Question

Cold water at \(10^{\circ} \mathrm{C}\) enters a water heater at the rate of \(0.02 \mathrm{m}^{3} / \mathrm{min}\) and leaves the water heater at \(50^{\circ} \mathrm{C}\). The water heater receives heat from a heat pump that receives heat from a heat source at \(0^{\circ} \mathrm{C}\). (a) Assuming the water to be an incompressible liquid that does not change phase during heat addition, determine the rate of heat supplied to the water, in \(\mathrm{kJ} / \mathrm{s}\) (b) Assuming the water heater acts as a heat sink having an average temperature of \(30^{\circ} \mathrm{C}\), determine the minimum power supplied to the heat pump, in \(\mathrm{kW}\)

Short answer

Answer: The rate of heat supplied to the water is 56.03 kJ/min and the minimum power supplied to the heat pump is 3.694 kW.

Step-by-step solution

(a) Calculate the mass flow rate of water entering the heater

To calculate the mass flow rate, we'll use the given volume flow rate and the density of water, assuming that the water's density does not change significantly over the temperature range in question. Water has approximately a constant density of 1000 kg/m^3. Mass flow rate = (volume flow rate) x (density of water) Mass flow rate = (0.02 m^3/min) x (1000 kg/m^3)

Convert the mass flow rate to kg/s

Now we need to convert the mass flow rate from kg/min to kg/s. Mass flow rate = (Mass flow rate in kg/min) x (1 min / 60 s) Mass flow rate = (20 kg/min) x (1 min / 60 s) = 0.333 kg/s

Calculate the rate of heat supplied to the water

Next, we'll calculate the rate of heat supplied to the water using the mass flow rate, specific heat of water, and the temperature difference. Rate of heat supplied = (mass flow rate) x (specific heat of water) x (temperature difference) The specific heat of water is about 4.186 kJ/(kg·°C). Rate of heat supplied = (0.333 kg/s) x (4.186 kJ/(kg·°C)) x (50°C - 10°C) Rate of heat supplied = 56.03 kJ/min

(b) Calculate the maximum COP of the heat pump

Now, we need to calculate the maximum COP of the heat pump based on given heat source and sink temperatures. COP (Coefficient of Performance) = T_sink / (T_sink - T_source) Where T_sink is the sink temperature (30°C) and T_source is the source temperature (0°C). To plug in these values, we need to convert them to Kelvin: T_sink = 30°C + 273.15 = 303.15 K T_source = 0°C + 273.15 = 273.15 K COP_max = 303.15 K / (303.15 K - 273.15 K) = 15.158

Calculate the minimum power supplied to the heat pump

Lastly, we'll calculate the minimum power supplied to the heat pump using the heat supply rate from part (a) and the maximum COP we just found: Power = (Rate of heat supplied) / (COP_max) Power = (56.03 kJ/min) / (15.158) Power = 3.694 kW. Thus, the rate of heat supplied to the water is 56.03 kJ/min and the minimum power supplied to the heat pump is 3.694 kW.

Key concepts

Mass Flow Rate Calculation

The mass flow rate is a fundamental concept in thermodynamics, reflecting the amount of mass passing through a given surface per unit time. It is a critical factor in systems involving fluid dynamics, such as heating and cooling processes.

In the exercise, the mass flow rate is calculated by multiplying the volume flow rate (given in cubic meters per minute) by the density of water, which is approximately 1000 kg/m³. This provides the mass flow rate in kilograms per minute, which is then converted to kilograms per second to match the standard SI units for subsequent calculations. This initial step is crucial because it allows us to determine how much water is being heated and at what pace.

Understanding the mass flow rate is not only essential for the problem at hand but also serves as a vital concept when we explore the workings of various industrial applications, ranging from HVAC systems to aerospace engineering.

Specific Heat of Water

Specific heat capacity, often referred to as specific heat, is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Water has a specific heat of approximately 4.186 kJ/(kg°C), which is relatively high compared to other substances. This high specific heat is why water is excellent for regulating temperatures, such as in our bodies or the environment.

In our textbook problem, the specific heat of water is crucial because it determines how much energy is needed to raise the temperature of the water flowing through the heater from 10°C to 50°C. The energy required, or the rate of heat supplied, is a product of the mass flow rate of water, the specific heat of water, and the temperature change.

When solving related heat transfer problems, keeping in mind the specific heat of involved substances is necessary to predict the thermal response accurately.

Heat Pump Coefficient of Performance

A heat pump's Coefficient of Performance (COP) is a measure of its efficiency. The COP is defined as the ratio of the heat provided to the heat source (the output) to the work or energy input required to transfer that heat. A higher COP means the heat pump is more efficient, as it can transfer more heat per unit of work input.

To calculate the COP of the heat pump in our exercise, we use the temperatures of the heat sink and source converted to Kelvin. This temperature conversion is important because the COP formula requires absolute temperature values for accuracy. The COP gives us an understanding of the theoretical maximum efficiency of the heat pump, which, in turn, allows us to deduce the minimum power that must be supplied to achieve the desired heat transfer rates for heating the water.

Temperature Conversion to Kelvin

Temperature can be measured using various scales, the Celsius scale and the Kelvin scale being the most common in scientific contexts. In the field of thermodynamics, Kelvin is the SI base unit for temperature, and it is preferred due to its fundamental relation to absolute zero—the point where no thermal energy exists.

To convert Celsius to Kelvin, one adds 273.15 to the Celsius temperature. This is precisely what we do in our exercise to find the COP of the heat pump. The temperature conversion ensures that we properly account for the absolute temperatures in our efficiency calculations. This conversion seems simple, but it is essential for any calculations involving thermodynamic cycles or principles.