Equation 1.41 is volume flow rate (i)=volume flow rate (o) + (dQ/dr)tank Q=volum

Question

Equation 1.41 is volume flow rate (i)=volume flow rate (o) + (dQ/dr)tank
Q=volume
Equation 1.44 is Outlet Volume flow rate= output gain×prediction of fluid level-h (t)

discharge various tanks with different geometrics. Consider such a tank as shown Figure 1.35, cylindrical. h, Figure 1.35 A cylindrical tank discharging at flow rate Q. a. Use the conservation of mass, Eq. (1.41), where 0 in this case, and Eq. (1.44) to derive a fist-order ODE for the fluid height at any time h). b. Solve the ODE with IC h(0) c. Plot ht) ver sus time in the interval 20, and determine how long it takes to empty the tank for A-/, ho =2m , and ko = 0.01-

Explanation / Answer

given m = rho * Q

where m = mass

Q = volume and rho = density

so, m' = rho*Q'

for incompressibel liquid rho is constant

Q'i = Q'o + (dQ/dt)tank

Qi' = inlet volume flow rate

Qo' = outlet fvolume flow rate

dq/dt is change of volume in the tank

5. a. given Qi' = 0

then

dQ/dt for tank = A*dh/dt ( where A is cross section area of the tank)

then

Qo' = -A*dh/dt

b. Qo'*dt = -Adh

now, Qo' = ko*sqroot(h(t))

ko*sqroot(h(t))*dt = -Adh

ko*dt = -A*h^-1/2*dh

integrating

ko*t = -2A*sqrt(h) + K

for t = 0, h = ho

hence

2Asqrt(ho) = K

hence

ko*t = -2Asqrt(h) + 2Asqrt(ho)

h = (sqrt(ho) - ko*t/2A)^2

c. A = pi/4

ho = 2

ko = 0.01

hence

h = (sqrt(2) - 0.02*t/pi)^2 = 0

sqrt(2) = 0.02t/pi

hence the tank empties at t = 222.144s

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