MECH 310 Exam 1, Fall 2009

MECH 310 Thermodynamics I November 20, 2006 American University of Beirut, Fall 2006 Handout # Quiz 1 Quiz 1 Problem 1 (50 points) A large rigid vessel of unknown total volume is divided into two compartments by a partition: The first compartment of volume 0.5 m3 initially contains saturated liquid- vapor mixture of water at 100 ◦C while the second compartment is evacuated. The water in the first compartment is heated while the partition is fixed (not allowed to move). In what follows, sketch all the processes on a p − v and T − v diagram. saturated liquid-vapor mixture initial state: 0.5 m3 100 oC evacuated unknown volume heat fixed partition Figure 1: Schematic for problem 1. (a) If at the end of the heating process, the final state of water is saturated liquid, what are the minimum and maximum possible values of the range of mass of the liquid water at the initial state? What is the corresponding range of the volume of liquid water at the initial state? (b) If at the end of the heating process, the final state of water is saturated vapor, what are the minimum and maximum possible values of the range of mass of the liquid water at the initial state? What is the corresponding range of the volume of liquid water at the initial state? If at the end of the heating process, the final state of water is compressed liquid water at 250 ◦C. At this point, the entire vessel is insulated and the partition separating the two compartments is removed allowing water to fill the entire vessel. At the end of this process, the water temperature dropped by 20 ◦C. (c) What is the initial volume of the vacuum compartment? 1 Problem 2 (50 points) Consider the piston-cylinder arrangement shown in the Figure. The cylinder contains 3.5 grams of air and the piston is tightly sealed so that air cannot escape to the environment. The piston mass is 1 kg and its diameter is 20 cm. Initially air inside the cylinder is in thermodynamic equilibrium and the piston is at an elevation of 10 cm relative to the bottom of the cylinder. Stops in the cylinder at an elevation of 15 cm relative to the cylinder bottom prevent the cylinder from upward motion beyond that elevation. 1 kJ of heat is slowly added via an electric resistance properly situated inside the bottom wall of the cylinder. (a) What are the initial pressure and temperature? (b) Determine the final pressure and temperature. (c) Determine the work done by the gas during the process and the work done against atmosphere. (d) Sketch the process on T − v and p − v diagrams. Assume the following: The piston is frictionless, air may be modeled as an ideal gas, and except for heat addition via the electric resistance no other forms of heat transfer take place across the boundary. Neglect changes in potential energy and kinetic energy of the air inside the cylinder. The ideal gas constant for air is R = 0.287 kJ/kg ◦K. State any additional assumptions you make. 10 cm 15 cm 20 cm Q = 1 kJ g atmosphere 1 bar Figure 2: Schematic for problem 2. 2 MECH 310 Thermodynamics I November 20, 2008 American University of Beirut, Fall 2008 Handout # Quiz 1 Quiz 1 Problem 1 (50 points) A large rigid vessel of unknown total volume is divided into two compartments by a partition: The first compartment of volume 0.5 m3 initially contains saturated liquid- vapor mixture of water at 100 ◦C while the second compartment is evacuated. The water in the first compartment is heated while the partition is fixed (not allowed to move). In what follows, sketch all the processes on a p − v and T − v diagram. saturated liquid-vapor mixture initial state: 0.5 m3 100 oC evacuated unknown volume heat fixed partition Figure 1: Schematic for problem 1. (a) If at the end of the heating process, the final state of water is saturated liquid, what are the minimum and maximum possible values of the range of mass of the liquid water at the initial state? What is the corresponding range of the volume of liquid water at the initial state? (b) If at the end of the heating process, the final state of water is saturated vapor, what are the minimum and maximum possible values of the range of mass of the liquid water at the initial state? What is the corresponding range of the volume of liquid water at the initial state? If at the end of the heating process, the final state of water is compressed liquid water at 250 ◦C. At this point, the entire vessel is insulated and the partition separating the two compartments is removed allowing water to fill the entire vessel. At the end of this process, the water temperature dropped by 20 ◦C. (c) What is the initial volume of the vacuum compartment? 