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Borgnakke Sonntag Fundamentals of Thermodynamics SOLUTION MANUAL CHAPTER 1 8e Updated June 2013  Borgnakke and Sonntag CONTENT CHAPTER 1 SUBSECTION...

Borgnakke Sonntag Fundamentals of Thermodynamics SOLUTION MANUAL CHAPTER 1 8e Updated June 2013  Borgnakke and Sonntag CONTENT CHAPTER 1 SUBSECTION PROB NO. Concept Problems 1-21 Properties, Units and Force 22-37 Specific Volume 38-44 Pressure 45-61 Manometers and Barometers 62-83 Energy and Temperature 84-95 Review problems 96-101 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 1.a Make a control volume around the turbine in the steam power plant in Fig. 1.2 and list the flows of mass and energy that are there. Solution: We see hot high pressure steam flowing in 1 at state 1 from the steam drum through a flow control (not shown). The steam leaves at a lower pressure to the condenser (heat WT exchanger) at state 2. A rotating shaft gives a rate of energy (power) to the electric generator set. 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 1.b Take a control volume around your kitchen refrigerator and indicate where the components shown in Figure 1.3 are located and show all flows of energy transfers. Solution: The valve and the. The black grille in Q leak Q cold line, the the back or at the evaporator, is bottom is the inside close to the condenser that inside wall and gives heat to the usually a small room air. blower distributes cold air from the. The compressor freezer box to the W sits at the bottom. refrigerator room. cb 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 1.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 1.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. ∆ρ = – ∆T/2 Since the specific volume is the inverse of the density v = 1/ρ it will increase. 1.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, ∆P, so the absolute pressure is found as P = Po + ∆P = 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 1.f Can I always neglect ∆P in the fluid above location A in figure 1.13? 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. 1.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: Box Since the left branch fluid surface Po feels 110 kPa and the right branch surface is at 100 kPa you must go further down to match the 110 kPa. H The right branch has a higher column of fluid. 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 1.1 Make a control volume around the whole power plant in Fig. 1.1 and with the help of Fig. 1.2 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: Boiler Smoke building stack Coal conveyor system Storage gypsum cb Coal storage flue gas Turbine house Dock Combustion air Flue gas Welectrical Underground power cable m Storage for later transport out: m District heating: Gypsum, fly ash, slag Coal Cold return m Hot water m 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 1.2 Make a control volume around the refrigerator in Fig. 1.3. Identify the mass flow of external air and show where you have significant heat transfer and where storage changes. The valve and the. The black grille in Q leak Q cold line, the the back or at the evaporator, is bottom is the inside close to the condenser that inside wall and gives heat to the usually a small room air. blower distributes cold air from the. The compressor freezer box to the W sits at the bottom. refrigerator room. cb The storage changes inside the box which is outside of the refrigeration cycle components of Fig. 1.3, when you put some warmer mass inside the refrigerator it is being cooled by the evaporator and the heat is leaving in the condenser. The condenser warms outside air so the air flow over the condenser line carries away some energy. If natural convection is not enough to do this a small fan is used to blow air in over the condenser (forced convection). Likewise the air being cooled by the evaporator is redistributed inside the refrigerator by a small fan and some ducts. Since the room is warmer than the inside of the refrigerator heat is transferred into the cold space through the sides and the seal around the door. Also when the door is opened warm air is pulled in and cold air comes out from the refrigerator giving a net energy transfer similar to a heat transfer. 