1-4 Review roblems 1-85 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. Assumptions 1 The cylinders of the lift are vertical. There are no leaks. Atmospheric pressure act on both sides, and thus it can be disregarded. Analysis Noting that pressure is force per unit area, the pressure on the smaller piston is determined from F1 m1g 1 A πd / 4 1 1 (5 kg)(9.81 m/s ) 1kN π (.1 m) / 4 1 kg m/s 5 kg F 1 1 cm Weight 5 kg 1. kn/m 1. ka From ascal s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from F m g (5 kg)(9.81 m/s ) 1kN 1 1. kn/m D 1. m A πd / 4 πd / 4 1 kg m/s Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of ascal s principle. F D 1-86 A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis The gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield WEIGTHS GAS m piston g (5 kg)(9.81 m/s ) 1kN atm 1 ka 95.66 kn/m 95.7 ka A (.1 m )/4 π 1 kg m/s The force balance when the weights are placed is used to determine the mass of the weights ( mpiston + mweights ) g atm + A (5 kg + mweights )(9.81m/s ) + 1kN ka 95.66 ka m 115. kg weights π (.1 m )/4 1 kg m/s A large mass is needed to double the pressure.
1-5 1-87 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. roperties The densities of air and mercury are given to be 1.15 kg/m and 1,6 kg/m. Analysis The local atmospheric pressure is determined from + ρgh atm plane 1kN 58 ka + (1.15 kg/m )(9.81m/s )( m) 1 kg m/s The atmospheric pressure may be expressed in mmhg as h Hg ρg atm 91.8 ka (1,6 kg/m )(9.81m/s 91.84 kn/m 1 a 1 mm 688 mmhg ) 1ka 1m 91.8 ka 1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 8-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation 6 W mg (8kg)(9.87. 1 z m/s Sea level: Denver: Mt. Ev.: 1N ) 1kg m/s (z m): W 8 (9.87-.x1-6 ) 8 9.87 784.6 N (z 161 m): W 8 (9.87-.x1-6 161) 8 9.8 784. N (z 8848 m): W 8 (9.87-.x1-6 8848) 8 9.778 78. N 1-89 A man is considering buying a 1-oz steak for $.15, or a -g steak for $.8. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be $.15 16 oz 1lbm 1 ounce steak: Unit Cost $9.6/kg 1 oz 1lbm.4559 kg gram steak: $.8 1 g Unit Cost $8.75/kg g 1kg Therefore, the steak at the international market is a better buy.
1-6 1-9 The thrust developed by the jet engine of a Boeing 777 is given to be 85, pounds. This thrust is to be expressed in N and kgf. Analysis Noting that 1 lbf 4.448 N and 1 kgf 9.81 N, the thrust developed can be expressed in two other units as Thrust in N: Thrust in kgf: 4.448 N Thrust (85, lbf ).78 1 1lbf Thrust (7.8 1 5 N) 1kgf 9.81N 5.85 1 4 N kgf 1-91E The efficiency of a refrigerator increases by % per C rise in the minimum temperature. This increase is to be expressed per F, K, and R rise in the minimum temperature. Analysis The magnitudes of 1 K and 1 C are identical, so are the magnitudes of 1 R and 1 F. Also, a change of 1 K or 1 C in temperature corresponds to a change of 1.8 R or 1.8 F. Therefore, the increase in efficiency is (a) % for each K rise in temperature, and (b), (c) /1.8 1.67% for each R or F rise in temperature. 1-9E The boiling temperature of water decreases by C for each 1 m rise in altitude. This decrease in temperature is to be expressed in F, K, and R. Analysis The magnitudes of 1 K and 1 C are identical, so are the magnitudes of 1 R and 1 F. Also, a change of 1 K or 1 C in temperature corresponds to a change of 1.8 R or 1.8 F. Therefore, the decrease in the boiling temperature is (a) K for each 1 m rise in altitude, and (b), (c) 1.8 5.4 F 5.4 R for each 1 m rise in altitude. 1-9E The average body temperature of a person rises by about C during strenuous exercise. This increase in temperature is to be expressed in F, K, and R. Analysis The magnitudes of 1 K and 1 C are identical, so are the magnitudes of 1 R and 1 F. Also, a change of 1 K or 1 C in temperature corresponds to a change of 1.8 R or 1.8 F. Therefore, the rise in the body temperature during strenuous exercise is (a) K (b) 1.8.6 F (c) 1.8.6 R