1 Problem 1 Solution Starting with a saturated liquid-vapor mixture (T1, v1, p1), an isochoric heating process may proceed into the subcooled liquid region via a saturated liquid state or into the superheated vapor via a saturated vapor state. Whether the first scenario takes place or the second depends on whether the initial state is on the left or right of the critical point, as shown in the figure. Let vc be the specific volume of the critical point, then if vf(T1) < v1 < vc ⇒ scenario (a) if vc < v1 < vg(T1) ⇒ scenario (b) The ranges of water volume required in parts (a) and (b) correspond to the above ranges of the specific volume. Noting that mf = m(1 − x) = V v � vg−v vfg � , then the required ranges for the mass of liquid are obtained if V vc �vg − vc vfg � < (mf)1 < V vf(T1) ⇒ scenario (a) if 0 < (mf)1 < V vc �vg − vc vfg � ⇒ scenario (b) The specific volume at the critical point may be obtained by observing that at the critical point vf = vg. From the saturation tables of water, vc = 0.003155 m3/kg. At T1 =100◦C, we have : vf(T1) = 0.0010434 m3/kg vg(T1) = 1.6729 m3/kg we get if 158.27 kg < (mf)1 < 479.2 kg ⇒ scenario (a) if 0 kg < (mf)1 < 158.27 kg kg ⇒ scenario (b) second part At the end of the heating process, the water is in the subcooled compressed liquid state. The internal energy is approximated as u2 = uf(T2) = uf(250C) = 1080.7 kJ/kg. Also v2 = vf(T2) = 0.0012509 m3/kg. The mass of water is then m = V2/v2 = 399.71 kg. Applying the first law between states 2 and 3, and noting that the vessel in insulated and that there are no changes in kinetic and potential energy, then ∆U + ∆ K.E. + ∆ P.E. = Q← + W ← ⇒ ∆U = 0 ⇒ ∆u = 0 So at state 3, we have u3 = 1080.7 kJ/kg and T3 = 250 − 20 = 230 C. This a saturated mixture state. From the tables and at T3 = 230 C, we have uf = 987.1 kJ/kg and ufg = 1616.3 kJ/kg, then x3 = 0.0579. The specific volume at state 3 is then v3 = vf + xvfg = 0.0012084 + 0.05791(0.071577 − 0.0012084) = 0.005283 m3/kg. The volume at state 3 is V3 = mv3 = 2.111 m3. So the volume of the initially evacuated compartment is 2.111 − 0.5 = 1.611 m3. 2 Problem 2 (50 points) Consider the piston-cylinder arrangement shown in the Figure. The cylinder contains 3.5 grams of air and the piston is tightly sealed so that air cannot escape to the environment. The piston mass is 1 kg and its diameter is 20 cm. Initially air inside the cylinder is in thermodynamic equilibrium and the piston is at an elevation of 10 cm relative to the bottom of the cylinder. Stops in the cylinder at an elevation of 15 cm relative to the cylinder bottom prevent the cylinder from upward motion beyond that elevation. 1 kJ of heat is slowly added via an electric resistance properly situated inside the bottom wall of the cylinder. (a) What are the initial pressure and temperature? (b) Determine the final pressure and temperature. (c) Determine the work done by the gas during the process and the work done against atmosphere. (d) Sketch the process on T − v and p − v diagrams. Assume the following: The piston is frictionless, air may be modeled as an ideal gas, and except for heat addition via the electric resistance no other forms of heat transfer take place across the boundary. Neglect changes in potential energy and kinetic energy of the air inside the cylinder. The ideal gas constant for air is R = 0.287 kJ/kg ◦K. State any additional assumptions you make. 10 cm 15 cm 20 cm Q = 1 kJ g atmosphere 1 bar Figure 2: Schematic for problem 2. 