The most important part of answers of this type of physics problems is the assumptions made to arrive at the answer. With a given answer, these assumptions are usually given, but not in the present case.
The given answer corresponds to the following assumptions:
1. D>>H leads us to assume that the hydrostatic head (H) is negligible => we do not need to know the variation of rate of discharge with respect to time.
2. Water has no friction (viscosity) leads us to model the problem as an agglomeration of water drops free falling without affecting each other, and will fill the siphon tube.
Volume of water = AH
rate of water exiting the drain tube at the outlet
=velocity of free-fall through a height of D
=√(2gD) [use kinematic equation below]
Volume of water drained per unit time
=velocity*area of siphone tube
=A'(√(2gD)
Time to empty=volume of water in tank ÷ volume of water drained per unit time
Kinematic equation for free-fall:
initial velocity v0=0
free-fall distance=D
g=acceleration due to gravity
V1²-V0²=2gD
=>
V1²=(2gD-v0²)
=>
V1=√(2gD)
a woman is draining her fish tank by siphoning the water into an outdoor drain the rectangular tank has footprint area A and depth H . the drain is located a distance D below the surface of the water in the tank , where D>>H . the cross section area of the siphon tube is A' model the water as no friction find 1) the initial speed of water out of the drain
show that the time interval need to empty the tank is
delta T = AH / A' * (2gD)^1/2
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