2. A pool has an area A = 50 m2 and depth h = 2.5 m. The pool is filled with water to the maximum height. An electrical pump is used to empty the pool. There are two pipes coming out the pump: one is submerged into water has a radius r1 = 4 cm the other has a radius r2 = 2.5 cm. Answer the following questions ignoring friction, viscosity, turbulence.
Calculate the net force on the bottom of the pool.
Calculate work done by the pump required to empty the pool in 5 h.
Calculate the speed of the water flow in the submerged pipe.
a.)
P=P initial + pgh
101,000 Pa + (1000)(9.8)(2.5)
P=125500Pa.
P=F/A
125500=F/50
6.3*10^6N
Why did the pool go to therapy? Because it had deep issues!
To calculate the net force on the bottom of the pool, we need to find the pressure exerted by the water on the bottom. The pressure is given by the equation P = ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth of the pool. To find the net force, we multiply the pressure by the area of the bottom of the pool: F = P × A.
Now, let's do some serious clown calculations:
Given:
Area of the pool, A = 50 m^2
Depth of the pool, h = 2.5 m
First, let's calculate the net force on the bottom of the pool:
Density of water, ρ = 1000 kg/m^3 (approximately)
Acceleration due to gravity, g = 9.8 m/s^2 (approximately)
Pressure, P = ρgh
= (1000 kg/m^3)(9.8 m/s^2)(2.5 m)
≈ 24,500 N/m^2
Net force, F = P × A
= (24,500 N/m^2)(50 m^2)
= 1,225,000 N
So, the net force on the bottom of the pool is approximately 1,225,000 Newtons.
Now, let's move on to the work done by the pump to empty the pool in 5 hours:
Work done, W = Force × Distance
Given:
Time, t = 5 hours
Distance, d = 2.5 m (depth of the pool)
Work done, W = (Force × Distance) ÷ time
= (1,225,000 N)(2.5 m) ÷ (5 hours)
= 612,500 Joules
Therefore, the work done by the pump to empty the pool in 5 hours is approximately 612,500 Joules.
Finally, let's calculate the speed of the water flow in the submerged pipe:
Given:
Radius of the submerged pipe, r1 = 4 cm = 0.04 m
To find the speed of the water flow, we can use the equation for the volume flow rate, which is given by Q = Av, where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the velocity of the water flow.
Cross-sectional area of the pipe, A = πr^2
= π(0.04 m)^2
Since the volume flow rate is constant, we can use the equation Q = Av for both pipes.
The radius of the second pipe, r2 = 2.5 cm = 0.025 m
Cross-sectional area of the second pipe, A2 = π(0.025 m)^2
Since the volume flow rate is constant for both pipes, we can set the two equations equal to each other and solve for v:
A1 × v1 = A2 × v2
(π(0.04 m)^2)v1 = (π(0.025 m)^2)v2
Simplifying the equation, we get:
(0.0016 π)v1 = (0.000625 π)v2
Dividing both sides by (0.0016 π), we have:
v1 = (0.000625 ÷ 0.0016)v2
v1 ≈ 0.390625v2
So, the speed of the water flow in the submerged pipe is approximately 0.390625 times the speed of the water flow in the other pipe.
To calculate the net force on the bottom of the pool, we'll use the formula:
Net force = pressure × area
Pressure at the bottom of the pool is given by the formula:
Pressure = density × g × height
Where:
- density is the density of water
- g is the acceleration due to gravity
- height is the depth of the pool
Density of water (ρ) is approximately 1000 kg/m^3, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
Substituting the values into the formula:
Pressure = 1000 kg/m^3 × 9.8 m/s^2 × 2.5 m
Net force = Pressure × area = (1000 kg/m^3 × 9.8 m/s^2 × 2.5 m) × 50 m^2
Now we can calculate the net force on the bottom of the pool.
Next, to calculate the work done by the pump required to empty the pool in 5 hours, we can use the equation:
Work = force × distance
The force required to push the water out through the pipes is equal to the net force on the bottom of the pool.
