Instructor: Hacker Engineering 232 Sample Exam 1 Solutions Answer Key
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Instructor: Hacker Engineering 232 Sample Exam 1 Solutions Answer Key
Instructor: Hacker Engineering 232 Sample Exam 1 Solutions Answer Key 1. a b c d ○ e 8. a b c a 2. ○ b c d e 9. a b c ○ a 3. ○ b c d e 10. a b c 4. a b c ○ d e 11. a b ○ 5. a b c ○ d e 12. a b 6. a b ○ c d e 13. a b ○ 7. a b c ○ d e 14. a b c c c c d ○ e 15. a b c ○ d e d e 16. a b c ○ d e d ○ e 17. a b ○ c d e 18. a b c ○ d e d e d ○ e 19. a b ○ c d e d e 20. a b c ○ d e d ○ e 21. a b c ○ d e Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.2 Chapter 1: Introduction to Thermodynamics Introduction to Thermodynamics Introductory Concepts and Definitions Problem 1. Consider a refrigerator in a kitchen. Take the refrigerator and everything in it to be our system. Which best describes the system’s surroundings? (a) All of the air in the kitchen. (b) Any one standing in the kitchen. (c) The air inside the refrigerator. *(d) Everything in the universe external to the system. Problem 2. Circle the true statement: *(a) a closed system is also called a control mass (b) a closed system is also called control volume (c) a closed system is an isolated system (d) an open system is also called control mass Problem 3. Which of the following is not an extensive property? *(a) Density (b) Volume (c) Mass (d) Energy (e) None of these Problem 4. Identify the SI of the expression P V , where P is pressure and V is volume. (a) Pa (b) Pa · m2 *(c) J (d) N (e) None of these Problem 5. Determine whether equations (i) and (ii) below are dimensionally consistent. 1 2y (ii) x = t v + at , (i) g = 2 t 2 where x = distance, y = distance, v = velocity, t = time, and a = acceleration. (a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent (b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent *(c) (i) and (ii) are both dimensionally consistent (d) (i) and (ii) are both dimensionally inconsistent Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.3 Static Fluids For all problems below assume the system is in static equilibrium. Problem 6. The mass of an unknown gas mixture in a room that is 5 m × 10 m × 10 m is known to be 500 kg. What is the specific volume of the gas? (a) 1 m3 *(b) 1 m3 /kg (c) 500 m3 (d) 500 m3 /kg (e) None of these Solution: v≡ length × width × height 5 · 10 · 10 m3 m3 volume = = =1 mass mass 500 kg kg Problem 7. A large storage tank is 10 m tall. The top of the storage tank is open to the atmosphere. What is the gage pressure at the bottom of the tank? Take the density of water to be ρ = 1000 mkg3 and g = 10 sm2 . (a) 1 kPa (b) 10 kPa 2 *(c) 10 kPa (d) 103 kPa (e) None of these Solution: Use the equation Pgage = ∆P = Pabs − Patm = ρgh = (103 m N kg )(10 2 )(10m) = 105 2 = 105 Pa = 102 kPa 3 m s m Problem 8. A submarine has a circular top hatch with a 1 m diameter. What is the force exerted on the hatch from the pressure due to the water above the hatch when the sub dives down to a depth of 200 m? Take the approximate density of sea water to be ρ = 1000 mkg3 and g = 10 sm2 . Note: we are ignoring atmospheric pressure Patm in this problem and we are assuming the submarine is a rigid body (i.e., we can ignore the pressure from inside the sub). (a) 2π × 105 N (b) π × 106 N (c) 2π × 106 N *(d) π2 × 106 N (e) None of these Solution: First we’ll find the pressure on the hatch, then we’ll use the definition of pressure to find the force Fhatch = Phatch Ahatch , where Fhatch is the force on the hatch from the pressure of the sea water above the hatch, Phatch is the pressure on the hatch, and Ahatch is the area of the hatch. Phatch = ρgh = (103 kg N m )(10 2 )(2 × 102 m) = 2 × 106 Pa = 2 × 106 2 . 3 m s m Using the definition of pressure we find: Fhatch = Phatch Ahatch = 2 (πrhatch ) Phatch 2 N 1 π =π m2 2 × 106 = × 106 N 2 2 m 2 Problem 9. The basic barometer can used to measure the height of a building. To do this a measurement is made at the bottom of the building at ground level and a second Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.4 measurement is taken at the top of the building on the roof. One can then relate these readings to pressure using the equation for a mercury-filled barometer: Patm (hHg ) = ρHg ghHg , where ρHg = 13, 600 kg/m3 is the density of mercury, g is the acceleration due to gravity, and hHg is the height of the mercury column in the barometer measured in millimeters. The approximate height of the building can then be found by making a few simplifying assumptions. For moderate sized buildings