Question

Consider the flow of saturated steam at \(270.1 \mathrm{kPa}\) that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter \(1.25 \mathrm{~cm}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) through each tube. The water enters the tubes of heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the fouling resistance for the steam and water may be taken as \(0.00015\) and \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Using the NTU method, determine \((a)\) effectiveness of the heat exchanger, \((b)\) length of the tube, and \((c)\) rate of steam condensation.

Short answer

Based on the step-by-step solution, we found the following results: (a) The effectiveness of the heat exchanger is approximately 0.712. (b) The length of the tubes is approximately 9.96 meters. (c) The rate of steam condensation is approximately 0.086 kg/s.

Step-by-step solution

Calculate the heat capacity rate of the water and steam

First, we need to determine the heat capacity rate of the water. The heat capacity rate of the water (\(\dot{C}_{w}\)) can be evaluated using the water flow rate (\(\dot{m}_{w}\)) and the specific heat (\(c_{p,w}\)) of the water at constant pressure: $$\dot{C}_{w}=\dot{m}_{w} c_{p,w}$$ The water flow rate through each tube is given as \(0.25 kg/s\). Since there are 4 tubes, the total water flow rate is: $$\dot{m}_{w}=0.25 kg/s \times 4 = 1 kg/s$$ The specific heat of the water can be approximated as \(4.18 kJ/kg \cdot K = 4180 W/kg \cdot K\). Now we can calculate the heat capacity rate of the water: $$\dot{C}_{w}= 1 kg/s \times 4180 W/kg \cdot K = 4180 W/K$$ Next, we need to determine the heat capacity rate of the steam (\(\dot{C}_{s}\)). We can use the enthalpy of saturated steam at the given pressure (\(270.1 kPa\)) and assuming a saturated liquid in the outlet. Use steam tables to get enthalpy values for saturated steam \(h_{v}\) and saturated liquid \(h_{f}\): $$h_{v}=2706.7 kJ/kg$$ $$h_{f}= 698.10 kJ/kg$$ The heat capacity rate of the steam can be calculated as: $$\dot{C}_{s}=\dot{m}_{s} \times (h_{v} - h_{f})$$ where \(\dot{m}_{s}\) is the steam mass flow rate. At this point, we still have an unknown variable, \(\dot{m}_{s}\), which we will determine in subsequent steps.

Determine the overall heat transfer coefficient

We are now going to determine the overall heat transfer coefficient for the heat exchanger (\(U_{o}\)). We are given the fouling resistance (\(R_{f, w}\) and \(R_{f, s}\)) and the convection heat transfer coefficient (\(h_{s}\)): $$\frac{1}{U_{o}}=\frac{1}{h_{s}}+R_{f, s}+R_{f, w}$$ Substituting the given values, we get: $$\frac{1}{U_{o}}=\frac{1}{1500} + 0.00015 + 0.0001$$ Now, we can solve for the overall heat transfer coefficient: $$U_{o}=\frac{1}{\frac{1}{1500}+0.00015+0.0001} \approx 1273.24 W/m^2 \cdot K$$

Calculate the NTU and the effectiveness of the heat exchanger

In order to determine the effectiveness of the heat exchanger using the NTU method, we first need to calculate the NTU (Number of Transfer Units). The expression for NTU is: $$\text{NTU}=\frac{U_{o}A}{\dot{C}_{min}}$$ In this case, the minimum heat capacity rate is the heat capacity rate of the water: $$\dot{C}_{min}=\dot{C}_{w}=4180 W/K$$ Since we don't have the value for the area (\(A\)) at this point, the NTU is unknown. We can, however, calculate the effectiveness-NTU. For a shell-and-tube heat exchanger with one shell pass and an arbitrary number of tube passes, the effectiveness-NTU relationship is given by: \(\epsilon = \frac{1 - e^{-\text{NTU}(1-\text{CR})}}{1-\text{CR}e^{-\text{NTU}(1-\text{CR})}}\) where \(\epsilon\) is the effectiveness and CR is the capacity rate ratio: $$\text{CR}=\frac{\dot{C}_{min}}{\dot{C}_{max}}$$ Since the heat capacity rate of the steam is yet to be determined, we can't calculate the effectiveness directly. We, however, have the inlet and outlet temperatures of water, and we can use the definition of effectiveness to find the relationship between \(\dot{m}_{s}\) and \(\epsilon\). So, $$\epsilon = \frac{\dot{Q}_{act}}{\dot{Q}_{max}}$$ Where \(\dot{Q}_{act}\) is the actual heat transfer rate and \(\dot{Q}_{max}\) is the maximum heat transfer rate. Rewriting the equation in terms of steam and water flow rates, $${\epsilon} = \frac{\dot{m}_{s}(h_v-h_f)}{\dot{m}_{w} C_{p,w}(T_{2w} - T_{1w})}$$ Where \(T_{1w}\) and \(T_{2w}\) are the inlet and outlet temperatures of the water, respectively. Substituting the known values, $$\epsilon=\frac{\dot{m}_s(2706.7-698.1)}{1\times 4180\times (60-20)}$$

Calculate the rate of heat transfer

Now, we can calculate the actual rate of heat transfer (\(\dot{Q}_{act}\)) using the outlet and inlet temperature of the water and the heat capacity rate of the water: $$\dot{Q}_{act}=\dot{C}_{w}(T_{2w} - T_{1w})$$ Substituting the known values, we get: $$\dot{Q}_{act}=4180 \times (60-20) = 167200 W$$

