Question

In a dairy plant, milk at \(4^{\circ} \mathrm{C}\) is pasteurized continuously at \(72^{\circ} \mathrm{C}\) at a rate of \(12 \mathrm{L} / \mathrm{s}\) for 24 hours a day and 365 days a year. The milk is heated to the pasteurizing temperature by hot water heated in a natural-gas-fired boiler that has an efficiency of 82 percent. The pasteurized milk is then cooled by cold water at \(18^{\circ} \mathrm{C}\) before it is finally refrigerated back to \(4^{\circ} \mathrm{C}\). To save energy and money, the plant installs a regenerator that has an effectiveness of 82 percent. If the cost of natural gas is \(\$ 1.30 /\) therm \((1 \text { therm }=105,500 \mathrm{kJ}),\) determine how much energy and money the regenerator will save this company per year and the annual reduction in entropy generation.

Short answer

In a milk pasteurization plant, the installation of a regenerator with an effectiveness of 82% can result in significant energy and cost savings, as well as a reduction in entropy generation. The annual energy savings amount to 25,949,983,296 kJ, while the annual cost savings amount to $320,904.87. Furthermore, the reduction in entropy generation amounts to 1,807,505.25 kJ/year K.

Step-by-step solution

Calculate the heat transfer rate without a regenerator

First, we need to determine the rate of heat transfer required to raise the temperature of the milk from \(4^{\circ}\mathrm{C}\) to \(72^{\circ}\mathrm{C}\). We can do this using the equation: $$Q_{milk} = \dot{m}_{milk} \times c_{p,milk} \times (T_{out} - T_{in})$$ where \(\dot{m}_{milk}\) is the mass flow rate of milk, \(c_{p,milk}\) is the specific heat of milk, and \(T_{out}\) and \(T_{in}\) are the outlet and inlet temperatures, respectively. To find the mass flow rate, we can use the given volumetric flow rate and the density of milk, which is approximately 1030 kg/m\(^3\). Since 1 L is equal to 0.001 m\(^3\), we can calculate the mass flow rate as follows. $$\dot{m}_{milk} = 12 \,\text{L/s} \times 0.001 \,\dfrac{\text{m}^3}{\text{L}} \times 1030 \,\dfrac{\text{kg}}{\text{m}^3} = 12.36 \,\text{kg/s}$$ Let's assume the specific heat of milk is approximately that of water, \(c_{p,milk} = 4.18 \,\text{kJ/kg}^\circ\text{C}\). Now, we can calculate the heat transfer rate. $$Q_{milk} = 12.36 \,\text{kg/s}\times 4.18\,\dfrac{\text{kJ}}{\text{kg}^\circ\text{C}} \times (72^\circ\text{C} - 4^\circ\text{C}) = 3466.97\,\text{kJ/s}$$

Calculate the heat transfer rate with a regenerator

Now, we need to determine the rate of heat transfer required when the regenerator is installed. Since the effectiveness of the regenerator is 82%, it will recover 82% of the waste heat from the pasteurized milk. The equation for the heat transfer rate with the regenerator is: $$Q_{regen} = Q_{milk} \times (1 - \text{effectiveness})$$ Substitute the given values to find \(Q_{regen}\). $$Q_{regen} = 3466.97 \,\text{kJ/s} \times (1 - 0.82) = 623.055 \,\text{kJ/s}$$

Calculate annual energy savings

Now, we can calculate the annual energy savings by comparing the heat transfer rates without and with the regenerator. $$E_{savings} = (Q_{milk} - Q_{regen}) \times \text{time}$$ Time = 24 hours/day × 365 days/year × 3600 seconds/hour $$E_{savings} = (3466.97 \,\text{kJ/s} - 623.055 \,\text{kJ/s})\times 24 \times 365 \times 3600 = 25949983296 \,\text{kJ/year}$$

Calculate annual cost savings

We can calculate the annual cost savings by converting the energy savings to therms and multiplying by the cost of natural gas. $$C_{savings} = \dfrac{E_{savings}}{1\,\text{therm}/105500\,\text{kJ}} \times \$1.30/\text{therm}$$ $$C_{savings} = \dfrac{25949983296\,\text{kJ/year}}{105500\,\text{kJ/therm}} \times \$1.30/\text{therm} = \$320,904.87\,\text{year}^{-1}$$

