An open-topped tank completely filled with water has a release valve near its bottom. The valve is 1.75m below the water surface. Water is released from the valve to power a turbine, which generates electricity. The area of the top of the tank, AT, is 11.5 times the cross-sectional area, AV, of the valve opening. Calculate the speed of the water as it exits the valve. Neglect friction and viscosity. In addition, calculate the speed of a drop of water released from rest at h=1.75m when it reaches the elevation of the valve.
To calculate the speed of the water as it exits the valve, we can use the principles of conservation of energy.
1. Determine the height difference (Δh) between the water surface and the valve.
- In this case, the height difference is given as 1.75m.
2. Calculate the potential energy difference (ΔPE) between the water surface and the valve.
- Potential energy difference can be calculated using the formula: ΔPE = m * g * Δh
- Here, m represents the mass of water and g is the acceleration due to gravity (approximately 9.8 m/s²).
- Since the tank is completely filled with water, the mass of water can be calculated by multiplying the density of water (ρ) with the volume (V). The density of water is approximately 1000 kg/m³.
- The volume can be calculated by multiplying the cross-sectional area (AV) of the valve opening with the height difference (Δh).
- So, m = ρ * V = ρ * AV * Δh.
3. Calculate the kinetic energy (KE) of the water when it reaches the valve.
- Since the tank is open-topped, the water has only potential energy at the surface and kinetic energy at the valve.
- The kinetic energy can be calculated using the formula: KE = 0.5 * m * v², where v represents the velocity of the water.
4. Equate the potential energy difference (ΔPE) to the kinetic energy (KE).
- ΔPE = KE => ρ * AV * Δh * g = 0.5 * ρ * AV * v²
- Canceling ρ and AV on both sides, we get: Δh * g = 0.5 * v²
- Rewrite the equation as: v = √(2 * g * Δh)
5. Substitute the values and calculate the velocity.
- v = √(2 * 9.8 * 1.75)
- v ≈ 6.15 m/s
Therefore, the speed of the water as it exits the valve is approximately 6.15 m/s.
To calculate the speed of a drop of water released from rest at a height of 1.75m when it reaches the elevation of the valve, we can use the principle of conservation of energy again.
1. Determine the potential energy (PE) of the water when it is at a height of 1.75m.
- PE = m * g * h
- Here, m represents the mass of the water and h is the height (1.75m).
- We can calculate the mass of the water using the same approach as before: m = ρ * V.
2. Calculate the kinetic energy (KE) of the water when it reaches the elevation of the valve.
- The initial potential energy (PE) is converted into kinetic energy (KE) as the water falls.
- PE = KE => m * g * h = 0.5 * m * v²
- Canceling m, we get: g * h = 0.5 * v²
- Rewrite the equation as: v = √(2 * g * h)
3. Substitute the values and calculate the velocity.
- v = √(2 * 9.8 * 1.75)
- v ≈ 6.15 m/s
Therefore, the speed of a drop of water released from rest at a height of 1.75m when it reaches the elevation of the valve is also approximately 6.15 m/s.