1) In 2011, the total of all residential customers in Halton Region used 54 540 000 m3 of water. Halton Region has a population of around 500 000 people and Milton (part of Halton Region) has a population of around 85 000 people. Assuming all people in Halton use the same amount of water, how much water was used in Milton in 2011?
2) 1 cm3 = 1 mL. The density of water is 1 g/cm3. Approximately what mass of water Milton use in 2011?
3) Most houses in Milton draw their water from the system that draws its water from Lake Ontario. Lak Ontario is 74 m above sea level and Milton is 221 m above sea level. The majority of the water in Milton passes through the water tower on Steeles Ave. It is 55 m from ground to top. In total, how much vertical distance must the water rise from Lake Ontario to the water tower?
4) How much gravitational potential energy does each kg of water gain travelling from Lake Ontario to the top of the water tower?
5) How much energy was needed to bring all the water used in Milton in 2011 from Lake Ontario to the water tower?
6) The energy used to lift the water comes from an electric water pump. Assuming the pump is 85% efficient at lifting the water, how much input electrical energy is needed for a year to lift the water?
7) Much of the energy used by the pump comes from natural gas fired plants (like the new one on Steeles Ave.). A typical efficiency for these power plants is 50%.
a. How much input energy (in the form of chemical energy) from natural gas is needed for a year to lift the water?
b. The chemical energy released from natural gas is around 35 MJ (Mega-Joules) per m3. What volume of gas needs to be burned to pump a year’s worth of water to Milton?
See previous post: Sat,11-15-14, 5:29 PM.
1) To calculate how much water was used in Milton in 2011, we need to determine the proportion of the Halton Region's population that resides in Milton. Since Milton has a population of approximately 85,000 people and Halton Region has a population of around 500,000 people, we can find the proportion by dividing the population of Milton by the population of Halton Region:
Proportion of Milton population = Population of Milton / Population of Halton Region
Proportion of Milton population = 85,000 / 500,000
By calculating this, we find that the proportion of Milton's population in Halton Region is 0.17 or 17%.
Next, we can calculate the water usage in Milton by multiplying the total water usage in Halton Region by the proportion of Milton's population:
Water usage in Milton = Total water usage in Halton Region * Proportion of Milton population
Water usage in Milton = 54,540,000 m3 * 0.17
By multiplying these values together, we find that approximately 9,268,800 m3 of water was used in Milton in 2011.
2) To calculate the mass of water used in Milton in 2011, we can use the density of water, which is 1 g/cm3 or 1 g/mL. Since 1 cm3 is equal to 1 mL, we can assume that 1 cm3 of water has a mass of 1 gram.
To convert the volume of water used in Milton (9,268,800 m3) to grams, we can multiply by 1,000,000 to convert from m3 to cm3:
Volume of water used in Milton (cm3) = Volume of water used in Milton (m3) * 1,000,000
Next, we multiply the volume of water used in Milton (cm3) by the density of water to get the mass of water used in Milton:
Mass of water used in Milton (grams) = Volume of water used in Milton (cm3) * Density of water (grams/cm3)
Since the density of water is 1 g/cm3, the mass of water used in Milton would be the same as the volume of water used in Milton, which is approximately 9,268,800 grams or 9,268.8 kilograms.
3) To find the total vertical distance the water must rise from Lake Ontario to the water tower in Milton, we need to subtract the elevation of Lake Ontario from the elevation of the water tower:
Total vertical distance = Elevation of water tower - Elevation of Lake Ontario
Total vertical distance = 221 m - 74 m
By subtracting these values, we find that the total vertical distance the water must rise is 147 meters.
4) The gravitational potential energy gained by each kilogram of water traveling from Lake Ontario to the top of the water tower can be calculated using the formula:
Gravitational potential energy = Mass of water (kg) * Gravitational acceleration (9.8 m/s2) * Vertical height (m)
Assuming the mass of water being lifted is 1 kg and the vertical height is 147 m (as calculated in the previous question), we can calculate the gravitational potential energy gained:
Gravitational potential energy = 1 kg * 9.8 m/s^2 * 147 m
By multiplying these values together, we find that each kilogram of water gains approximately 1,421.4 Joules (J) of gravitational potential energy.
5) To calculate the total energy needed to bring all the water used in Milton from Lake Ontario to the water tower, we multiply the gravitational potential energy gained per kilogram (1,421.4 J/kg) by the mass of water used in Milton (9,268.8 kg):
Total energy = Gravitational potential energy per kilogram * Mass of water used in Milton
Total energy = 1,421.4 J/kg * 9,268.8 kg
By multiplying these values together, we find that the total energy needed to lift the water is approximately 13,149,811.2 Joules (J).
6) To determine the input electrical energy needed to lift the water using an 85% efficient pump, we need to divide the total energy needed by the pump's efficiency:
Input electrical energy = Total energy needed / Pump efficiency
Input electrical energy = 13,149,811.2 J / 0.85
By dividing these values, we find that the input electrical energy needed to lift the water is approximately 15,470,363.76 Joules (J).
7a) To calculate the input energy in the form of chemical energy from natural gas needed to operate the pump, we need to divide the input electrical energy (15,470,363.76 J) by the efficiency of the power plant (50% or 0.5):
Input energy from natural gas = Input electrical energy / Power plant efficiency
Input energy from natural gas = 15,470,363.76 J / 0.5
By dividing these values, we find that the input energy from natural gas needed for a year is approximately 30,940,727.52 Joules (J).
7b) To determine the volume of gas that needs to be burned to pump a year's worth of water to Milton, we divide the input energy from natural gas (30,940,727.52 J) by the energy released per cubic meter of natural gas (35 MJ or 35,000,000 Joules):
Volume of gas needed = Input energy from natural gas / Energy released per cubic meter of gas
Volume of gas needed = 30,940,727.52 J / 35,000,000 J/m3
By dividing these values, we find that the volume of gas needed to pump a year's worth of water to Milton is approximately 0.883 cubic meters (m3).