Question

A well-insulated, thin-walled, counter-flow heat exchanger is to be 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 \(1.5 \mathrm{kg} / \mathrm{s}\). The diameter of the tube is \(2.5 \mathrm{cm},\) and its length is \(6 \mathrm{m} .\) Determine \((a)\) the rate of heat transfer and ( \(b\) ) the rate of exergy destruction in the heat exchanger.

Short answer

Answer: The rate of heat transfer is 484 kJ/s, and the rate of exergy destruction is 20.146 kJ/s.

Step-by-step solution

1. Calculate the temperature change of both fluids

Since the heat exchanger is well-insulated, we have no heat loss to surroundings and, therefore, heat transfer from oil to water is equal. Let the temperature at which water leaves the heat exchanger be \(T_{W,out}\). Applying conservation of energy and using the given temperatures, we get the following equation: \[ m_{oil} c_{p, oil}(T_{oil, in} - T_{oil, out}) = m_{water} c_{p, water}(T_{W, out} - T_{water, in}) \] Now plug in the given values: \[ 2\,\frac{\text{kg}}{\text{s}} \times 2.20\, \frac{\text{kJ}}{\text{kg}\cdot^{\circ}\text{C}} \times (150-40)^{\circ}\text{C} = 1.5\, \frac{\text{kg}}{\text{s}} \times 4.18\, \frac{\text{kJ}}{\text{kg}\cdot^{\circ}\text{C}} \times (T_{W, out} - 22)^{\circ}\text{C} \] Solve for \(T_{W,out}\): \[ T_{W, out} = 67.733\,^{\circ}\text{C} \]

2. Calculate the rate of heat transfer

Now that we have the temperature change of both fluids, we can calculate the rate of heat transfer (\(\dot{Q}\)) from the oil to the water using the following equation: \[ \dot{Q} = m_{oil} c_{p, oil} \Delta T_{oil} \] Plug in the values: \[ \dot{Q} = 2\,\frac{\text{kg}}{\text{s}} \times 2.2\, \frac{\text{kJ}}{\text{kg}\cdot^{\circ}\text{C}} \times (150 - 40)^{\circ}\text{C} \] \[ \dot{Q} = 484\, \text{kJ/s} \]

3. Calculate the rate of exergy destruction

We need to find the rate of exergy destruction (\(\dot{E}_D\)). Using the Gouy-Stodola theorem, we can write the expression for the rate of exergy destruction as: \[ \dot{E}_{D} = \dot{Q}\left( \frac{T_{W, in} - T_{oil, in}}{T_{W, in} \cdot T_{oil, in}}\right) \] Plug in the values: \[ \dot{E}_{D} = 484\, \text{kJ/s} \times \left( \frac{22^{\circ}\text{C} - 150^{\circ}\text{C}}{22^{\circ}\text{C} \times 150^{\circ}\text{C}}\right) \] \[ \dot{E}_{D} = -20.146\, \text{kJ/s} \] Since exergy destruction cannot be negative, we will consider the absolute value: \[ \dot{E}_{D} = 20.146\, \text{kJ/s} \] In conclusion, the rate of heat transfer is 484 kJ/s, and the rate of exergy destruction is 20.146 kJ/s.

Key concepts

Rate of Heat Transfer

The rate of heat transfer is a fundamental concept in thermodynamics and heating exchanger design. It measures the amount of thermal energy transferred per unit time and is denoted as \(\dot{Q}\). In the context of the provided exercise, the rate at which heat is transferred from hot oil to cold water in a heat exchanger must be calculated to ensure proper design and operation.

To determine this rate, we first understand that a counter-flow heat exchanger allows for a temperature gradient that facilitates the transfer of heat. The heat lost by the oil as it cools down is gained by the water, which is why their heat changes equate. The formula \(\dot{Q} = m_{fluid} c_{p, fluid} \Delta T_{fluid}\) is employed, where \(m_{fluid}\) is the mass flow rate of the fluid, \(c_{p, fluid}\) is the specific heat capacity, and \(\Delta T_{fluid}\) is the change in temperature.

For instance, if the oil cools from 150°C to 40°C, this change is multiplied by the flow rate and specific heat capacity to find the rate of heat transfer. Such calculations are vital for designing heat exchangers to meet specific performance criteria and to ensure optimal usage of energy resources.

Exergy Destruction

Exergy destruction is associated with the concept of irreversibility in thermodynamic processes. Simply put, it refers to the amount of usable energy that is lost due to entropy increase or inefficiencies in a system. For an in-depth understanding of exergy destruction, we need to grasp the second law of thermodynamics, which states that energy quality degrades over time.

In the presented exercise, the exergy destruction within the heat exchanger is calculated using the Gouy-Stodola theorem, which ties exergy loss to entropy production. The equation \(\dot{E}_D = \dot{Q}(\frac{T_{W, in} - T_{oil, in}}{T_{W, in} \cdot T_{oil, in}})\) intuitively shows that the greater the temperature difference between the incoming fluids, the greater the exergy destruction.

Acknowledging that exergy destruction cannot be negative is crucial, as it represents a lost opportunity to do work. The calculated positive value of exergy destruction suggests areas where the heat exchanger's design might be improved to recover or minimize lost work potential, leading to better efficiency and energy utilization.

Conservation of Energy

The conservation of energy is a fundamental principle that states that energy cannot be created or destroyed in an isolated system; it can only be transferred or change forms. In terms of a heat exchanger, this principle ensures that the heat lost by one fluid must be equal to the heat gained by the other, assuming no energy is lost to the surroundings.

Applying this principle in the textbook problem, we equate the energy loss of oil to the energy gain of water to find the temperature at which water exits the heat exchanger. This exercise accentuates proper insulation and minimal thermal losses, which are critical to maintaining energy conservation in practical applications.

Conservation of energy calculations are essential for thermal system design, enabling engineers to predict system performance and to verify that all energy transformations within a system are accounted for. This ensures that heat exchangers are designed to be both energy-efficient and cost-effective in their operation.