Question

Design a hydrocooling unit that can cool fruits and vegetables from 30 to \(5^{\circ} \mathrm{C}\) at a rate of \(20,000 \mathrm{kg} / \mathrm{h}\) under the following conditions: The unit will be of flood type, which will cool the products as they are conveyed into the channel filled with water. The products will be dropped into the channel filled with water at one end and be picked up at the other end. The channel can be as wide as \(3 \mathrm{m}\) and as high as \(90 \mathrm{cm} .\) The water is to be circulated and cooled by the evaporator section of a refrigeration system. The refrigerant temperature inside the coils is to be \(-2^{\circ} \mathrm{C}\), and the water temperature is not to drop below \(1^{\circ} \mathrm{C}\) and not to exceed \(6^{\circ} \mathrm{C}\) Assuming reasonable values for the average product density, specific heat, and porosity (the fraction of air volume in a box), recommend reasonable values for \((a)\) the water velocity through the channel and ( \(b\) ) the refrigeration capacity of the refrigeration system.

Short answer

Answer: The water velocity through the channel is approximately 0.205 m/s and the refrigeration capacity of the refrigeration system is 2000000 kJ/h.

Step-by-step solution

1. Calculate the cooling load

To calculate the cooling load, we need to determine the amount of heat that needs to be removed from the fruits and vegetables. We'll use the formula: \(Q = m \times C_p \times \Delta T\) Where: - \(Q\) is the heat that needs to be removed (in kJ) - \(m\) is the mass flow rate of the products (in kg/h) - \(C_p\) is the specific heat of the products (in kJ/kg·K) - \(\Delta T\) is the temperature change of the products (in K). For this problem, we have: - \(m = 20000 \, \mathrm{kg/h}\) - \(\Delta T = 30 \, ^{\circ} \mathrm{C} - 5 \, ^{\circ} \mathrm{C} = 25 \, ^{\circ} \mathrm{C} (25 \, \mathrm{K})\) Assuming a reasonable value for the specific heat of the products, we'll use: - \(C_p = 4 \, \mathrm{kJ/kg \cdot K}\) Now we can calculate the cooling load \(Q\): \(Q = 20000 \, \mathrm{kg/h} \times 4 \, \mathrm{kJ/kg \cdot K} \times 25 \, \mathrm{K} = 2000000 \, \mathrm{kJ/h}\)

2. Calculate the heat transfer area

To calculate the heat transfer area, we'll use the formula: \(A = \dfrac{Q}{h \times (T_{water} - T_{refrigerant})}\) Where: - \(A\) is the heat transfer area (in m²) - \(h\) is the heat transfer coefficient (in kW/m²·K) - \(T_{water}\) is the average water temperature (in K) - \(T_{refrigerant}\) is the refrigerant temperature (in K). Assuming a reasonable value for the heat transfer coefficient, we'll use: - \(h = 1 \, \mathrm{kW/m^2 \cdot K}\) The average water temperature is: - \(T_{water} = (1 + 6) / 2 = 3.5 \, ^{\circ} \mathrm{C} (276.65 \, \mathrm{K})\) The refrigerant temperature is: - \(T_{refrigerant} = -2 \, ^{\circ} \mathrm{C} (271.15 \, \mathrm{K})\) Now we can calculate the heat transfer area \(A\): \(A = \dfrac{2000000 \, \mathrm{kJ/h}}{ 1 \, \mathrm{kW/m^2 \cdot K} \times (276.65 \, \mathrm{K} - 271.15 \, \mathrm{K})} = 4000 \, \mathrm{m^2}\)

3. Calculate the water velocity

To calculate the water velocity \(v\), we'll use the formula: \(v = \dfrac{Q}{A}\) Where: - \(v\) is the water velocity (in m/s) - \(Q\) is the flow rate of water (in m³/s) - \(A\) is the channel cross-sectional area (in m²). The channel cross-sectional area is: - \(A = (3 \, \mathrm{m}) \times (0.9 \, \mathrm{m}) = 2.7 \, \mathrm{m^2}\) Now we can calculate the water velocity \(v\): \(v = \dfrac{2000000 \, \mathrm{kJ/h}}{2.7 \, \mathrm{m^2}} = 740.74 \, \mathrm{m/h} \approx 0.205 \, \mathrm{m/s}\)

4. Refrigeration capacity

To calculate the refrigeration capacity, we'll use the same formula as in step 1, but using the refrigerant temperature change instead of the product temperature change. Refrigeration capacity is equal to the cooling load, which we have already calculated: \(Q = 2000000 \, \mathrm{kJ/h}\) Thus, the refrigeration capacity of the system is 2000000 kJ/h. In conclusion, the recommended values are: a) Water velocity through the channel: \(0.205 \, \mathrm{m/s}\) b) Refrigeration capacity of the refrigeration system: \(2000000 \, \mathrm{kJ/h}\)

Key concepts

Cooling Load Calculation

Calculating the cooling load is a critical step in the design of a hydrocooling unit. It involves determining the amount of heat to be removed from the product—in this case, fruits and vegetables. This is a key aspect for students to understand as it impacts the efficiency and performance of the cooling system.

The formula used is:
\[ Q = m \times C_p \times \Delta T \]
where:

• \(Q\) is the total heat energy to be removed, measured in kilojoules (kJ),
• \(m\) is the mass flow rate of the product, indicating how much product is cooled per hour measured in kilograms per hour (kg/h),
• \(C_p\) is the specific heat capacity of the product, which is the amount of energy required to raise one kilogram of the product by one degree Celsius, and
• \(\Delta T\) is the change in temperature of the product from its initial to the desired final temperature.

It's important for students to use correct values for specific heat capacity and accurately calculate the change in temperature to ensure the cooling load is determined accurately. A clear understanding of these factors helps in making informed decisions about the design and capability of the cooling unit.

Heat Transfer Area

The heat transfer area is another fundamental concept in designing a hydrocooling unit. This area is defined by the surface over which heat removal occurs and directly correlates with the efficiency of heat exchange between the product and the cooling medium, typically water.

The formula to calculate the heat transfer area is:
\[ A = \dfrac{Q}{h \times (T_{water} - T_{refrigerant})} \]
where:

• \(A\) is the required heat transfer area in square meters (\(m^2\)),
• \(h\) is the heat transfer coefficient, which depends on the material and surface characteristics of the heat exchange surfaces,
• \(T_{water}\) is the average water temperature within the cooling system, and
• \(T_{refrigerant}\) is the temperature of the refrigerant used to cool the water.

A higher heat transfer coefficient or a larger temperature difference between water and refrigerant will allow for a smaller heat transfer area. Conversely, a lower heat transfer coefficient or a smaller temperature difference will require a larger area to achieve the same cooling effect. Students should comprehend the implications of these relationships when considering the overall design and cost of the hydrocooling unit.

Refrigeration Capacity

The refrigeration capacity is a measure of how much heat energy the refrigeration system can remove from the products per hour. It's essential for students to grasp this concept as it directly impacts the ability of the hydrocooling unit to maintain desired temperatures for the cooling process.

Refrigeration capacity is typically represented by the same formula used for cooling load (since these values are equal in a balanced system):
\[ Q = m \times C_p \times \Delta T \]
In the context of a hydrocooling unit, the refrigeration capacity must be sized to handle the total cooling load calculated previously. Students should understand that an undersized refrigeration system can lead to insufficient cooling, while an oversized system can be more costly and energy-inefficient.

For the exercise provided, the refrigeration capacity calculated matches the cooling load, ensuring that the system is correctly balanced. A clear understanding of these principles supports students in designing effective refrigeration systems for various applications.