Book "B": Worksheet 7: Water Heating Calculations - Answers and Commentary
Question 1
Calculate the time taken to heat the water in a hot water cylinder that is 2m tall and has a diameter of 750mm.
The cylinder has a 10kW element fitted and insulated to 85%.
Initial water temperature is 15°C and required temperature is 65°C.
Solution
The first step is to calculate the amount of energy required to heat the water.
The equation is
\( Q = m c \Delta T \)
where \( Q \) is the energy required (J), \( m \) is the mass of the substance (kg), \( c \) is the specific heat capacity (J.kg-1.°C-1), and \( \Delta T \) is the temperature change (°C).
For water, c = 4184 J.kg-1.°C-1.
Calculating the Mass of Water to be Heated \( m \)
First, the mass of water is calculated. The cylinder has a height of 2 m, and a diameter of 750 mm. The calculation is shown below. All dimensions are worked in metres.
\( V = \frac{\pi d^2 h}{4} = \frac{\pi \cdot 0.75^2 \cdot 2}{4} = \) 0.883573 m3 (6 sf).
The density of water is approximately 1000 kg.m-3. Therefore, the mass of water to be heated is 883.573 kg.
Calculating the Temperature Change \( \Delta T \)
The temperature change is the difference between the final temperature and the initial temperature. In this case, \( \Delta T = 65 - 15 = \) 50°C.
Calculating the Energy Required
The mass and temperature change can now be plugged into the energy equation.
\( Q = m c \Delta T = 883.573 \cdot 4184 \cdot 50 = \) 184843000 J (6 sf).
\( Q \) may also be expressed as 51.34528 kWh (6 sf). 1 kWh = 3600000 J.
Calculating the Electrical Energy Required
The cylinder has an electrical power input of 10 kW, and is "insulated to 85%". This will be assumed to mean an efficiency of 85% (i.e. \( \eta = 0.85 \)).
In other words, the "effective power" available for heating the water is assumed to be 8.5 kW (10 kW at 85% efficiency).
Note that 1 W = 1 J.s-1.
Calculating the Time Required - Using joules
The cylinder requires 184843000 J, with energy supplied at the effective rate of 8500 J.s-1.
This is enough information to calculate the heating time \( t \), using \( E = P t \).
\( E = P t \rightarrow t = \frac{E}{P} = \frac{184843000}{8500} = \) 21746.2 s (6 sf).
It is probably more convenient to work with a time in hours. The time in hours is 6.041 h (4 sf) #.
Calculating the Time Required - Using kilowatt-hours
The cylinder requires 51.34528 kWh, with energy supplied at the effective rate of 8.5 kW.
This is enough information to calculate the heating time \( t \), using \( E = P t \).
\( E = P t \rightarrow t = \frac{E}{P} = \frac{51.34528}{8.5} = \) 6.041 h (4 sf) #.
Question 2
A tank of water measures 1.5 m wide, 2 m high and 0.5 m deep.
The water in the tank needs to be heated from 25°C to 90°C in 21 hours. Efficiency is poor at 55%.
Calculate the most appropriate size element to carry out this task and the cost to heat this water if the tariff unit is 25 cents.
Calculate the current the element draws from the 230 volt supply.
Finally calculate the savings in cost over a year if a thermal blanket covers the tank that raises the efficiency to 90% and a whole tank of water is used 335 days of the year.
Solution
The first step is to calculate the amount of energy required to heat the water.
The equation is
\( Q = m c \Delta T \)
where \( Q \) is the energy required (J), \( m \) is the mass of the substance (kg), \( c \) is the specific heat capacity (J.kg-1.°C-1), and \( \Delta T \) is the temperature change (°C).
For water, c = 4184 J.kg-1.°C-1.
Calculating the Mass of Water to be Heated \( m \)
First, the mass of water is calculated. The tank is 1.5 m wide, 2 m high and 0.5 m deep. The calculation is shown below. All dimensions are worked in metres.
\( V = Whd = 1.5 \cdot 2 \cdot 0.5 = \) 1.5 m3.