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 1.3 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 1.4 A tray of liquid water is placed in a freezer where it cools from 20oC to -5oC. Show the energy flow(s) and storage and explain what changes. Inside the freezer box, the walls are very cold as they are the outside of the evaporator, or the air is cooled and a small fan moves the air around to redistribute the cold air to all the items stored in the freezer box. The fluid in the evaporator absorbs the energy and the fluid flows over to the compressor on its way around the cycle, see Fig. 1.3. As the water is cooled it eventually reaches the freezing point and ice starts to form. After a significant amount of energy is removed from the water it is turned completely into ice (at 0oC) and then cooled a little more to -5oC. The water has a negative energy storage and the energy is moved by the refrigerant fluid out of the evaporator into the compressor and then finally out of the condenser into the outside room air. ©C. Borgnakke 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 1.5 The overall 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 m msolid + mair ρ=V= V solid + 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 1.6 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 1.7 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 1.8 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 drops falling Cloud* * Steam (water vapor) cannot 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 1.8 What is the approximate mass of 1 L of gasoline? Of helium in a balloon at To, Po? Solution: Gasoline is a liquid slightly lighter than liquid water so its density is smaller than 1000 kg/m3. 1 L is 0.001 m3 which is a common volume used for food items. A more accurate density is listed in Table A.3 as 750 kg/m3 so the mass becomes m = ρ V = 750 kg/m3 × 0.001 m3 = 0.75 kg The helium is a gas highly sensitive to P and T, so its density is listed at the standard conditions (100 kPa, 25C) in Table A.5 as ρ = 0.1615 kg/m3, m = ρ V = 0.1615 kg/m3 × 0.001 m3 = 1.615 × 10-4 kg 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 1.9 Can you carry 1 m3 of liquid water? Solution: The density of liquid water is about 1000 kg/m3 from Figure 1.7, 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 1.10 A heavy refrigerator has four height-adjustable feet. 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 1.11 A swimming pool has an evenly distributed pressure at the bottom. Consider a stiff steel plate lying on the ground. Is the pressure below it just as evenly distributed? Solution: The pressure is force per unit area from page 13: P = F/A = mg/A The steel plate 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 actual plate area. Thus the local pressure at the contact locations is much larger than the average indicated 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. Steel plate Ground 1.12 What physically determines the variation of the atmospheric pressure with elevation? The total mass of the column of air over a unit area and the gravitation gives the force which per unit area is pressure. This is an integral of the density times gravitation over elevation as in Eq.1.4. To perform the integral the density and gravitation as a function of height (elevation) should be known. Later we will learn that air density is a function of temperature and pressure (and compositions if it varies). Standard curve fits are known that describes this variation and you can find tables with the information about a standard atmosphere. See problems 1.28, 1.64, and 1.95 for some examples. 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 1.13 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= P0 + ∆P = P0 + ρgH So they feel exactly the same pressure. 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 1.14 A manometer with water shows a ∆P of Po/20; what is the column height difference? Solution: ∆P = Po/20 = ρHg 101.3 × 1000 Pa H = Po/(20 ρ g) = 20 × 997 kg/m3 × 9.80665 m/s2 = 0.502 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 1.15 Does the pressure have to be uniform for equilibrium to exist? No. It depends on what causes a pressure difference. Think about the pressure increasing as you move down into the ocean, the water at different levels are in equilibrium. However if the pressure is different at nearby locations at same elevation in the water or in air that difference induces a motion of the fluid from the higher towards the lower pressure. The motion will persist as long as the pressure difference exist. 