3 Problem 2 Solution (a) The initial volume of air in the control mass is V1 = h1Ap where the piston area is Ap = πD2/4, where h1 = 10 cm , D = 20 cm, so that V1 = 0.00314159 m3. The initial pressure is obtained by performing force balance on the piston, so that p1 = pa + mpg Ap which for mp = 1 kg and g = 10 N/kg leads to p1 = 101318 pa. The initial temperature is then T1 = p1V1/mR = 316 K. (b) We do not know whether the piston hits the stops or not. We assume it does not hit the stops and we see what happens. Under this assumption we apply the first law of thermodynamics for the control mass between initial and final state, with neglecting changes in kinetic and potential energy, U2 − U1 = Q← + W ← ⇒ mcv(T2 − T1) = Q← − � 2 1 p dV since we assume that during the process the piston does not hit the stops and since the heat addition is slow then p = p1 during the process. Denoting h2 as the piston elevation at the end of the process, and using T2 = p2V2/(mR) = p1Aph2/(mR), then mcv �p1Aph2 mR − T1 � = Q← − pAp(h2 − h1) ⇒ h2 = Q← + mcvT1 + pAph1 pAp(cv/R + 1) We get h2 = 16.13 cm which is larger than 15 cm. So the piston hits the stops. Applying the first law again, but this time with piston hitting the stops. In this case p2 ̸= p1 and h2 = 15 cm and the only work done by the gas is when the piston is rising during which p = p1 so that W → = p1(V2 − V1), then mcv �p2V2 mR − T1 � = Q← − p1(V2 − V1) ⇒ p2 = mR V2 � T1 + Q← − p1(V2 − V1) mcv � = 138.68 kpa The final temperature is T2 = p2V2/(mR) = 650.6 K. (c) The work done by the gas during the process is W → = p1(V2 − V1) = 0.159 kJ. 4 MECH 310 Thermodynamics I November 20, 2007 American University of Beirut, Fall 2007 Handout # Quiz 1 Quiz 1 • This is a closed book exam. • You have 90 minutes. • Write your name and section number on both the question and answer sheets. • It is recommended that you read the whole exam before you start solving. Problem 1 (25 points) Two kilograms of water are contained within a piston-cylinder arrangement. The piston is weightless and undergoes frictionless motion. See Figure 1. At the initial state, there are equal masses of liquid and vapor initially. The system is (process 1) slowly heated to a position where the piston is locked, and then (process 2) the system is cooled to the saturated vapor state at 50 ◦C. The atmospheric pressure is pa = 101.4 kPa. (a) Is process 1 isothermal, isobaric, or isochoric? Why? What about process 2? why? (b) Draw the processes on a p − v diagram, and (c) Evaluate the work done during the process. H2O H2O vacuum H2O pa pa Figure 1: Schematic for problem 1. Problem 2 (25 points) Steam occupies one-half of the partitioned insulated chamber shown in Figure 2. The initial pressure and temperature of the steam are 0.3 MPa and 360 ◦C. The other half of the chamber contains a vacuum. The partition is removed, and the steam fills the entire volume at the end of the process. Determine the final temperature of the steam. 1 H2O H2O vacuum partition Figure 2: Schematic for problem 2. Problem 3 (25 points) The cylinder shown in Figure 3 contains 0.1 kg of water initially of volume 0.005 m3. The spring touches the piston but exerts no force at this initial state. The piston, assumed frictionless, rises owing to heat transfer to the water until the water is in the saturated vapor state. The spring constant is 149.1 kN/m, and the piston area is 0.02 m2, and the mass of the piston is 100 kg. The atmospheric pressure is pa = 100 kPa. (a) Determine the initial pressure and specific internal energy? (b) Determine the final temperature and pressure. (c) Evaluate the heat transfer H2O H2O vacuum partition H2O pa pa Figure 3: Schematic for problem 3. Problem 4 (25 points) Consider a copper sphere of mass m, radius R, constant specific heat c, initially (at time t = 0) of temperature T0. The sphere is then placed in a bath of flowing hot water maintained at temperature Tw > T0. At any time t, the temperature is assumed to be spatially uniform in the sphere. The convection heat transfer coefficient between the flowing water and the sphere is h. Assume that the work done in expanding the sphere is negligible and neglect changes in potential and kinetic energy. (a) What is the internal energy of the sphere as a function of given quantities and the sphere temperature T(t)? (b) What is the rate of heat gained by the sphere in terms of given quantities and T(t) ? (c) Using the first law of thermodynamics, find T(t). 