The density of water is approximately 1000 kg.m-3. Therefore, the mass of water to be heated is 1500 kg.
Calculating the Temperature Change \( \Delta T \)
The temperature change is the difference between the final temperature and the initial temperature. In this case, \( \Delta T = 90 - 25 = \) 65°C.
Calculating the Energy Required
The mass and temperature change can now be plugged into the energy equation.
\( Q = m c \Delta T = 1500 \cdot 4184 \cdot 65 = \) 407940000 J.
\( Q \) may also be expressed as 113.317 kWh (6 sf). 1 kWh = 3600000 J.
Calculating the Cylinder Power and Current - Using joules
The cylinder needs to be heated from 25°C to 90°C in 21 hours. The required rate of energy transfer to the water is calculated below. This is also the effective output power of the heating element.
\( P_{\mathrm{out}} = \frac{Q}{t} = \frac{407940000}{21 \cdot 3600} = \) 5396.03 J.s-1 or 5396.03 W (6 sf).
In other words, the effective output power of the element must be a minimum of 5396.03 W.
Calculating the Cylinder Power and Current - Using kilowatt-hours
The cylinder needs to be heated from 25°C to 90°C in 21 hours. The required rate of energy transfer to the water is calculated below. This is also the effective output power of the heating element.
\( P_{\mathrm{out}} = \frac{Q}{t} = \frac{113.317}{21} = \) 5.39603 kW or 5396.03 W (6 sf).
In other words, the effective output power of the element must be a minimum of 5396.03 W.
Calculating the Load Current and Cost at 55% Efficiency and a 25c tariff
The effective output power required is 5396.03 W, but the cylinder element is 55% efficient (\( \eta \) = 0.55). We need to calculate \( P_{\mathrm{in}} \).
\( P_{\mathrm{out}} = \eta P_{\mathrm{in}} \rightarrow P_{\mathrm{in}} = \frac{P_{\mathrm{in}}}{\eta} = \frac{5396.03}{0.55} \) = 9810.96 W (6 sf) or 9.81096 kW (6 sf).
The next size element up will be 10 kW.
The element is powered off 230 V. The load current can then be calculated.
\( I = \frac{P_{\mathrm{out}}}{V} = \frac{10000}{230} = \) 43.47 A (4 sf) #.
The amount of energy required from the element is calculated.
\( E_{\mathrm{out}} = \eta E_{\mathrm{in}} \rightarrow E_{\mathrm{in}} = \frac{E_{\mathrm{in}}}{\eta} = \frac{407940000}{0.55} \) = 741709000 J (6 sf) or 206.03 kWh (6 sf).
The element power is higher than what is required to achieve the heating in 21 hours. The operating time will be a little shorter. However, the amount of energy will remain the same.
The cost can be calculated using a tariff of $0.25 per kWh.
\( C = r E_{\mathrm{in}} = 0.25 \cdot 206.03 = \) $51.5075.
Calculating the Load Current and Cost at 90% Efficiency and a 25c tariff
The amount of energy required from the element is calculated.
\( E_{\mathrm{out}} = \eta E_{\mathrm{in}} \rightarrow E_{\mathrm{in}} = \frac{E_{\mathrm{in}}}{\eta} = \frac{407940000}{0.9} \) = 453267000 J (6 sf) or 125.907 kWh (6 sf).
The element power (10 kW as previously calculated) is higher than what is required to achieve the heating in 21 hours. The operating time will be a little shorter. However, the amount of energy will remain the same.
The cost can be calculated using a tariff of $0.25 per kWh.
\( C = r E_{\mathrm{in}} = 0.25 \cdot 125.907 = \) $31.4769.
Calculating the Difference in Yearly Running Costs
The cylinder is used 335 days a year to heat water in this manner.
The yearly cost at 55% efficiency is:
\( C_{55} = 51.5075 \cdot 335 = \) $17255.01 per year.
The yearly cost at 90% efficiency is:
\( C_{90} = 31.4769 \cdot 335 = \) $10544.76 per year.
The cost difference is $6710.25 per year #.