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 1.16 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 1.17 What is the lowest temperature in degrees Celsuis? In degrees 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 1.18 Convert the formula for water density in In-text Concept Question “d” 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 1.19 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 1.20 What is the main difference between the macroscopic kinetic energy in a motion like the blowing of wind versus the microscopic kinetic energy of individual molecules? Which one can you sense with your hand? Answer: The microscopic kinetic energy of individual molecules is too small for us to sense however when the combined action of billions (actually more like in the order of 1 E19) are added we get to the macroscopic magnitude we can sense. The wind velocity is the magnitude and direction of the averaged velocity over many molecules which we sense. The individual molecules are moving in a random motion (with zero average) on top of this mean (or average) motion. A characteristic velocity of this random motion is the speed of sound, around 340 m/s for atmospheric air and it changes with temperature. 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 1.21 How can you illustrate the binding energy between the three atoms in water as they sit in a tri-atomic water molecule. Hint: imagine what must happen to create three separate atoms. Answer: If you want to separate the atoms you must pull them apart. Since they are bound together with strong forces (like non-linear springs) you apply a force over a distance which is work (energy in transfer) to the system and you could end up with two hydrogen atoms and one oxygen atom far apart so they no longer have strong forces between them. If you do not do anything else the atoms will sooner or later recombine and release all the energy you put in and the energy will come out as radiation or given to other molecules by collision interactions. 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, Units, and 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 1.22 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: m 0.06 kg kg ρ = V = 0.000 075 3 = 800 m3 EA A EA m A AE A AE Intensive kg 1 m3 ρ = 800 3 ; A A v = = 0.001 25 kg ρ EA A A A EA A m A AE E E E T = 8°C; P = 101 kPa Extensive m = 60 g = 0.06 kg V = 75 cm3 = 0.075 L = 0.000 075 m3 A E A A E 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 1.23 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 A E A 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 1.24 A stainless steel storage tank contains 5 kg of oxygen gas and 7 kg of nitrogen gas. How many kmoles are in the tank? Table A.2: MO2 = 31.999 ; MN2 = 28.013 A A E A A E 5 nO2 = mO2 / MO2 = 31.999 = 0.15625 kmol A A E A A E A A E A A E 7 nO2 = mN2 / MN2 = 28.013 = 0.24988 kmol A A E A A E A A E A A E ntot = nO2 + nN2 = 0.15625 + 0.24988 = 0.40613 kmol A A E A A E A A E 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 1.25 A steel cylinder of mass 4 kg contains 4 L of liquid water at 25oC at 100 kPa. A E A 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 A E 4 kg Volume of steel: V = m/ρ = = 0.000 512 m3 E 7820 kg/m3 A A A E Density of water in Table A.4: ρ = 997 kg/m3 A E Mass of water: m = ρV = 997 kg/m3 × 0.004 m3 = 3.988 kg A E A A E A Total mass: m = msteel + mwater = 4 + 3.988 = 7.988 kg Total volume: V = Vsteel + Vwater = 0.000 512 + 0.004 = 0.004 512 m3 = 4.51 L A E A 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 1.26 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 A E A gravitational field? How much mass can a force of 1 N support? Solution: ma = 0 = ∑ F = F - mg F F = mg = 2 kg × 9.80665 m/s2 = 19.613 NA E A g m F = mg => F 1N m=g= = 0.102 kg 9.80665 m/s2 A A A A E E 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 1.27 An aluminum piston of 2.5 kg is in the standard gravitational field where a force of 25 N is applied vertically up. Find the acceleration of the piston. Solution: Fup = ma = F – mg F – mg F g a= m A =m–g A A A E E 25 N = 2.5 kg – 9.807 m/s2 A A A E E = 0.193 ms-2 A E F 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 1.28 When you move up from the surface of the earth the gravitation is reduced as g = 9.807 − 3.32 × 10-6 z, with z as the elevation in meters. How many percent is the weight of an airplane reduced when it cruises at 11 000 m? Solution: go= 9.807 ms-2 gH = 9.807 – 3.32 × 10-6 × 11 000 = 9.7705 ms-2 Wo = m go ; WH = m gH 9.7705 WH/ Wo = gH/ go = 9.807 = 0.9963 Reduction = 1 – 0.9963 = 0.0037 or 0.37% i.e. we can neglect that for most applications. 