Exercícios resolvidos, Notas de estudo de Mecatrônica

Baixe Exercícios resolvidos e outras Notas de estudo em PDF para Mecatrônica, somente na Docsity! SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 2 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG • BORGNAKKE • VAN WYLEN CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study Guide Problems 1-22 Properties and Units 23-26 Force and Energy 27-37 Specific Volume 38-43 Pressure 44-57 Manometers and Barometers 58-76 Temperature 77-80 Review Problems 81-86 Sonntag, Borgnakke and van Wylen Correspondence table CHAPTER 2 6th edition Sonntag/Borgnakke/Wylen The correspondence between the problem set in this sixth edition versus the problem set in the 5'th edition text. Problems that are new are marked new and those that are only slightly altered are marked as modified (mod). Study guide problems 2.1-2.22 and 2.23-2.26 are all new problems. New 5th Ed. New 5th Ed. New 5th Ed. 27 1 47 new 67 24 28 new 48 16 68 new 29 2 49 17 69 new 30 new 50 new 70 23 31 3 51 new 71 new 32 new 52 19 72 30 33 5 53 new 73 32 34 6 54 34 74 33 35 7 55 29 75 new 36 9 56 new 76 37 37 10 57 28 mod 77 27 38 12 58 new 78 new 39 new 59 20 79 38 40 new 60 26 80 new 41 new 61 new 81 31 42 11 62 21 82 new 43 13 63 new 83 22 44 new 64 new 84 35 45 18 65 15 85 36 46 14 66 new 86 new English Unit Problems New 5th Ed. SI New 5th Ed. SI 87 new - 97 43E 43 88 new 11 98 new 50 89 new 12 99 new 53 90 new 19 100 45E 70 91 new 20 101 46E 45 92 new 24 102 new 82 93 39E 33 103 48E 55 94 40E - 104 new 80 95 new 47 105 47E 77 96 42E 42 Design and Open ended problems 106-116 are from 5th edition problems 2.50- 2.60 Sonntag, Borgnakke and van Wylen 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: Welectrical 1 2 WT 1 3 Electric power gen. 5 4 6 7 Cooling by seawater Condensate to steam gen. cold Hot steam from generator cb The electrical power also leaves the C.V. to be used for lights, instruments and to charge the batteries. Sonntag, Borgnakke and van Wylen 2.4 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. Sonntag, Borgnakke and van Wylen 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 Q loss C B Sonntag, Borgnakke and van Wylen 2.8 Water in nature exist in different phases like 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) Sonntag, Borgnakke and van Wylen 2.9 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. Sonntag, Borgnakke and van Wylen 2.10 Density of fibers, rock wool insulation, foams and cotton is fairly low. Why is that? Solution: All these materials consists 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. Sonntag, Borgnakke and van Wylen 2.14 You dive 5 m down in the ocean. What is the absolute pressure there? Solution: The pressure difference for a column is from Eq.2.2 and the density of water is from Table A.4. ∆P = ρgH = 997 kg/m3 × 9.81 m/s2 × 5 m = 48 903 Pa = 48.903 kPa Pocean= P0 + ∆P = 101.325 + 48.903 = 150 kPa Sonntag, Borgnakke and van Wylen 2.15 What pressure difference does a 10 m column of atmospheric air show? Solution: The pressure difference for a column is from Eq.2.2 ∆P = ρgH So we need density of air from Fig.2.7, ρ = 1.2 kg/m3 ∆P = 1.2 kg/m3 × 9.81 ms-2 × 10 m = 117.7 Pa = 0.12 kPa Sonntag, Borgnakke and van Wylen 2.16 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? 