The system shown in Figure 1 consists of an upper tank full of water with two sp

Question

The system shown in Figure 1 consists of an upper tank full of water with two spigots that are initially closed. The spigots are mounted at the same height on the large tank at opposite sides from one another. Each spigot has a diameter of 0.023 ft. At time t=0 s, both spigots are opened and water discharges from each spigot at an average velocity of 4.1 ft/s. The water from the left spigot discharges into a cylindrical tank. The water from the right spigot discharges into a rectangular prismatic tank. The dimensions for each of these tanks are shown in Table A. Find the percent difference in water level height between the cylindrical and rectangular prism tanks at time t = 1x102 s, where: % diff=(height of water in cylindrical tank-height of water in rectangular prism tank)/(height of water in cylindrical tank)?100

Explanation / Answer

Flow rate in each tank Q = Area * Velocity = (/4*d^2)*V = (/4*0.023^2)*4.1 = 0.0017 ft^3/s

In time t = 1*10^2 - 0 = 100 s, volume of water into each tank = 0.0017*100 = 0.17 ft^3

We have, 0.17 = A_cyl*(h)_cyl and 0.17 = A_rect*(h)_rect

where, A_cyl and A_rect denote the cross-section area of cylindrical and rectangular tanks respectively and (h)_cyl, (h)_rect denote the rise in water heights in these tanks in 100 s.

A_cyl = *0.44^2 = 0.6079 ft^2

A_rect = 1.5*0.65 = 0.975 ft^2

(h)_cyl = 0.17/0.6079 = 0.28 ft

(h)_rect = 0.17/0.975 = 0.174 ft

% diff=(height of water in cylindrical tank-height of water in rectangular prism tank)/(height of water in cylindrical tank)×100

% diff = (0.28-0.174)/0.28*100 = 37.85 %

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