Determine the length of the tube

Now, we can determine the length of the tube (\(L\)) by using the heat transfer rate, the overall heat transfer coefficient, the temperature difference between the steam and water, and the area of the tube (\(A\)). The expression is: $$\dot{Q}_{act}=U_{o}A\Delta{T}_{lm}$$ Where \(\Delta{T}_{lm}\) is the log mean temperature difference, calculated as: $$\Delta{T}_{lm}=\frac{(\Delta{T}_1-\Delta{T}_2)}{\ln{(\Delta{T}_1/\Delta{T}_2)}}$$ In this case, \(\Delta{T}_1=T_{s}-T_{1w}\) and \(\Delta{T}_2=T_{s}-T_{2w}\). Assume \(T_s\) is the saturation temperature of steam at the given pressure, approximately \(134.2^{\circ}C\). Then, $$\Delta{T}_{lm}=\frac{(134.2-20)-(134.2-60)}{\ln{((134.2-20)/(134.2-60))}}\approx 69.77^{\circ}C$$ Now, we can solve for the area of the tube (\(A\)): $$A=\frac{\dot{Q}_{act}}{U_{o}\Delta{T}_{lm}}=\frac{167200}{1273.24\times 69.77} \approx 1.870 m^2$$ We can calculate the area of a single tube using the diameter of the tube \(D_{t}\). $$A_t=\pi D_{t}L$$ Since there are 4 tubes in total, the total area is: $$A=4A_t$$ Now we can solve for the length of the tube: $$L=\frac{A}{4\pi D_{t}}=\frac{1.870}{4\pi \times 0.0125}\approx 9.96 m$$

Calculate the rate of steam condensation

Finally, we can calculate the rate of steam condensation (\(\dot{m}_s\)) using the rate of heat transfer and the enthalpy difference between the steam and water: $$\dot{m}_s=\frac{\dot{Q}_{act}}{h_v-h_f}=\frac{167200}{2706.7-698.1}\approx 0.086 kg/s$$ Now, we can substitute this value back into the expression for the effectiveness to find the actual effectiveness: $$\epsilon=\frac{\dot{m}_s(2706.7-698.1)}{1\times 4180\times (60-20)}$$ $$\epsilon=\frac{0.086\times (2706.7-698.1)}{4180\times 40}\approx 0.712$$ In summary, the results are as follows: (a) The effectiveness of the heat exchanger is approximately \(0.712\). (b) The length of the tube is approximately \(9.96 m\). (c) The rate of steam condensation is approximately \(0.086 kg/s\).

Key concepts

Shell-and-Tube Heat Exchanger

A shell-and-tube heat exchanger is a common type of equipment used to transfer heat between two fluids. It consists of a cylindrical shell with several tubes running through it. One fluid flows through the tubes and the other flows around the tubes inside the shell, allowing for efficient heat exchange.
This configuration is ideal for applications where large volumes of fluid need to be processed because it offers a high surface area for heat transfer.

• The design consists of the **shell**, through which fluid, in this case, steam, flows, and the **tubes**, which carry another fluid, such as water.
• The **tubes** may be arranged in a series of passes, which can increase the efficiency of the heat exchange process.
• **Baffles** may be installed within the shell to direct the flow of fluid around the tubes, improving heat transfer.

Using tube passes and baffles enhances the heat transfer by maintaining the fluid turbulence. Understanding the basic mechanics of shell-and-tube heat exchangers is crucial for assessing their effectiveness in energy-intensive processes.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient is a measure of how effectively heat is conducted away from a surface through convection. It reflects the fluid properties, flow characteristics, and the nature of the surface material.
In the context of a shell-and-tube heat exchanger, this coefficient is significant because it impacts the overall heat transfer rate.

• There are two distinct heat transfer coefficients in a shell-and-tube heat exchanger: one for the fluid inside the tubes and another for the fluid outside (in the shell).
• In this exercise, the coefficient for steam is given as **1500 W/m²·K**, indicating efficient heat transfer from the steam to the tube walls.
• Improving the coefficient can be achieved by altering flow velocities, surface conditions, or fluid properties.

Higher coefficients lead to more efficient systems, reducing energy costs and improving performance. Engineers focus on optimizing this coefficient to ensure the heat exchanger operates at peak effectiveness.

Fouling Resistance

Fouling resistance is a parameter that represents the additional resistance to heat transfer due to the buildup of unwanted materials on the heat transfer surfaces. These materials can include minerals, biological materials, and other deposits.
In heat exchangers, fouling is a common issue, reducing effectiveness and requiring regular maintenance.

• The fouling resistance for **steam** and **water** in the exercise is given as **0.00015 m²·K/W** and **0.0001 m²·K/W**, respectively.
• This resistance adds to the overall thermal resistance, decreasing the amount of heat that can be transferred.
• Regular cleaning and monitoring are important to mitigate fouling impacts and maintain efficiency.

Understanding and accounting for fouling is essential for reliable operation and longevity of heat exchangers, helping minimize downtime and maintenance costs.

Log Mean Temperature Difference

The Log Mean Temperature Difference (LMTD) is a logarithmic average of the temperature difference between the hot and cold fluids across the heat exchanger. It provides a way to express the temperature driving force for heat transfer in shell-and-tube heat exchangers that minimizes errors.
Calculation of LMTD accounts for varying temperature differences along the length of the heat exchanger. For our scenario:

• **Temperature Difference 1**: Between the steam temperature and the cold water inlet.
• **Temperature Difference 2**: Between the steam temperature and the heated water outlet.
• The formula is: \[\Delta{T}_{lm} = \frac{\Delta{T}_1 - \Delta{T}_2}{\ln{(\Delta{T}_1/\Delta{T}_2)}}\]
• Using the values provided, this facilitates the determination of heat transfer rate over the course of the heat exchanger.

Accurate calculation of LMTD is vital to determine heat exchanger size and effectiveness, ensuring the device meets the required thermal performance specifications.