Page 1 Chapter 2 Properties of Fluids Chapter 2 PROPERTIES OF FLUIDS Density and Specific Gravity 2-1C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 2-2C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ? H2O = 1000 kg/m 3 ). That is, H2O / ? ? = SG . When specific gravity is known, density is determined from H2O ? ? × = SG . 2-3C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 2-4C R u is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = R u /M, where M is the molar mass of the gas. 2-5 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is R u = 8.314 kPa.m 3 /kmol.K. The molar mass of helium is 4.0 kg/kmol. Analysis The volume of the sphere is 3 3 3 m 113.1 m) (3 3 4 3 4 = = = p p r V He D = 6 m 20 °C 200 kPa Assuming ideal gas behavior, the mole numbers of He is determined from kmol 9.286 = · · = = K) K)(293 /kmol m kPa (8.314 ) m kPa)(113.1 (200 3 3 T R P N u V Then the mass of He can be determined from kg 37.1 = = = kg/kmol) kmol)(4.0 (9.286 NM m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-1 Page 2 Chapter 2 Properties of Fluids Chapter 2 PROPERTIES OF FLUIDS Density and Specific Gravity 2-1C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 2-2C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ? H2O = 1000 kg/m 3 ). That is, H2O / ? ? = SG . When specific gravity is known, density is determined from H2O ? ? × = SG . 2-3C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 2-4C R u is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = R u /M, where M is the molar mass of the gas. 2-5 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is R u = 8.314 kPa.m 3 /kmol.K. The molar mass of helium is 4.0 kg/kmol. Analysis The volume of the sphere is 3 3 3 m 113.1 m) (3 3 4 3 4 = = = p p r V He D = 6 m 20 °C 200 kPa Assuming ideal gas behavior, the mole numbers of He is determined from kmol 9.286 = · · = = K) K)(293 /kmol m kPa (8.314 ) m kPa)(113.1 (200 3 3 T R P N u V Then the mass of He can be determined from kg 37.1 = = = kg/kmol) kmol)(4.0 (9.286 NM m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-1 Chapter 2 Properties of Fluids 2-6 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted. "Given Data" {D=6"[m]"} {P=200"[kPa]"} T=20"[C]" P=100"[kPa]" R_u=8.314"[kJ/kmol*K]" "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3"[m^3]" m=N*MOLARMASS(Helium)"[kg]" D [m] m [kg] 0.5 0.01075 2.111 0.8095 3.722 4.437 5.333 13.05 6.944 28.81 8.556 53.88 10.17 90.41 11.78 140.6 13.39 206.5 15 290.4 0 2 4 6 8 10 12 14 0 100 200 300 400 500 D [m] m [kg] Mass of Helium in Balloon as function of Diameter P = 200 kPa m [kg] P = 100 kPa PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-2 Page 3 Chapter 2 Properties of Fluids Chapter 2 PROPERTIES OF FLUIDS Density and Specific Gravity 2-1C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 2-2C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ? H2O = 1000 kg/m 3 ). That is, H2O / ? ? = SG . When specific gravity is known, density is determined from H2O ? ? × = SG . 2-3C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 2-4C R u is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = R u /M, where M is the molar mass of the gas. 2-5 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is R u = 8.314 kPa.m 3 /kmol.K. The molar mass of helium is 4.0 kg/kmol. Analysis The volume of the sphere is 3 3 3 m 113.1 m) (3 3 4 3 4 = = = p p r V He D = 6 m 20 °C 200 kPa Assuming ideal gas behavior, the mole numbers of He is determined from kmol 9.286 = · · = = K) K)(293 /kmol m kPa (8.314 ) m kPa)(113.1 (200 3 3 T R P N u V Then the mass of He can be determined from kg 