Solution Manual For Fundamentals of Thermal-Fluid Sciences, 5th Edition

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1-1Solutions ManualforFundamentals of Thermal Fluid Sciences5th EditionYunus A. Çengel, John M. Cimbala, Robert H. TurnerChapter 1INTRODUCTION AND OVERVIEW

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1-2Thermodynamics, Heat Transfer, and Fluid Mechanics1-1COn a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclistpicks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.1-2CThere is no truth to his claim. It violates the second law of thermodynamics.1-3CA car going uphill without the engine running would increase the energy of the car, and thus it would be a violation ofthe first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble betweentwo marks of a horizontal water tube) it can shown that the road that looks uphill to the eye is actually downhill.1-4CThermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium stateto another. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution withinthe system at a specified time.1-5C(a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow isthe electric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference.1-6CHeat transfer is a non-equilibrium phenomena since in a system that is in equilibrium there can be no temperaturedifferences and thus no heat flow.1-7CNo, there cannot be any heat transfer between two bodies that are at the same temperature (regardless of pressure)since the driving force for heat transfer is temperature difference.1-8C Stressis defined as force per unit area, and is determined by dividing the force by the area upon which it acts. Thenormal component of a force acting on a surface per unit area is called thenormal stress, and the tangential component of aforce acting on a surface per unit area is calledshear stress. In a fluid, the normal stress is calledpressure.Mass, Force, and Units1-9CKg-mass is the mass unit in the SI system whereas kg-force is a force unit. 1-kg-force is the force required toaccelerate a 1-kg mass by 9.807 m/s2. In other words, the weight of 1-kg mass at sea level is 1 kg-force.1-10CIn this unit, the wordlightrefers to the speed of light. The light-year unit is then the product of a velocity and time.Hence, this product forms a distance dimension and unit.1-11CThere is no acceleration, thus the net force is zero in both cases.

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1-31-12The variation of gravitational acceleration above the sea level is given as a function of altitude. The height at which theweight of a body will decrease by 0.3% is to be determined.AnalysisThe weight of a body at the elevation z can be expressed asWmgmz( ..)9 8073 32106In our case,)807.9)((997.0997.0997.0)100/3.01(mmgWWWsssSubstituting,m8862zz)1032.3807.9()807.9(997.061-13The mass of an object is given. Its weight is to be determined.AnalysisApplying Newton's second law, the weight is determined to beN1920)m/s6.9)(kg200(2mgW1-14A plastic tank is filled with water. The weight of the combined system is to be determined.AssumptionsThe density of water is constant throughout.PropertiesThe density of water is given to be= 1000 kg/m3.AnalysisThe mass of the water in the tank and the total mass aremw=V=(1000 kg/m3)(0.2 m3) = 200 kgmtotal=mw+ mtank=200 + 3 = 203 kgThus,N1991m/skg1N1)m/skg)(9.81(20322mgW1-15EThe constant-pressure specific heat of air given in a specified unit is to be expressed in various units.AnalysisUsing proper unit conversions, the constant-pressure specific heat is determined in various units to beFBtu/lbm0.240Ckcal/kg0.240CJ/g1.005KkJ/kg1.005CkJ/kg4.1868FBtu/lbm1C)kJ/kg(1.005kJ4.1868kcal1C)kJ/kg(1.005g1000kg1kJ1J1000C)kJ/kg(1.005CkJ/kg1KkJ/kg1C)kJ/kg(1.005ppppccccSea levelz0mtank= 3kgV=0.2 m3H2O

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1-41-16A rock is thrown upward with a specified force. The acceleration of the rock is to be determined.AnalysisThe weight of the rock isN.3729m/skg1N1)m/skg)(9.79(322mgWThen the net force that acts on the rock isN6.17037.29020downupnetFFFFrom the Newton's second law, the acceleration of the rock becomes2m/s56.9N1m/skg1kg3N170.62mFaStone