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 1.29 A car rolls down a hill with a slope so the gravitational “pull” in the direction of motion is one tenth of the standard gravitational force (see Problem 1.26). If the car has a mass of 2500 kg find the acceleration. Solution: ma = ∑ F = mg / 10 a = mg / 10m = g/10 = 9.80665 (m/s2) / 10 = 0.981 m/s2 g This acceleration does not depend on the mass of the car. 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 1.30 A van is driven at 60 km/h and is brought to a full stop with constant deceleration in 5 seconds. If the total car and driver mass is 2075 kg find the necessary force. Solution: Acceleration is the time rate of change of velocity. dV 60 × 1000 a = dt = = 3.333 m/s2 3600 × 5 ma = ∑ F ; Fnet = ma = 2075 kg × 3.333 m/s2 = 6916 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 1.31 A 1500-kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed of 75 km/h. What are the force and total time required? Solution: dV ∆V a = dt = => ∆t ∆V (75 − 20) km/h × 1000 m/km ∆t = a = = 3.82 sec 3600 s/h × 4 m/s2 F = ma = 1500 kg × 4 m/s2 = 6000 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 1.32 On the moon the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is “weighed” with a beam balance on the surface on the moon. What is the expected reading? If this mass is weighed with a spring scale that reads correctly for standard gravity on earth (see Problem 1.26), what is the reading? Solution: Moon gravitation is: g = gearth/6 m m m Beam Balance Reading is 5 kg Spring Balance Reading is in kg units This is mass comparison Force comparison length ∝ F ∝ g 5 Reading will be kg 6 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 1.33 The elevator in a hotel has a mass of 750 kg, and it carries six people with a total mass of 450 kg. How much force should the cable pull up with to have an acceleration of 1 m/s2 in the upwards direction? Solution: The total mass moves upwards with an F acceleration plus the gravitations acts with a force pointing down. ma = ∑ F = F – mg F = ma + mg = m(a + g) = (750 + 450) kg × (1 + 9.81) m/s2 = 12 972 N 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 1.34 One of the people in the previous problem weighs 80 kg standing still. How much weight does this person feel when the elevator starts moving? Solution: The equation of motion is ma = ∑ F = F – mg so the force from the floor becomes F = ma + mg = m(a + g) = 80 kg × (1 + 9.81) m/s2 = 864.8 N = x kg × 9.81 m/s2 Solve for x x = 864.8 N/ 9.81 m/s2 = 88.15 kg The person then feels like having a mass of 88 kg instead of 80 kg. The weight is really force so to compare to standard mass we should use kp. So in this example the person is experiencing a force of 88 kp instead of the normal 80 kp. 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 1.35 A bottle of 12 kg steel has 1.75 kmole of liquid propane. It accelerates horizontal with 3 m/s2, what is the needed force? Solution: The molecular weight for propane is M = 44.094 from Table A.2. The force must accelerate both the container mass and the propane mass. m = msteel + mpropane = 12 + (1.75 × 44.094) = 90.645 kg ma = ∑ F ⇒ F = ma = 90.645 kg × 3 m/s2 = 271.9 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 1.36 Some steel beams with a total mass of 700 kg are raised by a crane with an acceleration of 2 m/s2 relative to the ground at a location where the local gravitational acceleration is 9.5 m/s2. Find the required force. Solution: F = ma = Fup − mg Fup = ma + mg = 700 kg ( 2 + 9.5 ) m/s2 = 80 500 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 Specific Volume 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 1.37 A 1 m3 container is filled with 400 kg of granite stone, 200 kg dry sand and 0.2 m3 of liquid 25°C water. Use properties from tables A.3 and A.4. Find the average specific volume and density of the masses when you