Exam (elaborations) TEST BANK FOR Fundamentals of Thermodynamics 7th Edition By Claus Borgnakke, Richard E. Sonntag (Solution Manual)

Exam (elaborations) TEST BANK FOR Fundamentals of Thermodynamics 7th Edition By Claus Borgnakke, Richard E. Sonntag (Solution Manual) Borgnakke and Sonntag CONTENT SUBSECTION PROB NO. Concept Problems 1-18 Properties and Units 19-22 Force and Energy 23-34 Specific Volume 35-40 Pressure 41-56 Manometers and Barometers 57-77 Temperature 78-83 Review problems 84-89 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag In-Text Concept Questions Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.a Make a control volume around the turbine in the steam power plant in Fig. 1.1 and list the flows of mass and energy that are there. Solution: We see hot high pressure steam flowing in at state 1 from the steam drum through a flow control (not shown). The steam leaves at a lower pressure to the condenser (heat exchanger) at state 2. A rotating shaft gives a rate of energy (power) to the electric generator set. W T 1 2 2.b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.6 are located and show all flows of energy transfer. Solution: The valve and the cold line, the evaporator, is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room. cb W . Q . Q leak The black grille in the back or at the bottom is the condenser that gives heat to the room air. The compressor sits at the bottom. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.c Why do people float high in the water when swimming in the Dead Sea as compared with swimming in a fresh water lake? As the dead sea is very salty its density is higher than fresh water density. The buoyancy effect gives a force up that equals the weight of the displaced water. Since density is higher the displaced volume is smaller for the same force. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.d Density of liquid water is ρ = 1008 – T/2 [kg/m3] with T in oC. If the temperature increases, what happens to the density and specific volume? Solution: The density is seen to decrease as the temperature increases. Δρ = – ΔT/2 Since the specific volume is the inverse of the density v = 1/ρ it will increase. 2.e A car tire gauge indicates 195 kPa; what is the air pressure inside? The pressure you read on the gauge is a gauge pressure, ΔP, so the absolute pressure is found as ` P = Po + ΔP = 101 + 195 = 296 kPa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.f Can I always neglect ΔP in the fluid above location A in figure 2.12? What does that depend on? If the fluid density above A is low relative to the manometer fluid then you neglect the pressure variation above position A, say the fluid is a gas like air and the manometer fluid is like liquid water. However, if the fluid above A has a density of the same order of magnitude as the manometer fluid then the pressure variation with elevation is as large as in the manometer fluid and it must be accounted for. 2.g A U tube manometer has the left branch connected to a box with a pressure of 110 kPa and the right branch open. Which side has a higher column of fluid? Solution: Since the left branch fluid surface feels 110 kPa and the right branch surface is at 100 kPa you must go further down to match the 110 kPa. The right branch has a higher column of fluid. cb Po Box H Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Concept Problems Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.1 Make a control volume around the whole power plant in Fig. 1.2 and with the help of Fig. 1.1 list what flows of mass and energy are in or out and any storage of energy. Make sure you know what is inside and what is outside your chosen C.V. Solution: Smoke stack Boiler building Coal conveyor system Dock Turbine house Storage gypsum Coal storage flue gas cb Underground power cable W electrical Hot water District heating m Coal m m Flue gas Storage for later Gypsum, fly ash, slag transport out: Cold return m m Combustion air Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2.2 Take a control volume around the rocket engine in Fig. 1.12. Identify the mass flows and where you have significant kinetic energy and where storage changes. We have storage in both tanks are reduced, mass flows out with modest velocities. Energy conversion in the combustion process. gas at high pressure expands towards lower pressure outside and thus accelerates to high velocity with significant kinetic energy flowing out. Control and mixing Combustion Oxydizer Fuel Nozzle High speed flow out Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2.3 Make a control volume that includes the steam flow around in the main turbine loop in the nuclear propulsion system in Fig.1.3. Identify mass flows (hot or cold) and energy transfers that enter or leave the C.V. Solution: The electrical power also leaves the C.V. to be used for lights, instruments and to charge the batteries. 