Question

A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

Short answer

Based on the given information, the analysis and calculations show that fouling has indeed affected the overall heat transfer coefficient in the double pipe counterflow heat exchanger. The magnitude of the fouling resistance has been determined to be approximately \(0.00049 \mathrm{~m^2 \cdot K/W}\).

Step-by-step solution

Calculate the heat transfer from oil

To calculate the heat transfer from oil, we will use the formula: \(\dot{Q}=\dot{m} \cdot c_{p} \cdot \Delta T\) Where \(\dot{Q}\) is the heat transfer, \(\dot{m}\) is the mass flow rate, \(c_{p}\) is the specific heat, and \(\Delta T\) is the temperature difference. For the engine oil, we have: \(\dot{m}_{oil} = 2.5 \thinspace \mathrm{kg/s}\), \(\Delta T_{oil} = (55-25)^{\circ} \mathrm{C}=30^{\circ}\mathrm{C}\), Assuming the specific heat of oil, \(c_{p_{oil}} = 2000 \thinspace \mathrm{J/kg \cdot K}\) (typical value), we can calculate the heat transfer from oil: \(\dot{Q}_{oil} = 2.5 \mathrm{\thinspace kg/s} \times 2000 \mathrm{~J/kg \cdot K} \times 30 \mathrm{K} = 150000 \mathrm{~W}\)

Calculate the temperature difference for water

Next, we'll use the heat transfer from oil to determine the temperature difference for the water: \(\dot{Q}_{water} = \dot{m}_{water} \cdot c_{p_{water}} \cdot \Delta T_{water}\) Where \(\dot{m}_{water} = 1.75 \thinspace \mathrm{kg/s}\) and \(c_{p_{water}} = 4206 \thinspace \mathrm{J/kg \cdot K}\). We can rearrange the equation to solve for the temperature difference for the water: \(\Delta T_{water} = \frac{\dot{Q}_{water}}{\dot{m}_{water} \cdot c_{p_{water}}} = \frac{150000 \mathrm{~W}}{1.75 \mathrm{\thinspace kg/s} \times 4206 \mathrm{~J/kg \cdot K}} \approx 20.09 \mathrm{K}\) Since water enters the heat exchanger at \(100^{\circ}\mathrm{C}\), its exit temperature will be approximately \(79.91^{\circ}\mathrm{C}\).

Calculate the Log Mean Temperature Difference (LMTD)

To determine the LMTD, we will use the following formula: \(LMTD=\frac{\Delta T_{1}-\Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}}}\) Where \(\Delta T_{1}\) represents the temperature difference between the hot and cold fluids at one end of the heat exchanger and \(\Delta T_{2}\) represents the temperature difference at the other end. For our heat exchanger, we have \(\Delta T_{1} = 100^{\circ}\mathrm{C} - 25^{\circ}\mathrm{C} = 75^{\circ}\mathrm{C}\) and \(\Delta T_{2}= 79.91^{\circ}\mathrm{C} - 55^{\circ}\mathrm{C} = 24.91^{\circ}\mathrm{C}\). Thus, we can calculate the LMTD: \(LMTD=\frac{75 \mathrm{K}-24.91 \mathrm{K}}{\ln \frac{75 \mathrm{K}}{24.91 \mathrm{K}}} \approx 46.59 \mathrm{K}\)

Calculate the expected surface area without fouling

Now that we have the LMTD, we can calculate the surface area required to transfer the heat using the designed overall heat transfer coefficient, \(U = 450 \mathrm{\thinspace W/m^2 \cdot K}\): \(A = \frac{\dot{Q}_{oil}}{U \cdot LMTD} = \frac{150000 \mathrm{~W}}{450 \mathrm{\thinspace W/m^2 \cdot K} \times 46.59 \mathrm{K}} \approx 7.03 \mathrm{\thinspace m^2}\)

Determine if fouling is present and calculate the fouling resistance

Since the calculated surface area (7.03 \(\mathrm{m^2}\)) is less than the given surface area (7.5 \(\mathrm{m^2}\)), fouling is likely present in the heat exchanger. To determine the magnitude of the fouling resistance, we'll use the following equation: \(R_{f} = \frac{1}{U_{actual}\cdot A} - \frac{1}{U_{design}\cdot A}\) Where \(U_{actual}\) is the actual overall heat transfer coefficient after fouling and \(A\) is the given surface area. First, we need to find the actual overall heat transfer coefficient: \(\dot{Q} = U_{actual} \cdot A \cdot LMTD\) \(U_{actual} = \frac{\dot{Q}}{A \cdot LMTD} = \frac{150000 \mathrm{~W}}{7.5 \mathrm{\thinspace m^2} \times 46.59 \mathrm{K}} \approx 431.20 \mathrm{\thinspace W/m^2 \cdot K}\) Now, we can calculate the fouling resistance: \(R_{f} = \frac{1}{431.20 \mathrm{\thinspace W/m^2 \cdot K} \cdot 7.5 \mathrm{\thinspace m^2}} - \frac{1}{450 \mathrm{\thinspace W/m^2 \cdot K} \cdot 7.5 \mathrm{\thinspace m^2}} \approx 0.00049 \mathrm{\thinspace m^2 \cdot K/W}\) So, fouling has affected the overall heat transfer coefficient, and the magnitude of the fouling resistance is approximately \(0.00049 \mathrm{\thinspace m^2 \cdot K/W}\).

