/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A liquefied natural gas (LNG) re... [FREE SOLUTION] | 91Ó°ÊÓ

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A liquefied natural gas (LNG) regasification facility utilizes a vertical heat exchanger or vaporizer that consists of a shell with a single-pass tube bundle used to convert the fuel to its vapor form for subsequent delivery through a land-based pipeline. Pressurized LNG is off-loaded from an oceangoing tanker to the bottom of the vaporizer at \(T_{c, i}=-155^{\circ} \mathrm{C}\) and \(\dot{m}_{\mathrm{LNG}}=150 \mathrm{~kg} / \mathrm{s}\) and flows through the shell. The pressurized LNG has a vaporization temperature of \(T_{f}=-75^{\circ} \mathrm{C}\) and specific heat \(c_{p l}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The specific heat of the vaporized natural gas is \(c_{p, v}=2210 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) while the gas has a latent heat of vaporization of \(h_{f g}=575 \mathrm{~kJ} / \mathrm{kg}\). The LNG is heated with seawater flowing through the tubes, also introduced at the bottom of the vaporizer, that is available at \(T_{h, i}=20^{\circ} \mathrm{C}\) with a specific heat of \(c_{\mu \mathrm{Sw}}=3985 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). If the gas is to leave the vaporizer at \(T_{c o}=8^{\circ} \mathrm{C}\) and the seawater is to exit the device at \(T_{\text {hot }}=10^{\circ} \mathrm{C}\), determine the required vaporizer heat transfer area. Hint: Divide the vaporizer into three sections, as shown in the schematic, with \(U_{\mathrm{A}}=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(U_{\mathrm{B}}=260 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(U_{\mathrm{C}}=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
The required vaporizer heat transfer area is calculated in six steps: determine the heat required for each section of the vaporizer, calculate the heat capacity rate and mass flow rate of seawater, find the logarithmic mean temperature difference (LMTD) for each section, compute the heat transfer area for each section, and finally sum up the heat transfer areas of all sections. With all given parameters and equations in place, the total heat transfer area required for the vaporizer is given by: \(A_{total} = A_A + A_B + A_C\)

Step by step solution

01

Calculate the heat required for each section of the vaporizer

To find the required heat transfer area, we need to determine the amount of heat that would be transferred in each section of the vaporizer. We can calculate this by using the mass flow rate, specific heat values, and the temperature differences. Section A: Heating the LNG from -155°C to -75°C \(Q_A = \dot{m}_{LNG} \cdot c_{pl} \cdot (T_f - T_{c,i})\) Section B: Vaporizing the LNG \(Q_B = \dot{m}_{LNG} \cdot h_{fg}\) Section C: Heating the vaporized gas to the desired exit temperature \(Q_C = \dot{m}_{LNG} \cdot c_{pv} \cdot (T_{co} - T_f)\)
02

Calculate the heat capacity rate for the seawater

Now, we have to find the heat capacity rate of the seawater, which is the product of the mass flow rate of seawater and its specific heat. \(\dot{C}_{Sw} = \dot{m}_{Sw} \cdot c_{pSw}\)
03

Calculate the mass flow rate of seawater

Since the seawater is the heating medium, the heat transferred to the LNG and vapor must equal the heat absorbed by the seawater. We can use this fact to determine the mass flow rate of seawater. \(Q_A + Q_B + Q_C = \dot{C}_{Sw} \cdot (T_{hot} - T_{h,i})\) Solve for \(\dot{m}_{Sw}\)
04

Calculate the logarithmic mean temperature difference (LMTD) for each section

LMTD is used to characterize the average temperature difference between the hot and cold fluids in the heat exchanger for each section. LMTD for Section A: \(T_{h,i-A} = T_{h,i}\) \(T_{c,i-A} = T_{c,i}\) \(T_{h,o-A} = T_{h,i} - Q_A/\dot{C}_{Sw}\) \(T_{c,o-A} = T_f\) \( \Delta T_{lm-A} = \frac{(T_{h,i-A} - T_{c,o-A}) - (T_{h,o-A} - T_{c,i-A})}{\ln[(T_{h,i-A} - T_{c,o-A})/(T_{h,o-A} - T_{c,i-A})]}\) LMTD for Section B: \(T_{h,i-B} = T_{h,o-A}\) \(T_{c,i-B} = T_f\) \(T_{h,o-B} = T_{h,i-B} - Q_B/\dot{C}_{Sw}\) \(T_{c,o-B} = T_f\) \( \Delta T_{lm-B} = \frac{(T_{h,i-B} - T_{c,o-B}) - (T_{h,o-B} - T_{c,i-B})}{\ln[(T_{h,i-B} - T_{c,o-B})/(T_{h,o-B} - T_{c,i-B})]}\) LMTD for Section C: \(T_{h,i-C} = T_{h,o-B}\) \(T_{c,i-C} = T_f\) \(T_{h,o-C} = T_{hot}\) \(T_{c,o-C} = T_{co}\) \( \Delta T_{lm-C} = \frac{(T_{h,i-C} - T_{c,o-C}) - (T_{h,o-C} - T_{c,i-C})}{\ln[(T_{h,i-C} - T_{c,o-C})/(T_{h,o-C} - T_{c,i-C})]}\)
05