1-7 1-94E Hyperthermia of 5 C is considered fatal. This fatal level temperature change of body temperature is to be expressed in F, K, and R. Analysis The magnitudes of 1 K and 1 C are identical, so are the magnitudes of 1 R and 1 F. Also, a change of 1 K or 1 C in temperature corresponds to a change of 1.8 R or 1.8 F. Therefore, the fatal level of hypothermia is (a) 5 K (b) 5 1.8 9 F (c) 5 1.8 9 R 1-95E A house is losing heat at a rate of 45 kj/h per C temperature difference between the indoor and the outdoor temperatures. The rate of heat loss is to be expressed per F, K, and R of temperature difference between the indoor and the outdoor temperatures. Analysis The magnitudes of 1 K and 1 C are identical, so are the magnitudes of 1 R and 1 F. Also, a change of 1 K or 1 C in temperature corresponds to a change of 1.8 R or 1.8 F. Therefore, the rate of heat loss from the house is (a) 45 kj/h per K difference in temperature, and (b), (c) 45/1.8 5 kj/h per R or F rise in temperature. 1-96 The average temperature of the atmosphere is expressed as T atm 88.15 6.5z where z is altitude in km. The temperature outside an airplane cruising at 1, m is to be determined. Analysis Using the relation given, the average temperature of the atmosphere at an altitude of 1, m is determined to be T atm 88.15-6.5z 88.15-6.5 1 1.15 K - 6 C Discussion This is the average temperature. The actual temperature at different times can be different. 1-97 A new Smith absolute temperature scale is proposed, and a value of 1 S is assigned to the boiling point of water. The ice point on this scale, and its relation to the Kelvin scale are to be determined. Analysis All linear absolute temperature scales read zero at absolute zero pressure, and are constant multiples of each other. For example, T(R) 1.8 T(K). That is, multiplying a temperature value in K by 1.8 will give the same temperature in R. K S The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only. The boiling temperature of water in the Kelvin and the Smith scales are 15.15 K and 1 K, respectively. Therefore, these two temperature scales are related to each other by 1 TS ( ) TK ( ) 6799T. ( K ) 7.15 The ice point of water on the Smith scale is 7.15 1 T(S) ice.6799 T(K) ice.6799 7.15 7. S
1-8 1-98E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are 1 mph 1.69 km/h and T( F) 1.8T( C) +. The first thought that comes to mind is to replace T( F) in the equation by its equivalent 1.8T( C) +, and V in mph by 1.69 km/h, which is the regular way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.69 to convert it to the desired unit of mph. That is, or T ( F) 914. [ 914. T ( F)][. 475. ( V / 169. ) +. 4 V / 169. ] equiv ambient T ( F) 914. [ 914. T ( F)][. 475. 16V +. 4 V ] equiv ambient where V is in km/h. Now the problem reduces to converting a temperature in F to a temperature in C, using the proper convection relation: 1. 8T ( C) + 914. [ 914. ( 18. T ( C) + )][. 475. 16V +. 4 V ] equiv which simplifies to ambient T ( C). (. T )(. 475. 16V +. 4 V ) equiv ambient where the ambient air temperature is in C.
