Question

Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), where it is heated by a hot water stream \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(70^{\circ} \mathrm{C}\) at a rate of \(1 \mathrm{~kg} / \mathrm{s}\). Determine the maximum heat transfer rate and the outlet temperatures of both fluids for that case.

Short answer

Answer: The maximum heat transfer rate is 209500 J/s, with the outlet temperature of the hot water stream being 20°C and the outlet temperature of the air being 43.28°C.

Step-by-step solution

Use the conservation of energy principle

Since energy is conserved, for the maximum heat transfer rate, all the energy lost by the hot water stream must be equal to the energy gained by the air stream. Let \(Q\) be the heat transfer rate, \(m\) be the mass flow rate, and \(c_p\) be the specific heat capacity. The energy change for the hot water stream can be represented as follows: \((m_{hot} \cdot c_{p_{hot}}) (T_{in_{hot}} - T_{out_{hot}}) = Q\) Simultaneously, the energy change for the air can be represented as: \(Q = (m_{air} \cdot c_{p_{air}}) (T_{out_{air}} - T_{in_{air}})\)

Find the maximum heat transfer rate, Q

Since we want the maximum heat transfer rate, the outlet temperature of the hot water stream will be equal to the inlet temperature of the air. Therefore, \(T_{out_{hot}} = T_{in_{air}}\). Plugging in the provided values and solving for \(Q\), we get \(Q = (m_{hot} \cdot c_{p_{hot}}) (T_{in_{hot}} - T_{in_{air}})\) \(Q = (1\,\mathrm{kg/s}) \cdot (4190\,\mathrm{J/kg\cdot K}) \cdot (70\,^{\circ}\mathrm{C} - 20\,^{\circ}\mathrm{C})\) \(Q = 209500\,\mathrm{J/s}\)

Calculate the outlet temperature of the air

Now, we will use the heat transfer rate and the air stream's energy change equation to find the outlet temperature of the air: \(Q = (m_{air} \cdot c_{p_{air}}) (T_{out_{air}} - T_{in_{air}})\) Rearranging the equation, we get: \(T_{out_{air}} = T_{in_{air}} + \frac{Q}{(m_{air} \cdot c_{p_{air}})}\) Plugging in the provided values, we get: \(T_{out_{air}} = 20\,^{\circ}\mathrm{C} + \frac{209500\,\mathrm{J/s}}{(3\,\mathrm{kg/s}) \cdot (1005\,\mathrm{J/kg\cdot K})}\) \(T_{out_{air}} = 20\,^{\circ}\mathrm{C} + 23.28\,^{\circ}\mathrm{C}\) \(T_{out_{air}} = 43.28\,^{\circ}\mathrm{C}\)

Present the final results

We calculated the maximum heat transfer rate, Q, to be \(209500\,\mathrm{J/s}\) and the outlet temperatures for both fluids: - Outlet temperature of the hot water stream: \(T_{out_{hot}} = 20\,^{\circ}\mathrm{C}\) - Outlet temperature of the air: \(T_{out_{air}} = 43.28\,^{\circ}\mathrm{C}\)

Key concepts

Cross-flow Heat Exchanger

A cross-flow heat exchanger is a device designed to transfer heat between two fluid streams that flow perpendicular to each other. This unique arrangement allows for efficient heat transfer, even when the temperature difference between the two streams is not significant.
In cross-flow heat exchangers, one fluid typically flows through tubes, while the other flows across the tubes' outer surfaces. This configuration encourages maximum heat exchange surface contact.
The temperature profiles of both streams in a cross-flow heat exchanger are typically different, with a characteristic that neither fluid ever fully reaches the inlet temperature of the other. This makes it increasingly efficient in systems where a moderate temperature change is required.

• Air and water often serve as the fluid streams in cross-flow heat exchangers.
• Their perpendicular flow pattern optimizes the heat transfer rate.

By understanding how these devices function, engineers can design systems that utilize the benefits of cross-flow heat transfers, maximizing efficiency in heating or cooling applications.

Conservation of Energy

The principle of conservation of energy is fundamental to solving heat transfer problems. It states that energy cannot be created or destroyed in an isolated system. Instead, energy can only be transferred from one form to another.
In the context of a heat exchanger, this means the heat lost by one fluid must equal the heat gained by the other fluid. This balance is crucial for designing and analyzing any heating or cooling process.
Applying this principle to a heat exchange scenario helps you calculate the heat transfer rate. By leveraging the known mass flow rates and specific heat capacities, the exchange rate equation can be dynamically balanced.

• Heat loss in one fluid equals heat gain in the other.
• This concept ensures energy efficiency and accuracy in design.

Understanding conservation of energy is essential for troubleshooting heat systems and ensuring they perform to expectations.

Specific Heat Capacity

Specific heat capacity, often denoted as \(c_p\), is a measure of how much heat energy a substance can store per unit mass and per degree Kelvin or Celsius. It is a crucial factor in thermal calculations, especially in heat exchangers.
Air and water, for instance, have different specific heat capacities. In the context of the exercise, air has a specific heat capacity of 1005 J/kg·K, while water's is higher, at 4190 J/kg·K. This difference plays a significant role in how these substances absorb and lose heat.
Specific heat capacity directly influences how quickly a substance can heat up or cool down. For example, a substance with a high specific heat capacity (like water) can absorb more heat without a significant increase in temperature.

• Determines a substance's ability to hold heat.
• Critical for calculating temperature changes in heat exchange processes.

By understanding specific heat capacity, you can better predict temperature shifts and energy requirements in thermal systems.

Outlet Temperature Calculation

Calculating outlet temperatures in a heat exchanger involves balancing the energy exchanges between the incoming and outgoing streams using the equations derived from conservation of energy.
In practical terms, this means applying the equation for energy balance for the heat exchange process. It involves the mass flow rate, specific heat capacity, and the difference between inlet and outlet temperatures.
In the provided exercise, the formula used ensures that the calculations consider the specific properties of the materials involved, such as the specific heat capacities of air and water.

• Outlet temperature for the hot fluid is calculated under the assumption of maximum heat transfer.
• The outlet temperature for the cold fluid is derived using the calculated heat transfer rate and the properties of the cold fluid.

Developing proficiency in these calculations is important for engineers who need to design systems where precise temperature control is critical.