Question

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools \(5000 \mathrm{~kg} / \mathrm{h}\) of air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) which enters the air handler at \(25^{\circ} \mathrm{C}\). If these streams are in counter-flow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Short answer

Answer: (c) \(19^{\circ} \mathrm{C}\)

Step-by-step solution

Recall energy balance equation

The energy balance equation states that the energy entering or leaving a system must be equal to the energy entering or leaving it. For this problem, we can apply an energy balance on the air side, assuming no additional energy is exchanged with the surroundings: \(\dot{m}_{air} c_{p, air} (T_{in, air} - T_{out, air}) = \dot{m}_{water} c_{p, water} (T_{in, water} - T_{out, water})\) where \(\dot{m}_{air}\) and \(\dot{m}_{water}\) are the mass flow rates of the air and water respectively; \(c_{p, air}\) and \(c_{p, water}\) are the specific heat capacities of air and water, respectively; and \(T_{in, air}\), \(T_{out, air}\), \(T_{in, water}\), and \(T_{out, water}\) are the inlet and outlet temperatures of the air and water, respectively.

Plug in the given values and solve for the unknown

The mass flow rates of air and water are given, as well as their respective specific heat capacities and inlet and outlet temperatures. Using these values and plugging them into the energy balance equation, we have: \((5000 \mathrm{~kg} / \mathrm{h}) (1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(25^{\circ} \mathrm{C} - T_{out, air}) = (1000 \mathrm{~kg} / \mathrm{h}) (4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(5^{\circ} \mathrm{C} - 12^{\circ} \mathrm{C})\) Now we can solve for the unknown temperature \(T_{out, air}\): \(5000(25 - T_{out, air}) = -1000(4.2)(5 - 12)\) \(125000 - 5000 T_{out, air} = 29400\) \(-5000 T_{out, air} = -95500\) \(T_{out, air} = 19.1^{\circ} \mathrm{C}\) Since the air is cooled in a heat exchanger, it is not possible to obtain a lower temperature at the air outlet because the water temperature is fixed. Therefore, the minimum temperature at the air outlet is (c) \(19^{\circ} \mathrm{C}\).

Key concepts

Energy Balance

The concept of energy balance is fundamental in understanding heat exchangers. It helps us track and equate energy entering and leaving a system, ensuring conservation of energy. In the context of a heat exchanger, energy balance ensures that the heat lost by one fluid is completely gained by the other, assuming no energy is exchanged with the surroundings.
For this problem, we applied the energy balance on the air and water flowing through the heat exchanger. The equation used was:

• Air Side: \(\dot{m}_{\text{air}} \cdot c_{p, \text{air}} \cdot (T_{\text{in, air}} - T_{\text{out, air}})\)
• Water Side: \(\dot{m}_{\text{water}} \cdot c_{p, \text{water}} \cdot (T_{\text{in, water}} - T_{\text{out, water}})\)

The equation equates these two expressions to conserve energy: \(\dot{m}_{\text{air}} \cdot c_{p, \text{air}} \cdot (T_{\text{in, air}} - T_{\text{out, air}}) = \dot{m}_{\text{water}} \cdot c_{p, \text{water}} \cdot (T_{\text{in, water}} - T_{\text{out, water}})\).
Using this balance, one can solve for unknown temperatures or flow rates.

Specific Heat Capacity

Specific heat capacity is a crucial property that measures how much energy is needed to change the temperature of a substance. It’s denoted by the symbol \(c_p\) and gives the energy required per unit mass per degree change in temperature.
In this problem:

• Water has a specific heat capacity of \(4.2 \text{ kJ} / \text{kg} \cdot \text{K}\), meaning it requires 4.2 kJ to raise the temperature of 1 kg of water by 1 K.
• Air has a specific heat capacity of \(1.0 \text{ kJ} / \text{kg} \cdot \text{K}\), which means it needs 1.0 kJ to increase the temperature of 1 kg of air by 1 K.

The different specific heat capacities for air and water dictate how much energy each of these fluids can carry with a given temperature change. This difference is why the flow rate and specific heat work together to determine the outlet temperatures.

Counter-Flow Heat Exchanger

The counter-flow heat exchanger is a design that maximizes heat transfer efficiency. In this setup, fluids flow in opposite directions. Imagine two parallel paths, one carrying cold fluid and the other hot fluid, but in the opposite direction.
This arrangement enhances the temperature gradient between the fluids, allowing more effective energy transfer. The counter-flow design offers superior thermal performance compared to parallel flow arrangements.
In this exercise, the chilled water flows in one direction while the air flows in the opposite direction. This configuration helps the air be gradually cooled to its minimum possible temperature as it encounters increasingly colder water.
The advantage of the counter-flow heat exchanger is showcased by achieving the minimum outlet temperature for the air, demonstrating its high-efficiency principle.

Mass Flow Rate

The mass flow rate is a measure of the mass of fluid passing through a point per unit time, typically expressed in kg/h. It's a critical factor in determining the overall heat transfer in a heat exchanger.
For our specific problem:

• The flow rate of the chilled water is \(1000 \text{ kg/h}\).
• The air's mass flow rate is \(5000 \text{ kg/h}\).

These rates give us the total mass of air and water that flows through the heat exchanger, influencing the heat exchange process. The higher mass flow rate of air means more energy is required to change its temperature compared to the chilled water.
In the heat exchanger calculation, the mass flow rate is multiplied by specific heat and temperature changes, which directly impacts the heat transfer capacity and the temperature outcome at the exit point of each fluid.