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 Sonntag, Borgnakke and van Wylen 2.19 What is a temperature of –5oC in degrees Kelvin? Solution: The offset from Celsius to Kelvin is 273.15 K, so we get TK = TC + 273.15 = -5 + 273.15 = 268.15 K Sonntag, Borgnakke and van Wylen 2.20 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 Sonntag, Borgnakke and van Wylen 2.21 Density of liquid water is ρ = 1008 – T/2 [kg/m3] with T in oC. If the temperature increases 10oC how much deeper does a 1 m layer of water become? Solution: The density change for a change in temperature of 10oC becomes ∆ρ = – ∆T/2 = –5 kg/m3 from an ambient density of ρ = 1008 – T/2 = 1008 – 25/2 = 995.5 kg/m3 Assume the area is the same and the mass is the same m = ρV = ρAH, then we have ∆m = 0 = V∆ρ + ρ∆V ⇒ ∆V = - V∆ρ/ρ and the change in the height is ∆H = ∆V A = H∆V V = -H∆ρ ρ = -1 × (-5) 995.5 = 0.005 m barely measurable. Sonntag, Borgnakke and van Wylen 2.24 An apple “weighs” 80 g and has a volume of 100 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 = 0.08 0.0001 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 = 80 g = 0.08 kg V =100 cm3 = 0.1 L = 0.0001 m3 Sonntag, Borgnakke and van Wylen 2.25 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 Sonntag, Borgnakke and van Wylen 2.26 A pressurized steel bottle is charged with 5 kg of oxygen gas and 7 kg of nitrogen gas. How many kmoles are in the bottle? Table A2 : MO2 = 31.999 ; MN2 = 28.013 nO2 = mO2 / MO2 = 5 31.999 = 0.15625 kmol nO2 = mN2 / MN2 = 7 28.013 = 0.24988 kmol ntot = nO2 + nN2 = 0.15625 + 0.24988 = 0.406 kmol Sonntag, Borgnakke and van Wylen 2.29 A model car rolls down an incline with a slope so the gravitational “pull” in the direction of motion is one third of the standard gravitational force (see Problem 2.1). If the car has a mass of 0.45 kg find the acceleration. Solution: ma = ∑ F = mg / 3 a = mg / 3m = g/3 = 9.80665 / 3 = 3.27 m/s2 g This acceleration does not depend on the mass of the model car. Sonntag, Borgnakke and van Wylen 2.30 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 WH/Wo = gH/go = 9.7705 9.807 = 0.9963 Reduction = 1 – 0.9963 = 0.0037 or 0.37% i.e. we can neglect that for most application Sonntag, Borgnakke and van Wylen 2.31 A car drives 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 1075 kg find the necessary force. Solution: Acceleration is the time rate of change of velocity. a = dV dt = 60 × 1000 3600 × 5 = 3.333 m/s 2 ma = ∑ F ; Fnet = ma = 1075 kg × 3.333 m/s2 = 3583 N Sonntag, Borgnakke and van Wylen 2.34 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What force is needed and what is the final velocity? Solution: Constant acceleration can be integrated to get velocity. a = dV dt => ∫ dV = ∫ a dt => ∆V = a ∆t ∆V = a ∆t = 3 m/s2 × 10 s = 30 m/s => V = 30 m/s F = ma = 950 kg × 3 m/s2 = 2850 N F Sonntag, Borgnakke and van Wylen 2.35 A 15 kg steel container has 1.75 kilomoles of liquid propane inside. A force of 2 kN now accelerates this system. What is the acceleration? 