1-9 1-99E EES roblem 1-98E is reconsidered. The equivalent wind-chill temperatures in F as a function of wind velocity in the range of 4 mph to 1 mph for the ambient temperatures of, 4, and 6 F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Obtain V and T_ambient from the Diagram Window" {T_ambient1 V} V_usemax(V,4) T_equiv91.4-(91.4-T_ambient)*(.475 -.*V_use +.4*sqrt(V_use)) "The parametric table was used to generate the plot, Fill in values for T_ambient and V (use Alter Values under Tables menu) then use F to solve table. lot the first 1 rows and then overlay the second ten, and so on. lace the text on the plot using Add Text under the lot menu." T equiv T ambient V [mph] [F] [F] -5-5 1-46 - 1-4 -15 1-4 -1 1-7 -5 1-1 1-15 5 1-9 1 1-15 1 1-75 -5-68 - -61-15 -5-1 -46-5 -9-5 -5 1-18 15-11 -87-5 -79 - -7-15 -64-1 -56-5 -49-41 5-1 -6 15-18 -9-5 4-85 - 4-77 -15 4-69 -1 4-61 -5 4-54 4-46 5 4-8 1 4-15 4-4 T WindChill ] [F u iv T eq 1-1 - - -4-5 -6-7 Wind Chill Temperature Wind speed 1 mph mph mph 4 mph -8 - - -1 1 6 5 4 1-1 - T ambient T amb 6F T amb 4F T amb F 4 6 8 1 V [mph]
1-4 1-1 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). The weight of the duct and the air in is negligible. roperties The density of air is given to be ρ 1. kg/m. We take the density of water to be 1 kg/m. D 15 cm Analysis Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy L m F force exerted by water. The volume of the underground section of B the duct is V AL ( π D / 4) L [ π (.15 m) /4]( m).5 m Then the buoyancy force becomes 1kN F B ρ gv (1 kg/m )(9.81 m/s )(.5 m ).46 kn 1 kg m/s Discussion The upward force exerted by water on the duct is.46 kn, which is equivalent to the weight of a mass of 5 kg. Therefore, this force must be treated seriously. 1-11 A helium balloon tied to the ground carries people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. roperties The density of air is given to be ρ 1.16 kg/m. The density of helium gas is 1/7 th of this. Analysis The buoyancy force acting on the balloon is V balloon F B The total mass is 4πr / 4π(5 m) / 5.6 m ρ air gv balloon 1 N (1.16 kg/m )(9.81m/s )(5.6 m ) 5958 N 1 kg m/s m m He total The total weight is 1.16 ρ HeV kg/m (5.6 m ) 86.8 kg 7 m + m 86.8 + 7 6.8 kg He people 1 N W m g (6.8 kg)(9.81 m/s ) 1 kg m/s Thus the net force acting on the balloon is F W 5958 5 7 N total Fnet B Then the acceleration becomes F a m net total 7 N 1kg m/s 6.8 kg 1 N 16.5 m/s 5 N m 14 kg Helium balloon
1-41 1-1 EES roblem 1-11 is reconsidered. The effect of the number of people carried in the balloon on acceleration is to be investigated. Acceleration is to be plotted against the number of people, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given Data:" rho_air1.16"[kg/m^]" "density of air" g9.87"[m/s^]" d_balloon1"[m]" m_1person7"[kg]" {Noeople } "Data suppied in arametric Table" "Calculated values:" rho_herho_air/7"[kg/m^]" "density of helium" r_balloond_balloon/"[m]" V_balloon4*pi*r_balloon^/"[m^]" m_peoplenoeople*m_1person"[kg]" m_herho_he*v_balloon"[kg]" m_totalm_he+m_people"[kg]" "The total weight of balloon and people is:" W_totalm_total*g"[N]" "The buoyancy force acting on the balloon, F_b, is equal to the weight of the air displaced by the balloon." F_brho_air*V_balloon*g"[N]" "From the free body diagram of the balloon, the balancing vertical forces must equal the product of the total mass and the vertical acceleration:" F_b- W_totalm_total*a_up A up [m/s ] Noeople 8.19 1 16.46 1.6 6.44 4.81 5 1.947 6.54 7 -.597 8-1.497 9 -.6 1 a up [m/s^] 5 15 1 5-5 1 4 5 6 7 8 9 1 Noeople
1-4 1-1 A balloon is filled with helium gas. The maximum amount of load the balloon can carry is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. roperties The density of air is given to be ρ 1.16 kg/m. The density of helium gas is 1/7th of this. Analysis In the limiting case, the net force acting on the balloon will be zero. That is, the buoyancy force and the weight will balance each other: W mg F Thus, m m F 5958 N 9.81 m/s B total people g B 67. kg m m 67. 86.8 5.5 kg total He Helium balloon m 1-14E The pressure in a steam boiler is given in kgf/cm. It is to be expressed in psi, ka, atm, and bars. Analysis We note that 1 atm 1. kgf/cm, 1 atm 14.696 psi, 1 atm 11.5 ka, and 1 atm 1.15 bar (inner cover page of text). Then the desired conversions become: In atm: In psi: In ka: In bars: (9 kgf/cm (9 kgf/cm (9 kgf/cm (9 kgf/cm 1 atm ) 1. kgf/cm 1 atm ) 1. kgf/cm 1 atm ) 1. kgf/cm 1 atm ) 1. kgf/cm 89.4 atm 14.696 psi 19 psi 1 atm 11.5 ka 9 ka 1 atm 1.15 bar 9. 1 atm Discussion Note that the units atm, kgf/cm, and bar are almost identical to each other. bar