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1-51-17Problem 1-16 is reconsidered. The entire software solution is to be printed out, including the numerical resultswith proper units.AnalysisThe problem is solved using EES, and the solution is given below."The weight of the rock is"W=m*gm=3 [kg]g=9.79 [m/s2]"The force balance on the rock yields the net force acting on the rock as"F_up=200 [N]F_net = F_up - F_downF_down=W"The acceleration of the rock is determined from Newton's second law."F_net=m*a"To Run the program, press F2 or select Solve from the Calculate menu."SOLUTIONa=56.88 [m/s^2]F_down=29.37 [N]F_net=170.6 [N]F_up=200 [N]g=9.79 [m/s2]m=3 [kg]W=29.37 [N]m [kg]a[m/s2]12345678910190.290.2156.8840.2130.2123.5418.7815.2112.4310.211234567891004080120160200m [kg]a [m/s2]

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1-61-18A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and kJ areto be determined.AnalysisThe resistance heater consumes electric energy at a rate of 4 kW or 4 kJ/s. Then the total amount of electric energyused in 3 hours becomesTotal energy = (Energy per unit time)(Time interval)= (4 kW)(3 h)=12 kWhNoting that 1 kWh = (1 kJ/s)(3600 s) = 3600 kJ,Total energy = (12 kWh)(3600 kJ/kWh)=43,200 kJDiscussionNote kW is a unit for power whereas kWh is a unit for energy.1-19EAn astronaut took his scales with him to space. It is to be determined how much he will weigh on the spring andbeam scales in space.Analysis(a) A spring scale measures weight, which is the local gravitational force applied on a body:lbf25.522ft/slbm32.2lbf1)ft/slbm)(5.48(150mgW(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The beam scalewill read what it reads on earth,W150 lbf1-20A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to beobtained for the filling time.AssumptionsGasoline is an incompressible substance and the flow rate is constant.AnalysisThe filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unitof time is „seconds‟. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds.Putting the given information into perspective, we havet[s]V[L], andV[L/s}It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate.Therefore, the desired relation isVtVDiscussionNote that thisapproach may not work for cases that involve dimensionless (and thus unitless) quantities.

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1-71-21A pool is to be filled with water using a hose. Based on unit considerations, a relation is to be obtained for the volumeof the pool.AssumptionsWater is an incompressible substance and the average flow velocity is constant.AnalysisThe pool volume depends on the filling time, the cross-sectional area which depends on hose diameter, and flowvelocity. Also, we know that the unit of volume is m3. Therefore, the independent quantities should be arranged such thatwe end up with the unit of seconds. Putting the given information into perspective, we haveV[m3]is a function oft[s],D[m], andV[m/s}It is obvious that the only wayto end up with the unit “m3” for volume is to multiply the quantitiestandVwith the squareofD. Therefore, the desired relation isV=CD2Vtwhere the constant of proportionality is obtained for a round hose, namely,C=π/4 so thatV= (D2/4)Vt.DiscussionNote that the values of dimensionless constants of proportionality cannot be determined with this approach.Review Problems1-22The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at differentlocations are to be determined.AnalysisThe weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into therelation226m/skg1N1)m/s103.32kg)(9.807(80zmgWSea level:(z = 0 m): W = 80(9.807-3.32x10-60) = 809.807 =784.6 NDenver:(z = 1610 m): W = 80(9.807-3.32x10-61610) = 809.802 =784.2 NMt. Ev.:(z = 8848 m): W = 80(9.807-3.32x10-68848) =809.778 =782.2 N1-23EA man is considering buying a 12-oz steak for $3.15, or a 300-g steak for $2.95. The steak that is a better buy is tobe determined.AssumptionsThe steaks are of identical quality.AnalysisTo make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kgas the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be12 ounce steak:$9.26/kgkg0.45359lbm1lbm1oz16oz12$3.15=CostUnit300 gram steak:$9.83/kgkg1g1000g300$2.95=CostUnitTherefore, the steak at the traditional market is a better buy.

