The below are some of the examples worked out to calculate specific weight, specific volume, specific gravity and density of substance(matter) at given conditions:
Problem 1:
At temperature of 100° F and 120 psi (pressure per square inch) absolute, calculation of above properties for methane (CH4):
solution:
By using equation of state = PV= nRT ( R= 96.3 ft/°R)
specific weight = P/RT = n/V = 120 × 144 / [96.3 (460+100)] = 0.320 lb/ft³ (note ° R= 460 + ° F)
density = specific weight / gravity = 0.320/32.2 = 0.00994 slug/ft³
specific volume = 1/density = 1/0.00994 = 101 ft³/slug
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Problem 2:
For oil of volume 6 m³ and weighs 47 kN
solution:
specific weight = 47 kN / 6 m³ = 7.833 kN/m³
density =7833/9.81 = 798 kg/m³
specific gravity = specific weight of oil /specific weight of water = 7.833/9.79 =0.800
Note: specific gravity of liquid is the ratio of specific weight of particular liquid to specific of water at same conditions, and it specific gravity as no units it has only a numerical value.
specific gravity of substance is calculated as: = weight of substance / weight of equal volume of water
= specific weight of substance / specific weight of water
= density of substance / density of water
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Problem 3:
At 90° F and 300psi absolute, find gas constant and density for a gas which has 11.4 ft³/ lb.
solution:
Data given is the condition of gas i.e temperature and pressure and specific volume
R= P/(specific volume . T) = [(30.0 × 144)11.4]/460+90 = 89.5 ft/° R
density = specific weight / gravity = 1 / ( volume × gravity ) = 1 / (11.4 × 32.2) = 0.00272 slugs/ft³
Problem 1:
At temperature of 100° F and 120 psi (pressure per square inch) absolute, calculation of above properties for methane (CH4):
solution:
By using equation of state = PV= nRT ( R= 96.3 ft/°R)
specific weight = P/RT = n/V = 120 × 144 / [96.3 (460+100)] = 0.320 lb/ft³ (note ° R= 460 + ° F)
density = specific weight / gravity = 0.320/32.2 = 0.00994 slug/ft³
specific volume = 1/density = 1/0.00994 = 101 ft³/slug
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Problem 2:
For oil of volume 6 m³ and weighs 47 kN
solution:
specific weight = 47 kN / 6 m³ = 7.833 kN/m³
density =7833/9.81 = 798 kg/m³
specific gravity = specific weight of oil /specific weight of water = 7.833/9.79 =0.800
Note: specific gravity of liquid is the ratio of specific weight of particular liquid to specific of water at same conditions, and it specific gravity as no units it has only a numerical value.
specific gravity of substance is calculated as: = weight of substance / weight of equal volume of water
= specific weight of substance / specific weight of water
= density of substance / density of water
_____________________________________________________________________________________
Problem 3:
At 90° F and 300psi absolute, find gas constant and density for a gas which has 11.4 ft³/ lb.
solution:
Data given is the condition of gas i.e temperature and pressure and specific volume
R= P/(specific volume . T) = [(30.0 × 144)11.4]/460+90 = 89.5 ft/° R
density = specific weight / gravity = 1 / ( volume × gravity ) = 1 / (11.4 × 32.2) = 0.00272 slugs/ft³