)/vfg = 0.597. We can then get u1 = uf + xufg = 418.94 + 0.597(2506.5 − 418.94) = 1665.2 kJ/kg (a) For an isobaric process, p = p1, and in particular the final state is saturated va- por at 100 C and 1 atm, so that v2 = vg = 1.673 m3/kg. The work done is W → = p(V2 − V1) = p(mv2 − V1) = 101.4(1.673 − 1) = 68.24 kJ. The change in internal en- ergy is ∆U = m(u2 − u1) = 1(2506.5 − 1665.2) = 841.28 kJ. The heat added is then Q← = ∆U + W → = 909.52 kJ. To maintain a constant pressure, the piston must be free to move with a negligible acceleration. (b) Same as for part (a). Why? (c) Final state is saturated vapor with vg = 1m3/kg. From the tables T2 = 116 C, p2 = 175 kPa and u2 = 2524 kJ/kg. Since both mass and specific volume are constant during this process, the volume must be constant which requires the piston to be held stationary. The produced work is zero W → = 0 and the heat added to the water is Q← = ∆U = 1(2524 − 1665.2) = 860 kJ. Problem 2: Solution Applying first law for control mass consisting of water inside the container, and noting that the volume is constant (p−v work is zero), no heat is transferred across the boundary, and electric work crosses the boundary ∆U = W ← electrical (1) or m(u2 − u1) = IV t Initial state is saturated liquid/vapor at 100 kPa so that u1 = (uf + xufg)1 = 417.36 + x1(2506.1 − 417.3) = 417.36 + 2088.8x1. Final state is saturated vapor, with v2 = v1 = (vf +xvfg)1 = 0.0010431+x1(1.694−0.0010431) = 0.001+1.693x1. So the first law looks like 5(ug(vg = 0.001 + 1.693x1) − (417.36 + 2088.8x1)) = 8(110)(600)/1000 or 2088.8x1 = (ug(vg = 0.001 + 1.693x1) − 522.96 Let’s start with an initial guess x1 = 0.5, then vg = 0.001 + 1.693x1 = 0.8475 m3/kg, then u2 = 2533 kJ/kg, updating x1 = (2533 − 552.96)/2088.8 = 0.95. Then vg = 0.001 + 1.693x1 = 1.606 m3/kg, then u2 = 2505 kJ/kg. Updating x1 = (2505 − 552.96)/2088.8 = 0.934. 4 Problem 3: Solution (a) Ideal gas law pV = mRT for each compartment yields pA1 = 430.5 kPa and pB1 = 143.5 kPa. (b) Force balance on the piston, assuming zero acceleration, yields pA = pB + k(l0 − l) A so that l0 − l = A(pA − pB)/k which for the initial state gives l0 − l = 7.2 cm. Since l = VB1/A, we get l0 = 31.2 cm. (c) Force balance on the piston yields dpA = dpB − k A2dVB ⇒ (pA − pB) = (pA − pB)1 − k A2(VB − VB1) (2) First law for compartments A and B respectively ∆UA = Q← A − W → A ∆UB = −W → B or mAcv(T2 − T1) = Q← A − � VA2 VA1 pA dVA mBcv(T2 − T1) = − � VB2 VB1 pB dVB Noting that dVA = −dVB, and adding the two equations, we get (mA + mB)cv(T2 − T1) = Q← A + � VB2 VB1 (pA − pB) dVB So the net work output from air in both compartments is W → A+B = − � VB2 VB1 (pA − pB) dVB = − � VB2 VB1 � (pA − pB)1 − k A2(VB − VB1) � dVB = 0.1235 kJ Ideal gas law at state 2, RT2 = pA2VA2 mA = pB2VB2 mB ⇒ pA2(V − VB2) mA = pB2VB2 mB ⇒ pA2 = pB2 VB2 V − VB2 mA mB leading to pA2 = 2.444pB2. Substituting in Equation (2), pA2 − pB2 = (pA − pB)1 − k A2(VB2 − VB1) = 367 kPa resulting in pB2 = 254 kPa and pA2 = 621 kPa. Using ideal gas law, we get T2 = 487 K. Finally first law, yields Q← A = (mA + mB)cv(T2 − T1) + W → A+B = 4.16 kJ 5