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 = 15 + (1.75 × 44.094) = 92.165 kg ma = ∑ F ⇒ a = ∑ F / m a = 2000 N 92.165 kg = 21.7 m/s 2 Sonntag, Borgnakke and van Wylen 2.36 A bucket of concrete of total mass 200 kg is 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 = 200 ( 2 + 9.5 ) = 2300 N g F up Sonntag, Borgnakke and van Wylen 2.39 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 ρ = m V = 1.6 1.25 = 1.28 kg/m 3 A B cb Sonntag, Borgnakke and van Wylen 2.40 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 × 997 = 199.4 kg mTOT = mstone + msand + mliq = 400 + 200 + 199.4 = 799.4 kg Vstone = mv = m/ρ = 400/ 2750 = 0.1455 m 3 Vsand = mv = m/ρ = 200/ 1500 = 0.1333 m 3 VTOT = Vstone + Vsand + Vliq = 0.1455 + 0.1333 + 0.2 = 0.4788 m3







































































































































































































































































v = VTOT / mTOT = 0.4788/799.4 = 0.000599 m 3/kg ρ = 1/v = mTOT/VTOT = 799.4/0.4788 = 1669.6 kg/m 3 Sonntag, Borgnakke and van Wylen 2.41 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 and use air density of 1.1 kg/m3. Find the average specific volume and density of the 1 m3 volume. Solution: Specific volume and density are ratios of total mass and total volume. Vstone = mv = m/ρ = 400/ 2750 = 0.1455 m 3 Vsand = mv = m/ρ = 200/ 1500 = 0.1333 m 3 Vair = VTOT − Vstone − Vsand − Vliq = 1− 0.1455 − 0.1333 − 0.2 = 0.5212 m3



























































































































































































































mair = Vair/vair = Vair ρair = 0.5212 × 1.1 = 0.573 kg mliq = Vliq/vliq = Vliq ρliq = 0.2 × 997 = 199.4 kg mTOT = mstone + msand + mliq + mair = 400 + 200 + 199.4 + 0.573 ≈ 800 kg v = VTOT / mTOT = 1/800 = 0.00125 m 3/kg ρ = 1/v = mTOT/VTOT = 800/1 = 800 kg/m 3 Sonntag, Borgnakke and van Wylen Pressure 2.44 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 r2 = π D2/4 PA = P π D2/4 = mg ⇒ D2 = 4mg P π D = 2 mg Pπ = 2 850 × 9.807 500 π × 1000 = 0.146 m Sonntag, Borgnakke and van Wylen 2.45 A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg resting on the stops, as shown in Fig. P2.45. 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 + P0A Now solve for P (divide by 1000 to convert to kPa for 2nd term) P = P0 + mpg A = 100 kPa + 100 × 9.80665 0.01 × 1000 = 100 kPa + 98.07 kPa = 198 kPa Water cb Sonntag, Borgnakke and van Wylen 2.46 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: F↑ = PA = F↓ = P0A + mpg; P0 = 1 bar = 100 kPa A = (π/4) D2 = (π/4) × 0.1252 = 0.01227 m2 cb g Po mp = (P − P0) A g = ( 1500 − 100 ) × 1000 × 0.01227 9.80665 = 1752 kg Sonntag, Borgnakke and van Wylen 2.49 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 40 0 ma = (7000 - 101 ) kPa × π × ( 0.152 / 4 ) m2 - 5 × 9.807 × 0.6428 N = 121.9 kN - 31.52 N = 121.87 kN a = F m = 121.87 kN 5 kg = 24 374 m/s 2 Sonntag, Borgnakke and van Wylen 2.50 A large exhaust fan in a laboratory room keeps the pressure inside at 10 cm water relative vacuum 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 = 2049 N Table A.1: 1 m H2O is 9.80638 kPa and kPa is kN/m2. Sonntag, Borgnakke and van Wylen 2.51 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) light oil Solution: Table A.4: ρH2O = 997 kg/m3; ρGlyc = 1260 kg/m3; ρOil = 910 kg/m3 ∆P = ρgh P = Ptop + ∆P a) ∆P = ρgh = 997× 9.807× 5 = 