Explain how the LMTD method can be used to determine the heat transfer surface area of a multipass shelland-tube heat exchanger when all the necessary information, including the outlet temperatures, is given.

Short Answer

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Explain how to use the LMTD method to determine the heat transfer surface area of a multipass shell-and-tube heat exchanger when all necessary information, including outlet temperatures, is given.

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01

Understanding the LMTD Method

The LMTD method is a widely used approach for determining the heat transfer surface area of heat exchangers. With inlet and outlet temperatures given for both hot and cold fluids, as well as an overall heat transfer coefficient (U) and the heat transfer rate (Q), we can determine the required heat transfer surface area (A) using this method.
02

Calculate the Heat Transfer Rate (Q)

Calculate the heat transfer rate (Q) using the given outlet temperatures and the mass flow rates and specific heat capacities of both hot and cold fluids: Q = m_c * C_p,c * (T_c,out - T_c,in) = m_h * C_p,h * (T_h,in - T_h,out) where: - Q: heat transfer rate (W) - m_c: mass flow rate of the cold fluid (kg/s) - C_p,c: specific heat capacity of the cold fluid (J/kg*K) - T_c,out: outlet temperature of the cold fluid (K) - T_c,in: inlet temperature of the cold fluid (K) - m_h: mass flow rate of the hot fluid (kg/s) - C_p,h: specific heat capacity of the hot fluid (J/kg*K) - T_h,in: inlet temperature of the hot fluid (K) - T_h,out: outlet temperature of the hot fluid (K)
03

Calculate Temperature Differences

Calculate the temperature differences at both ends of the exchanger: ΔT_1 = T_h,in - T_c,out ΔT_2 = T_h,out - T_c,in
04

Calculate the LMTD

Calculate the logarithmic mean temperature difference (LMTD) using the temperature differences obtained in the previous step: LMTD = (ΔT_1 - ΔT_2) / ln(ΔT_1 / ΔT_2)
05

Calculate the Overall Heat Transfer Coefficient (U)

The overall heat transfer coefficient (U) needs to be given for the shell-and-tube heat exchanger. This value depends on the design and material properties of the heat exchanger.
06

Calculate the Heat Transfer Surface Area (A)

Finally, calculate the heat transfer surface area (A) using the LMTD method with the overall heat transfer coefficient (U), heat transfer rate (Q), and LMTD values: A = Q / (U * LMTD) The result will give you the heat transfer surface area (A) of the multipass shell-and-tube heat exchanger.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Exchanger Design
When designing a heat exchanger, engineers must take into account various crucial factors to ensure efficient heat transfer between two fluids. The choice of heat exchanger type, such as a shell-and-tube, and its configurations like the number of tube passes, plays a pivotal role in determining how the heat will be exchanged. The material of construction must be selected based on the working fluids to resist corrosion and withstand high temperatures.

In addition to material considerations, the design process also involves the sizing of the heat exchanger. This requires determining the heat transfer surface area, which is affected by the flow rates of the fluids, their specific heat capacities, and the temperature differences. The LMTD method offers a systematic way to calculate the necessary area for a desired heat transfer rate, making it foundational in the design of a multipass shell-and-tube heat exchanger.
Heat Transfer Rate Calculation
The heat transfer rate, denoted as Q, represents the amount of heat energy transferred between the hot and cold fluids in the heat exchanger per unit time, and it's typically expressed in watts (W). To calculate Q, we utilize the mass flow rates, specific heat capacities of the fluids involved, and their respective inlet and outlet temperatures.

The formula for the hot and cold sides must be equated as they should transfer an equal amount of heat energy under steady-state conditions:
  • For the cold fluid:
\(Q = m_c \times C_{p,c} \times (T_{c,out} - T_{c,in})\)
  • For the hot fluid:
\(Q = m_h \times C_{p,h} \times (T_{h,in} - T_{h,out})\)

This balance is essential to ensure that the energy gained by the cold fluid is equal to the energy lost by the hot fluid, which is the principle of conservation of energy in heat exchanger operations.
Logarithmic Mean Temperature Difference
The Logarithmic Mean Temperature Difference (LMTD) is a critical value in heat exchanger calculations that accounts for the variation in temperature difference between the hot and cold fluids across the length of the exchanger. It is particularly useful when there is a significant temperature change and provides a more accurate representation of the average temperature gradient than a simple arithmetic mean would.

The LMTD is calculated using the temperature differences at the two ends of the exchanger (\( \Delta T_1 \) and \( \Delta T_2 \)), as follows:
  • At one end (where the hot fluid enters): \( \Delta T_1 = T_{h,in} - T_{c,out} \)
  • At the other end (where the hot fluid exits): \( \Delta T_2 = T_{h,out} - T_{c,in} \)
\[ LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})} \]
This logarithmic approach ensures that the temperature gradient's variance is considered, leading to a more optimal and efficient heat exchanger design.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, symbolized as U, quantifies the heat transfer capacity of the heat exchanger per unit area per unit temperature difference. It is a measure of the heat exchanger’s ability to conduct and transfer heat and is influenced by material properties, surface condition, fluid velocity, and temperature. The higher the U value, the more effective the heat exchanger is at transferring heat.

In the LMTD method, once the LMTD and the heat transfer rate (Q) are known, the overall heat transfer coefficient (U) can be used to find the necessary heat transfer surface area (A):
  • \( A = \frac{Q}{U \times LMTD} \)
The U value is essential for ensuring that the heat exchanger design meets the required thermal performance and that the fluids achieve the desired outlet temperatures. An accurately determined U value can prevent oversizing or undersizing the heat exchanger, which could lead to inefficiency or excessive costs.

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Most popular questions from this chapter

The radiator in an automobile is a cross-flow heat exchanger \(\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\) that uses air \(\left(c_{p}=1.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) to cool the engine-coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(30^{\circ} \mathrm{C}\) air through this radiator at a rate of \(10 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, the effectiveness of the radiator is \(0.4\). Determine \((a)\) the outlet temperature of the air and (b) the rate of heat transfer between the two fluids.

Consider a shell and tube heat exchanger in a milk be heated from \(20^{\circ} \mathrm{C}\) by hot water initially at \(140^{\circ} \mathrm{C}\) and flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The milk flows through 30 thin-walled tubes with an inside diameter of \(20 \mathrm{~mm}\) with each tube making 10 passes through the shell. The average convective heat transfer coefficients on the milk and water side are \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. In order to complete the pasteurizing process and hence restrict the microbial growth in the milk, it is required to have the exit temperature of milk attain at least \(70^{\circ} \mathrm{C}\). As a design engineer, your job is to decide upon the shell width (tube length in each pass) so that the milk exit temperature of \(70^{\circ} \mathrm{C}\) can be achieved. One of the design requirements is that the exit temperature of hot water should be at least \(10^{\circ} \mathrm{C}\) higher than the exit temperature of milk.

Can the temperature of the cold fluid rise above the inlet temperature of the hot fluid at any location in a heat exchanger? Explain.

Can the temperature of the hot fluid drop below the inlet temperature of the cold fluid at any location in a heat exchanger? Explain.

Ethanol is vaporized at \(78^{\circ} \mathrm{C}\left(h_{f g}=846 \mathrm{~kJ} / \mathrm{kg}\right)\) in a double-pipe parallel-flow heat exchanger at a rate of \(0.03 \mathrm{~kg} / \mathrm{s}\) by hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(120^{\circ} \mathrm{C}\). If the heat transfer surface area and the overall heat transfer coefficients are \(6.2 \mathrm{~m}^{2}\) and \(320 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the outlet temperature and the mass flow rate of oil using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

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