A vertical cylindrical tank of liquid has diameter of 4 ft and is supported on l

Question

A vertical cylindrical tank of liquid has diameter of 4 ft and is supported on load cells that can read the mass of the contents of the tank. When the height in the tank is exactly 5 ft the load cells read 5884 Ib_n. The material in the tank is an aqueous salt solution and has a specific gravity, SG, that varies with the salt concentration according to the relationship: SG = l + 0.03S. Where S is the mass percentage of salt. The molar mast of S is 225. Determine the following: The volume of material in the tank (ft^3) The specific gravity of the material in the tank The gauge pressure at the bottom of the tank (in psig) The mass percentage of the salt in the tank The number of Ib moles of S in the tank If I said the concentration of SO_2 in the air was 0.0002 %- would this be a molar concentration (mole fraction), mass concentration (mass fraction)? Pick one.

Explanation / Answer

a)volume of material in the tank= surface area of cylindrical base*height of liquid=r^2*h

r=radius of tank=diameter/2=4 ft/2=2 ft

volume of material in the tank=3.14*(2ft)^2 *5ft=62.8 ft3

b)SG=1+0.03S

S=mass % of salt

Mass of water in the tank=density*volume=62.31 lb/ft3*62.8 ft3=3913.0680 lbs

Mass of salt=5884 lbs-3913.068=1970.932

S=1970.932 lbs/3913.0680*100=50.368%

SG=1+0.03*50.368=2.511

c)gauge pressure=force/area^2=mg/r^2=5884lbs*(32.174 ft/s2)/3.14*(2ft)^2=15072.597 lb/ft^2=15072.597 lbs/(12in)^2=104.671 lb/in^2=104.671 psig

1ft=12inch

g=acceleration due to gravity

d) S=mass % of salt

Mass of water in the tank=density*volume=62.31 lb/ft3*62.8 ft3=3913.0680 lbs

Mass of salt=5884 lbs-3913.068=1970.932

S=1970.932 lbs/3913.0680*100=50.368%

e)no of lb moles of salt=mass of S/molar mass of S=1970.932 lbs/(225 lb/mol)=8.760 lbmol or mol

2) mass fraction=mass of SO2/total mass of air*100

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