Question

A shell-and-tube heat exchanger is used for heating \(10 \mathrm{~kg} / \mathrm{s}\) of oil \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(46^{\circ} \mathrm{C}\). The heat exchanger has 1 -shell pass and 6-tube passes. Water enters the shell side at \(80^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). The overall heat transfer coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the rate of heat transfer and the heat transfer area.

Short answer

Based on the given information and calculations, the rate of heat transfer (Q) in the shell-and-tube heat exchanger is approximately 420,000 W, and the required heat transfer area (A) is approximately 14.19 m².

Step-by-step solution

Calculate the heat transfer rate (Q)

To compute the heat transfer rate, apply the formula \(Q=m_{o} \cdot c_{p} \cdot \Delta T_{o}\), where \(\Delta T_{o}\) is the temperature difference of the oil between the inlet and outlet. \(\Delta T_{o} = T_{2} - T_{1} = 46^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 21^{\circ} \mathrm{C}\) \(Q = 10\ \mathrm{kg/s} \cdot 2.0\ \mathrm{kJ/(kg\cdot K)} \cdot 21\ \mathrm{K}\) Since \(1\ \mathrm{kJ/s} = 1000\ \mathrm{W}\), we can convert the units: \(Q = 10 \cdot 2.0 \cdot 21 \cdot 1000\ \mathrm{W} = 420,000\ \mathrm{W}\)

Calculate temperature differences (ΔT)

Calculate the temperature difference between inlet and outlet temperatures on both sides: \(\Delta T_{1} = T_{3} - T_{1} = 80^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 55^{\circ} \mathrm{C}\) \(\Delta T_{2} = T_{4} - T_{2} = 60^{\circ} \mathrm{C} - 46^{\circ} \mathrm{C} = 14^{\circ} \mathrm{C}\)

Calculate Log Mean Temperature Difference (LMTD)

To compute the LMTD, use the formula \(\mathrm{LMTD} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\): \(\mathrm{LMTD} = \frac{55^{\circ} \mathrm{C} - 14^{\circ} \mathrm{C}}{\ln(\frac{55^{\circ} \mathrm{C}}{14^{\circ} \mathrm{C}})} = \frac{41}{\ln(\frac{55}{14})} \approx 29.59^{\circ} \mathrm{C}\)

Calculate the heat transfer area (A)

To calculate the heat transfer area, use the formula \(A=\frac{Q}{U \cdot \mathrm{LMTD}}\): \(A = \frac{420,000\ \mathrm{W}}{1000\ \mathrm{W/m^{2}\cdot K} \cdot 29.59^{\circ} \mathrm{C}} \approx 14.19\ \mathrm{m^{2}}\) The rate of heat transfer is approximately \(420,000\ \mathrm{W}\), and the heat transfer area required is approximately \(14.19\ \mathrm{m^{2}}\).

Key concepts

Shell-and-tube heat exchanger

A shell-and-tube heat exchanger is a device commonly used for heat transfer between two fluids. In this configuration, one fluid flows through the tubes while the other flows around the outside in the shell, allowing for efficient heat exchange. This design is highly versatile and can handle various fluids, temperatures, and pressures, making it one of the most popular types of heat exchangers.

Shell-and-tube heat exchangers are particularly effective in industries like oil refineries, chemical plants, and power stations due to their durability and high heat transfer efficiency. This specific system typically consists of a bundle of tubes housed within a cylindrical shell. One fluid runs through the tubes and gathers or releases heat, while the second fluid flows over the tubes inside the shell to either collect or provide heat.

This type of heat exchanger's performance relies heavily on the arrangement of the tubes. The specific problem describes a setup with 6 tube passes and 1 shell pass, indicating a certain complexity and efficiency in heat transfer across the surface area available.

Log Mean Temperature Difference (LMTD)

The Log Mean Temperature Difference (LMTD) is a crucial factor in calculating the heat transfer in a heat exchanger. LMTD allows for a more accurate estimation of the temperature driving force by considering the varying temperature differences at different points in the exchanger.

LMTD is particularly useful for heat exchangers with changing temperatures across their length, like the shell-and-tube configuration. It provides a steady value that effectively represents the temperature difference needed for calculations.

To calculate the LMTD, first determine the temperature differences at the inlet and outlet on both the hot and cold sides. Use the formula: \[ \mathrm{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \] This formula computes the logarithmic mean of these temperature differences, simplifying the design and analysis of heat exchangers.

Heat transfer coefficient

The heat transfer coefficient is essential for estimating the heat exchange efficiency between the fluids in a heat exchanger. It represents how well heat is transferred from one medium to another, considering the thermal conductivity of the materials, surface area, and fluid flows.

In this problem, the given overall heat transfer coefficient is \(1000\ \mathrm{W/m^2\cdot K}\). It is critical to recognize this as an aggregation of several factors, including convection between the fluid and the tube wall, conduction through the tube wall, and convection into the other fluid.

Higher heat transfer coefficients correlate with better thermal performance. It influences the rate of heat transfer and is particularly important during design adjustments and scaling, impacting overall system efficiency and required heat transfer area.

Heat transfer area calculation

Calculating the heat transfer area is pivotal for designing effective heat exchangers. It determines how much surface area the fluids need to exchange heat optimally. This calculation helps ensure that the exchanger can adequately manage the load with the given flows and thermal conditions.

To find the heat transfer area, use the formula: \[ A = \frac{Q}{U \cdot \mathrm{LMTD}} \] Where \( Q \) is the heat transfer rate, \( U \) is the overall heat transfer coefficient, and LMTD is the log mean temperature difference.

In this example, the area calculated is approximately \( 14.19\ \mathrm{m^2} \). This result is integral to ensuring the heat exchanger meets operational requirements by accommodating enough surface area for the given conditions. Understanding and designing around this value helps in creating efficient, scalable heat-exchanging solutions.