48887.9 Pa P = 101 + 48.99 = 149.9 kPa b) ∆P = ρgh = 1260× 9.807× 5 = 61784 Pa P = 101 + 61.8 = 162.8 kPa c) ∆P = ρgh = 910× 9.807× 5 = 44622 Pa P = 101 + 44.6 = 145.6 kPa Sonntag, Borgnakke and van Wylen 2.54 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 you feel at each place? Solution: ∆P = ρgh Pocean= P0 + ∆P = 1025 × 100 + 1000 × 9.81 × 15 = 2.4965 × 105 Pa = 250 kPa Phill = P0 - ∆P = 1025 × 100 - 1.18 × 9.81 × 250 = 0.99606 × 105 Pa = 99.61 kPa Sonntag, Borgnakke and van Wylen 2.55 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 P = P0 + mpg A = 101.325 + 5 × 25 1000 × 0.0015 kPa kg m/s2 Pa m2 = 184.7 kPa gasg Po Sonntag, Borgnakke and van Wylen 2.56 A steel tank of cross sectional area 3 m2 and 16 m tall weighs 10 000 kg and it is open at the top. 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 the atmospheric pressure. F↑ = PA = (ρoceangh + P0)A F↓ = (mtank + mconcrete)g + P0A The force balance becomes Air Ocean Concrete 10 m 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 net force up is the weight (mg) of this mass called bouyancy, P0 cancel. Sonntag, Borgnakke and van Wylen 2.59 A differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel. Solution: Convert all pressures to units of kPa. Pgauge = 1.25 MPa = 1250 kPa; P0 = 0.96 bar = 96 kPa P = Pgauge + P0 = 1250 + 96 = 1346 kPa Sonntag, Borgnakke and van Wylen 2.60 Two vertical cylindrical storage tanks are full of liquid water, density 1000 kg/m3, the top open to the atmoshere. One is 10 m tall, 2 m diameter, the other is 2.5 m tall with diameter 4 m. What is the total force from the bottom of each tank to the water and what is the pressure at the bottom of each tank? Solution: VA = H × πD2 × (1 / 4) = 10 × π × 22 × ( 1 / 4) = 31.416 m3 VB = H × πD2 × (1 / 4) = 2.5 × π × 42 × ( 1 / 4) = 31.416 m3 Tanks have the same volume, so same mass of water gives gravitational force F = mg = ρ V g = 1000 × 31.416 × 9.80665 = 308 086 N this is the force the legs have to supply (assuming Po below the bottom). Tanks have total force up from bottom as Ftot A = F + PoA = 308 086 + 101325 × 3.1416 = 626 408 N Ftot B = F + PoA = 308 086 + 101325 × 12.5664 = 1 581 374 N Pbot = Po + ρ H g Pbot A = 101 + (1000 × 10 × 9.80665 / 1000) = 199 kPa Pbot B = 101 + (1000 × 2.5 × 9.80665 / 1000) = 125.5 kPa Po Po g m m cb A B Sonntag, Borgnakke and van Wylen 2.61 Blue manometer fluid of density 925 kg/m3 shows a column height difference of 6 cm vacuum with one end attached to a pipe and the other open to P0 = 101 kPa. What is the absolute pressure in the pipe? Solution: Since the manometer shows a vacuum we have PPIPE = P0 - ∆P ∆P = ρgh = 925 × 9.807 × 0.06 = 544.3 Pa = 0.544 kPa PPIPE = 101 – 0.544 = 100.46 kPa cb Po Pipe Sonntag, Borgnakke and van Wylen 2.64 A submarine maintains 101 kPa inside it and it dives 240 m down in the ocean having an average density of 1030 kg/m3. What is the pressure difference between the inside and the outside of the submarine hull? Solution: Assume the atmosphere over the ocean is at 101 kPa, then ∆P is from the 240 m column water. ∆P = ρLg = (1030 kg/m3 × 240 m × 9.807 m/s2) / 1000 = 2424 kPa Sonntag, Borgnakke and van Wylen 2.65 A barometer to measure absolute pressure shows a mercury column height of 725 mm. The temperature is such that the density of the mercury