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1-81-24EThe mass of a substance is given. Its weight is to be determined in various units.AnalysisApplying Newton's second law, the weight is determined in various units to beN9.8122m/skg1N1)m/skg)(9.81(1mgWkN0.0098122m/skg1000kN1)m/skg)(9.81(1mgW2m/skg1)m/skg)(9.81(12mgWkgf1N9.81kgf1m/skg1N1)m/skg)(9.81(122mgW2ft/slbm71)ft/s(32.2kg1lbm2.205kg)(12mgWlbf2.2122ft/slbm32.2lbf1)ft/s(32.2kg1lbm2.205kg)(1mgW1-25The flow of air through a wind turbine is considered. Based on unit considerations, a proportionality relation is to beobtained for the mass flow rate of air through the blades.AssumptionsWind approaches the turbine blades with a uniform velocity.AnalysisThe mass flow rate depends on the air density, average wind velocity, and the cross-sectional area which dependson hose diameter. Also, the unit of mass flow ratemis kg/s. Therefore, the independent quantities should be arranged suchthat we end up with the proper unit. Putting the given information into perspective, we havem[kg/s]is a function of[kg/m3],D[m], andV[m/s}It is obvious that the only way to end up with the unit “kg/s” for mass flow rate is to multiply the quantitiesandVwiththe square ofD. Therefore, the desired proportionality relation is2is proportional tomD Vor,VDCm2where the constant of proportionality isC=π/4 so thatVDm)4/(2DiscussionNote that the dimensionless constants of proportionality cannot be determined with this approach.

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1-91-26A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient, the air density, the carvelocity, and the frontal area of the car.AnalysisThe drag force depends on a dimensionless drag coefficient, the air density, the car velocity, and the frontal area.Also, the unit of forceFis newton N, which is equivalent to kgm/s2. Therefore, the independent quantities should bearranged such that we end up with the unit kgm/s2for the drag force. Putting the given information into perspective, wehaveFD[ kgm/s2]CDrag[],Afront[m2],[kg/m3], andV[m/s]It is obvious that the only way to end up with the unit “kgm/s2” for drag force is to multiply mass with the square of thevelocity and the fontal area, with the drag coefficient serving as the constant of proportionality. Therefore, the desiredrelation is2frontDragVACFDDiscussionNote that this approach is not sensitive to dimensionless quantities, and thus a strong reasoning is required.1-27CDespite the convenience and capability the engineering software packages offer, they are still just tools, and they willnot replace the traditional engineering courses. They will simply cause a shift in emphasis in the course material frommathematics to physics. They are of great value in engineering practice, however, as engineers today rely on softwarepackages for solving large and complex problems in a short time, and perform optimization studies efficiently.1-28 Design and Essay Problems

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2-1Solutions ManualforFundamentals of Thermal Fluid Sciences5th EditionYunus A. Çengel, John M. Cimbala, Robert H. TurnerChapter 2BASIC CONCEPTS OF THERMODYNAMICS

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2-2Systems, Properties, State, and Processes2-1CThe radiator should be analyzed as an open system since mass is crossing the boundaries of the system.2-2CThe system is taken as the air contained in the piston-cylinder device. This system is a closed or fixed mass systemsince no mass enters or leaves it.2-3CA can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the system.2-4CIntensive properties do not depend on the size (extent) of the system but extensive properties do.2-5CIf we were to divide the system into smaller portions, the weight of each portion would also be smaller. Hence, theweight is anextensive property.2-6CYes, because temperature and pressure are two independent properties and the air in an isolated room is a simplecompressible system.2-7CIf we were to divide this system in half, both the volume and the number of moles contained in each half would beone-half that of the original system. The molar specific volume of the original system isNVvand the molar specific volume of one of the smaller systems isNN/VVv2/2which is the same as that of the original system. The molar specific volume is then anintensive property.2-8CA process during which a system remains almost in equilibrium at all times is called a quasi-equilibrium process.Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum andthe work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibriumprocesses.2-9CA process during which the temperature remains constant is called isothermal; a process during which the pressureremains constant is called isobaric; and a process during which the volume remains constant is called isochoric.2-10CThe pressure and temperature of the water are normally used to describe the state. Chemical composition, surfacetension coefficient, and other properties may be required in some cases.As the water cools, its pressure remains fixed. This cooling process is then an isobaric process.

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2-32-11CWhen analyzing the acceleration of gases as they flow through a nozzle, the proper choice for the system is thevolume within the nozzle, bounded by the entire inner surface of the nozzle and the inlet and outlet cross-sections. This is acontrol volume since mass crosses the boundary.2-12CThespecific gravity,orrelative density,and is defined as the ratio of the density of a substance to the density ofsome standard substance at a specified temperature(usually water at 4°C, for whichH2O= 1000 kg/m3). That is,H2O/SG. When specific gravity is known, density is determined fromH2OSG.

