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 did Milton use in 2011
3) Most houses in Milton draw their water from the municipal system that draws its water from Lake Ontario. Lake 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?
202
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?
c. Use the internet to find the cost/m3 of natural gas in Ontario. Be sure to record your source. What is the cost to Milton to pump a year’s worth of water? Use the cost you think is appropriate.
1. V = (85,000/500,000) * 54,540,000 m^3
2. Dw = 1g/cm^3 = 1,000kg/m^3 = Density
of water.
Mass(kg) = V * 1000
3. h1 = 74 m, h2 = 221, h3 = 55
h = (h2-h1) + h3.
4. PE(gained) = mg*h - Mg*h1
PE/Mass = PE gained per kg of water.
1) To determine how much water was used in Milton in 2011, we need to find the proportion of Milton's population to the total population of Halton Region.
Population of Milton: 85,000 people
Population of Halton Region: 500,000 people
Proportion of Milton's population to Halton Region's population: 85,000 / 500,000 = 0.17
Next, we can find the amount of water used in Milton by multiplying the total water used in Halton Region by the proportion:
Amount of water used in Milton = 54,540,000 m3 * 0.17 = 9,257,800 m3
Therefore, approximately 9,257,800 m3 of water was used in Milton in 2011.
2) Since 1 cm3 is equal to 1 mL and the density of water is 1 g/cm3, we can directly convert the volume of water used in Milton to mass in grams.
Mass of water used in Milton = 9,257,800 m3 * 1 g/cm3 = 9,257,800 grams
Therefore, approximately 9,257,800 grams (or 9,257.8 kg) of water was used in Milton in 2011.
3) To find the total vertical distance the water must rise from Lake Ontario to the water tower, we need to calculate the difference in elevations.
Elevation of Lake Ontario: 74 m
Elevation of water tower: 221 m
Height of water tower: 55 m
Total vertical distance = (Elevation of water tower + Height of water tower) - Elevation of Lake Ontario
Total vertical distance = (221 m + 55 m) - 74 m
Total vertical distance = 202 m
Therefore, the total vertical distance the water must rise from Lake Ontario to the water tower is 202 meters.
4) Gravitational potential energy can be calculated using the formula:
Gravitational potential energy = mass * acceleration due to gravity * height
Since the height in this case is the total vertical distance the water must rise (202 m) and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the gravitational potential energy per kilogram of water:
Gravitational potential energy per kg of water = 1 kg * 9.8 m/s^2 * 202 m
Gravitational potential energy per kg of water = 1,993.6 Joules
Therefore, each kilogram of water gains approximately 1,993.6 Joules of gravitational potential energy when traveling from Lake Ontario to the top of the water tower.
5) To determine the total energy needed to bring all the water used in Milton in 2011 from Lake Ontario to the water tower, we multiply the total mass of water used in Milton (9,257.8 kg) by the gravitational potential energy per kg of water (1,993.6 Joules).
Total energy needed = mass of water used * gravitational potential energy per kg
Total energy needed = 9,257.8 kg * 1,993.6 Joules/kg
Total energy needed = 18,484,908.48 Joules
Therefore, approximately 18,484,908.48 Joules of energy were needed to bring all the water used in Milton in 2011 from Lake Ontario to the water tower.
6) To calculate the input electrical energy needed for a year to lift the water, we need to consider the efficiency of the electric water pump.
Assuming the pump is 85% efficient, the input electrical energy needed can be calculated as:
Input electrical energy needed = Total energy needed / Pump efficiency
Input electrical energy needed = 18,484,908.48 Joules / 0.85
Therefore, the input electrical energy needed for a year to lift the water is approximately 21,816,364.69 Joules.
7) a) Since the pump is powered by natural gas-fired power plants, we need to consider the efficiency of these plants. Assuming a typical efficiency of 50% for natural gas-fired power plants, we can calculate the input energy needed from natural gas for a year to lift the water.
Input energy needed from natural gas = Input electrical energy needed / Power plant efficiency
Input energy needed from natural gas = 21,816,364.69 Joules / 0.5
Therefore, the input energy needed from natural gas for a year to lift the water is approximately 43,632,729.38 Joules.
b) To determine the volume of gas that needs to be burned to pump a year's worth of water to Milton, we need the energy released per cubic meter of natural gas.
Energy released per m3 of natural gas = 35 MJ
Volume of gas burned = Input energy needed from natural gas / Energy released per m3
Volume of gas burned = 43,632,729.38 Joules / (35,000,000 Joules/m3)
Therefore, the volume of gas that needs to be burned to pump a year's worth of water to Milton is approximately 1.25 m3.
c) Please provide an appropriate cost/m3 of natural gas in Ontario and I will help you calculate the cost to Milton to pump a year's worth of water.