1-4 1-15 A barometer is used to measure the altitude of a plane relative to the ground. The barometric readings at the ground and in the plane are given. The altitude of the plane is to be determined. Assumptions The variation of air density with altitude is negligible. roperties The densities of air and mercury are given to be ρ 1. kg/m and ρ 1,6 kg/m. Analysis Atmospheric pressures at the location of the plane and the ground level are plane ( ρ gh) plane 1 N 1 ka (1,6 kg/m )(9.81 m/s )(.69 m) 1 kg m/s 1 N/m 9.6 ka ground ( ρ gh) ground 1 N (1,6 kg/m )(9.81 m/s )(.75 m) 1 kg m/s 1.46 ka 1 ka 1 N/m Taking an air column between the airplane and the ground and writing a force balance per unit base area, we obtain W / A 1 N (1. kg/m )(9.81 m/s )( h) 1 kg m/s It yields h 714 m which is also the altitude of the airplane. air ( ρ gh) 1 ka 1 N/m air ground ground plane plane (1.46 9.6) ka h Sea level 1-16 A 1-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. roperties The density of water is given to be ρ 1 kg/m. The specific gravity of oil is given to be.85. Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water, ρ SG ρ HO (.85)(1 kg/m ) 85 kg/m The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, total + ( ρgh) + ( ρgh) Oil SG.85 oil water oil water [(85 kg/m )(9.81 m/s )(5 m) + (1 kg/m )(9.81 m/s )(5 m) ] 9.7 ka Water 1 ka 1 N/m h 1 m
1-44 1-17 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 5 ka. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder. atm Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield W A A It yields mg ( atm atm ) A ( m)(9.81 m/s ) (5 1 ka)( 1 m 45.9 kg 4 1 kg/m s m ) 1ka W mg 1-18 The gage pressure in a pressure cooker is maintained constant at 1 ka by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. Analysis Atmospheric pressure is acting on all surfaces of the petcock, which atm balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣF y ) yields W gage m g gage A.48 kg A (1 ka)(4 1 9.81 m/s 6 m ) 1 kg/m s 1 ka W mg 1-19 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is measured. It is to be determined how high the water will rise in the tube. roperties The density of water is given to be ρ 1 kg/m. Analysis The pressure at the bottom of the tube can be expressed as + ( ρ gh Solving for h, h ρ g atm ) tube atm (115 9) ka (1 kg/m )(9.81 m/s.4 m 1 kg m/s ) 1 N 1 N/m 1 ka Water atm 9 ka h
1-45 1-11 The average atmospheric pressure is given as atm 11.5(1.56z) where z is the altitude in km. The atmospheric pressures at various locations are to be determined. Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation atm 115. ( 1. 56z) 5. 56 Atlanta: (z.6 km): atm 11.5(1 -.56.6) 5.56 97.7 ka Denver: (z 1.61 km): atm 11.5(1 -.56 1.61) 5.56 8.4 ka M. City: (z.9 km): atm 11.5(1 -.56.9) 5.56 76.5 ka Mt. Ev.: (z 8.848 km): atm 11.5(1 -.56 8.848) 5.56 1.4 ka 5.56 1-111 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. roperties The density of the liquid is given to be ρ.81 kg/l 81 kg/m. Analysis The gage pressure in the duct is determined from gage abs atm ρgh 1 N (81 kg/m )(9.81 m/s )(.8 m) 66 a 1 kg m/s The length of the differential fluid column is L h / sinθ (8 cm) / sin 5 1.9 cm 1 a 1 N/m Fresh Water Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability. L 5 8 cm 1-11E Equal volumes of water and oil are poured into a U-tube from different arms, and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same. The excess pressure applied on the oil side is to be determined. Assumptions 1 Both water and oil are incompressible substances. Oil does not mix with water. The cross-sectional area of the U-tube is constant. roperties The density of oil is given to be ρ oil 49. lbm/ft. We take the density of water to be ρ w 6.4 lbm/ft. Analysis Noting that the pressure of both the water and the oil is the same at the contact surface, the pressure at this surface can be expressed as + ρ gh + ρ gh contact blow a a Noting that h a h w and rearranging, gage,blow blow atm ( ρ w ρ oil ) gh 1 lbf (6.4-49. lbm/ft )(. ft/s )(/1 ft).7 psi. lbm ft/s atm w w in Water 1 ft 144 in Blown air Discussion When the person stops blowing, the oil will rise and some water will flow into the right arm. It can be shown that when the curvature effects of the tube are disregarded, the differential height of water will be.7 in to balance -in of oil. Oil