exclude air mass and volume. Solution: Specific volume and density are ratios of total mass and total volume. mliq = Vliq/vliq = Vliq ρliq = 0.2 m3 × 997 kg/m3 = 199.4 kg mTOT = mstone + msand + mliq = 400 + 200 + 199.4 = 799.4 kg Vstone = mv = m/ρ = 400 kg/ 2750 kg/m3 = 0.1455 m3 Vsand = mv = m/ρ = 200/ 1500 = 0.1333 m3 VTOT = Vstone + Vsand + Vliq = 0.1455 + 0.1333 + 0.2 = 0.4788 m3 v = VTOT / mTOT = 0.4788/799.4 = 0.000599 m3/kg ρ = 1/v = mTOT/VTOT = 799.4/0.4788 = 1669.6 kg/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 1.38 A power plant that separates carbon-dioxide from the exhaust gases compresses it to a density of 110 kg/m3 and stores it in an un-minable coal seam with a porous volume of 100 000 m3. Find the mass they can store. Solution: m = ρ V = 110 kg/m3 × 100 000 m3 = 11 × 10 6 kg Comment: Just to put this in perspective a power plant that generates 2000 MW by burning coal would make about 20 million tons of carbon-dioxide a year. That is 2000 times the above mass so it is nearly impossible to store all the carbon-dioxide being produced. 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 1.39 A 15-kg steel gas tank holds 300 L of liquid gasoline, having a density of 800 kg/m3. If the system is decelerated with 2g what is the needed force? Solution: m = mtank + mgasoline = 15 kg + 0.3 m3 × 800 kg/m3 = 255 kg cb F = ma = 255 kg × 2 × 9.81 m/s2 = 5003 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 1.40 A 5 m3 container is filled with 900 kg of granite (density 2400 kg/m3 ) and the rest of the volume is air with density 1.15 kg/m3. Find the mass of air and the overall (average) specific volume. Solution: mgranite mair = ρ V = ρair ( Vtot − ) ρ 900 = 1.15 [ 5 - 2400 ] = 1.15 × 4.625 = 5.32 kg V 5 v = m = 900 + 5.32 = 0.005 52 m3/kg Comment: Because the air and the granite are not mixed or evenly distributed in the container the overall specific volume or density does not have much meaning. 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 1.41 A tank has two rooms separated by a membrane. Room A has 1 kg air and volume 0.5 m3, room B has 0.75 m3 air with density 0.8 kg/m3. The membrane is broken and the air comes to a uniform state. Find the final density of the air. Solution: Density is mass per unit volume m = mA + mB = mA + ρBVB = 1 + 0.8 × 0.75 = 1.6 kg V = VA + VB = 0.5 + 0.75 = 1.25 m3 A B m 1.6 kg ρ = V = 1.25 3= 1.28 kg/m3 m cb 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 1.42 One kilogram of diatomic oxygen (O2 molecular weight 32) is contained in a 500- L tank. Find the specific volume on both a mass and mole basis (v and v ). Solution: From the definition of the specific volume V 0.5 m3 v = m = 1 kg = 0.5 m3/kg V V n m/M = M v = 32 kg/kmol × 0.5 m /kg = 16 m /kmol v = = 3 3 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 Pressure 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 1.43 A 5000-kg elephant has a cross sectional area of 0.02 m2 on each foot. Assuming an even distribution, what is the pressure under its feet? Force balance: ma = 0 = PA – mg P = mg/A = 5000 kg × 9.81 m/s2 /(4 × 0.02 m2) = 613 125 Pa = 613 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 1.44 A valve in a cylinder has a cross sectional area of 11 cm2 with a pressure of 735 kPa inside the cylinder and 99 kPa outside. How large a force is needed to open the valve? Fnet = PinA – PoutA = (735 – 99) kPa × 11 cm2 Pcyl = 6996 kPa cm2 kN cb = 6996 × EA 2 × 10-4 m2 A A E A A E m A AE = 700 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 1.45 The hydraulic lift in an auto-repair shop has a cylinder diameter of 0.2 m. To what pressure should the hydraulic fluid be pumped to lift 40 kg of piston/arms and 700 kg of a car? Solution: Force acting on the mass by the gravitational field F↓ = ma = mg = 740 × 9.80665 = 7256.9 N = 7.257 kN Force balance: F↑ = ( P - P0 ) A = F↓ A A E => P = P0 + F↓ / A A = π D2 (1 / 4) = 0.031416 m2 7.257 kN P = 101 kPa + = 332 kPa 0.031416 m2 EA A A AE 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 1.46 A hydraulic lift has a maximum fluid pressure of 500 kPa. What should the piston-cylinder diameter be so it can lift a mass of 850 kg? Solution: With the piston at rest the static force balance is F↑ = P A = F↓ = mg A = π r 2 = π D 2 /4 A E A A E A 4mg PA = P π D2/4 = mg ⇒ D2 = Pπ E E A A A A A A E mg 850 kg × 9.807 m/s2 D=2 =2 = 