MECH 310 Exam 1 with Solution, Fall 2008

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