A bathtub is a rectangular shape 3 feet long, 1.5 feet wide and 1 foot high. The

Question

A bathtub is a rectangular shape 3 feet long, 1.5 feet wide and 1 foot high. The inflow when the faucet is on is a constant rate of 3 gallons per minute. The exit flow is calculated from a simplified Bernoulli’s equation as a function of the height of the water in the tub through a 1/2-inch diameter drain. (Note: there are 7.5 gallons in a cubic foot. We are assuming the viscosity of water does not affect the drain outflow rate!) Height of water in tub (h) = Gallons / (7.5 gal/ft^^3 * TubArea) Where: h= height of water in tub in feet TubArea = 3ft*1.5 ft^^2 = 4.5 ft^^2 Simplified: h = Gallons in Tub * 0.03 (ft) Outflow equation Outflow(exit) = Velocity(exit) * Area of Drain = SQRT(2*G*h) * PipeArea / CONVERSION measured in gallons/minute Where: G= 32.2 ft/sec^^2 Gravity coefficient h= height of water in tub in feet PipeArea = 3.14*(0.25/12)(0.25/12) ft^^2 = 0.0013626 ft^^2 CONVERSION = ft^^3/gallon (1/7.5)*min/sec(1/60) = 1/450 = 0.00222 Simplified: Outflow(gallons/minute) = 4.91 SQRT(h) (h in feet) OR Outflow(gallons/minute) = 0.85 SQRT(TubGallons)

1) With the drain closed (Outflow = 0.0 gallons/minute), how long does it take for the tub to fill to its full 1-foot level capacity with a fully open faucet?

2) If the Inflow faucet was full on and the drain was left open, what is the steady state height that the tub equalizes at and how long does it take to reach steady state?

3) If the tub starts filled at its 1-foot rim and the drain is opened, how long does it take the tub to drain to the 1-inch (.0833 ft) level?

it is required to use anylogic application, but i don't know if it can solved without it.

Explanation / Answer

(1) VOLUME OFTUB =3 ft X 1.5 ft x 1ft = 4.5 cubic feet = 7.5 x 4.5 gallon as 1 cubic feet = 7.5 gallon Now Rate of infow 3 Gallon/minute hence required time = ( 7.5 x 4.5 ) /3 minutes =11.25 minutes

(2) FOR STEADY STATE HEIGHT IN FLOW AND OUT FLOW MUST BE SAME

OUT FLOW IS GIVEN AS 4.91 X SQRT (h) GALLON/MINUTE HENCE FOR STEADY HEIGHT THIS MUST

BE EQUAL TO IN FLOW. i..e. 4.91 x SQRT (H) = 3 i.e SQRT (h) = 3 / 4.91 = 0.611 , i.e h = (0.611)2

i.e.  h = 0.3733 ft

TO REACH THIS STATE TIME REQUIRED = ( VOLUME OF WATER) / INFLOW

= ( 3.5 ft x 1.5 ft x 0.3733 ft x 7.5 ) /3 (converting in gallon)

= 4.9 minutes

(3) TUB IS FULL AND ONLY OUT FLOW IS OPEN THEN REQUIRED TIME

= (3.5 X 1.5 X(11 x 0.0833) X7.5) / ( 4.91 X 0.542) ( here `h` should be taken as average of (1ft + 1 inch) /2 as out flow depends on height of water in tub = 0.542ft)

= 36.08 / 3.615 = 9.98 minutes ,after this height of water in tub will be one inch

IF INFLOW   

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