Question:

A container is filled with a viscous fluid to a height of 2 meters. The fluid has a density of 1000 kg/m^3. A small hole is made at the bottom of the container, and the fluid begins to flow out. Assuming the fluid is incompressible and neglecting any effects of air resistance, answer the following:

(a) Determine the pressure at the bottom of the container.

(b) Calculate the speed of the fluid as it exits the container.

(c) Find the flow rate of the fluid leaving the container.

Answer:

(a) To determine the pressure at the bottom of the container, we can use the concept of hydrostatic pressure. The pressure at a certain depth in a fluid is given by the formula:

P = P₀ + ρgh

where: P = pressure at certain depth P₀ = atmospheric pressure (assumed to be 1 atm) ρ = density of the fluid g = acceleration due to gravity h = height or depth of the fluid

Substituting the given values:

P = 1 atm + (1000 kg/m³)(9.8 m/s²)(2 m)

P = 1 atm + 19600 Pa

P ≈ 1 atm + 19600 Pa

P ≈ 2 atm

Therefore, the pressure at the bottom of the container is approximately 2 atmospheres.

(b) To calculate the speed of the fluid as it exits the container, we can use the principle of conservation of energy. The potential energy (PE) of the fluid at the top of the container is converted into kinetic energy (KE) as it exits.

Using the equation:

PE = KE + PE₂

where: PE = potential energy KE = kinetic energy PE₂ = potential energy at the exit point (which is negligible since it's at the same height as the exit)

We can equate the two energies:

(mgh) = (1/2)mv²

Cancelling out mass (m) from both sides and rearranging the equation:

v² = 2gh

v = √(2gh)

Substituting the given values:

v = √(2)(9.8 m/s²)(2 m)

v ≈ √(39.2 m²/s²)

v ≈ 6.27 m/s

Therefore, the speed of the fluid as it exits the container is approximately 6.27 m/s.

(c) To find the flow rate of the fluid leaving the container, we can use the equation for volume flow rate:

Q = Av

where: Q = flow rate A = cross-sectional area of the hole v = speed of the fluid

The cross-sectional area of the hole is assumed to be small and considered as a point, thus the equation can be simplified as:

Q = Av

Q = πr²v

where: r = radius of the hole

Substituting the given values:

Q = π(0.01 m)²(6.27 m/s)

Q ≈ π(0.0001 m²)(6.27 m/s)

Q ≈ 0.000393 m³/s

Therefore, the flow rate of the fluid leaving the container is approximately 0.000393 cubic meters per second.