Question

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes of a heat exchanger with 2 -shell passes and 23 -tube passes at \(14^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\), while hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell at \(200^{\circ} \mathrm{C}\) at the same mass flow rate. The overall heat transfer coefficient based on the outer surface of the tube is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area on that side is \(20 \mathrm{~m}^{2}\). Determine the rate of heat transfer using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Short answer

Question: Determine the rate of heat transfer in a heat exchanger using (a) the LMTD method and (b) the NTU-effectiveness method, given the following information: 1. Cold water properties: cp = 4180 J/kgK 2. Cold water inlet temperature: T_ci = 14°C 3. Cold water mass flow rate: m_c = 3 kg/s 4. Hot oil properties: cp = 2200 J/kgK 5. Hot oil inlet temperature: T_hi = 200°C 6. Hot oil mass flow rate: m_h = 3 kg/s 7. Number of shell and tube passes 8. Overall heat transfer coefficient: U = 300 W/m²K 9. Heat transfer surface area: A = 20 m² Calculate the rate of heat transfer for both methods and compare the results.

Step-by-step solution

1. Determine the log mean temperature difference (LMTD)

First, we need to find the temperature differences at the inlet and outlet ends of the heat exchanger. We can find these by applying the energy balance equation to the cold water side: $$Q = m_cc_p(T_{co} - T_{ci}) $$ Since we don't have the outlet temperatures yet, we can find an initial LMTD assuming counter-current flow (where the hot fluid leaves at the opposite end of the cold fluid entrance). To find this initial LMTD, we use the following formula: $$LMTD_{initial} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$$ where - \(\Delta T_1 = T_{hi} - T_{ci}\) - \(\Delta T_2 = T_{ho} - T_{co}\) We will be revising the LMTD with the information on shell and tube passes later on.

2. Determine heat capacity rates

Next, we need to find the heat capacity rates for the cold water (C_c) and hot oil (C_h) using the following formulas: $$C_c = m_cc_p$$ $$C_h = m_hc_p$$

3. Calculate the initial heat exchanger effectiveness

With the heat capacity rates, we can calculate the initial heat exchanger effectiveness (ε_initial) using the counter-current flow formula: $$\varepsilon_{initial} = \frac{1 - e^{\frac{-NTU(1+CR)}{1+CR}}}{1 + CR}$$ where - \(NTU = UA / C_{min}\) (Number of Transfer Units) - \(C_{min} = \min(C_c, C_h)\) (Minimum heat capacity rate) - \(CR = \frac{C_{min}}{C_{max}}\) (Capacity rate ratio)

4. Revise the LMTD and find outlet temperatures

Since the heat exchanger has multiple shell and tube passes, we need to revise the initial LMTD to account for these. We can use the following formula (for 2-shell passes and an even number of tube passes): $$LMTD_{actual} = LMTD_{initial}\cdot F$$ where \(F\) is the correction factor, which can be found using the number of tube passes and the heat exchanger configuration. With the actual LMTD, we can find the actual heat transfer rate (Q_actual): $$Q_{actual} = UA\cdot LMTD_{actual}$$ Finally, we can find the outlet temperatures for the cold water (T_co) and hot oil (T_ho) using the energy balance equations: $$T_{co} = T_{ci} + \frac{Q_{actual}}{m_cc_p}$$ $$T_{ho} = T_{hi} - \frac{Q_{actual}}{m_hc_p}$$

5. Determine the rate of heat transfer using the ε-NTU method

With the outlet temperatures and heat exchanger effectiveness, we can determine the rate of heat transfer (Q) using the ε-NTU method formula: $$Q = \varepsilon_{actual}C_{min}(T_{hi} - T_{ci})$$ Now, you can compare the rate of heat transfer using the LMTD and ε-NTU methods. Keep in mind that the values obtained from both methods should be similar.

Key concepts

LMTD Method

When it comes to understanding heat transfer in heat exchangers, the LMTD method, or Log Mean Temperature Difference method, is an essential concept. The LMTD is a way to calculate the average temperature difference between the hot and cold fluids across the heat exchanger. This average is logarithmic because of the changing temperature difference as the fluids move through the exchanger. The LMTD represents the driving force that allows heat transfer to occur.