Calculate the required heat transfer area for each section

With the overall heat transfer coefficients (U) and LMTD calculated in previous steps, we can calculate the required heat transfer area for each section of the vaporizer using the equation: \(A = \frac{Q}{U \cdot \Delta T_{lm}}\) \(A_A = \frac{Q_A}{U_A \cdot \Delta T_{lm-A}}\) \(A_B = \frac{Q_B}{U_B \cdot \Delta T_{lm-B}}\) \(A_C = \frac{Q_C}{U_C \cdot \Delta T_{lm-C}}\)
06

Calculate the total heat transfer area

Finally, sum up the heat transfer areas of the three sections to find the total heat transfer area required for the vaporizer. \(A_{total} = A_A + A_B + A_C\)

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Key Concepts

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

LNG Regasification
Liquefied natural gas (LNG) regasification is the process of converting LNG back into its gaseous state for use or transportation. During this process, LNG is typically warmed and vaporized in a device called a vaporizer or heat exchanger. This involves transferring heat from a warmer fluid, often seawater in coastal facilities, to the cold LNG.

This heat exchange is crucial because LNG is stored at extremely low temperatures, around -160°C, to keep it in liquid form and minimize volume for transport. By regasifying LNG, it becomes feasible to transport natural gas through pipelines to reach users who need it for residential, industrial, or commercial energy consumption.

Understanding LNG regasification is key for the wider distribution of natural gas, a cleaner-burning fossil fuel relative to coal and oil.
Heat Transfer Area
The heat transfer area is one of the most important aspects of designing a heat exchanger, such as the one used in LNG regasification. It determines the surface over which heat exchange occurs between two fluids at different temperatures. For LNG vaporizers, this is where the colder LNG receives heat from the warmer seawater to transform from liquid to gas.

Calculating the heat transfer area involves understanding how much heat needs to be transferred. This is calculated based on the mass flow rate of the LNG, its specific heat capacities, and the temperature difference it must overcome.

A larger heat transfer area can handle larger quantities of heat transfer, which is crucial in efficiently regasifying LNG at high flow rates.

Hence, optimizing the heat transfer area is vital for ensuring that the LNG can be efficiently converted while ensuring the vaporizer operates well within its design parameters.
Logarithmic Mean Temperature Difference (LMTD)
The Logarithmic Mean Temperature Difference (LMTD) is a key calculation in designing efficient heat exchangers. It serves as a simplified method to estimate the thermal driving force needed for heat exchange processes.

LMTD considers both the inlet and outlet temperatures of the hot and cold fluids and provides a mean temperature difference that fuels the heat exchange. It's important because the temperature differences aren't always linear, especially in counterflow or cross-flow heat exchangers.

By using LMTD, engineers can design heat exchangers, like those used in LNG regasification, more effectively. The heat exchanger's size and efficiency are directly influenced by this value, making it pivotal in determining how much heat can be effectively transferred.

An accurate LMTD ensures optimized regasification rates, aligning with operational needs while minimizing energy wastage.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient (U) is another essential parameter in the design and operation of heat exchangers. It represents the heat transfer ability of the materials and configurations used in the exchanger, accounting for all forms of resistance to heat flow.

This coefficient includes the conductance of the heat exchanger materials, the resistance offered by the boundary layers of the fluids, and any fouling factors that might reduce heat transfer efficiency.

In an LNG regasification context, U values can vary for different sections of the vaporizer, reflecting changes in structure or material optimal for either phase of LNG. For example, U might be higher in areas where vaporization requires more intense heat exchange conditions.

Optimizing the overall heat transfer coefficient enhances the performance and energy efficiency of the heat exchanger, ensuring maximum conversion with minimal energy input.

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Most popular questions from this chapter

The oil in an engine is cooled by air in a cross-flow heat exchanger where both fluids are unmixed. Atmospheric air enters at \(30^{\circ} \mathrm{C}\) and \(0.53 \mathrm{~kg} / \mathrm{s}\). Oil at \(0.026 \mathrm{~kg} / \mathrm{s}\) enters at \(75^{\circ} \mathrm{C}\) and flows through a tube of 10-mm diameter. Assuming fully developed flow and constant wall heat flux, estimate the oil- side heat transfer coefficient. If the overall convection coefficient is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total heat transfer area is \(1 \mathrm{~m}^{2}\), determine the effectiveness. What is the exit temperature of the oil?

