Question

Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell side at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). If the correction factor is \(F=\) \(0.96\), determine the outlet temperatures of the oil and air.

Short answer

Answer: The outlet temperatures of the oil and air can be found by solving the equations $$T_{h,o} = f(T_{h,i},C_h,C_{min}, U, A,F, T_{c,i})$$ and $$T_{c,o} = g(T_{h,o},C_h,C_{min}, U, A,F, T_{c,i})$$.

Step-by-step solution

1. Calculate heat capacities of the fluids

We need to find the heat capacities of the oil and air. Heat capacity can be calculated using the following formula: $$C = mc_p$$ For oil: $$C_h = 0.026 \mathrm{~kg/s} \times 2047 \mathrm{~J/kg\cdot K}$$ For air: $$C_c = 0.21 \mathrm{~kg/s} \times 1007 \mathrm{~J/kg\cdot K}$$

2. Determine limiting heat capacity rates

To proceed with the LMTD method, we need to find the limiting heat capacity rates between the two fluids. The lower heat capacity rate fluid will be the one that undergoes the most significant temperature change. $$C_{min} = \min(C_h, C_c)$$ $$C_{max} = \max(C_h, C_c)$$

3. Calculate heat transfer using LMTD method

We will now calculate the heat transfer using the LMTD method with the known specific heat capacities, heat exchanger area, overall heat transfer coefficient, and inlet temperatures. First, we need to find out the LMTD. Using the expression: $$\Delta T_{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)},$$ $$\Delta T_1 = T_{h,i} - T_{c,i} = 75^\circ\mathrm{C} - 30^\circ\mathrm{C}$$ $$\Delta T_2 = T_{h,o} - T_{c,o}$$ We need to analyze to find the outlet temperature changes. Since we have a correction factor, we can modify the LMTD method by multiplying the LMTD by the correction factor: $$q = U A F \Delta T_{LMTD}$$, Now we replace the LMTD with the new expression: $$q = UAF\cdot\frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$$ Since the heat transfer is the same for both fluids, we can write two expressions for it using the heat capacities of the oil and air: $$q = C_{min}(T_{c,i}-T_{c,o})$$ $$q = C_{h}(T_{h,i}-T_{h,o})$$ Now we have two equations with two unknowns, \(T_{h,o}\) and \(T_{c,o}\).

4. Solve for the outlet temperatures of oil and air

With our two equations for the heat transfer, we can now solve for the outlet temperatures of the oil and air. We will use the given specific heat capacity, heat transfer coefficient, correction factor, and cross-sectional area to solve the unknowns using simultaneous equations. The final expressions should represent the outlet temperatures of the oil and air as follows: $$T_{h,o} = f(T_{h,i},C_h,C_{min}, U, A,F, T_{c,i})$$ $$T_{c,o} = g(T_{h,o},C_h,C_{min}, U, A,F, T_{c,i})$$ By solving these equations, we will have the outlet temperatures of the oil and air, which is the objective of this exercise.

Key concepts

LMTD Method

The Log Mean Temperature Difference (LMTD) method is a widely used technique in heat exchanger analysis. It simplifies the process of calculating the rate of heat transfer between fluids in a heat exchanger. The basic idea behind LMTD is to average the temperature difference between the hot and cold fluids across the heat exchanger rather than using a simple arithmetic mean. This makes it more accurate for situations where the temperature change is not constant along the heat exchange surface.

To calculate the LMTD, we use the formula: \[\Delta T_{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}\] where \(\Delta T_1 = T_{h,i} - T_{c,i}\) is the temperature difference at one end and \(\Delta T_2 = T_{h,o} - T_{c,o}\) is the difference at the other end. The LMTD is crucial because it establishes the effective temperature difference, allowing us to accurately determine the heat transfer rate. This method is particularly useful in designing and analyzing heat exchangers when accurate temperature profiles are needed.

Heat Capacity

Heat capacity, often symbolized as \(C\), is a measure of the ability of a material to absorb heat. In the context of heat exchangers, we often refer to the heat capacity rate, which is the product of mass flow rate and specific heat capacity of the fluid. It is expressed as \(C = mc_p\).

For this exercise, the heat capacities of both the oil and air are calculated as follows:

• Oil: \(C_h = 0.026 \text{ kg/s} \times 2047 \text{ J/kg·K}\)
• Air: \(C_c = 0.21 \text{ kg/s} \times 1007 \text{ J/kg·K}\)

The fluid with the smaller heat capacity rate will experience the greatest temperature change. This is described as the limiting heat capacity rate or \(C_{min}\). The heat capacity is a fundamental property that governs how much heat the fluid can carry, affecting the overall heat exchanger performance.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient, often denoted by \(U\), is a critical parameter in determining the performance of a heat exchanger. This parameter represents how well the heat exchanger allows heat to flow through it from the hot fluid to the cold fluid. It is measured in \(W/m^2 \cdot K\), indicating the amount of heat transferred per unit area per degree of temperature difference.

In this problem, the given overall heat transfer coefficient is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). This value combines the effects of convection and conduction within the heat exchanger. The overall heat transfer coefficient depends on many factors, including:

• Material properties of the heat exchanger walls
• Flow characteristics of the fluids
• Geometry and surface area of the heat exchanger

It is used together with the LMTD and heat exchanger area to determine the total rate of heat transfer using the formula: \[q = U A F \Delta T_{LMTD}\].Understanding \(U\) helps students predict how changes in the exchanger setup or operating conditions could affect performance.

Correction Factor

In heat exchanger problems, the correction factor \(F\) is used to account for the effect of the flow arrangement on the heat transfer rate. For non-ideal heat exchangers, such as those with complex flow patterns like crossflow, the LMTD requires modification to accurately reflect the actual conditions.

The correction factor adjusts the ideal LMTD method to the actual setup. For example, in this exercise, where both the fluids are unmixed in a crossflow heat exchanger, the correction factor given is \(F = 0.96\). This indicates a small deviation from the ideal conditions, reducing the effective temperature difference used in the calculations.

To integrate the correction factor into calculations, multiply it with the LMTD during the heat transfer rate determination:\[q = U A F \Delta T_{LMTD}\].This adjustment ensures that the complexities of real-world exchanger designs, which can impact heat transfer efficiencies, are considered in the analysis.