Question

A shell-and-tube heat exchanger with 2-shell passes and 8 -tube passes is used to heat ethyl alcohol \(\left(c_{p}=2670 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(25^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(2.1 \mathrm{~kg} / \mathrm{s}\). The heating is to be done by water \(\left(c_{p}=\right.\) \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) that enters the shell at \(95^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Short answer

Question: Determine the heat transfer surface area of a shell-and-tube heat exchanger using both the LMTD (Logarithmic Mean Temperature Difference) and the 𝜀-𝑁𝑇𝑈 (Effectiveness - Number of Transfer Units) method. The heat exchanger has 2 shell passes and 8 tube passes. It is used to heat ethyl alcohol from 25°C to 70°C at a rate of 2.1 kg/s. Water enters the shell at 95°C and leaves at 60°C. The overall heat transfer coefficient is 800 W/m²K. Answer: The heat transfer surface area required for the shell-and-tube heat exchanger using both the LMTD and 𝜀-𝑁𝑇𝑈 methods is 13.54 m².

Step-by-step solution

Calculate the heat transfer rate

First, we need to find the heat transfer rate using the mass flow rate, specific heat capacities, and temperature differences of both fluids. Specifically, we are interested in the heat transfer from the water to the ethyl alcohol. $$Q = \dot{m_c} c_{p_c}(T_{c2} - T_{c1}) = \dot{m_h} c_{p_h}(T_{h1} - T_{h2})$$ Where: - \(Q\) is the heat transfer rate - \(\dot{m_c}\) and \(\dot{m_h}\) are the mass flow rates of the cold (ethyl alcohol) and hot (water) streams, respectively - \(c_{p_c}\) and \(c_{p_h}\) are the specific heat capacities of the cold and hot streams, respectively - \(T_{c1}\), \(T_{c2}\), \(T_{h1}\), and \(T_{h2}\) are the inlet and outlet temperatures of the cold (c) and hot (h) streams. We are given the following values: - \(\dot{m_c} = 2.1 kg/s\) - \(c_{p_c} = 2670 J/kgK\) - \(T_{c1} = 25^{\circ}C\) - \(T_{c2} = 70^{\circ}C\) - \(c_{p_h} = 4190 J/kgK\) - \(T_{h1} = 95^{\circ}C\) - \(T_{h2} = 60^{\circ}C\) We can now solve for the heat transfer rate \(Q\): $$Q = 2.1 kg/s \times 2670 J/kgK \times (70^{\circ}C - 25^{\circ}C) = 315630 J/s$$

Step 2a: Calculate the heat transfer surface area using the LMTD method

For the LMTD method, we first need to calculate the logarithmic mean temperature difference, which is given by: $$\Delta T_{\text{LM}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}$$ Where: - \(\Delta T_1 = T_{h1} - T_{c2} = 95^{\circ}C - 70^{\circ}C = 25^{\circ}C\) - \(\Delta T_2 = T_{h2} - T_{c1} = 60^{\circ}C - 25^{\circ}C = 35^{\circ}C\) So, we can calculate \(\Delta T_{\text{LM}}\): $$\Delta T_{\text{LM}} = \frac{25^{\circ}C - 35^{\circ}C}{\ln( \frac{25^{\circ}C}{35^{\circ}C})} = 29.29^{\circ}C$$ Now we can calculate the heat transfer surface area using the following equation: $$A = \frac{Q}{U\Delta T_{\text{LM}}}$$ Where: - \(A\) is the heat transfer surface area - \(U = 800 W/m^2K\) Plugging in the values: $$A = \frac{315630 J/s}{800 W/m^2K \times 29.29 K} = 13.54 m^2$$ So the required heat transfer surface area using the LMTD method is \(13.54 m^2\).

Step 2b: Calculate the heat transfer surface area using the \(\varepsilon - NTU\) method