0.146 m Pπ 500 kPa × π × 1000 (Pa/kPa) A EA A EA E 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 1.47 A laboratory room keeps a vacuum of 0.1 kPa. What net force does that put on the door of size 2 m by 1 m? Solution: The net force on the door is the difference between the forces on the two sides as the pressure times the area F = Poutside A – Pinside A = ∆P A = 0.1 kPa × 2 m × 1 m = 200 N Remember that kPa is kN/m2. A E A Pabs = Po - ∆P ∆P = 0.1 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 1.48 A vertical hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 1500 kPa. Solution: Force balance: Po g F↑ = PA = F↓ = P0A + mpg; A A E A A E P0 = 1 bar = 100 kPa A A E cb A = (π/4) D2 = (π/4) × 0.1252 = 0.01227 m2 A E A A E A A E A 0.01227 m2 mp = (P − P0) = ( 1500 − 100 ) kPa × 1000 Pa/kPa × g 9.80665 m/s2 A A A A A A A A E E E E E = 1752 kg 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 1.49 A 75-kg human footprint is 0.05 m2 when the human is wearing boots. Suppose A E A you want to walk on snow that can at most support an extra 3 kPa; what should the total snowshoe area be? Force balance: ma = 0 = PA – mg mg 75 kg × 9.81 m/s2 A= P = = 0.245 m2 E 3 kPa A A A A A E E E 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 1.50 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg A E A resting on the stops, as shown in Fig. P1.50. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston? Solution: The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface. Force balance: F↑ = F↓ = PA = mpg + P0AA A E A A E Now solve for P (divide by 1000 to convert to kPa for 2nd term) mpg 100 × 9.80665 P = P0 + A = 100 kPa + kPa 0.01 × 1000 A A A A A A E E E E cb = 100 kPa + 98.07 kPa = 198 kPa Water 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 1.51 A large exhaust fan in a laboratory room keeps the pressure inside at 10 cm water vacuum relative to the hallway. What is the net force on the door measuring 1.9 m by 1.1 m? Solution: The net force on the door is the difference between the forces on the two sides as the pressure times the area F = Poutside A – Pinside A = ∆P × A = 10 cm H2O × 1.9 m × 1.1 m = 0.10 × 9.80638 kPa × 2.09 m2 A E A = 2049 N Table A.1: 1 m H2O is 9.80638 kPa and kPa is kN/m2. A E A 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 1.52 A tornado rips off a 100 m2 roof with a mass of 1000 kg. What is the minimum A E A vacuum pressure needed to do that if we neglect the anchoring forces? Solution: The net force on the roof is the difference between the forces on the two sides as the pressure times the area F = Pinside A – PoutsideA = ∆P A That force must overcome the gravitation mg, so the balance is ∆P A = mg ∆P = mg/A = (1000 kg × 9.807 m/s2 )/100 m2 = 98 Pa = 0.098 kPa A E A A E A Remember that kPa is kN/m2. A E A 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 1.53 A 5-kg cannon-ball acts as a piston in a cylinder with a diameter of 0.15 m. As the gun-powder is burned a pressure of 7 MPa is created in the gas behind the ball. What is the acceleration of the ball if the cylinder (cannon) is pointing horizontally? Solution: The cannon ball has 101 kPa on the side facing the atmosphere. ma = F = P1 × A − P0 × A = (P1 − P0 ) × A = (7000 – 101) kPa × π ( 0.152 /4 ) m2 = 121.9 kN F 121.9 kN a = m = 5 kg = 24 380 m/s2 A A A A E E 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 1.54 Repeat the previous problem for a cylinder (cannon) pointing 40 degrees up relative to the horizontal direction. Solution: ma = F = ( P1 - P0 ) A - mg sin 400 ma = (7000 - 101 ) kPa × π × ( 0.152 / 4 ) m2 - 5 × 9.807 × 0.6428 N = 121.9 kN - 31.52 N = 121.87 kN F 121.87 kN a= m = = 24 374 m/s2 5 kg A A A A E E 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 1.55 A 2.5 m tall steel cylinder has a cross sectional area of 1.5 m2. At the bottom with A E A a height of 0.5 m is liquid water on top of which is a 1 m high layer of gasoline. This is shown in Fig. P1.55. The gasoline surface is exposed to atmospheric air at 101 kPa. What is the highest pressure in the water? Solution: The pressure in the fluid goes up with the depth as Air P = Ptop + ∆P = Ptop + ρgh A A A A Gasoline E E 1m and since we have two fluid layers we get 0.5 m P = Ptop + [(ρh)gasoline + (ρh)water] g A A A A A A Water E E E The densities from Table A.4 are: ρgasoline = 750 kg/m3; A E A A E A ρwater = 