This question is incomplete, the complete question is;
A crude oil storage tank initially contains 1000 m³ of crude oil. Oil is pumped into the tank through a pipe at a rate of 2 m³/min and out of the tank at a velocity of 1.5 m/s through another pipe having a diameter of 0.15m. The crude oil has a specific volume of 0.0015 m³/kg .
Determine : a) The mass of oil in the tank, in kg, after 24 hours b) The volume of oil in the tank, in m³, at that time ( after 24hrs )
Answer:
a) The mass of oil in the tank after 24hrs is 1059852.6667 kg
b) The volume of oil in the tank, in m³, at that time is 1589.779 m³
Explanation:
Given that;
volume of tank v = 1000 m³
Discharge (AV)₁ = 2 m³/min
specific volume (v) = 0.0015 m³/kg
a)
for one inlet, one exit control volume, we have]
d(mc[tex]_v[/tex])/dt = "m[tex]_i[/tex] - "m[tex]_e[/tex]
"m[tex]_i[/tex] = (AV)₁ / v
"m[tex]_i[/tex] = 2 / 0.0015
"m[tex]_i[/tex] = 1333.33 kg/min = ( 1333.33 kg/min × 60 ) = 80,000 kg/hr
"m[tex]_e[/tex] = AV₂/v
AREA A = πr² = π(0.15m/2)²
so
"m[tex]_e[/tex] = ( π(0.15m/2)² × 1.5 m/s ) / 0.0015m³/kg
"m[tex]_e[/tex] = 17.6714 kg/s = (17.6714 kg/s × 60 × 60 ) = 63,617.25 kg/hr
d(m[tex]_{cv}[/tex])/dt = 80000 - 63,617.25 = 16382.75 kg/hr
now after 24 hours, the mass of the crude oil tank is;
mc[tex]_v[/tex]( 24 ) = ( 16382.75 × 24 ) + mass initially contained
so initial mass m[tex]_i[/tex] = volume / specific volume
initial mass m[tex]_i[/tex] = V / v = 1000 / 0.0015 = 666,666.6667 kg
Total mass = m[tex]_i[/tex] + mc[tex]_v[/tex]
Total mass = 666,666.6667 kg + (16382.75 × 24)
Total mass = 1059852.6667 kg
Therefore, The mass of oil in the tank after 24hrs is 1059852.6667 kg
b)
Volume will be;
V = total mass × density
V = 1059852.6667 kg × 0.0015 m³/kg
V = 1589.779 m³
Therefore, The volume of oil in the tank, in m³, at that time is 1589.779 m³