1-46 1-11 It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height, and a certain gage pressure at the arm level is needed for sufficient flow rate. The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined. Assumptions 1 The IV fluid is incompressible. The IV bottle is open to the atmosphere. roperties The density of the IV fluid is given to be ρ 1 kg/m. Analysis (a) Noting that the IV fluid and the blood pressures balance each other when the bottle is 1. m above the arm level, the gage pressure of the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 1. m, gage, arm abs atm ρgharm-bottle 1 kn 1 ka (1 kg/m )(9.81 m/s )(1. m) 1. ka 1 kg m/s 1 kn/m (b) To provide a gage pressure of ka at the arm level, the height of the bottle from the arm level is again determined from ρgh to be h arm-bottle gage, arm ρg gage, arm arm-bottle IV Bottle ka 1 kg m/s 1 kn/m. m (1 kg/m )(9.81 m/s ) 1 kn 1 ka Discussion Note that the height of the reservoir can be used to control flow rates in gravity driven flows. When there is flow, the pressure drop in the tube due to friction should also be considered. This will result in raising the bottle a little higher to overcome pressure drop. atm 1. m
1-47 1-114 A gasoline line is connected to a pressure gage through a double-u manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Assumptions 1 All the liquids are incompressible. The effect of air column on pressure is negligible. roperties The specific gravities of oil, mercury, and gasoline are given to be.79, 1.6, and.7, respectively. We take the density of water to be ρ w 1 kg/m. Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρ gh terms until we reach the gasoline pipe, and setting the result equal to gasoline gives Air Water gage 7 ka Oil 45 cm cm 5 cm 1 cm Mercury Gasoline ρ ghw + ρ gh ρ gh ρ gh gage w oil oil Hg Hg gasoline gasoline gasoline Rearranging, ρ g hw SG h + SG h + SG Substituting, gasoline gasoline gage w 7 ka - (1 kg/m ( oil oil Hg Hg gasolinehgasoline )(9.81m/s )[(.45 m).79(.5 m) + 1.6(.1 m) +.7(. m)] 1kN 1ka 1 kg m/s 1kN/m 54.6 ka Therefore, the pressure in the gasoline pipe is 15.4 ka lower than the pressure reading of the pressure gage. Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several fluids. )
1-48 1-115 A gasoline line is connected to a pressure gage through a double-u manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Assumptions 1 All the liquids are incompressible. The effect of air column on pressure is negligible. roperties The specific gravities of oil, mercury, and gasoline are given to be.79, 1.6, and.7, respectively. We take the density of water to be ρ w 1 kg/m. Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρ gh terms until we reach the gasoline pipe, and setting the result equal to gasoline gives Air Water gage 18 ka Oil 45 cm cm 5 cm 1 cm Mercury Gasoline gage ρ w ghw + ρ oil ghoil ρ Hg ghhg ρ gasoline ghgasoline Rearranging, ρ g hw SG h + SG h + SG Substituting, gasoline gasoline gage 1kN 1 kg m/s 164.6 ka w 18 ka - (1 kg/m gasoline ( oil oil Hg Hg gasolinehgasoline 1ka 1kN/m )(9.87 m/s )[(.45 m).79(.5 m) + 1.6(.1 m) +.7(. m)] Therefore, the pressure in the gasoline pipe is 15.4 ka lower than the pressure reading of the pressure gage. Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several fluids. )
1-49 1-116E A water pipe is connected to a double-u manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Assumptions 1 All the liquids are incompressible. The solubility of the liquids in each other is negligible. roperties The specific gravities of mercury and oil are given to be 1.6 and.8, respectively. We take the density of water to be ρ w 6.4 lbm/ft. Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ρ gh terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to atm gives 5 in Water 6 in Oil 4 in 15 in Oil ρ gh + ρ gh ρ gh ρ gh water pipe water water oil oil Hg Hg oil oil atm Mercury Solving for water pipe, atm + ρ g h SG h + SG h + SG h ) Substituting, water pipe water pipe water 14.psia + (6.4lbm/ft 1 lbf +.8(4/1 ft)]. psia. lbm ft/s ( water oil oil Hg Hg oil oil )(. ft/s )[(5/1 ft).8(6/1 ft) + 1.6(15/1 ft) 1 ft 144 in Therefore, the absolute pressure in the water pipe is. psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.