The formula for calculating the LMTD is given by:
\[\begin{equation}LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}\end{equation}\]
where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences at the two ends of the heat exchanger. To apply the formula, you need to know the inlet and outlet temperatures of both the hot and cold fluids. If these are not directly given, you might have to use energy balances to find them.

One important aspect in heat exchanger design, reflected in this exercise, is accounting for multiple shell and tube passes. These complicate the simple LMTD calculation because they change the flow arrangement and, consequently, the temperature profiles of the fluids. Correction factors, often found from specialized charts or calculations based on the heat exchanger's geometry and flow arrangement, are used to adjust the initial LMTD to a more accurate value that accounts for these complexities.

ε-NTU Method

Another fundamental approach to analyze heat exchanger performance is the ε-NTU method. This method is particularly useful when the inlet or outlet temperatures are unknown and involves calculating the heat exchanger's effectiveness (ε), which is the ratio of the actual heat transfer to the maximum possible heat transfer. By defining the Number of Transfer Units (NTU) and capacity rate ratio (CR), this method can often be more versatile than the LMTD method, especially for complex heat exchanger configurations.

The NTU is defined as:\[\begin{equation} NTU = \frac{UA}{C_{min}}\end{equation}\]
This represents the size of the heat exchanger relative to the heat capacity rate of the working fluid. The heat exchanger effectiveness (ε) can be found using relation charts or formulas specific to the flow arrangement. For counter-current flow, a common arrangement, the effectiveness is calculated as:\[\begin{equation}\varepsilon = \frac{1 - e^{\frac{-NTU(1+CR)}{1+CR}}}{1 + CR}\end{equation}\]
Once the effectiveness is determined, the actual rate of heat transfer is found by multiplying the effectiveness by the minimum heat capacity rate and the temperature difference between the hot and cold fluids at the inlet.

Understanding the ε-NTU method is crucial for students as it applies to various real-world scenarios beyond textbook examples, where explicit inlet and outlet temperatures might not be readily available. Practical applications include designing new heat exchangers or troubleshooting existing ones.

Heat Exchanger Effectiveness

Concept of Effectiveness

Heat exchanger effectiveness is a dimensionless parameter measuring the thermal performance of a heat exchanger. It's the ratio of actual heat transfer to the maximum theoretical heat transfer if the cold fluid were to reach the inlet temperature of the hot fluid. To put it simply, it's a measure of how well the heat exchanger is doing its job.

In the context of our exercise, the effectiveness is influenced by various factors such as the flow arrangement, the number of shell and tube passes, and the properties of the fluids involved. Now, let's examine how this concept applies to our problem:
\[\begin{equation}\varepsilon_{actual} = \frac{Q_{actual}}{C_{min}(T_{hi} - T_{ci})}\end{equation}\]
Here, \(Q_{actual}\) is the rate of heat transfer calculated using the detailed methods mentioned earlier, and \(C_{min}\) is the minimum heat capacity rate between the hot and cold fluids. By increasing the effectiveness, a heat exchanger can transfer more heat without necessarily increasing the size or cost significantly.

Design Implications

Designing for high effectiveness means balancing material and operating costs with the thermal demands of the system. It's important for students to not only understand the calculation but also the practical significance in engineering design. Maximizing effectiveness without oversizing the exchanger is a common challenge for engineers.

Log Mean Temperature Difference

The log mean temperature difference (LMTD) is an essential term in the field of heat transfer, particularly when dealing with heat exchangers. In our exercise, the LMTD provides the needed temperature gradient that promotes heat exchange. Essentially, it's this gradient that is 'pushing' the heat from the hot fluid to the cold fluid.

Correctly calculating the LMTD is important because heat transfer equipment is often rated based on an LMTD condition; hence, this method is frequently used in industry. The formula for the LMTD is designed to give a single value for temperature difference that would result in the same heat transfer as the varying temperature difference that actually exists in a heat exchanger.

Understanding the log mean in LMTD is also important because it correctly averages the temperature difference when there is a large variation between the values of \( \Delta T_1 \) and \( \Delta T_2 \). A simple arithmetic mean would not accurately represent the driving force for heat transfer since the rate is not linear. It’s the logarithmic nature of the mean that ensures we account for the heat transfer efficiency at varying temperatures along the exchanger’s length. This detail is particularly important to bring out in educational resources to ensure students grasp the practical rather than just theoretical aspects of heat transfer.