a house at the bottom of a hill is fed by a full tank of water 5m deep and connected to the house by a pipe that is 110m long at an angle of 58 from the horizontal.
a) determine the water gauge pressure at the house
b) how high could the water shoot if it came vertically out of a broken pipe in front of the house?
a) gage pressure = added pressure above atmospheric p = (water density) g H,
where
H = 5 m + 110 sin 58
= water level elevation above house ground level
b) same as the top level of the water in the tank
thankyou!
To calculate the water gauge pressure at the house, we need to use the formula for gauge pressure:
Gauge pressure = density of fluid × gravitational acceleration × height
a) Determine the water gauge pressure at the house:
Given:
Height of the water tank (h) = 5m
Density of water (ρ) = 1000 kg/m^3 (approximately)
Gravitational acceleration (g) = 9.8 m/s^2 (approximately)
Gauge pressure = ρ × g × h
Gauge pressure = 1000 kg/m^3 × 9.8 m/s^2 × 5 m
Gauge pressure = 49,000 N/m^2 (or Pascals, Pa)
Therefore, the water gauge pressure at the house is 49,000 N/m^2 or 49 kPa.
b) To determine how high the water could shoot if it came vertically out of a broken pipe in front of the house, we can use the principle of conservation of energy. The potential energy of the water at the bottom of the hill is converted to kinetic energy as it shoots out of the broken pipe.
Using the equation for kinetic energy:
Kinetic energy = 1/2 × mass of water × velocity^2
Assuming there is no energy loss due to friction or other factors, we can equate the initial potential energy of the water to its final kinetic energy:
Potential energy = Kinetic energy
m × g × h = 1/2 × m × v^2
Here, m is the mass of the water and v is the velocity of the water when it shoots out of the pipe.
Since we're looking for height, we need to solve for h:
h = (1/2 × v^2) / g
To determine the value of v, we can use trigonometry:
sin(angle) = opposite/hypotenuse
sin(58°) = height/110m
height = sin(58°) × 110m
Now, we can substitute the values in the equation for h:
h = (1/2 × (sin(58°) × 110m)^2) / 9.8 m/s^2
Calculating the value of h will give us the maximum height the water could shoot vertically out of the broken pipe.
To determine the water gauge pressure at the house, we can use the concept of hydrostatic pressure.
a) Water gauge pressure is the pressure exerted by a fluid at a specific point due to the weight of the fluid above it. In this case, the water in the tank creates pressure at the house.
First, we need to calculate the vertical distance from the bottom of the tank to the house. Since the tank is 5 meters deep and connected to the house by a pipe that is 110 meters long at an angle of 58 degrees from the horizontal, we can use trigonometry to find the vertical distance.
Vertical distance = 5m * sin(58 degrees)
Vertical distance = 4.08 meters (approx.)
Now, we need to convert the vertical distance into pressure. The pressure due to the weight of water is given by the formula:
Pressure = Density * g * Height
The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is 9.8 m/s².
Pressure = 1000 kg/m³ * 9.8 m/s² * 4.08 meters
Pressure ≈ 39816 Pascal (approx.)
Therefore, the water gauge pressure at the house is approximately 39816 Pascal.
b) To determine how high the water could shoot if it came vertically out of a broken pipe in front of the house, we can use the principle of conservation of energy.
The energy at the bottom of the hill is entirely in the form of potential energy, which is given by:
Potential energy = m * g * h
Since we know the density of water and the gravitational acceleration, we can calculate the mass of the water in the tank:
Mass = Density * Volume = Density * Area * Height
The area can be calculated using the formula for the area of a circle:
Area = π * r²
Since the pipe is connected to the house, it has the same diameter as the tank, which is equal to twice the radius.
Area = π * (2 * radius)²
Area ≈ π * (2 * 2.5m)²
Area ≈ 39.25 m² (approx.)
Now we can calculate the mass of water in the tank:
Mass = Density * Area * Height
Mass = 1000 kg/m³ * 39.25 m² * 5m
Mass ≈ 196250 kg (approx.)
Finally, we can calculate the height the water could shoot if it came vertically out of the broken pipe:
Potential energy = Mass * g * Height
Potential energy = 196250 kg * 9.8 m/s² * h
Let's assume the water shoots vertically upwards with a velocity of zero at the point of ejection, so all the potential energy is converted into kinetic energy.
Potential energy = Kinetic energy
1/2 * Mass * Velocity² = Mass * g * h
Canceling the mass:
1/2 * Velocity² = g * h
Rearranging the equation:
Velocity² = 2 * g * h
Velocity = sqrt(2 * g * h)
Plugging in the values:
Velocity ≈ sqrt(2 * 9.8 m/s² * h)
We assume the water shoots vertically, so the angle of projection is 90 degrees. From basic principles of projectile motion, we know that the maximum height reached by a projectile thrown vertically is given by:
Maximum height = (Velocity * sin(θ))² / (2 * g)
Plugging in the values:
Maximum height ≈ (sqrt(2 * 9.8 m/s² * h) * sin(90 degrees))² / (2 * 9.8 m/s²)
Since sin(90 degrees) = 1, we can simplify the equation:
Maximum height ≈ (sqrt(2 * 9.8 m/s² * h))² / (2 * 9.8 m/s²)
Maximum height ≈ (2 * 9.8 m/s² * h) / (2 * 9.8 m/s²)
Maximum height = h
Therefore, the maximum height the water could shoot if it came vertically out of a broken pipe in front of the house would be equal to the height of the tank, which is 5 meters.