37.1 = = = kg/kmol) kmol)(4.0 (9.286 NM m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-1 Chapter 2 Properties of Fluids 2-6 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted. "Given Data" {D=6"[m]"} {P=200"[kPa]"} T=20"[C]" P=100"[kPa]" R_u=8.314"[kJ/kmol*K]" "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3"[m^3]" m=N*MOLARMASS(Helium)"[kg]" D [m] m [kg] 0.5 0.01075 2.111 0.8095 3.722 4.437 5.333 13.05 6.944 28.81 8.556 53.88 10.17 90.41 11.78 140.6 13.39 206.5 15 290.4 0 2 4 6 8 10 12 14 0 100 200 300 400 500 D [m] m [kg] Mass of Helium in Balloon as function of Diameter P = 200 kPa m [kg] P = 100 kPa PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-2 Chapter 2 Properties of Fluids 2-7 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa ·m 3 /kg ·K. Analysis Initially, the absolute pressure in the tire is PP P gatm 1 =+ = + = 210 100 310 kPa Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from kPa 336 kPa) (310 K 298 K 323 1 1 2 2 2 2 2 1 1 1 = = = ?? ? = P T T P T P T P V V Tire 25 °C 210 kPa Thus the pressure rise is kPa 26 = - = - = ? 310 336 1 2 P P P The amount of air that needs to be bled off to restore pressure to its original value is kg 0.0070 = - = - = ? = · · = = = · · = = 0.0836 0.0906 kg 0.0836 K) K)(323 /kg m kPa (0.287 ) m kPa)(0.025 (310 kg 0.0906 K) K)(298 /kg m kPa (0.287 ) m kPa)(0.025 (310 2 1 3 3 2 2 2 3 3 1 1 1 m m m RT P m RT P m V V 2-8E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.3704 psia ·ft 3 /lbm ·R. Tire 0.53 ft 3 90 °F 20 psia Analysis The initial and final absolute pressures in the tire are P 1 = P g1 + P atm = 20 + 14.6 = 34.6 psia P 2 = P g2 + P atm = 30 + 14.6 = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is lbm 0.0900 R) R)(550 /lbm ft psia (0.3704 ) ft psia)(0.53 (34.6 3 3 1 1 1 = · · = = RT P m V Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes lbm 0.1160 R) R)(550 /lbm ft psia (0.3704 ) ft psia)(0.53 (44.6 3 3 2 2 2 = · · = = RT P m V Thus the amount of air that needs to be added is lbm 0.0260 = - = - = ? 0.0900 0.1160 1 2 m m m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-3 Page 4 Chapter 2 Properties of Fluids Chapter 2 PROPERTIES OF FLUIDS Density and Specific Gravity 2-1C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 2-2C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ? H2O = 1000 kg/m 3 ). That is, H2O / ? ? = SG . When specific gravity is known, density is determined from H2O ? ? × = SG . 2-3C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 2-4C R u is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = R u /M, where M is the molar mass of the gas. 2-5 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is R u = 8.314 kPa.m 3 /kmol.K. The molar mass of helium is 4.0 kg/kmol. Analysis The volume of the sphere is 3 3 3 m 113.1 m) (3 3 4 3 4 = = = p p r V He D = 6 m 20 °C 200 kPa Assuming ideal gas behavior, the mole numbers of He is determined from kmol 9.286 = · · = = K) K)(293 /kmol m kPa (8.314 ) m kPa)(113.1 (200 3 3 T R P N u V Then the mass of He can be determined from kg 37.1 = = = kg/kmol) kmol)(4.0 (9.286 NM m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-1 Chapter 2 Properties of Fluids 2-6 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted. "Given Data" {D=6"[m]"} {P=200"[kPa]"} T=20"[C]" P=100"[kPa]" R_u=8.314"[kJ/kmol*K]" "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3"[m^3]" m=N*MOLARMASS(Helium)"[kg]" D [m] m [kg] 0.5 0.01075 2.111 0.8095 3.722 4.437 5.333 13.05 6.944 28.81 8.556 53.88 10.17 90.41 11.78 140.6 13.39 206.5 15 290.4 0 2 4 6 8 10 12 14 0 100 200 300 400 500 D [m] m [kg] Mass of Helium in Balloon as function of Diameter P = 200 kPa m [kg] P = 100 kPa PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-2 Chapter 2 Properties of Fluids 2-7 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa ·m 3 /kg ·K. Analysis Initially, the absolute pressure in the tire is PP P gatm 1 =+ = + = 210 100 310 kPa Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from kPa 336 kPa) (310 K 298 K 323 1 1 2 2 2 2 2 1 1 1 = = = ?? ? = P T T P T P T P V V Tire 25 °C 210 kPa Thus the pressure rise is kPa 26 = - = - = ? 