Question

Water flows through a shower head steadily at a rate of \(10 \mathrm{L} / \mathrm{min}\). An electric resistance heater placed in the water pipe heats the water from 16 to \(43^{\circ} \mathrm{C}\). Taking the density of water to be \(1 \mathrm{kg} / \mathrm{L},\) determine the electric power input to the heater, in \(\mathrm{kW}\). In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of \(39^{\circ} \mathrm{C}\) through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water), determine the electric power input required in this case. If the price of the electric energy is 11.5 e \(/ \mathrm{kWh}\), determine how much money is saved during a 10 -min shower as a result of installing this heat exchanger.

Short answer

Answer: €1,083.3

Step-by-step solution

1. Calculate the mass flow rate

We are given the volume flow rate, which is \(10\,\mathrm{L/min}\). To convert this value to mass flow rate, we multiply the volume flow rate by the density of water, which is \(1\,\mathrm{kg/L}\). Mass flow rate = \(10\,\mathrm{L/min} * 1\,\mathrm{kg/L} = 10\,\mathrm{kg/min}\).

2. Calculate the initial electric power input

We can use the power formula: \(P = m * c * \Delta T\). For water, the specific heat capacity \(c = 4.18\,\mathrm{kJ/kgK}\). The temperature difference \(\Delta T = 43 - 16 = 27^{\circ}\mathrm{C}\). Thus, the initial electric power input is: \(P = 10\,\mathrm{kg/min} * 4.18\,\mathrm{kJ/kgK} * 27\,\mathrm{K} = 1130.6\,\mathrm{kJ/min}\). Convert this value to \(\mathrm{kW}\): \(P = 1130.6\,\mathrm{kJ/min} * \dfrac{1\,\mathrm{kW}}{60\,\mathrm{kJ/s}} = 18.84\,\mathrm{kW}\).

3. Calculate the power input with the heat exchanger

The heat exchanger has an effectiveness of 0.50, meaning it recovers half of the energy that can possibly be transferred. Since the drained warm water has a temperature of \(39^{\circ}\mathrm{C}\), the temperature difference is reduced to \(27 * 0.5 = 13.5^{\circ}\mathrm{C}\). Calculate the power input using this reduced temperature difference: \(P_{ex} = 10\,\mathrm{kg/min} * 4.18\,\mathrm{kJ/kgK} * 13.5\,\mathrm{K} = 565.3\,\mathrm{kJ/min}\). Convert this value to \(\mathrm{kW}\): \(P_{ex} = 565.3\,\mathrm{kJ/min} * \dfrac{1\,\mathrm{kW}}{60\,\mathrm{kJ/s}} = 9.42\,\mathrm{kW}\).

4. Calculate the energy savings in a 10-minute shower

Compare the power input with and without the heat exchanger: \(P_{savings} = 18.84\,\mathrm{kW} - 9.42\,\mathrm{kW} = 9.42\,\mathrm{kW}\). Calculate the energy savings in a 10-minute shower: \(E_{savings} = 9.42\,\mathrm{kW} * 10\,\mathrm{min} = 94.2\,\mathrm{kWh}\).

5. Calculate the money saved during a 10-minute shower

Use the given price of electric energy: \(11.5\,\mathrm{e/kWh}\). Calculate the money saved for a 10-minute shower: Money saved = \(94.2\,\mathrm{kWh} * 11.5\,\mathrm{e/kWh} = 1,083.3\,\mathrm{e}\).