Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.

Repeat the calculation for the time until overflow for the case of water entering a 4.00-m3 tank at a rate of 6.83 kg/s and withdrawn at a rate of 3.50 kg/s. The tank is initially two thirds full.

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mtosi17 mtosi17 · Answer 1 · 2019-10-01T04:38:25+02:00

Answer:

(a) The time until overflow is 649 s

(b) The time until overflow is 355 s

Explanation:

The volume as a function of time can be expressed as

[tex]V(t) = V_0+(q_i-q_o)*t[/tex]

If the tank is initially half full, V(0) = V0 = 4/2 = 2 m3.

With ρ=1000 kg/m3, the volume flows are

Flow in = 6.33 kg/s * 0.001 m3/kg = 0.00633 m3/s

Flow out = 3.25 kg/s * 0.001 m3/kg = 0.00325 m3/s

The time until overflow (V(t)=4 m3) is

[tex]V(t) = 2+(0.00633 - 0.00325)*t=2+0.00308*t=4\\\\t=(4-2)/0.00308 = 649.4 s=11 min[/tex]

If the flows are

Flow in = 6.83 kg/s * 0.001 m3/kg = 0.00683 m3/s

Flow out = 3.50 kg/s * 0.001 m3/kg = 0.0035 m3/s

And the tank is initially 2/3 full (V(0)=2.67 m3)

The time until overflow is

[tex]V(t) = 2.67+(0.00683 - 0.00350)*t=2.67+0.00375*t=4\\\\t=(4-2,67)/0.00375 = 354.67 s =6 min[/tex]