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2-42-13The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation ofdensity with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphereusing the correlation is to be estimated.Assumptions1Atmospheric air behaves as an ideal gas.2The earth is perfectly sphere with a radius of 6377 km, and thethickness of the atmosphere is 25 km.PropertiesThe density data are given in tabular form asr, kmz, km, kg/m3637701.225637811.112637921.007638030.9093638140.8194638250.7364638360.6601638580.52586387100.41356392150.19486397200.088916402250.04008AnalysisUsing EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table fromTables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit”to get curve fit window. Then specify 2ndorder polynomial and enter/edit equation. The results are:(z) = a +bz+cz2= 1.20252–0.101674z+ 0.0022375z2for the unit of kg/m3,(or,(z) = (1.20252–0.101674z+ 0.0022375z2)109for the unit of kg/km3)wherezis the vertical distance from the earth surface at sea level. Atz= 7 km, the equation would give= 0.60 kg/m3.(b)The mass of atmosphere can be evaluated by integration to be5/4/)2(3/)2(2/)2(4)2)((4)(4)(5403200200202020202020chhcrbhcrbrahbrarhardzzzrrczbzadzzrczbzadVmhzhzVwherer0= 6377 km is the radius of the earth,h= 25 km is the thickness of the atmosphere, anda= 1.20252,b= -0.101674, andc= 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 109for thedensity unity kg/km3, the mass of the atmosphere is determined to bem=5.0921018kgDiscussionPerforming the analysis with excel would yield exactly the same results.EES Solution for final result:a=1.2025166;b=-0.10167c=0.0022375;r=6377;h=25m=4*pi*(a*r^2*h+r*(2*a+b*r)*h^2/2+(a+2*b*r+c*r^2)*h^3/3+(b+2*c*r)*h^4/4+c*h^5/5)*1E+9

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2-5Temperature2-14CThey are Celsius (C) and kelvin (K) in the SI, and fahrenheit (F) and rankine (R) in the English system.2-15CProbably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid.If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the samerate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate.2-16CTwo systems having different temperatures and energy contents are brought in contact. The direction of heat transferis to be determined.AnalysisHeat transfer occurs from warmer to cooler objects. Therefore, heat will be transferred from system B to system Auntil both systems reach the same temperature.2-17A temperature is given inC. It is to be expressed in K.AnalysisThe Kelvin scale is related to Celsius scale byT(K] =T(C) + 273Thus,T(K] = 37C + 273 =310 K2-18EThe temperature of air given inC unit is to be converted toF and R unit.AnalysisUsing the conversion relations between the various temperature scales,R762F302460302460)F()R(32)150)(8.1(32)C(8.1)F(TTTT2-19A temperature change is given inC. It is to be expressed in K.AnalysisThis problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus,T(K] =T(C) =70K2-20EThe flash point temperature of engine oil given inF unit is to be converted to K and R units.AnalysisUsing the conversion relations between the various temperature scales,K457R8238.18231.8)R()K(460363460)F()R(TTTT

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2-62-21EThe temperature of ambient air given inC unit is to be converted toF, K and R units.AnalysisUsing the conversion relations between the various temperature scales,R419.67K233.15F4067.4594015.2734032)8.1)(40(C40TTT2-22EA temperature change is given inF. It is to be expressed inC, K, and R.AnalysisThis problem deals with temperature changes, which are identical in Rankine and Fahrenheit scales. Thus,T(R) =T(°F) = 45RThe temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in Fahrenheit andRankine scales byT(K) =T(R)/1.8 = 45/1.8 =25 KandT(C) =T(K) =25CPressure, Manometer, and Barometer2-23CThe atmospheric pressure, which is the external pressure exerted on the skin, decreases with increasing elevation.Therefore, the pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins andthe air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose toburst, causing bleeding. The shortness of breath is caused by the lower air density at higher elevations, and thus loweramount of oxygen per unit volume.2-24CThe blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to thebody. For a constant volume of blood to be discharged by the heart, the blood pressure must increase to overcome theincreased resistance to flow.2-25CNo, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is thegagepressurethat doubles when the depth is doubled.2-26CPascal’s principlestates thatthe pressureapplied to a confined fluid increases the pressure throughout by the sameamount. This is a consequence of the pressure in a fluid remaining constant in the horizontal direction. An example ofPascal’s principle is the operation of the hydraulic car jack.2-27CThe density of air at sea level is higher than the density of air on top of a high mountain. Therefore, the volume flowrates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will behigher.

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