2 MECH 310 Thermodynamics I November 20, 2008 American University of Beirut, Fall 2008 Handout # Quiz 1 Quiz 1 Solution • This is a closed book 90 minutes exam. You allowed to bring one A4 hand written cheat sheet which will be collected at end of exam. It is recommended that you read the whole exam before you start solving. • Write your name and section number on both the question and answer sheets. • Clearly identify your control mass / control volume. Problem 1 (25 points) Consider the counter-flow heat exchanger shown in Figure 1 operating at steady condi- tions. Heat is exchanged between two streams of fluid moving opposite to each other. The mass flowrate of the hot fluid is ˙mh = 10 kg/s and that of the cold fluid is ˙mc = 5 kg/s. The inlet and outlet specific enthalpies of the hot fluid are hh,1 = 3000 kJ/kg and hh,2 = 2500 kJ/kg. The inlet specific enthalpy of the cold fluid is hc,1 = 1500 kJ/kg. In this problem changes in kinetic and potential energies are neglected. (a) Assuming that the heat exchanger is perfectly insulated, what is the outlet specific enthalpy of the cold fluid hc,2? (b) What is the rate of entropy transport into the heat exchanger across its boundary due to heat and work? (c) Assuming specific entropy is known at both inlets and both outlets, sh,1, sh2, sc,1, sc,2, are given. What is the rate of entropy generation by irreversibility ˙Sirr ? Is ˙Sirr positive, negative, zero? Explain. cold stream hot stream hh,1, sh,1 hh,2, sh,2 hc,1, sc,1 hc,2, sc,2 , ˙mh , ˙mc control volume Figure 1: Schematic for problem 1. Problem 1 Solution (a) First law for a control volume at steady state, insulated Q← = 0, and in absence of all kinds of work, and neglecting changes in kinetic and potential energy ˙mc(hc,2 − hc,1) + ˙mh(hh,2 − hh,1) = 0 ⇒ hc,2 = hc,1 + ˙mh ˙mc (hh,1 − hh,2) ⇒ hc,2 = 1500 + 2(3000 − 2500) = 2500 kJ/kg 1 (b) Entropy transport due to work is zero. so ˙S← W = 0 ALWAYS. Since there is not heat transfer, entropy transport is due to heat zero; ˙S← Q = 0. (c) Entropy balance ˙mc(sc,2 − sc,1) + ˙mh(sh,2 − sh,1) = ˙Sirr It must be positive because the process is irreversible due to the heat transfer from the hot stream to the cold stream. In order to reverse this heat transfer, we must put work, which permanently changes the environment. Problem 2 (25 points) Consider the power generation engine shown in Figure 2. The engine operates in a ther- modynamic cycle. (Thermodynamic cycle: final state is identical to initial state.) During this cycle, the engine produces work W → cycle while receiving heat Q← H for a high-temperature reservoir. The entropy transported into the engine due to heat transfer Q← H is S← H . (a) Relate W → cycle to Q← H . (b) By performing entropy balance, show that the above cyclic process is impossible. (c) Explain that the only way the entropy balance for the cyclic engine is satisfied, the engine must reject entropy S→ to the surrounding. How much should S→ be? How can it be practically done ? cyclic engine high temperature reservoir Q← H W → control mass Figure 2: Schematic for problem 2. Problem 2 Solution (a) For an engine operating in a cycle ∆E = 0 since E2 = E1, then W → = Q←. (b) Entropy balance S2 = S1, so that S2 − S1 = S← QH + Sirr ⇒ 0 = S← QH + Sirr Since both terms on the right hand side of the above equation are positive, their sum cannot be zero. 2 (c) We need to remove entropy by rejecting heat Q→ L to the surrounding so that entropy balance becomes 0 = S← QH − S→ QL + Sirr ⇒ S→ QL = S← QH + Sirr It is impossible for a cyclic engine to produce work while interacting with only one thermal reservoir. Problem 3 (25 points) We consider isothermal expansion of 10 kg of air in the the piston cylinder arrangement of Figure 3. Air is modeled as an ideal gas with R = 0.287 kJ/kg.K. During the process from initial (equilibrium) state 1 to final (equilibrium) state 2, heat is added to the gas by conduction across a wall of thickness h = 0.5 cm, area A = 0.1 m2, and thermal conductivity k = 50 W/(m.K). On the gas side of the conducting wall, the gas temperature is T = 50 ◦C and on the outer side of the wall, the temperature is To = 100 ◦C . (a) If heat is conducted steadily through the wall into the gas from time t1 = 0 to times t2 = 10 sec, what is the heat gained by gas Q←? [Hint: conduction heat flux is q = −k∇T and assume linear temperature distribution on conducting wall.] (b) Can we neglect changes in kinetic energy and changes in potential energy, i.e. KE2- KE1 and PE2-PE1? Why? (c) What is the work done by the gas during the expansion process? If the initial volume is V1 = 0.01 m3, what is the final volume V2, temperature T2, and pressure p2? piston conduction heat transfer Q← ideal gas Temperature T Temperature To h area A control mass Figure 3: Schematic for problem 3. Problem 3 Solution (a) The heat gain per unit area per unit time is q = k(To − Ti)/h = 50(100 − 50)/0.005 = 500 kJ/m2.s. The rate of heat gain is then qA = 500×0.1 = 50 kW. The heat transferred over the period of 10 s is then Q← = qA(t2 − t1) = 500 kJ. (b) The changes in potential and kinetic energy may be neglected if |∆K.E.