Key concepts

Fouling Resistance

Fouling is a common issue in heat exchangers, where unwanted deposits build up on the heat transfer surfaces. This accumulation can significantly deteriorate the heat exchanger's efficiency by adding an extra layer that heat must travel through. This extra passage created by fouling is called fouling resistance and is represented by the symbol \( R_f \).

Fouling resistance reflects the impact of these deposits on the thermal resistance. Essentially, instead of heat exchanging directly between the fluids, it must also penetrate the unwanted layer, which acts as an insulative barrier.

When calculating the fouling resistance, you compare the heat exchanger's performance with and without fouling. It's measured using the formula:

• \( R_f = \left( \frac{1}{U_{actual} \cdot A} - \frac{1}{U_{design} \cdot A} \right) \) Where \( U_{actual} \) and \( U_{design} \) are the actual and expected overall heat transfer coefficients, respectively, and \( A \) is the surface area.

Recognizing and quantifying fouling within a heat exchanger is crucial for maintenance and ensuring optimal performance.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, denoted as \( U \), is a key parameter in evaluating a heat exchanger's efficiency. It factors in all modes of heat transfer, including conduction, convection, and potentially radiation, across the different layers of materials and fluids involved.

This coefficient provides a comprehensive view of the heat exchanger's ability to transfer heat per unit area and temperature difference between fluids. It's measured in \( \mathrm{W/m^2 \cdot K} \). A higher \( U \) value indicates a more efficient heat exchanger.

To determine the impact of fouling on \( U \), we compare the actual \( U \) value from operational data to the designed \( U \). Any significant decrease usually points to fouling or other efficiency issues.

• Calculation: \( U_{actual} = \frac{\dot{Q}}{A \cdot LMTD} \) This approach uses the measured heat transfer rate \( \dot{Q} \), the surface area \( A \), and the Log Mean Temperature Difference \( LMTD \).

Log Mean Temperature Difference (LMTD)

The Log Mean Temperature Difference, or LMTD, is a vital concept in the heat transfer calculation for heat exchangers. It averages the temperature difference between the fluids at the inlet and outlet, providing a more accurate measure of the effective temperature driving force. This factor is crucial because, in practice, temperature differences change along the length of the exchanger.

LMTD is calculated using the formula:

• \( LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln\frac{\Delta T_1}{\Delta T_2}} \) Where \( \Delta T_1 \) and \( \Delta T_2 \) are the temperature differences between the hot and cold fluids at the respective ends of the exchanger.

The log mean approach prevents the gross overestimation or underestimation of the heat transfer rate that can occur if using a simple arithmetic average. LMTD is essential for determining realistic and efficient design and operational strategies for heat exchangers.

Counterflow Heat Exchanger

A counterflow heat exchanger is a type of heat exchanger where the fluids flow in opposite directions. This arrangement is often more efficient than parallel flow systems, as it allows for a larger mean temperature difference between the fluids over the length of the exchanger.

In this setup, the hot and cold fluids enter and exit the exchanger at different ends. The advantage is that the outlet temperature of the cold fluid can approach the inlet temperature of the hot fluid, maximizing heat transfer rate. This configuration is beneficial for systems where achieving a high degree of temperature change in the fluids is desired.

The counterflow design can be readily identified by drawing temperature profiles for the fluids involved. It offers several efficiency benefits:

• Enhanced heat recovery potential.
• Capability to reach higher exit temperatures.
• More uniform temperature changes compared to parallel flow designs.

This makes counterflow heat exchangers particularly suitable for processes where precise temperature control is critical.

Heat Transfer Calculation

Heat transfer calculations are at the core of designing and operating heat exchangers. The principal aim is to ensure that the exchanger meets the thermal duties required.

The basic equation for heat transfer in exchangers is:

• \( \dot{Q} = \dot{m} \cdot c_p \cdot \Delta T \) Where \( \dot{Q} \) is the heat transfer rate, \( \dot{m} \) is the mass flow rate, \( c_p \) is the specific heat capacity, and \( \Delta T \) is the temperature change of the fluid.

In practical applications, this equation helps determine how much heat is transferred from one fluid to another. It aids in selecting the right heat exchanger design and operational parameters to match real-world thermal loads.

Simple solutions often assume constant temperature differences to simplify the calculations. However, more detailed examination includes temperature variations across the exchanger and the effects of fouling, which must be monitored and managed to maintain efficiency.