In open heart surgery under hypothermic conditions, the patient's blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube, counterflow heat exchanger of length \(0.5 \mathrm{~m}\) be used for this purpose, with the thin-walled inner tube having a diameter of \(55 \mathrm{~mm}\). The specific heat of the blood is \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If water at \(T_{h j}=60^{\circ} \mathrm{C}\) and \(\dot{m}_{h}=0.10 \mathrm{~kg} / \mathrm{s}\) is used to heat blood entering the exchanger at \(T_{c A}=18^{\circ} \mathrm{C}\) and \(\dot{m}_{c}=0.05 \mathrm{~kg} / \mathrm{s}\), what is the temperature of the blood leaving the exchanger? The overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) The surgeon may wish to control the heat rate \(q\) and the outlet temperature \(T_{c, 0}\) of the blood by altering the flow rate and/or inlet temperature of the water during the rewarming process. To assist in the development of an appropriate controller for the prescribed values of \(\hat{m}_{c}\) and \(T_{c \jmath}\), compute and plot \(q\) and \(T_{c, \rho}\) as a function of \(\dot{m}_{h}\) for \(0.05 \leq \dot{m}_{\mathrm{h}} \leq 0.20 \mathrm{~kg} / \mathrm{s}\) and values of \(T_{h, l}=50,60\), and \(70^{\circ} \mathrm{C}\). Since the dominant influence on the overall heat transfer coefficient is associated with the blood flow conditions, the value of \(U\) may be assumed to remain at \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Should certain operating conditions be excluded?

The chief engineer at a university that is constructing a large number of new student dormitories decides to install a counterflow concentric tube heat exchanger on each of the dormitory shower drains. The thinwalled copper drains are of diameter \(D_{i}=50 \mathrm{~mm}\). Wastewater from the shower enters the heat exchanger at \(T_{h, i}=38^{\circ} \mathrm{C}\) while fresh water enters the dormitory at \(T_{c, l}=10^{\circ} \mathrm{C}\). The wastewater flows down the vertical wall of the drain in a thin, falling \(f\) m , providing \(h_{h}=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the annular gap is \(d=10 \mathrm{~mm}\), the heat exchanger length is \(L=1 \mathrm{~m}\), and the water flow rate is \(\dot{m}=10 \mathrm{~kg} / \mathrm{min}\), determine the heat transfer rate and the outlet temperature of the warmed fresh water. (b) If a helical spring is installed in the annular gap so the fresh water is forced to follow a spiral path from the inlet to the fresh water outlet, resulting in \(h_{c}=9050 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer rate and the outlet temperature of the fresh water. (c) Based on the result for part (b), calculate the daily savings if 15,000 students each take a 10 -minute shower per day and the cost of water heating is \(\$ 0.07 / \mathrm{kW} \cdot \mathrm{h}\).

A shell-and-tube heat exchanger consisting of one shell pass and two tube passes is used to transfer heat from an ethylene glycol-water solution (shell side) supplied from a rooftop solar collector to pure water (tube side) used for household purposes. The tubes are of inner and outer diameters \(D_{i}=3.6 \mathrm{~mm}\) and \(D_{o}=3.8 \mathrm{~mm}\), respectively. Each of the 100 tubes is \(0.8 \mathrm{~m}\) long ( \(0.4 \mathrm{~m}\) per pass), and the heat transfer coefficient associated with the ethylene glycol-water mixture is \(h_{o}=11,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) For pure copper tubes, calculate the heat transfer rate from the ethylene glycol-water solution \(\left(\dot{m}=2.5 \mathrm{~kg} / \mathrm{s}, T_{h, i}=80^{\circ} \mathrm{C}\right)\) to the pure water \((\dot{m}=\) \(2.5 \mathrm{~kg} / \mathrm{s}, T_{c, i}=20^{\circ} \mathrm{C}\) ). Determine the outlet temperatures of both streams of fluid. The density and specific heat of the ethylene glycol-water mixture are \(1040 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3660 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) It is proposed to replace the copper tube bundle with a bundle composed of high-temperature nylon tubes of the same diameter and tube wall thickness. The nylon is characterized by a thermal conductivity of \(k_{n}=0.31 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the tube length required to transfer the same amount of energy as in part (a).

As part of a senior project, a student was given the assignment to design a heat exchanger that meets the following specifications: \begin{tabular}{lccc} \hline & \(\dot{m}(\mathrm{~kg} / \mathrm{s})\) & \(T_{m, i}\left({ }^{\circ} \mathrm{C}\right)\) & \(T_{m, \theta}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline Hot water & 28 & 90 & \(-\) \\ Cold water & 27 & 34 & 60 \\ \hline \end{tabular} Like many real-world situations, the customer hasn't revealed, or doesn't know, additional requirements that would allow you to proceed directly to a final configuration. At the outset, it is helpful to make a first-cut design based upon simplifying assumptions, which can be evaluated to determine what additional requirements and trade-offs should be considered by the customer. (a) Design a heat exchanger to meet the foregoing specifications. List and explain your assumptions. Hint: Begin by finding the required value for \(U A\) and using representative values of \(U\) to determine \(A\). (b) Evaluate your design by identifying what features and configurations could be explored with your customer in order to develop more complete specifications.

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