a reasonable assumption for the average density of air under fair weather conditions is approximately ρair ≈ 1.18 Kg/m3 . If the barometric readings at the top and bottom of a certain building are 730 mmHg and 755 mmHg respectively, what is the height of the building in meters? (a) 50 m (b) 120 m *(c) 280 m (d) 320 m (e) None of these Solution: See handwritten solutions Temperature Use the equations TC = TK − 273 and TF = 59 TC + 32 to answer the following two questions. Round your answers to the nearest integer. Problem 10. Use the equation TF = 95 TC + 32 to convert 98◦ F to ◦ C. (a) 32◦ C (b) 208◦ C (c) 20◦ C *(d) 37◦ C (e) None of these 5 5 5 5 5 Solution: TC = (TF −32) = (98−32)◦ C = ·66◦ C = ·22◦ C = (35+ )◦ C ≈ 37◦ C 9 9 9 3 3 Chapter 2: The first law of thermodynamics Work, Power, Heat, and Energy Problem 11. A skyrocket is fired upward. As it rises, the work done by gravity on the rocket is: (a) positive *(b) negative (c) zero Solution: The rocket is moving upward; gravity is pulling downward. Since the force is in the opposite direction of the motion, the work done by the force is negative. Problem 12. How much power, in kilowatts, must be developed by the electric motor of a 1600-kg car moving at 26 m/s on a level road if the forces of resistance total 720 N? Round your answer to the nearest tenth of a kilowatt. (a) 95.2 kW (c) 133.2 kW (e) None of these (b) 36.3 kW *(d) 18.7 kW Solution: Use the equation for power as a product of force times velocity: P = F v. Then P = Fmotor · vcar = (720N)(26m/s) = 18.7 kW. To find Fmotor , draw a free-body diagram and apply Newton’s second law together with the fact that the car is moving at a constant speed (so it’s not accelerating) to arrive at Fmotor = fk . See my solutions on the web for the details. Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.5 Problem 13. Ricky, the 5 kg snowboarding raccoon, is on his snowboard atop a frictionless ice-covered quarterpipe with radius 3 meters (see figure at right). If he starts from rest at the top of the quarter-pipe, what is his speed at the bottom of the quarter-pipe? For ease of calculation, assume that g = 10 m/s2 . Round your answer to the nearest 0.1 m/s. (a) 7.0 m/s (c) 8.5 m/s (e) None of these *(b) (d) x Raccoon: 5 kg 3m 3m 7.7 m/s 9.3 m/s Solution: We are given: m = 5 Kg, R = 3 m, vi = 0 m/s, and g = 10 m/s2 . We want: vf =? We assume: no friction, hence the mechanical energy of the system is conserved. Since the system, Ricky and the earth, has no external forces on it, such as friction, we can apply the conservation of mechanical energy: MEi = MEf . Now, the initial potential due to gravity is Ugrav = mgh = mgR and the initial kinetic energy is Ki = 0 since Ricky is at rest. The final P.E. is zero, since Ricky is now at ground level h = 0. The final K.E. is Kf = 12 mvf2 . Using the definition of mechanical energy: ME = PE + KE gives 1 mgR = PEi + KEi = MEi = MEf = PEf + KEf = mvf2 2 ÷1m 2 −−−− −−→ vf2 = 2Rg √ p 1/2 −−−−−−→ vf = 2Rg = (2)(3 m)(10 m/s2 ) = 7.7 m/s Notice that since R = h, this is the same speed that Ricky would have if he had jumped straight down (go look back at our 1-D kinematic problems). The only difference is the direction of his speed would have been different. If he had jumped straight down, then his velocity would also have been straight down, and into the ground would go Ricky; but the quarter pipe redirected his motion to being tangent to the ground, while preserving his speed. So instead of busting his head, and winding up dead; Ricky sped off to work, where he was a clerk. Were it not for conservation of mechanical energy he would have been late for his job, and replaced by his competitor Rob. Problem 14. If heat is added to a system and the temperature of a system increases, without knowing anything else, which form of energy will be definitely increase? (a) The kinetic energy of the system (b) The potential energy of the system (c) The work done by the system *(d) The internal energy of the system Solution: If the temperature increased, then the internal energy must have increased. That’s all that we can say for sure. The only way the system’s kinetic energy could be increased is if the bulk motion of system were increased, and we have no such information. A similar statement goes for the potential of the system. We also don’t know if the system does any work. Unless the system were a gas in a frictionless piston-cylinder mechanism, we’d have no way of knowing if the system did work owing to it’s increase in internal energy. Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.6 Problem 15. (The sign convention for heat transfer and work) Consider a closed system. Suppose that heat Q is added to the system and work W is done on the system. Use the standard sign conventions for heat and work transfer given in class (and in the book) to determine which of the following combinations is correct. Circle the correct answer. (a) W > 0 and Q > 0 (b) W > 0 and Q < 0 *(c) W < 0 and Q > 0 (d) W < 0 and Q < 0 Solution: By convention: W > 0 if the system does work on the surroundings (environment) and Q > 0 if heat is added to the system. The sign convention for thermodynamics is different from the sign convention for physics. In physics, if energy flows into the system, then the energy is added to the system and the system’s energy increases. This is defined to be the positive direction for energy transfer for both heat and work. The positive direction in thermodynamics differs from physics. The positive direction for heat flow in thermodynamics is identical to the positive direction in physics, but the positive direction in thermodynamics for work is the opposite direction as that for physics. The positive direction for energy transfer in thermodynamics comes from thermo’s history in the study of the steam engine. Thus the positive direction is the natural direction for the flow of energy in a steam engine. Heat is added to the cylinder full of steam (the system), hence Q > 0 if steam is added to the system. However, the hot steam is used to push a piston, so the work is done by the system on the piston (not part of the system). Therefore, W > 0 if the work is done by the system, not on the system. Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.7 Work done on and by a thermodynamic system For the following problems use the general formula for work done by a thermodynamic system: Z Vf P (V ) dV (0.1) W = Vi where W = work, Vi = initial volume, Vf = final volume, and P (V ) = pressure. Notice that, in general, pressure will be a function of volume. All of these problems will involve questions for a gas contained within a closed frictionless piston-cyclinder assembly. Recall: a polytropic process is one that satisfies P V n = constant for some value of n and for all time during the process. Problem 16. A gas is contained within a closed frictionless piston-cyclinder assembly. The system undergoes a constant-pressure process at 5 bar beginning at an initial volume Vi = 1 m3 and ending in a final volume Vf = 3 m3 . Using the standard sign convention for work, determine how much work was done by the system. (a) 10 kJ (b) 100 kJ *(c) 1000 kJ (d) 200 kJ (e) None of these Solution: Since the process is an isobaric process (constant pressure) we have P (V ) = P0 for all time t ∈ [ti , tf ]. The integral for work becomes: Z Vf W = P (V ) dV Vi Z Vf = P0 dV (since P (V ) = P0 ) Vi Z Vf = P0 dV Vi = P0 (Vf − Vi ) = P0 ∆V Now ∆V = 3 m3 − 1 m3 = 2 m3 , and P0 = 5 bar = 5 × 105 Pa = 5 × 105 N . Thus m2 5 N W = P0 ∆V = 5 × 10 2 (2 m3 ) = 106 N · m = 106 J = 1000 kJ m Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.8 Problem 17. A vertical, frictionless piston-cylinder device contains a gas that is in equilibrium with the weight of the piston balanced by the internal pressure for the gas. The mass of the piston is 4 kg. The atmospheric pressure outside the cylinder is Patm = 105 Pa. The area of the piston is A = 20 cm2 . Determine the pressure of the gas contained in the cylinder Pgas . Take the magnitude of gravity g = 10 m/s2 . (a) 102 kPa *(b) 120 kPa (c) 10,020 Pa (d) 100,200 Pa (e) None of these Solution: Since the piston is in mechanical equilibrium, the net force on the piston must be zero. The upward force is supplied by the gas inside the piston-cylinder device, and the downward for on the piston comes from the pull of gravity and the atmospheric pressure. Let m be the mass of the piston, A be the surface area of the face of the piston, Fgas be the force of the gas, and Fatm be the force due to atmospheric pressure. Then, the force balance becomes: ÷A −−−−−→ Fgas = Fatm + mg mg Pgas = Patm + A (4kg)(10 m/s2 ) 5 = 10 Pa + 2 m2 103 = 105 Pa + 2 × 104 Pa = (100 + 20) kPa = 120 kPa Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.9 Heat transfer Modes: Conduction, Radiation, and Convection You may find at least one of the following formulas useful (in all formulas below A is area and T is temperature): ∆T • For heat transfer by conduction use Fourier’s Law: Q˙ x = ±κ A , where κ is L the thermal conductivity coefficient, L is the thickness, and ∆T is the temperature difference. • For heat