Question

A thin-walled double-pipe counter-flow heat exchanger is used to cool oil \(\left(c_{p}=2.20 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from 150 to \(40^{\circ} \mathrm{C}\) at a rate of \(2 \mathrm{kg} / \mathrm{s}\) by water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(1.5 \mathrm{kg} / \mathrm{s}\). Determine the rate of heat transfer in the heat exchanger and the exit temperature of water.

Short answer

Answer: The rate of heat transfer in the heat exchanger is 484 kJ/s, and the exit temperature of the water is approximately 45.8°C.

Step-by-step solution

Determine the heat transfer rate

From energy balance equation, the heat transfer rate (Q) can be calculated as follows: $$ Q = m_{oil}c_{p_{oil}}(T_{oil_{in}} - T_{oil_{out}}) $$ where: - \(m_{oil} = 2\,\text{kg/s}\) is the mass flow rate of oil - \(c_{p_{oil}} = 2.20\,\text{kJ/kg °C}\) is the specific heat capacity of oil - \(T_{oil_{in}} = 150^\circ\,\text{C}\) is the initial temperature of oil - \(T_{oil_{out}} = 40^\circ\,\text{C}\) is the final temperature of oil Now, plug the values into the formula: $$ Q = 2\,\text{kg/s}\times 2.20\,\text{kJ/kg °C} \times 110 ^\circ\,\text{C} $$ Calculating the heat transfer rate, Q: $$ Q = 484\,\text{kJ/s} $$

Calculate the heat gained by the water

Using the same energy balance equation for the water: $$ Q = m_{water}c_{p_{water}}\Delta T_{water} $$ where: - \(m_{water} = 1.5\,\text{kg/s}\) is the mass flow rate of water - \(c_{p_{water}} = 4.18\,\text{kJ/kg °C}\) is the specific heat capacity of water - \(\Delta T_{water} = T_{water_{out}} - T_{water_{in}}\) We know that the heat gained by the water should be equal to the heat lost by the oil. Thus, $$ 484\,\text{kJ/s} = 1.5\,\text{kg/s}\times 4.18\,\text{kJ/kg °C}\times (T_{water_{out}} - 22^\circ\,\text{C}) $$

Calculate the exit temperature of water

Now, isolate \(T_{water_{out}}\) to calculate the exit temperature of water: $$ T_{water_{out}} = \frac{484\,\text{kJ/s}}{1.5\,\text{kg/s}\times 4.18\,\text{kJ/kg °C}} + 22^\circ\,\text{C} $$ Calculating the exit temperature of water: $$ T_{water_{out}} \approx 45.8^\circ\,\text{C} $$ Therefore, the rate of heat transfer in the heat exchanger is \(484\,\text{kJ/s}\), and the exit temperature of the water is approximately \(45.8^\circ\,\text{C}\).

Key concepts

Counter-flow Heat Exchanger

In a counter-flow heat exchanger, two fluids flow in opposite directions, which provides a more uniform temperature gradient and increases the heat transfer efficiency compared to parallel flow heat exchangers. It’s especially effective for transferring heat between fluids with a small temperature difference. The effectiveness of counter-flow design allows for the hotter fluid to continue transferring heat until the end of the exchanger, while in parallel flow, the heat transfer would diminish as the fluids approach the same temperature.

The thin-walled double-pipe configuration mentioned in the exercise is a common type of counter-flow heat exchanger. This setup maximizes the heat transferred from the hot oil to the cooler water due to the larger temperature difference at any given point along the length of the exchanger. The geometry ensures close contact between the pipes, thus enhancing the efficient use of space and materials.

Specific Heat Capacity

Specific heat capacity, represented by the symbol \(c_p\), is a measure of how much heat energy is required to raise the temperature of one kilogram of a substance by one degree Celsius. Expressed in units of kJ/kg°C, this property is intrinsic to the material and plays a vital role in calculations involving heat transfer.

For example, in our exercise, the oil and water have different specific heat capacities (2.20 kJ/kg°C for oil and 4.18 kJ/kg°C for water), meaning water requires almost twice as much heat energy to increase its temperature by the same amount as oil would. Understanding specific heat capacity is crucial for determining the amount of energy involved in heating or cooling processes.

Energy Balance Equation

The energy balance equation is a mathematical statement of the conservation of energy principle applied to heat transfer processes. It states that heat lost by one fluid must equal the heat gained by another in a closed system, assuming no heat is lost to the surroundings.

In the provided solution, the energy balance equation is used twice: first, to calculate the rate of heat transfer from the oil, and second, to find out the heat gained by the water, symbolized by Q. By setting the heat lost by oil equal to the heat gained by water, we can solve for unknown variables such as the exit temperature of the water. This equation is fundamental in determining the thermal performance of heat exchangers.

Mass Flow Rate

Mass flow rate is the amount of mass flowing through a cross-section per unit time, typically measured in kg/s. It is an essential factor in heat exchanger calculations because the amount of heat transferred is proportional to the mass flow rate of the fluid and its specific heat capacity.

In the given exercise, the mass flow rates of the oil and water are provided, which are crucial for computing the heat transfer rate (Q). By knowing the mass flow rates, 2 kg/s for oil and 1.5 kg/s for water, we can apply them along with specific heat capacities to find the rate of heat exchange and the resulting temperature changes. This variable directly influences how quickly a fluid will heat up or cool down within the heat exchanger.