Question

A heat pump receives heat from a lake that has an average winter time temperature of \(6^{\circ} \mathrm{C}\) and supplies heat into a house having an average temperature of \(23^{\circ} \mathrm{C}\). (a) If the house loses heat to the atmosphere at the rate of \(52,000 \mathrm{kJ} / \mathrm{h},\) determine the minimum power supplied to the heat pump, in \(\mathrm{kW}\) (b) A heat exchanger is used to transfer the energy from the lake water to the heat pump. If the lake water temperature decreases by \(5^{\circ} \mathrm{C}\) as it flows through the lake water-to-heat pump heat exchanger, determine the minimum mass flow rate of lake water, in \(\mathrm{kg} / \mathrm{s}\). Neglect the effect of the lake water pump.

Short answer

Answer: The minimum power supplied to the heat pump is 1.76 kW, and the minimum mass flow rate of lake water is 0.61 kg/s.

Step-by-step solution

Calculate the coefficient of performance (COP) of the heat pump

We will calculate the COP for a heat pump, using its formula: $$ COP = \frac{Q_H}{W}$$ COP is the coefficient of performance, \(Q_H\) is the heat transfer to the house, and \(W\) is the work input. For a heat pump, we also have the following relationship: $$Q_{H} = Q_{L} + W$$ Where \(Q_L\) is the heat extracted from the lake. To find the minimum power supplied, we need to use the Carnot COP, which can be related to temperatures: $$COP_{Carnot} = \frac{T_H}{T_H - T_L}$$ Where \(T_H\) is the temperature of the house (in Kelvin) and \(T_L\) is the temperature of the lake (in Kelvin). First, let's convert the given temperatures to Kelvin: $$T_H = 23 + 273.15 = 296.15 K$$ $$T_L = 6 + 273.15 = 279.15 K$$ Now, we can calculate the Carnot COP: $$COP_{Carnot} = \frac{296.15}{296.15 - 279.15} = 8.12$$

Calculate the minimum power supplied to the heat pump

Now that we have the COP, we can calculate the required work input W. Using the definition of COP and given house heat loss rate, $$W = \frac{Q_H}{COP_{Carnot}}$$ $$W = \frac{52000 \mathrm{kJ\ /h}}{8.12}$$ Let's convert the answer to kilowatts: $$W = \frac{52000 \mathrm{kJ/h}}{8.12\times 3600 \mathrm{s/h}} = 1.76 \mathrm{kW}$$ The minimum power supplied to the heat pump is 1.76 kW.

Calculate the minimum mass flow rate

We are given that the lake water temperature decreases by \(5^{\circ} \mathrm{C}\) as it flows through the heat exchanger. To calculate the minimum mass flow rate, we will use the formula for heat transfer in the heat exchanger: $$Q_L = mc_p\Delta T$$ where \(Q_L\) is the heat extracted from the lake, \(m\) is the mass flow rate of the lake water, \(c_p\) is the specific heat of water (which is \(4.18 \mathrm{kJ/kg\cdot K}\)), and \(\Delta T\) is the temperature change of the lake water. First, let's find \(Q_L\) using the house heat loss rate \(Q_H\) and the work input \(W\), $$Q_L = Q_H - W \times 3600 \mathrm{s}$$ $$Q_L = 52000\mathrm{kJ/h} - 1.76\mathrm{kW} \times 3600 \mathrm{s/h} = 52000-6336\mathrm{kJ/h}=45664\mathrm{kJ/h}$$ Now, we can use this value in our formula to calculate the mass flow rate \(m\), We can rearrange the formula: $$m = \frac{Q_L}{c_p\Delta T}$$ The temperature change \(\Delta T\) is given as \(5^{\circ} \mathrm{C}\). Plugging in the values: $$m = \frac{45664\mathrm{kJ/h}}{4.18 \mathrm{kJ/kg\cdot K} \times 5^{\circ} \mathrm{C}}$$ Let's convert the answer to \(\mathrm{kg/s}\): $$m = \frac{45664 / 3600 \mathrm{kJ/s}}{4.18 \times 5 \mathrm{kJ/kg}}= 0.61 \mathrm{kg/s}$$ The minimum mass flow rate of lake water is 0.61 kg/s.

Key concepts

Carnot Coefficient of Performance

The Carnot coefficient of performance (COP) is a theoretical maximum COP of a heat pump or refrigerator operating between two constant temperature reservoirs. It's defined by the temperatures of the hot and cold reservoirs. For a heat pump, the Carnot COP is given by the formula:
\[COP_{Carnot} = \frac{T_H}{T_H - T_L}\]
Where \(T_H\) and \(T_L\) are the absolute temperatures (in Kelvin) of the hot and cold reservoirs, respectively. It paints a picture of an ideal heat pump that would work without losing any energy to friction, electrical resistance or other inefficiencies. In real scenarios, no heat pump can achieve the Carnot COP, but it serves as a benchmark to measure how a real system compares to the perfect one.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. In the context of a heat pump, heat transfer occurs from the cooler lake to the warmer house. The basic formula for calculating this energy transfer during a temperature change is:
\[Q = mc_p\Delta T\]
The symbol \(Q\) represents the amount of heat transferred, \(m\) is the mass of the substance, \(c_p\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Heat pumps leverage this concept to maintain the desired temperature inside homes, working against the natural flow of heat from warm to cool areas.

Mass Flow Rate Calculation

The mass flow rate is a measure of how much mass moves through a given surface per time period, often indicated in units like \(\mathrm{kg/s}\) or \(\mathrm{kg/h}\). In thermal systems, determining the mass flow rate is crucial for understanding how quickly heat can be transferred using a fluid medium. The mass flow rate can be computed when we know how much heat needs to be moved and the properties of the fluid:
\[m = \frac{Q}{c_p\Delta T}\]
This calculation aids in designing systems, such as those in the exercise where the movement of lake water through a heat exchanger is analyzed, to ensure they meet the thermal requirements for heating or cooling spaces.

Specific Heat

Specific heat, denoted as \(c_p\), is the amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius (or Kelvin). Water has a high specific heat capacity, meaning it can hold a substantial amount of heat energy, which makes it a good medium for transferring heat in systems like heat pumps. In the exercise, the specific heat of water (4.18 \(\mathrm{kJ/kg\cdot K}\)) is a key factor in determining the mass flow rate of the lake water through the heat exchanger, as it influences the amount of heat water can carry per degree of temperature change.