transfer by radiation use Stefan-Boltzmann Law: 4 4 ], where 0 ≤ ε ≤ 1 is the emissivity and σ = − Tsurroundings Q˙ e = ±εσA[Tbody −8 W 5.67 × 10 m2 ·K4 is the Stefan-Boltmann constant. • For heat transfer by heat transfer by free and forced convection use Newton’s Law of cooling: Q˙ c = ±hA[Tbody − Tsurroundings ], where h is the heat transfer coefficient. Each formula contains a ± and it is your job to determine the correct sign for the problem at hand. In general, the sign depends on the direction of the energy transfer with respect to the chosen system. Problem 18. A large 1-cm thick plate glass window in a restaurant has a thermal W . This coefficient is valid over a temperature range of 280 - 320 conductivity of 23 m·K kelvin. The outside temperature is a sweltering 313 K (104◦ F), while the restaurant is kept at a comfortable temperature of 293 K (68◦ F). Assuming steady-state conditions, it can be shown that the temperature profile through the glass is linear. Assuming a linear profile, and taking the inside of the restaurant as our system, what is the rate of heat transfer via conduction through an area 1-m2 area of the plate glass? (a) 1 kW (b) 3/2 kW *(c) 3 kW (d) 7 kW (e) None of these Solution: Since energy flows from hot to cold, energy is flowing into the restaurant. That is, energy is flowing into the system. Since heat is defined to be positive if energy flows into the system, and since ∆t > 0, it follows that Q˙ x > 0. If we take ∆T = Tout −Tin , then ∆T > 0. Thus the correct form of the equation is T − T out in > 0. Q˙ x = κA L We are given the following information: W κ = 23 m·K A = 1 m 2 given: L = 1 cm = .01 m ∆T = Tout − Tin = (313 − 293) K = 20 K Q˙ x = 3 W 2 m·K 2 1m 2 · 10 K 1 m 102 = 3 × 103 W = 3 kW Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.10 Power and Refrigeration Cycles Problem 19. Smokey the choo-choo train is an old steam engine with a heat-transfer input of Qin = 100 kJ and a heat-transfer output of Qout = 70 kJ. What is the thermal W of the power cycle? efficiency η = Qcycle in (a) .7 *(b) .3 (c) .7 kJ (d) .3 kJ (e) None of these Solution: Using an energy balance we see that Qin = Wcycle + Qout −Qout −−−−−−−→ Wcycle = Qin − Qout Qout 7 3 Wcycle =1− =1− = ⇒ η= Qin Qin 10 10 Problem 20. Only one of the following statements is true. For the figure below draw arrows that indicate the positive direction for Qin , Qout , and Wcycle for a power cycle, then use your graph to determine the true statement. Hot Body System Cold Body Figure 1: Power Cycle Template (a) Heat flows from the cold body into the system and the “exaust” exits into to the hot body; the system uses the input heat to do work. (b) Heat flows from the cold body into the system and the “exaust” exits into to the hot body; the system uses the input work to extract heat. *(c) Heat flows from the hot body into the system and the “unused portion” exits into to the cold body; the system uses the input heat to do work. (d) Heat flows from the hot body into the system and the “unused portion” exits into to the cold body; the system uses the input work to extract heat. Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.11 Problem 21. A gas is contained within a closed frictionless piston-cyclinder assembly. Take the gas to be the system under investigation. In a careful experiment heat is added to the system in such a way that the system remains in quasi-equilibrium and the pressure satisfies a polytropic process of the form P V n = constant over the time that the experiment is conducted. Take n > 0 to be known. The gas expands from an initial volume Vi to a final volume Vf with Vi < Vf . The initial pressure was measured to be Pi and the final pressure was measured to be Pf . Derive a formula and use it to determine the work done by the system if Vi = 1 m3 , Pi = 100 kPa, Pf = 0.5 Pi and n = 0.5. Circle your answer. (a) 50 J (b) -50 J *(c) 200 kJ (d) -200 kJ (e) None of these Solution: See handwritten notes on the website for the derivation of the formulas: n Pf Vi n n = (∗) ⇒ Pi Vi = Pf Vf Vf Pi " (n−1) # Vi Pi Vi 1− , for n 6= 1 . (∗∗) Wsys = n−1 Vf Rewriting equation (∗) as Vi Vf (n−1) = Pf Pi (n−1) n and substituting it into (∗∗) yields Wsys " # (n−1) Pi Vi Pf n = 1− n−1 Pi (∗ ∗ ∗) Using the initial conditions: Vi = 1 m3 and Pi = 100 kPa (Pi Vi = 100kJ), and the parameter values: Pf = 0.5 Pi , n = 0.5 yields " (−1) # 1 = 200 kJ . Wsys = −2 · 100 kJ 1 − 2 Notice that the work is positive. This agrees with our sign convention, since the closed system must do work on the environment in order to expand the volume.