1-5 1-117 The temperature of the atmosphere varies with altitude z as T T βz, while the gravitational acceleration varies by g ( z) g /(1 + z / 6,7,). Relations for the variation of pressure in atmosphere are to be obtained (a) by ignoring and (b) by considering the variation of g with altitude. Assumptions The air in the troposphere behaves as an ideal gas. Analysis (a) ressure change across a differential fluid layer of thickness dz in the vertical z direction is d ρgdz From the ideal gas relation, the air density can be expressed as ρ. Then, RT R( T βz) d gdz R( T βz) Separating variables and integrating from z where to z z where, d z gdz R T β ) ( z erforming the integrations. g T βz ln ln Rβ T Rearranging, the desired relation for atmospheric pressure for the case of constant g becomes g βr βz 1 T (b) When the variation of g with altitude is considered, the procedure remains the same but the expressions become more complicated, g d dz R( T βz) (1 + z / 6,7, ) Separating variables and integrating from z where to z z where, d dz z R( T βz)(1 + z / 6,7,) erforming the integrations, ln g z g 1 1 1+ kz ln Rβ (1 + kt / β )(1 + kz) (1 + kt / β ) T βz where R 87 J/kg K 87 m /s K is the gas constant of air. After some manipulations, we obtain g exp R( β + kt 1 1 + ) 1+ 1/ kz 1+ kt 1+ kz ln / β 1 βz / T where T 88.15 K, β.65 K/m, g 9.87 m/s, k 1/6,7, m -1, and z is the elevation in m.. Discussion When performing the integration in part (b), the following expression from integral tables is used, together with a transformation of variable x T βz, dx 1 1 a + bx ln x( a + bx) a( a + bx) a x Also, for z 11, m, for example, the relations in (a) and (b) give.6 and.69 ka, respectively.
1-51 1-118 The variation of pressure with density in a thick gas layer is given. A relation is to be obtained for pressure as a function of elevation z. Assumptions The property relation Cρ n is valid over the entire region considered. Analysis The pressure change across a differential fluid layer of thickness dz in the vertical z direction is given as, d ρgdz n n n / Also, the relation Cρ can be expressed as C / ρ ρ, and thus ρ ρ Substituting, 1/ n ( / ) d gρ ( / ) 1/ n dz n ρ Separating variables and integrating from z where C to z z where, 1/ n ( / ) d ρ g erforming the integrations. Solving for, 1/ n+ 1 ( / ) 1/ n + 1 n 1 ρ gz 1 n z dz ρ gz n /( n 1) ( n 1) / n n 1 ρ gz 1 n which is the desired relation. Discussion The final result could be expressed in various forms. The form given is very convenient for calculations as it facilitates unit cancellations and reduces the chance of error.