310 336 1 2 P P P The amount of air that needs to be bled off to restore pressure to its original value is kg 0.0070 = - = - = ? = · · = = = · · = = 0.0836 0.0906 kg 0.0836 K) K)(323 /kg m kPa (0.287 ) m kPa)(0.025 (310 kg 0.0906 K) K)(298 /kg m kPa (0.287 ) m kPa)(0.025 (310 2 1 3 3 2 2 2 3 3 1 1 1 m m m RT P m RT P m V V 2-8E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.3704 psia ·ft 3 /lbm ·R. Tire 0.53 ft 3 90 °F 20 psia Analysis The initial and final absolute pressures in the tire are P 1 = P g1 + P atm = 20 + 14.6 = 34.6 psia P 2 = P g2 + P atm = 30 + 14.6 = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is lbm 0.0900 R) R)(550 /lbm ft psia (0.3704 ) ft psia)(0.53 (34.6 3 3 1 1 1 = · · = = RT P m V Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes lbm 0.1160 R) R)(550 /lbm ft psia (0.3704 ) ft psia)(0.53 (44.6 3 3 2 2 2 = · · = = RT P m V Thus the amount of air that needs to be added is lbm 0.0260 = - = - = ? 0.0900 0.1160 1 2 m m m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-3 Chapter 2 Properties of Fluids 2-9E A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. v Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant. Properties The gas constant of air is R = 0.3704 psia ·ft 3 /lbm ·R. Analysis Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be lbm 33.73 R) R)(550 /lbm ft psia (0.3704 ) ft 3 psia)(196. (35 ft 196.3 psia 20 R) R)(530 /lbm ft psia 4 lbm)(0.370 (20 3 3 2 2 2 3 3 1 1 1 = · · = = = · · = = RT P m P RT m V V Air, 20 lbm 20 psia 70 °F Thus the amount of air added is lbm 13.7 = - = - = ? 20.0 33.73 1 2 m m m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-4 Page 5 Chapter 2 Properties of Fluids Chapter 2 PROPERTIES OF FLUIDS Density and Specific Gravity 2-1C Intensive properties do not depend on the size (extent) of the system but extensive properties do. 2-2C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ? H2O = 1000 kg/m 3 ). That is, H2O / ? ? = SG . When specific gravity is known, density is determined from H2O ? ? × = SG . 2-3C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure. 2-4C R u is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases. These two are related to each other by R = R u /M, where M is the molar mass of the gas. 2-5 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas. Properties The universal gas constant is R u = 8.314 kPa.m 3 /kmol.K. The molar mass of helium is 4.0 kg/kmol. Analysis The volume of the sphere is 3 3 3 m 113.1 m) (3 3 4 3 4 = = = p p r V He D = 6 m 20 °C 200 kPa Assuming ideal gas behavior, the mole numbers of He is determined from kmol 9.286 = · · = = K) K)(293 /kmol m kPa (8.314 ) m kPa)(113.1 (200 3 3 T R P N u V Then the mass of He can be determined from kg 37.1 = = = kg/kmol) kmol)(4.0 (9.286 NM m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-1 Chapter 2 Properties of Fluids 2-6 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. The effect of the balloon diameter on the mass of helium is to be investigated, and the results are to be tabulated and plotted. "Given