a homeowner wishes to drain her swimming pool by siphoning the water, whose depth is h, into a nearby gully a distance H below it,where H is much larger than h.she uses pipe of cross-sectional area a, and the pool has surface area A.how long does it take to empty the pool if h=2 m, H=20 m ,A=50 m2 ,a=5 cm2?

Question

a homeowner wishes to drain her swimming pool by siphoning the water, whose depth is h, into a nearby gully a distance H below it,where H is much larger than h.she uses pipe of cross-sectional area a, and the pool has surface area A.how long does it take to empty the pool if h=2 m, H=20 m ,A=50 m2 ,a=5 cm2?

Answer 1

To calculate the time it takes to empty the pool through siphoning, we need to consider the flow rate of the water. The flow rate can be determined using the principles of fluid dynamics.

The flow rate (Q) is given by the equation:

Q = A * v

where Q is the flow rate, A is the cross-sectional area of the pipe, and v is the velocity of the water.

To find the velocity of the water, we can use Bernoulli's equation, which relates the pressure, velocity, and height of the water in two different points. In this case, we can use it to relate the pressure at the water surface in the pool to the pressure at the surface of the gully.

Bernoulli's equation is:

P + 1/2 * ρ * v^2 + ρ * g * h = constant

where P is the pressure, ρ is the density of water, v is the velocity of the water, g is the acceleration due to gravity, and h is the height of the water.

Since the two points we are considering are open to the atmosphere, the pressure terms cancel out, and we can simplify the equation to:

1/2 * ρ * v^2 + ρ * g * h = 0

To solve for the velocity (v), we can rearrange the equation:

v = sqrt(2 * g * h)

Now that we have the velocity, we can calculate the flow rate (Q):

Q = A * sqrt(2 * g * h)

Substituting the given values:

A = 50 m^2 (surface area of the pool)
a = 5 cm^2 (cross-sectional area of the pipe) = 0.0005 m^2
h = 2 m (depth of the water)
H = 20 m (height of the gully)
g = 9.8 m/s^2 (acceleration due to gravity)

First, we need to convert the cross-sectional area of the pipe to square meters:

A = a * (50 / 100^2) = 0.0005 * 0.5 = 0.00025 m^2

Now we can calculate the flow rate:

Q = A * sqrt(2 * g * h) = 0.00025 * sqrt(2 * 9.8 * 2) = 0.01 m^3/s

Finally, to find the time it takes to empty the pool, we divide the volume of the pool (V) by the flow rate:

V = A * h = 50 * 2 = 100 m^3

Time = V / Q = 100 / 0.01 = 10,000 seconds

Therefore, it will take approximately 10,000 seconds to empty the pool by siphoning.