4) The tank shown has a bottom area A. A pressure source p̂s = ps(t) + pa is connected through a

Step 1: The pressure difference across the valve $R_1$ is $p_s - p_a$ and the flow rate through

4) The tank shown has a bottom area A. A pressure source p̂s...

4) The tank shown has a bottom area A. A pressure source \hat{p}_s = p_s(t) + p_a is connected through a linear valve resistance, R?, to the bottom of the tank. Another linear valve resistance, R?, allows the fluid to exit the tank. (a) Assume R? = R? = R, and develop a model of h, where \hat{h} = h(t). What is the time constant of the model? (b) Right before both valves open simultaneously to allow fluid in and out of the tank, the height of the fluid in the tank is h?. If p?(t) = P_u s(t), determine the total response of the system for when both valves are open and fluid flows in and out of the tank, i.e., h(t). (20 points)

Transcript

00:01 Liquid system which is consist of liquid system which is consist of consist of two terms two terms which have cross sectional are cross sectional are a1 and a2 respectively and a2 respectively and the given system is a non -interactive system which means that height as to t in tan second will not depend on h1t in tan1.

01:06 Delta h is equal to hat responsible for inlet flow rate through tan1 and and flow through volume delta h is equal to has a head response.

01:24 Responsible for inlet flow rate, flow rate, flow rate through tank 1, tank 1 and flow through volume we have this relation like this for f is directly proportional to delta delta h and delta h is head difference and f is flow rate through the volume.

02:03 So f1 is equal to delta h, f2 is equal to h1 t r2 and f3 is equal to h2 t upon r3 and in point a mass balance from cos tan.

02:27 So total mass in, total mass out.

02:30 Total mass in is equal to total mass out.

02:37 Total mass accumulated.

02:43 So let rho be the density of the fluid.

02:47 So rho f1 is equal to rho f2 is equal to t upon tt because of volume.

02:58 So, rho f1 is equal to rho f2 is equal to p upon dt.

03:12 So, rho will be cancelled on each side because rho is constant.

03:18 So, f1 minus f2 is equal to p upon dt into h1 for cross sectional area is constant.

03:32 So delta h upon r1 is equal to h1t r2 is equal to a into dh1t upon tt.

03:46 And mass balance for tank laplace transform from equation first...

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