Calculate the annual reduction in entropy generation

In order to calculate the annual reduction in entropy generation, we first need to find the rate of entropy generation without and with the regenerator. Entropy generation can be calculated using the equation: $$\Delta S = \dfrac{Q}{T}$$ We can find the entropy generation rates for the milk heating and cooling processes without the regenerator, and then find the difference with the process when the regenerator is installed. $$\Delta S_{heating,wo} = \dfrac{Q_{milk}}{T_{out}}$$ $$\Delta S_{cooling,wo} = \dfrac{Q_{milk}}{T_{in}}$$ $$\Delta S_{regen} = \dfrac{Q_{regen}}{T_{out}}$$ Now we can find the entropy generation rates with and without the regenerator. $$\Delta S_{wo} = \Delta S_{heating,wo} + \Delta S_{cooling,wo}$$ $$\Delta S_{w} = \Delta S_{heating,wo} + \Delta S_{regen}$$ $$\Delta S_{reduction} = \Delta S_{wo} - \Delta S_{w}$$ Substitute the given temperatures of 277.15 K and 301.15 K for \(T_{in}\) and \(T_{out}\), respectively. $$\Delta S_{reduction} = \left[\dfrac{3466.97\,\text{kJ/s}}{301.15\,\text{K}} - \dfrac{3466.97\,\text{kJ/s}}{277.15\,\text{K}}\right] - \left[\dfrac{623.055\,\text{kJ/s}}{301.15\,\text{K}}\right]$$ $$\Delta S_{reduction} = 0.0573\,\text{kJ/s}\, \text{K}$$ To get the annual reduction in entropy generation, we multiply the entropy generation rate by time. $$\Delta S_{annual} = \Delta S_{reduction} \times 24 \times 365 \times 3600 = 1807505.25\,\text{kJ/year}\, \text{K}$$

Summary

By installing a regenerator with an effectiveness of 82%, the dairy plant can achieve annual energy savings of 25949983296 kJ, cost savings of $320,904.87, and a reduction in entropy generation of 1807505.25 kJ/year K.

Key concepts

Heat Transfer Rate

Understanding the heat transfer rate is crucial when analyzing processes like pasteurization in a dairy plant. The heat transfer rate quantifies how fast thermal energy is exchanged between the milk and the heating medium, in this case, hot water. It's typically measured in kilojoules per second (kJ/s), also known as kilowatts (kW).

In our exercise, we found that without a regenerator, the heat transfer rate needed to bring the milk to pasteurizing temperature is quite high. When the regenerator is installed, it significantly reduces the heat transfer required from the boiler, because it recovers some of the energy from the pasteurized milk to preheat the incoming cold milk.

With an effectiveness of 82%, we calculate the new reduced heat transfer rate, showing a substantial decrease in the energy needed, which directly correlates to energy savings. In real-world applications, understanding and optimizing heat transfer rates can lead to significant cost savings and more efficient use of resources.

Energy Efficiency

Energy efficiency is a term that describes getting the same functional output (like heating milk to pasteurization temperatures) with less energy input. It's a measure that is key in reducing operating costs and environmental impact. The efficiency of a natural-gas-fired boiler, for example, determines how much of the energy content of the gas is actually converted into usable heat.

In our exercise, the regenerator improves the plant's energy efficiency by reusing waste heat. This action exemplifies a principle in thermodynamics: to maximize efficiency, minimize waste. By recuperating otherwise lost energy, the plant uses less gas and therefore spends less money on fuel. The calculation of cost savings, derived from the energy efficiency improvement, emphasizes the direct economic benefits of investing in energy-efficient technologies.

Another layer to consider is environmental impact—higher energy efficiency generally means less fossil fuel usage and lower greenhouse gas emissions, aligning economic and ecological interests.

Entropy Generation

Entropy generation is a thermodynamic concept that measures the irreversible dispersal of energy in a system. It's a profound concept as it relates to the Second Law of Thermodynamics, which states that entropy within an isolated system always increases over time. Entropy generation is typically seen as a bad player in thermal systems as it's associated with energy loss.

In the exercise, the entropy generation before the installation of the regenerator is higher than after. This means that without the regenerator, more energy becomes unavailable for useful work because it’s dissipated in the form of heat to the surroundings. A reduction in entropy generation, as calculated after the regenerator's installation, indicates a more efficient process with fewer irreversible losses.

From an engineering standpoint, reducing entropy generation equates to designing processes that are closer to ideal, reversible cycles. This is an important indicator of a system's thermodynamic performance and places a spotlight on the importance of reducing waste to increase the sustainability of thermal systems.