1 Hot steam from generator 1 c Electric power gen. W T cb Welectrical 3 2 5 4 Condensate to steam gen. cold 7 6 Cooling by seawater Borgnakke and Sonntag 2.4 Separate the list P, F, V, v, ρ, T, a, m, L, t, and V into intensive, extensive, and non- properties. Solution: Intensive properties are independent upon mass: P, v, ρ, T Extensive properties scales with mass: V, m Non-properties: F, a, L, t, V Comment: You could claim that acceleration a and velocity V are physical properties for the dynamic motion of the mass, but not thermal properties. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.5 An electric dip heater is put into a cup of water and heats it from 20oC to 80oC. Show the energy flow(s) and storage and explain what changes. Solution: Electric power is converted in the heater element (an electric resistor) so it becomes hot and gives energy by heat transfer to the water. The water heats up and thus stores energy and as it is warmer than the cup material it heats the cup which also stores some energy. The cup being warmer than the air gives a smaller amount of energy (a rate) to the air as a heat loss. Welectric Qloss C B Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.6 Water in nature exists in different phases such as solid, liquid and vapor (gas). Indicate the relative magnitude of density and specific volume for the three phases. Solution: Values are indicated in Figure 2.7 as density for common substances. More accurate values are found in Tables A.3, A.4 and A.5 Water as solid (ice) has density of around 900 kg/m3 Water as liquid has density of around 1000 kg/m3 Water as vapor has density of around 1 kg/m3 (sensitive to P and T) Ice cube Liquid drop Cloud* * Steam (water vapor) can not be seen what you see are tiny drops suspended in air from which we infer that there was some water vapor before it condensed. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.7 Is density a unique measure of mass distribution in a volume? Does it vary? If so, on what kind of scale (distance)? Solution: Density is an average of mass per unit volume and we sense if it is not evenly distributed by holding a mass that is more heavy in one side than the other. Through the volume of the same substance (say air in a room) density varies only little from one location to another on scales of meter, cm or mm. If the volume you look at has different substances (air and the furniture in the room) then it can change abruptly as you look at a small volume of air next to a volume of hardwood. Finally if we look at very small scales on the order of the size of atoms the density can vary infinitely, since the mass (electrons, neutrons and positrons) occupy very little volume relative to all the empty space between them. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.8 Density of fibers, rock wool insulation, foams and cotton is fairly low. Why is that? Solution: All these materials consist of some solid substance and mainly air or other gas. The volume of fibers (clothes) and rockwool that is solid substance is low relative to the total volume that includes air. The overall density is ρ = m V = msolid + mair Vsolid + Vair where most of the mass is the solid and most of the volume is air. If you talk about the density of the solid only, it is high. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.9 How much mass is there approximately in 1 L of engine oil? Atmospheric air? Solution: A volume of 1 L equals 0.001 m3, see Table A.1. From Table A.4 the density is 885 kg/m3 so we get m = ρV = 885 kg/m3 × 0.001 m3 = 0.885 kg For the air we see in Figure 2.7 that density is about 1 kg/m3 so we get m = ρV = 1 kg/m3 × 0.001 m3 = 0.001 kg A more accurate value from Table A.5 is ρ = 1.17 kg/m3 at 100 kPa, 25oC. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.10 Can you carry 1 m3 of liquid water? Solution: The density of liquid water is about 1000 kg/m3 from Figure 2.7, see also Table A.3. Therefore the mass in one cubic meter is m = ρV = 1000 kg/m3 × 1 m3 = 1000 kg and we can not carry that in the standard gravitational field. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.11 A heavy cabinet has four adjustable feet on it. What feature of the feet will ensure that they do not make dents in the floor? Answer: The area that is in contact with the floor supports the total mass in the gravitational field. F = PA = mg so for a given mass the smaller the area is the larger the pressure becomes. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.12 The pressure at the bottom of a swimming pool is evenly distributed. Suppose we look at a cast iron plate of 7272 kg lying on the ground with an area of 100 m2. What is the average pressure below that? Is it just as evenly distributed as the pressure at the bottom of the pool? Solution: The pressure is force per unit area from page 25: P = F/A = mg/A = 7272 kg × (9.81 m/s2) / 100 m2 = 713.4 Pa The iron plate being cast can be reasonable plane and flat, but it is stiff and rigid. However, the ground is usually uneven so the contact between the plate and the ground is made over an area much smaller than the 100 m2. Thus the local pressure at the contact locations is much larger than the quoted value above. The pressure at the bottom of the swimming pool is very even due to the ability of the fluid (water) to have full contact with the bottom by deforming itself. This is the main difference between a fluid behavior and a solid behavior. Iron plate Ground Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2.13 Two divers swim at 20 m depth. One of them swims right in under a supertanker; the other stays away from the tanker. Who feels a greater pressure? Solution: Each one feels the local pressure which is the static pressure only a function of depth. Pocean= P 0 + ΔP = P 0 + ρgH So they feel exactly the same pressure. Borgnakke and Sonntag 2.14 A manometer with water shows a ΔP of Po/10; what is the column height difference? Solution: ΔP = Po/10 = ρHg H = Po/(10 ρ g) = 101.3 × 1000 Pa 10 × 997 kg/m3 × 9.80665 m/s2 = 1.036 m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.15 A water skier does not sink too far down in the water if the speed is high enough. What makes that situation different from our static pressure calculations? The water pressure right under the ski is not a static pressure but a static plus dynamic pressure that pushes the water away from the ski. The faster you go, the smaller amount of water is displaced but at a higher velocity. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.16 What is the smallest temperature in degrees Celsuis you can have? Kelvin? Solution: The lowest temperature is absolute zero which is at zero degrees Kelvin at which point the temperature in Celsius is negative TK = 0 K = −273.15 oC Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.17 Convert the formula for water density in In-text Concept Question “e” to be for T in degrees Kelvin. Solution: ρ = 1008 – TC/2 [kg/m3] We need to express degrees Celsius in degrees Kelvin TC = TK – 273.15 and substitute into formula ρ = 1008 – TC/2 = 1008 – (TK – 273.15)/2 = 1144.6 – TK/2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.18 A thermometer that indicates the temperature with a liquid column has a bulb with a larger volume of liquid, why is that? The expansion of the liquid volume with temperature is rather small so by having a larger volume expand with all the volume increase showing in the very small diameter column of fluid greatly increases the signal that can be read. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Properties and units Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.19 An apple “weighs” 60 g and has a volume of 75 cm3 in a refrigerator at 8oC. What is the apple density? List three intensive and two extensive properties of the apple. Solution: ρ = m V = 0.06 0.000 075 kg m3 = 800 kg m3 Intensive ρ = 800 kg m3 ; v = 1 ρ = 0.001 25 m3 kg T = 8°C; P = 101 kPa Extensive m = 60 g = 0.06 kg V = 75 cm3 = 0.075 L = 0.000 075 m3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.20 A steel cylinder of mass 2 kg contains 4 L of liquid water at 25oC at 200 kPa. Find the total mass and volume of the system. List two extensive and three intensive properties of the water Solution: Density of steel in Table A.3: ρ = 7820 kg/m3 Volume of steel: V = m/ρ = 2 kg 7820 kg/m3 = 0.000 256 m3 Density of water in Table A.4: ρ = 997 kg/m3 Mass of water: m = ρV = 997 kg/m3 ×0.004 m3 = 3.988 kg Total mass: m = msteel + mwater = 2 + 3.988 = 5.988 kg Total volume: V = Vsteel + Vwater = 0.000 256 + 0.004 = 0.004 256 m3 = 4.26 L Extensive properties: m, V Intensive properties: ρ (or v = 1/ρ), T, P Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.21 A storage tank of stainless steel contains 7 kg of oxygen gas and 5 kg of nitrogen gas. How many kmoles are in the tank? Table A.2: MO2 = 31.999 ; MN2 = 28.013 nO2 = mO2 / MO2 = 7 31.999 = 0.21876 kmol nO2 = mN2 / MN2 = 5 28.013 = 0.17848 kmol ntot = nO2 + nN2 = 0.21876 + 0.17848 = 0.3972 kmol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.22 One kilopond (1 kp) is the weight of 1 kg in the standard gravitational field. How many Newtons (N) is that? F = ma = mg 1 kp = 1 kg × 9.807 m/s2 = 9.807 N Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Force and Energy Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag 2.23 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gravitational field? How much mass can a force of 1 N support? Solution: ma = 0 = Σ F = F - mg F = mg = 2 kg × 9.80665 m/s2 = 19.613 N F = mg = m = F g = 1 N 9.80665 m/s2 = 0.102 kg m F g Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which this textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Borgnakke and Sonntag Excerpts from this work may be reproduced by instructors for distribution