Please explain how to get answers thanks 2. Consider the 1-m2 storage tank shown
ID: 3279856 • Letter: P
Question
Please explain how to get answers thanks 2. Consider the 1-m2 storage tank shown in Figure 4.49. Initially, the density of the liquid in the tank is 1000 kg/m3. At a particular instant, the inlet valve is opened and a fluid of density 900 kg/m3 flows into the tank at a rate of 100 L/min. At this same instant, the outlet valve is opened and adjusted such that the liquid level in the tank is maintained at a constant elevation. A mixer is also turned on to ensure that the liquid in the tank is well mixed. After the valves are open and the mixer is turned on, how long wil it take the density of the liquid in the tank to decline from 1000 kg/m3 to 950 kg/m2? If the system is left operating long enough, what will be the ultimate density of the iquid in the tank? Mixer Storage tank 100 /min 900 kg/m , 1 m3 ANS: 6.92 min, density 900 kg/m
Explanation / Answer
volume of storage tank V = 1m^3
intiail density of liquid inside = rho1 = 1000 kg/m^3
density of input liquid, = rho2 = 900 kg/m^3
input rate = 100 L/min = r
so as the elevation of the liquid in the tank does not change
input flow volume rate = output flow volume rate
also, at time t, let density of the liquid in the tank be rho
mass of the liquid inside tank, m = rho*V
then after time dt
input mass = r*rho2*dt
output mass = r*rho*dt
new mass left = rho*V + r*rho2*dt - r*rho*dt
rho(t + dt) = rho + r*rho2*dt/V - r*rho*dt/V
rho(t + dt) - rho(t) = d(rho) = r[rho2 - rho]dt/V
d(rho)/dt = r[rho2 - rho]/V
d(rho)/(rho - rho2) = -r*dt/V
integrating from rho = rho1 to rho = rho
ln([rho - rho2]/[rho1 - rho2]) = -rt/V
given rho = 950
rho1 = 1000
so ln([950 - 900]/[1000 - 900]) = -100*10^-3*t/1
t = 6.9314 minutes
also, e^(-rt/V) = (rho - rho2)/(rho1 - rho2)
when t-> infinity
e^(-rt/V) = 0
rho = rho2
so ultimate density rho = 900 kg/m^3