997 kg/m3 A E A A E 9.807 P = 101 kPa + [750 × 1 + 997 × 0.5] kg/m2 × 1000 (m/s2) (kPa/Pa) A A E = 113.2 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 1.56 An underwater buoy is anchored at the seabed with a cable, and it contains a total mass of 250 kg. What should the volume be so that the cable holds it down with a force of 1000 N? Solution: We need to do a force balance on the system at rest and the combined pressure over the buoy surface is the buoyancy (lift) equal to the “weight” of the displaced water volume ma = 0 = mH2Og – mg – F = ρH2OVg – mg – F V = (mg + F)/ ρH2Og = (m + F/g)/ ρH2O = (250 kg + 1000 N/9.81 m/s2) / 997 kg/m3 A E A A E A = 0.353 m3 A E 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 1.57 At the beach, atmospheric pressure is 1025 mbar. You dive 15 m down in the ocean and you later climb a hill up to 250 m elevation. Assume the density of water is about 1000 kg/m3 and the density of air is 1.18 kg/m3. What pressure do A E A A E A you feel at each place? Solution: ∆P = ρgh, Units from A.1: 1 mbar = 100 Pa (1 bar = 100 kPa). Pocean= P0 + ∆P = 1025 × 100 Pa + 1000 kg/m3 × 9.81 m/s2 × 15 m A E A A A E A E A A E A = 2.4965 × 105 Pa = 250 kPa A E A Phill = P0 - ∆P = 1025 × 100 Pa - 1.18 kg/m3 × 9.81 m/s2 × 250 m A A E A A E A E A A E A = 0.99606 × 105 Pa = 99.61 kPa A E A 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 1.58 What is the pressure at the bottom of a 5 m tall column of fluid with atmospheric pressure 101 kPa on the top surface if the fluid is a) water at 20°C b) glycerine 25°C or c) gasoline 25°C Solution: Table A.4: ρH2O = 997 kg/m3; A E A ρGlyc = 1260 kg/m3; A E A ρgasoline = 750 kg/m3A E ∆P = ρgh P = Ptop + ∆P A A E a) ∆P = ρgh = 997× 9.807× 5 = 48 888 Pa P = 101 + 48.99 = 149.9 kPa b) ∆P = ρgh = 1260× 9.807× 5 = 61 784 Pa P = 101 + 61.8 = 162.8 kPa c) ∆P = ρgh = 750× 9.807× 5 = 36 776 Pa P = 101 + 36.8 = 137.8 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 1.59 A steel tank of cross sectional area 3 m2 and 16 m tall weighs 10 000 kg and it is A E A open at the top, as shown in Fig. P1.59. We want to float it in the ocean so it sticks 10 m straight down by pouring concrete into the bottom of it. How much concrete should I put in? Solution: The force up on the tank is from the water pressure at the bottom times its area. The force down is the gravitation times mass and Air the atmospheric pressure. Ocean 10 m F↑ = PA = (ρoceangh + P0)A F↓ = (mtank + mconcrete)g + P0A Concrete The force balance becomes F↑ = F↓ = (ρoceangh + P0)A = (mtank + mconcrete)g + P0A Solve for the mass of concrete mconcrete = (ρoceanhA - mtank) = 997 × 10 × 3 – 10 000 = 19 910 kg Notice: The first term is the mass of the displaced ocean water. The force up is the weight (mg) of this mass called buoyancy which balances with gravitation and the force from P0 cancel. 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 1.60 A piston, mp= 5 kg, is fitted in a cylinder, A = 15 cm2, that contains a gas. The setup is in a centrifuge that creates an acceleration of 25 m/s2 in the direction of piston motion towards the gas. Assuming standard atmospheric pressure outside the cylinder, find the gas pressure. Solution: Force balance: F↑ = F↓ = P0A + mpg = PA A A A A Po E E mpg P = P0 + A g gas A A A A E E E 5 × 25 kPa kg m/s2 = 101.325 + 1000 × 0.0015 P a m2 A A A E E E = 184.7 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 1.61 Liquid water with density ρ is filled on top of a thin piston in a cylinder with cross-sectional area A and total height H, as shown in Fig. P1.61. Air is let in under the piston so it pushes up, spilling the water over the edge. Derive the formula for the air pressure as a function of piston elevation from the bottom, h. Solution: Force balance P0 Piston: F↑ = F↓ P PA = P0A + mH Og A A 2 E H P = P0 + mH Og/A A A P0 2 E h h, V air P = P0 + (H − h)ρg 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 Manometers and Barometers 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 1.62 A probe is lowered 16 m into a lake. Find the absolute pressure there? Solution: The pressure difference for a column is from Eq.1.2 and the density of water is from Table A.4. ∆P = ρgH = 997 kg/m3 × 9.81 m/s2 × 16 m A E A = 156 489 Pa = 156.489 kPa Pocean = P0 + ∆P A E A A A E = 101.325 + 156.489 = 257.8 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 Son