| << Q← 3 and |∆P.E.| << Q←. Some of you stated that they should be less that ∆U, which is zero in our problem, the process being isothermal. So this last statement is wrong. For equilibrium or quasi-equilibrium initial and final states, the gas velocities are small, and when multiplied by the gas density, which is small, yield small |∆K.E.|. Note that for (exact) equilibrium states, K.E.1 = 0 and K.E.2 = 0. As for changes in potential energy |∆P.E.| = ρg(V2 − V1) which for small ρ and |V2 − V1| may be neglected with respect to Q←. (c) First law of thermodynamics for a control mass between state 1 and state 2 is E2 − E1 = Q← + W ← Since the process is isothermal U2 − U1 = 0, neglecting changes in kinetic and potential energy, we get W → = Q← = 500 kJ. The work done by a system during an isothermal process is W → = mRT ln V2 V1 ⇒ V2 = V1e W → mRT = 0.01 × 1.715 = 0.0175 m3. Ideal has law leads to p2 = 54055 pa. Problem 4 (35 points) In this problem, we investigate the thermodynamics of friction within a fluid. In par- ticular, we consider a layer of lubricant (oil) of viscosity µ and density ρ. The oil layer separates a fixed wall (which could be the cylinder wall) from a moving wall (which could be the piston wall). The problem is steady and the upper wall velocity is V , and the oil layer thickness is h. The velocity distribution inside the oil may be obtained from fluid mechanics to be u(y) = V y h. (a) The friction (viscous) force per unit area acting on (the lower side of) the moving wall is τw = −µ (∂u/∂y)y=h. In which direction does this force pull the upper wall with respect to it motion? Explain the molecular origins of this force. (b) What is the rate of work done on the oil assuming that the dimensions of the oil layer in the x and z are respectively 1 m and 1 m. (c) If the oil enters the left side (i.e. at x = 0) at temperature T, what is its temperature at exit (i.e. at x = 1 m). Assume the oil layer to be thermally insulated at the upper and lower wall sides. Assume also that the specific heat of oil, cv, does not change over the ranges of temperatures involved in the problem. (d) Given initial and final equilibrium states, is the work done by the friction force path- dependent? Support your answer by one or more scenarios. (e) Is this process reversible? Explain. Problem 4 Solution (a) Friction pulls the wall in a direction opposite to its motion. Friction is due to the tangential component of momentum transfer between adjacent layers of molecules that due to their temperature are randomly oscillating about a mean position. (b) The rate work done on oil per unit area is given by ˙W A = 1 A ∂ ∂t � F · dx = � F A · dV Notice that the force per unit area on the oil is opposite to the force per unit area on the upper wall (action/reaction) so that F/A = −τwˆx. The fluid velocity at y = h is V = V ˆx 4 piston moving wall fixed wall oil, viscosity µ density ρ wall velocity V x y oil velocity V y h h control volume Figure 4: Schematic for problem 4. so that ˙W A = −τwV = µV �∂u ∂y � y=h = µV 2 h The rate of work done on the oil is then ˙W ← = ˙W A A = µV 2A h where A = 1 m2. (c) For the control volume chosen to contain the oil of thickness h and extending from x = 0 to x = 1 m, applying the first law for a control volume and noting that the flow is steady, and noting that changes in kinetic energy and potential energy between inlet and outlet is zero, then ˙W ← = ˙m(h2 − h1) where ˙m = ρV h/2. Considering oil to be compressed liquid for which h ≃ h(T), then h2 − h1 = cp(T2 − T1) so that T2 = T1 + 2 ˙W ← ρV hcp = T1 + 2 µV ρcph2 where it was assumed that cp does not change over the range [T1, T2]. (d) The work done by friction on the oil is path dependent. We could replace the work effect by heat transfer into the oil that yields the same temperature rise. In the first scenario ˙W ← = µV 2/h, Q← = 0 and in the second scenario ˙W ← = 0, Q← = µV 2/h. In both cases we have the same inlet and outlet states. (e) The process is irreversible because if we isolate the system, which amounts to stopping the upper wall, there will internal heat transfer in the oil until a final uniform temperature is reached at equilibrium. 5