Data" {D=6"[m]"} {P=200"[kPa]"} T=20"[C]" P=100"[kPa]" R_u=8.314"[kJ/kmol*K]" "Solution" P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3"[m^3]" m=N*MOLARMASS(Helium)"[kg]" D [m] m [kg] 0.5 0.01075 2.111 0.8095 3.722 4.437 5.333 13.05 6.944 28.81 8.556 53.88 10.17 90.41 11.78 140.6 13.39 206.5 15 290.4 0 2 4 6 8 10 12 14 0 100 200 300 400 500 D [m] m [kg] Mass of Helium in Balloon as function of Diameter P = 200 kPa m [kg] P = 100 kPa PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-2 Chapter 2 Properties of Fluids 2-7 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa ·m 3 /kg ·K. Analysis Initially, the absolute pressure in the tire is PP P gatm 1 =+ = + = 210 100 310 kPa Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from kPa 336 kPa) (310 K 298 K 323 1 1 2 2 2 2 2 1 1 1 = = = ?? ? = P T T P T P T P V V Tire 25 °C 210 kPa Thus the pressure rise is kPa 26 = - = - = ? 310 336 1 2 P P P The amount of air that needs to be bled off to restore pressure to its original value is kg 0.0070 = - = - = ? = · · = = = · · = = 0.0836 0.0906 kg 0.0836 K) K)(323 /kg m kPa (0.287 ) m kPa)(0.025 (310 kg 0.0906 K) K)(298 /kg m kPa (0.287 ) m kPa)(0.025 (310 2 1 3 3 2 2 2 3 3 1 1 1 m m m RT P m RT P m V V 2-8E An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.3704 psia ·ft 3 /lbm ·R. Tire 0.53 ft 3 90 °F 20 psia Analysis The initial and final absolute pressures in the tire are P 1 = P g1 + P atm = 20 + 14.6 = 34.6 psia P 2 = P g2 + P atm = 30 + 14.6 = 44.6 psia Treating air as an ideal gas, the initial mass in the tire is lbm 0.0900 R) R)(550 /lbm ft psia (0.3704 ) ft psia)(0.53 (34.6 3 3 1 1 1 = · · = = RT P m V Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes lbm 0.1160 R) R)(550 /lbm ft psia (0.3704 ) ft psia)(0.53 (44.6 3 3 2 2 2 = · · = = RT P m V Thus the amount of air that needs to be added is lbm 0.0260 = - = - = ? 0.0900 0.1160 1 2 m m m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-3 Chapter 2 Properties of Fluids 2-9E A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. v Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant. Properties The gas constant of air is R = 0.3704 psia ·ft 3 /lbm ·R. Analysis Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be lbm 33.73 R) R)(550 /lbm ft psia (0.3704 ) ft 3 psia)(196. (35 ft 196.3 psia 20 R) R)(530 /lbm ft psia 4 lbm)(0.370 (20 3 3 2 2 2 3 3 1 1 1 = · · = = = · · = = RT P m P RT m V V Air, 20 lbm 20 psia 70 °F Thus the amount of air added is lbm 13.7 = - = - = ? 20.0 33.73 1 2 m m m PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-4 Chapter 2 Properties of Fluids 2-10 The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. v Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km. Properties The density data are given in tabular form as r, km z, km ?, kg/m 3 6377 0 1.225 6378 1 1.112 6379 2 1.007 6380 3 0.9093 6381 4 0.8194 6382 5 0.7364 6383 6 0.6601 6385 8 0.5258 6387 10 0.4135 6392 15 0.1948 6397 20 0.08891 6402 25 0.04008 Analysis Using EES, (1) Define a trivial function rho= a+z in equation window, (2) select new parametric table from Tables, 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 2 nd order polynomial and enter/edit equation. The results are: 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 1.2 1.4 z, km ?, kg/m 3 ?(z) = a + bz + cz 2 = 1.20252 – 0.101674z + 0.0022375z 2 for the unit of kg/m 3 , (or, ?(z) = (1.20252 – 0.101674z + 0.0022375z 2 ) ×10 9 for the unit of kg/km 3 ) where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give ? = 0.60 kg/m 3 . PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-5Read More

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!