1-5 1-119 A pressure transducers is used to measure pressure by generating analogue signals, and it is to be calibrated by measuring both the pressure and the electric current simultaneously for various settings, and the results are tabulated. A calibration curve in the form of ai + b is to be obtained, and the pressure corresponding to a signal of 1 ma is to be calculated. Assumptions Mercury is an incompressible liquid. roperties The specific gravity of mercury is given to be 1.56, and thus its density is 1,56 kg/m. Analysis For a given differential height, the pressure can be calculated from ρ g h For h 8. mm.8 m, for example, 1kN 1.56(1 kg/m )(9.81m/s )(.8 m) 1 kg m/s Repeating the calculations and tabulating, we have 1ka 1kN/m.75 ka h(mm) 8. 181.5 97.8 41.1 765.9 17 1149 16 1458 156 (ka).7 4.14 9.61 54.95 11.9 16.6 15.8 181. 19.9 4. I (ma) 4.1 5.78 6.97 8.15 11.76 14.4 15.68 17.86 18.84 19.64 A plot of versus I is given below. It is clear that the pressure varies linearly with the current, and using EES, the best curve fit is obtained to be 1.I - 51. (ka) for 4. 1 I 19. 64. Multimeter For I 1 ma, for example, we would get 79. ka. 5 ressure transducer 18 Valve, ka 15 9 ressurized Air, 45 Rigid container h Manometer 4 6 8 1 1 14 16 18 I, ma Mercury SG1.5 6 Discussion Note that the calibration relation is valid in the specified range of currents or pressures.
1-5 Fundamentals of Engineering (FE) Exam roblems 1-1 Consider a fish swimming 5 m below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of 45 m below the free surface is (a) 9 a (b) 98 a (c) 5, a (d) 9, a (e) 441, a Answer (d) 9, a Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1 "kg/m" g9.81 "m/s" z15 "m" z45 "m" DELTArho*g*(z-z1) "a" "Some Wrong Solutions with Common Mistakes:" W1_rho*g*(z-z1)/1 "dividing by 1" W_rho*g*(z1+z) "adding depts instead of subtracting" W_rho*(z1+z) "not using g" W4_rho*g*(+z) "ignoring z1" 1-11 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96. and 98. ka. If the density of air is 1. kg/m, the height of the building is (a) 17 m (b) m (c) 17 m (d) 4 m (e) 5 m Answer (d) 4 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1. "kg/m" g9.81 "m/s" 196 "ka" 98 "ka" DELTA-1 "ka" DELTArho*g*h/1 "ka" "Some Wrong Solutions with Common Mistakes:" DELTArho*W1_h/1 "not using g" DELTAg*W_h/1 "not using rho" rho*g*w_h/1 "ignoring 1" 1rho*g*W4_h/1 "ignoring "
1-54 1-1 An apple loses 4.5 kj of heat as it cools per C drop in its temperature. The amount of heat loss from the apple per F drop in its temperature is (a) 1.5 kj (b).5 kj (c) 5. kj (d) 8.1 kj (e) 4.1 kj Answer (b).5 kj Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_perC4.5 "kj" Q_perFQ_perC/1.8 "kj" "Some Wrong Solutions with Common Mistakes:" W1_QQ_perC*1.8 "multiplying instead of dividing" W_QQ_perC "setting them equal to each other" 1-1 Consider a -m deep swimming pool. The pressure difference between the top and bottom of the pool is (a) 1. ka (b) 19.6 ka (c) 8.1 ka (d) 5.8 ka (e) ka Answer (b) 19.6 ka Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1 "kg/m^" g9.81 "m/s" z1 "m" z "m" DELTArho*g*(z-z1)/1 "ka" "Some Wrong Solutions with Common Mistakes:" W1_rho*(z1+z)/1 "not using g" W_rho*g*(z-z1)/ "taking half of z" W_rho*g*(z-z1) "not dividing by 1"
1-55 1-14 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is (a) 1 lbf (b) 9.81 lbf (c). lbf (d).1 lbf (e).1 lbf Answer (a) 1 lbf Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m1 "lbm" g. "ft/s" Wm*g/. "lbf" "Some Wrong Solutions with Common Mistakes:" gsi9.81 "m/s" W1_W m*gsi "Using wrong conversion" W_W m*g "Using wrong conversion" W_W m/gsi "Using wrong conversion" W4_W m/g "Using wrong conversion" 1-15 During a heating process, the temperature of an object rises by C. This temperature rise is equivalent to a temperature rise of (a) F (b) 5 F (c) 6 K (d) 6 R (e) 9 K Answer (d) 6 R Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_inC "C" T_inRT_inC*1.8 "R" "Some Wrong Solutions with Common Mistakes:" W1_TinFT_inC "F, setting C and F equal to each other" W_TinFT_inC*1.8+ "F, converting to F " W_TinK1.8*T_inC "K, wrong conversion from C to K" W4_TinKT_inC+7 "K, converting to K" 1-16 1-19 Design, Essay, and Experiment roblems