Question

Определить внешнее поверхностное давление p в закрытом резервуаре с водой (плотность воды ρ = 1000 кг/м3), при котором будет обеспечен расход воды Q = 100 л/с, вытекающий через цилиндрический насадок диаметром d = 0,1 м, расположенный на глубине H = 2 м под уровнем воды.

Answer 1

Q=V/t=Svt/t=Sv;
v=12.74 m/s;
Bernoulli
P+pgh=pv^2/2;
P=61 146.5 Pa

Answer 2

Calculation of External Surface Pressure in a Closed Water Reservoir

To determine the external surface pressure p in a closed reservoir with water, we need to consider the flow rate Q of water and the cylindrical nozzle diameter d located at a depth H below the water level.

The external surface pressure can be calculated using the following formula:

p = (ρ * g * H) + (ρ * Q^2) / (2 * A^2)

Where: - p is the external surface pressure - ρ is the density of water (ρ = 1000 kg/m^3) - g is the acceleration due to gravity (approximately 9.8 m/s^2) - H is the depth of the nozzle below the water level (H = 2 m) - Q is the flow rate of water (Q = 100 L/s = 0.1 m^3/s) - A is the cross-sectional area of the nozzle (A = π * (d/2)^2)

Let's calculate the external surface pressure using the given values:

1. Calculate the cross-sectional area of the nozzle: - A = π * (d/2)^2 - Substituting the given diameter d = 0.1 m: - A = π * (0.1/2)^2 = 0.00785 m^2

2. Calculate the external surface pressure: - p = (ρ * g * H) + (ρ * Q^2) / (2 * A^2) - Substituting the given values: - p = (1000 kg/m^3 * 9.8 m/s^2 * 2 m) + (1000 kg/m^3 * (0.1 m^3/s)^2) / (2 * (0.00785 m^2)^2) - p ≈ 19600 Pa

Therefore, the external surface pressure in the closed water reservoir, which ensures a water flow rate of 100 L/s through a cylindrical nozzle with a diameter of 0.1 m located at a depth of 2 m below the water level, is approximately 19600 Pa.

Please note that the above calculation assumes ideal conditions and neglects factors such as friction and turbulence.

Answer 3

Calculation of External Surface Pressure in a Closed Reservoir

To determine the external surface pressure p in a closed reservoir with water, we can use the following formula:

p = ρgh

Where: - p is the external surface pressure - ρ is the density of water (ρ = 1000 kg/m³) - g is the acceleration due to gravity (approximately 9.8 m/s²) - h is the depth of the cylindrical nozzle below the water level

In this case, the depth h is given as 2 m.

Let's calculate the external surface pressure using the given values:

p = (1000 kg/m³) * (9.8 m/s²) * (2 m)

Calculating this expression, we find:

p ≈ 19,600 Pa

Therefore, the external surface pressure in the closed reservoir with water is approximately 19,600 Pascal (Pa).

Please note that the above calculation assumes that the water is incompressible and there are no other external forces acting on the reservoir.

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