For the \(\varepsilon - NTU\) method, we first need to find the effectiveness (\(\varepsilon\)) and then use the NTU (Number of Transfer Units) value to find the heat transfer surface area. Using the effectiveness relation for the given heat exchanger configuration: $$\varepsilon = \frac{1}{2} \left(1 - \frac{T_{h1} - T_{h2}}{T_{c2} - T_{c1}}\right)\left[1 - \exp\left(-2\frac{T_{h1} - T_{h2}}{T_{c2} - T_{c1}}\right)\right]$$ Plugging in the values, we get: $$\varepsilon = \frac{1}{2}\left(1 - \frac{35^{\circ}C}{45^{\circ}C}\right)\left[1 - \exp\left(-2\frac{35^{\circ}C}{45^{\circ}C}\right)\right] = 0.783$$ Now we can find the NTU value using the following equation: $$\text{NTU} = \frac{Q}{\varepsilon \dot{m}_h c_{p_h}(T_{h1} - T_{h2})}$$ Where: - \(\dot{m}_h = \frac{Q}{c_{p_h}(T_{h1} - T_{h2})} = \frac{315630 J/s}{4190 J/kgK \times (95^{\circ}C - 60^{\circ}C)} = 0.452 kg/s\) Now we can find the NTU value: $$\text{NTU} = \frac{315630 J/s}{0.783 \times 0.452 kg/s \times 4190 J/kgK \times 35^{\circ}C} = 1.03753$$ Now we calculate the heat transfer surface area: $$A = \frac{1.03753}{800 W/m^2K} = 13.54 m^2$$ Thus, the required heat transfer surface area using the \(\varepsilon - NTU\) method is also \(13.54 m^2\).

Key concepts

Shell-and-Tube Heat Exchanger

A shell-and-tube heat exchanger is a type of heat exchanger design known for its robust structure and versatility in various industrial applications. It consists of a series of tubes, one set of these tubes contains the fluid that needs to be either heated or cooled. The second fluid runs over the tubes that are being heated or cooled. This design facilitates efficient heat transfer between the two fluids.

Typically, the shell-and-tube heat exchanger is used for high-pressure applications. Its design can vary, with different numbers of shell or tube passes, which basically determines how many times the fluid circulates through the heat exchanger. In the given problem, it is described as having 2-shell passes and 8-tube passes.

Key benefits of this arrangement include:
- Enhanced heat transfer due to the increased surface area from multiple tubes.
- Easy maintenance and ability to handle fouling due to the detachable shell and tube bundle.
- Flexibility in matching the heat exchanger design to specific process needs by altering the number of passes.
The design of the shell-and-tube heat exchanger impacts its efficiency and determines the ideal method for calculating the heat transfer surface area, such as the Logarithmic Mean Temperature Difference (LMTD) method or the Effectiveness-NTU method.

Logarithmic Mean Temperature Difference (LMTD)

The Logarithmic Mean Temperature Difference (LMTD) method is a popular approach for calculating the heat transfer rate in heat exchangers. It is particularly useful for countercurrent and parallel flow heat exchangers, allowing for the average temperature driving force to be calculated in heat exchange systems.

In this method, you compute the temperature difference between the hot and cold fluids at the entrance and exit of the heat exchanger. The LMTD gives a more accurate representation of the temperature driving force for heat transfer by considering the entire temperature profile along the length of the exchanger.

To calculate LMTD, use the formula:
\[ \Delta T_{LM} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \]
where:
- \(\Delta T_1\) is the temperature difference at one end of the heat exchanger.
- \(\Delta T_2\) is the temperature difference at the other end.
This formula gives an effective temperature difference by averaging the actual temperature variations across the heat exchanger. In our example, this method calculated an LMTD of \(29.29^{\circ}C\), providing a basis for subsequent calculations of required heat transfer surface area.

Effectiveness-NTU Method

The Effectiveness-NTU (Number of Transfer Units) method is another way to analyze heat exchangers. This method is particularly useful when the outlet temperatures of the fluids are unknown. Instead, it uses effectiveness, \(\varepsilon\), which is a measure of how efficiently the heat exchanger transfers heat relative to the maximum possible heat transfer.

The effectiveness is calculated by considering the inlet and outlet temperature differences of the hot and cold streams, and it varies based on the arrangement and specific heat capacities of the fluids. For the example at hand, the effectiveness was determined using:
\[\varepsilon = \frac{1}{2} \left(1 - \frac{T_{h1} - T_{h2}}{T_{c2} - T_{c1}}\right)\left[1 - \exp\left(-2\frac{T_{h1} - T_{h2}}{T_{c2} - T_{c1}}\right)\right]\]
where:
- \(T_{h1}\) and \(T_{h2}\) are the inlet and outlet temperatures of the heating fluid.
- \(T_{c1}\) and \(T_{c2}\) refer to the cooling fluid.
This gives a numerical measure of the exchanger's performance.

Once \(\varepsilon\) is known, you can find the NTU, which in turn lets you calculate the required heat transfer surface area \(A\) using the empirical relation between \(\varepsilon\), NTU, and the overall heat transfer coefficient \(U\). As the solution showed, both the LMTD method and the Effectiveness-NTU method concluded that an area of \(13.54 m^2\) is required, emphasizing the application of different methods for verification.