is 13 550 kg/m3. Find the ambient pressure. Solution: Hg : L = 725 mm = 0.725 m; ρ = 13 550 kg/m3 The external pressure P balances the column of height L so from Fig.2.10 P = ρ L g = 13 550 kg/m3 × 9.80665 m/s2 × 0.725 m × 10-3 kPa/Pa = 96.34 kPa Sonntag, Borgnakke and van Wylen 2.66 An absolute pressure gauge attached to a steel cylinder shows 135 kPa. We want to attach a manometer using liquid water a day that Patm = 101 kPa. How high a fluid level difference must we plan for? Solution: Since the manometer shows a pressure difference we have ∆P = PCYL - Patm = ρ L g L = ∆P / ρg = (135 – 101) kPa 997 kg m-3 × 10 × 9.807 m/s2 1000 Pa kPa = 3.467 m H Sonntag, Borgnakke and van Wylen 2.69 Assume we use a pressure gauge to measure the air pressure at street level and at the roof of a tall building. If the pressure difference can be determined with an accuracy of 1 mbar (0.001 bar) what uncertainty in the height estimate does that corresponds to? Solution: ρair = 1.169 kg/m 3 from Table A.5 ∆P = 0.001 bar = 100 Pa L = ∆P ρg = 100 1.169 × 9.807 = 8.72 m Sonntag, Borgnakke and van Wylen 2.70 A U-tube manometer filled with water, density 1000 kg/m3, shows a height difference of 25 cm. What is the gauge pressure? If the right branch is tilted to make an angle of 30° with the horizontal, as shown in Fig. P2.70, what should the length of the column in the tilted tube be relative to the U-tube? Solution: Same height in the two sides in the direction of g. ∆P = F/A = mg/A = Vρg/A = hρg = 0.25 × 1000 × 9.807 = 2452.5 Pa = 2.45 kPa h = H × sin 30° ⇒ H = h/sin 30° = 2h = 50 cm 30 o H h Sonntag, Borgnakke and van Wylen 2.71 A barometer measures 760 mmHg at street level and 735 mmHg on top of a building. How tall is the building if we assume air density of 1.15 kg/m3? Solution: ∆P = ρgH H = ∆P/ρg = 760 – 735 1.15 × 9.807 mmHg kg/m2s2 133.32 Pa mmHg = 295 m Sonntag, Borgnakke and van Wylen 2.74 Two hydraulic piston/cylinders are of same size and setup as in Problem 2.73, but with negligible piston masses. A single point force of 250 N presses down on piston A. Find the needed extra force on piston B so that none of the pistons have to move. Solution: AA = 75 cm 2 ; AB = 25 cm 2 No motion in connecting pipe: PA = PB Forces on pistons balance Po Po cb A B FBFA PA = P0 + FA / AA = PB = P0 + FB / AB FB = FA × AB AA = 250 × 25 75 = 83.33 N Sonntag, Borgnakke and van Wylen 2.75 A pipe flowing light oil has a manometer attached as shown in Fig. P2.75. What is the absolute pressure in the pipe flow? Solution: Table A.3: ρoil = 910 kg/m 3; ρwater = 997 kg/m 3 PBOT = P0 + ρwater g Htot = P0 + 997 × 9.807 × 0.8 = Po + 7822 Pa PPIPE = PBOT – ρwater g H1 – ρoil g H2 = PBOT – 997 × 9.807 × 0.1 – 910 × 9.807 × 0.2 = PBOT – 977.7 Pa – 1784.9 Pa PPIPE = Po + (7822 – 977.7 – 1784.9) Pa = Po + 5059.4 Pa = 101.325 + 5.06 = 106.4 kPa Sonntag, Borgnakke and van Wylen 2.76 Two cylinders are filled with liquid water, ρ = 1000 kg/m3, and connected by a line with a closed valve. A has 100 kg and B has 500 kg of water, their cross- sectional areas are AA = 0.1 m 2 and AB = 0.25 m 2 and the height h is 1 m. Find the pressure on each side of the valve. The valve is opened and water flows to an equilibrium. Find the final pressure at the valve location. Solution: VA = vH2OmA = mA/ρ = 0.1 = AAhA => hA = 1 m VB = vH2OmB = mB/ρ = 0.5 = ABhB => hB = 2 m PVB = P0 + ρg(hB+H) = 101325 + 1000 × 9.81 × 3 = 130 755 Pa PVA = P0 + ρghA = 101325 + 1000 × 9.81 × 1 = 111 135 Pa Equilibrium: same height over valve in both Vtot = VA + VB = h2AA + (h2 - H)AB ⇒ h2 = hAAA + (hB+H)AB AA + AB = 2.43 m PV2 = P0 + ρgh2 = 101.325 + (1000 × 9.81 × 2.43)/1000 = 125.2 kPa Sonntag, Borgnakke and van Wylen 2.79 Using the freezing and boiling point temperatures for water in both Celsius and Fahrenheit scales, develop a conversion formula between the scales. Find the conversion formula between Kelvin and Rankine temperature scales. Solution: TFreezing = 0 oC = 32 F; TBoiling = 100 oC = 212 F ∆T = 100 oC = 180 F ⇒ ToC = (TF - 32)/1.8 or TF = 1.8 ToC + 32 For the absolute K & R scales both are zero at absolute zero. TR = 1.8 × TK Sonntag, Borgnakke and van Wylen 2.80 The atmosphere becomes colder at higher elevation. As an average the standard atmospheric absolute temperature can be expressed as Tatm = 288 - 6.5 × 10 −3 z, where z is the elevation in meters. How cold is it outside an airplane cruising at 12 000 m expressed in Kelvin and in Celsius? Solution: For an elevation of z = 12 000 m we get Tatm = 288 - 6.5 × 10 −3 z = 210 K To express that in degrees Celsius we get TC = T – 273.15 = −63.15 oC Sonntag, Borgnakke and van Wylen Review Problems 2.81 Repeat problem 2.72 if the flow inside the apparatus is liquid water, ρ ≅ 1000 kg/m3, instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns. Solution: Balance forces in the manometer: P P 1 . 2 · h h 1 2 H (H - h2) - (H - h1) = ∆hHg = h1 - h2 P1A + ρH2Oh1gA + ρHg(H - h1)gA = P2A + ρH2Oh2gA + ρHg(H - h2)gA ⇒ P1 - P2 = ρH2O(h2 - h1)g + ρHg(h1 - h2)g P1 - P2 = ρHg∆hHgg - ρH2O∆hHgg = 13600 × 0.2 × 9.5 - 1000 × 0.2 × 9.5 = 25840 - 1900 = 23940 Pa = 23.94 kPa Sonntag, Borgnakke and van Wylen 2.84 In the city water tower, water is pumped up to a level 25 m above ground in a pressurized tank with air at 125 kPa over the water surface. This is illustrated in Fig. P2.84. Assuming the water density is 1000 kg/m3 and standard gravity, find the pressure required to pump more water in at ground level. Solution: ∆P = ρ L g = 1000 kg/m3 × 25 m × 9.807 m/s2 = 245 175 Pa = 245.2 kPa Pbottom = Ptop + ∆P = 125 + 245.2 = 370 kPa cb Sonntag, Borgnakke and van Wylen 2.85 Two cylinders are connected by a piston as shown in Fig. P2.85. Cylinder A is used as a hydraulic lift and pumped up to 500 kPa. The piston mass is 25 kg and there is standard gravity. What is the gas pressure in cylinder B? Solution: Force balance for the piston: PBAB + mpg + P0(AA - AB) = PAAA AA = (π/4)0.1 2 = 0.00785 m2; AB = (π/4)0.025 2 = 0.000 491 m2 PBAB = PAAA - mpg - P0(AA - AB) = 500× 0.00785 - (25 × 9.807/1000) - 100 (0.00785 - 0.000 491) = 2.944 kN PB = 2.944/0.000 491 = 5996 kPa = 6.0 MPa P B GAS A Oil Po cb Sonntag, Borgnakke and van Wylen 2.86 A dam retains a lake 6 m deep. To construct a gate in the dam we need to know the net horizontal force on a 5 m wide and 6 m tall port section that then replaces a 5 m section of the dam. Find the net horizontal force from the water on one side and air on the other side of the port. Solution: Pbot = P0 + ∆P ∆P = ρgh = 997× 9.807× 6 = 58 665 Pa = 58.66 kPa Neglect ∆P in air Fnet = Fright – Fleft = Pavg A - P0A Pavg = P0 + 0.5 ∆P Since a linear pressure variation with depth. Fnet = (P0 + 0.5 ∆P)A - P0A = 0.5 ∆P A = 0.5 × 58.66 × 5 × 6 = 880 kN F F left righ t Sonntag, Borgnakke and Wylen Concept Problems 2.87E A mass of 2 lbm has acceleration of 5 ft/s2, what is the needed force in lbf? Solution: Newtons 2nd law: F = ma F = ma = 2 lbm × 5 ft/s2 = 10 lbm ft/s2 = 10 32.174 lbf = 0.31 lbf 2.88E How much mass is in 0.25 gallon of liquid mercury (Hg)? Atmospheric air? Solution: A volume of 1 gal equals 231 in3, see Table A.1. From Figure 2.7 the density is in the range of 10 000 kg/m3 = 624.28 lbm/ft3, so we get m = ρV = 624.3 lbm/ft3 × 0.25 × (231/123 ) ft3 = 20.86 lbm A more accurate value from Table F.3 is ρ = 848 lbm/ft3. For the air we see in Figure 2.7 that density is about 1 kg/m3 = 0.06243 lbm/ft3 so we get m = ρV = 0.06243 lbm/ft3 × 0.25 × (231/123 ) ft3 = 0.00209 lbm A more accurate value from Table F.4 is ρ = 0.073 lbm/ft3 at 77 F, 1 atm. Sonntag, Borgnakke and Wylen 2.89E Can you easily carry a one gallon bar of solid gold? Solution: The density of solid gold is about 1205 lbm/ft3 from Table F.2, we could also have read Figure 2.7 and converted the units. V = 1 gal = 231 in3 = 231 × 12-3 ft3 = 0.13368 ft3 Therefore the mass in one gallon is m = ρV = 1205 lbm/ft3 × 0.13368 ft3 = 161 lbm and some people can just about carry that in the standard gravitational field. 2.90E What is the temperature of –5F in degrees Rankine? Solution: The offset from Fahrenheit to Rankine is 459.67 R, so we get TR = TF + 459.67 = -5 + 459.67 = 454.7 R 2.91E What is the smallest temperature in degrees Fahrenheit you can have? Rankine? Solution: The lowest temperature is absolute zero which is at zero degrees Rankine at which point the temperature in Fahrenheit is negative TR = 0 R = −459.67 F Sonntag, Borgnakke and Wylen Properties and Units 2.92E An apple weighs 0.2 lbm and has a volume of 6 in3 in a refrigerator at 38 F. What is the apple density? List three intensive and two extensive properties for the apple. Solution: ρ = m V = 0.2 6 lbm in3 = 0.0333 lbm in3 = 57.6 lbm ft3 Intensive ρ = 57.6 lbm ft3 ; v = 1 ρ = 0.0174 ft3 lbm T = 38 F; P = 14.696 lbf/in2 Extensive m = 0.2 lbm V = 6 in3 = 0.026 gal = 0.00347 ft3 Sonntag, Borgnakke and Wylen 2.95E A valve in a cylinder has a cross sectional area of 2 in2 with a pressure of 100 psia inside the cylinder and 14.7 psia outside. How large a force is needed to open the valve? Solution: Fnet = PinA – PoutA = (100 – 14.7) psia × 2 in2 = 170.6 (lbf/in2) × in2 = 170.6 lbf cb Pcyl Sonntag, Borgnakke and Wylen 2.96E One pound-mass of diatomic oxygen (O2 molecular weight 32) is contained in a 100-gal tank. Find the specific volume on both a mass and mole basis (v and v ). Solution: V = 231 in3 = (231 / 123) ft3 = 0.1337 ft3 conversion seen in Table A.1 This is based on the definition of the specific volume v = V/m = 0.1337 ft3/1 lbm = 0.1337 ft3/lbm v̄ = V/n = V m/M = Mv = 32 × 0.1337 = 4.278 ft 3/lbmol Sonntag, Borgnakke and Wylen Pressure 2.97E A 30-lbm steel gas tank holds 10 ft3 of liquid gasoline, having a density of 50 lbm/ft3. What force is needed to accelerate this combined system at a rate of 15 ft/s2? Solution: m = mtank + mgasoline = 30 lbm + 10 ft3 × 50 lbm/ft3 = 530 lbm cb F = ma = (530 lbm × 15 ft/s2) / (32.174 lbm ft/s2 lbf) = 247.1 lbf