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

To find the initial speed of water out of the drain, we can apply Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant for an ideal, incompressible, and non-viscous fluid.

Let's consider two points: Point 1 - at the surface of the water in the tank, and Point 2 - at the drain, where the water is flowing out.

At Point 1:
- Pressure energy: P1 - atmospheric pressure (assumed to be constant)
- Kinetic energy: 1/2 * ρ * v1^2 (where v1 is the velocity of the water at Point 1)
- Potential energy: ρ * g * H (where H is the depth of the tank)

At Point 2:
- Pressure energy: P2 - atmospheric pressure (assumed to be constant)
- Kinetic energy: 1/2 * ρ * v2^2 (where v2 is the velocity of the water at Point 2)
- Potential energy: ρ * g * (H - D) (where D is the distance of the drain below the surface)

Since we are assuming no friction, the energy terms cancel each other out, and we can equate the kinetic energy terms at both points:

1/2 * ρ * v1^2 = 1/2 * ρ * v2^2

Simplifying, we get:

v1^2 = v2^2

Taking the square root of both sides, we have:

v1 = v2

This tells us that the initial speed of the water out of the drain is equal to the speed of the water at the surface of the tank.

Now, let's calculate the time interval needed to empty the tank (delta T). We can use Torricelli's Law, which is a special case of Bernoulli's equation for a fluid flowing out of an orifice.

Torricelli's Law states that the speed of efflux (v2) is given by:

v2 = (2gD)^1/2

where g is the acceleration due to gravity.

Now, considering that the volume of water in the tank (V) is equal to A * H (footprint area multiplied by depth), and the volume flow rate (Q) is equal to the product of the cross-sectional area of the siphon tube (A') and the speed of efflux (v2), we can write:

V = Q * delta T

Substituting the values, we get:

A * H = A' * v2 * delta T

Rearranging the equation, we find:

delta T = (A * H) / (A' * v2)

Substituting the value of v2 from Torricelli's Law, we have:

delta T = (A * H) / (A' * (2gD)^1/2)

This is the expression for the time interval needed to empty the tank (delta T), as required.

Please note that this analysis assumes ideal conditions, neglecting factors such as friction and viscosity.

To find the initial speed of water out of the drain, we can use Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid remains constant.

In our case, the pressure energy and the potential energy are negligible compared to the kinetic energy, so we can ignore them. This allows us to simplify Bernoulli's equation to:

1/2 ρ v^2 = ρgh

where v is the velocity of the water at the surface of the tank, ρ is the density of water, g is the acceleration due to gravity, and h is the height of the water in the tank.

Since we are interested in finding the initial speed of water out of the drain, we can assume that the velocity of water at the surface of the tank is negligible.

1/2 ρ v^2 = ρgh

v^2 = 2gh

v = √(2gh)

Now, let's find the time interval required to empty the tank.

The volume of water in the tank is given by V = AH, where A is the footprint area of the tank and H is the depth of the tank.

The volumetric flow rate can be calculated as Q = A'v, where A' is the cross-sectional area of the siphon tube.

The time interval required to empty the tank, ΔT, can be found by dividing the volume of water in the tank by the volumetric flow rate, i.e., ΔT = V / Q.

Substituting the values for V, Q, and v, we get:

ΔT = (AH) / (A'v)

ΔT = (AH) / (A' √(2gh))

ΔT = AH / A' * (2gD)^1/2

Therefore, the time interval required to empty the tank is ΔT = AH / A' * (2gD)^1/2.

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)