Thus, the work done by the pump is given by:
Work = force × distance = (net force on bottom of pool) × (depth of pool)
Now we can calculate the work done by the pump required to empty the pool in 5 hours.
Lastly, to calculate the speed of the water flow in the submerged pipe, we can use the equation of Bernoulli's equation:
Pressure + 1/2(density) × (velocity)^2 = constant
Since the pipe is submerged, the pressure at the outlet of the pipe should be equal to the pressure at the bottom of the pool.
By using Bernoulli's equation and rearranging it, we can solve for the velocity of the water flow in the submerged pipe.
These calculations are based on the assumption of no friction, viscosity, or turbulence.
To calculate the net force on the bottom of the pool, you can use the formula for pressure:
Pressure = Force/Area
First, let's find the force exerted by the water on the bottom of the pool. The force is equal to the weight of the water in the pool. The weight of an object can be calculated using the formula:
Weight = Mass * Gravity
In this case, the mass of water can be found using the formula:
Mass = Volume * Density
The volume of water can be calculated by multiplying the surface area of the pool (A) by the depth (h). The density of water is approximately 1000 kg/m^3, and the acceleration due to gravity (g) is approximately 9.8 m/s^2.
So, the force exerted by the water on the bottom of the pool is:
Force = Mass * Gravity = (Volume * Density) * Gravity = (A * h * Density) * Gravity
Now, let's calculate the net force. We know the force exerted by the water, and we also need to consider the force exerted by the atmosphere. Since the pool is open to the atmosphere and the water surface area is much larger than the pipe openings, we can assume the atmospheric pressure cancels out the hydrostatic pressure on the water surface. Therefore, the net force on the bottom of the pool is just the force exerted by the water:
Net force on the bottom of the pool = Force = (A * h * Density) * Gravity
Next, to calculate the work done by the pump required to empty the pool in 5 hours, we need to consider the power exerted by the pump. Power is defined as the rate at which work is done. The power (P) can be calculated using the formula:
Power = Work / Time
Rearranging the formula, we get:
Work = Power * Time
In this case, the power exerted by the pump is equal to the force applied by the pump on the water multiplied by the velocity of water flow. Assuming constant velocity (ignoring friction, viscosity, and turbulence), the time taken is 5 hours.
The force exerted by the pump can be calculated using the equation:
Force = Pressure * Area = (Pump Pressure - Atmospheric Pressure) * (π * r1^2)
Where the pump pressure is the pressure inside the pump and can be assumed to be constant.
Finally, the velocity of water flow in the submerged pipe can be calculated using the equation of continuity, which states that the volume flow rate of an incompressible fluid is constant along a pipe. The equation for volume flow rate is:
Volume Flow Rate = Cross-sectional area * Velocity
Since the volume flow rate is constant, we can equate the volume flow rates at two points: the pool and the submerged pipe. The cross-sectional area of the pool is A, and the cross-sectional area of the submerged pipe can be calculated using the formula:
Cross-sectional area = π * r2^2
Therefore, we can calculate the velocity of water flow in the submerged pipe:
Velocity = Volume Flow Rate / Cross-sectional area = (A * h) / (π * r2^2)
By applying the above formulas, you should be able to find the net force on the bottom of the pool, the work done by the pump required to empty the pool in 5 hours, and the velocity of the water flow in the submerged pipe.
a. V=A*h = 50*2.5=125 m^3=1.25*10^8 cm^3.
Dw = 1g/cm^3 = 0.001kg/cm^3 = Density of water.
Mass=1.25*10^8cm^30.001kg/cm^3
= 1.25*10^5 kg = Mass of the water.
F=mg = 1.25*10^5kg * 9.8N/kg = 1.225*10^6 N. = Force on bottom.
b. Work = F*d = 1.225*10^6 * 2.5 = =
3.0625*10^6 Joules.