91Ó°ÊÓ

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).

Step by step solution

01

Find Temperature Distribution for Original Configuration

First, we will find the temperature distribution for the original configuration. The total thermal resistance across the layers can be given as: \[R_{total} = R_{in}+R_{iw}+R_{insulation}+R_{ow}+R_{out}\] The thermal resistances for convection at the inside and outside surfaces of the pipe are given by: \[R_{in} = \frac{1}{h_{i}A_{i}}\] and \[R_{out} = \frac{1}{h_{o}A_{o}}\] Here, \(A_{i} = \pi D_{i, 1}\) is the surface area on the inside and \(A_{o} = \pi D_{o, 2}\) is the surface area on the outside. The thermal resistances for both inner and outer pipe wall can be given by: \[R_{iw} = R_{ow} = \frac{t_i}{k_sA_{wall}}\] Here, \(A_{wall} = \frac{(D_{o, 1} + D_{i, 1})}{2}\) is the average of inside and outside wall surface areas of the pipe wall. The thermal resistance for the polyurethane insulation can be given by: \[R_{insulation} = \frac{t_p}{k_pA_p}\] Here, \(A_p = \frac{(D_{i, 2} + D_{o, 2})}{2}\) is the average of inside and outside surface areas of the insulation layer. Once all the resistance values are calculated, we can find the relationship between the temperature difference of the oil and the heat transfer rate as follows: \[Q = \frac{(T_{oil}-T_{\infty, o})}{R_{total}}\] Where the goal is to calculate \(T_{oil}\) such that \(T_{p, max} = 70^{\circ} C\).

02

Calculate Maximum Allowable Crude Oil Temperature

With the given values for the convection coefficients, diameters, thicknesses, and thermal conductivities, we can plug them into the above equations, determine the total thermal resistance, and then find the heat transfer rate. We can now calculate \(T_{p, max}\) using the total thermal resistance (considering only polyurethane insulation layer) and the maximum allowable polyurethane temperature: \[T_{oil} - T_{\infty, o} = (T_{p, max} - T_{\infty, o}) R_{total}\] Solving for the maximum allowable crude oil temperature, \(T_{oil}\), we obtain the result for part (a).

03

Enhance Configuration with an Aerogel layer

For the second part of the question, a layer of aerogel insulation is added to enhance the performance of the pipe-in-pipe device, which means modifying the \(R_{total}\) by considering the aerogel's thermal resistance. To account for this, we will consider a new total thermal resistance \(R'_{total}\) which includes the aerogel insulation layer: \[R'_{total} = R_{in}+R_{iw}+R_{a}+R_{insulation}+R_{ow}+R_{out}\] The thermal resistance for the aerogel layer can be given as: \[R_{a} = \frac{t_a}{k_aA_a}\] Here, \(A_a = \frac{D_{i, 2} + D_{i, 2} + 2 t_a}{2}\) is the average of the aerogel's inside and outside surface areas.

04

Calculate Maximum Allowable Crude Oil Temperature for Enhanced Configuration

Similar to Step 2, we can calculate the heat transfer rate and the maximum allowable crude oil temperature, holding \(T_{p, max} = 70^{\circ} C\), with the added aerogel layer: \[T_{oil}' - T_{\infty, o} = (T_{p, max} - T_{\infty, o}) R'_{total}\] Solving for the new maximum allowable crude oil temperature, \(T_{oil}'\), we obtain the result for part (b).

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Key Concepts

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

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics that involves the movement of heat energy from a region of high temperature to a region of lower temperature. It occurs through three primary mechanisms: conduction, convection, and radiation. In the context of our exercise, the focus is on conduction and convection, which play critical roles in the thermal management of crude oil transportation.

In conduction, heat is transferred through a solid material due to a temperature gradient, where the heat flow is directly proportional to the thermal conductivity of the material and the temperature difference across it. On the other hand, convection involves the transfer of heat through a fluid, which is caused by the movement of the fluid itself, often induced by differences in temperature and density within the fluid. The effectiveness of convection is characterized by the convection heat transfer coefficient.

Conduction and Convection

The interplay between conduction and convection is crucial in systems where heat transfer efficiency is of utmost importance. In the textbook exercise, the oil inside the pipe and the surrounding ocean water are involved in convective heat transfer with the steel pipe and insulation material. The conduction occurs through the steel pipe and insulation material, constituting layers with different thermal resistances.

A good understanding of these processes helps in calculating and maximizing the efficiency of heat transfer systems. For example, the inner steel pipe transfers heat to the polyurethane foam by conduction while the outer surface loses heat to the ocean water through convection. Balancing these mechanisms ensures the crude oil remains at an optimal temperature to reduce viscosity for efficient pumping while not exceeding the maximum service temperature of the insulation material.

Thermal Conductivity

Thermal conductivity, represented by the symbol \(k\), is a measure of a material's ability to conduct heat. It is defined as the quantity of heat, in watts, transmitted through a cube of material 1 meter on each side in a second, for a temperature difference of 1 Kelvins between the sides. Materials with high thermal conductivity, such as steel \(\left(k_s = 35\mathrm{~W/m\cdot K}\right)\), quickly transfer heat, whereas materials with low thermal conductivity, like polyurethane foam \(\left(k_p = 0.075\mathrm{~W/m\cdot K}\right)\), are better at insulating and reduce the rate of heat loss.

In thermal resistance calculations, the thermal conductivity directly affects the resistance of the material layer. The higher the thermal conductivity, the lower the thermal resistance, meaning heat can pass through more easily. This is an important factor when designing insulation for systems like the pipe-in-pipe configuration for transporting crude oil.

Insulation Materials

Insulation materials are crucial for managing and controlling heat transfer in various applications. These materials, such as polyurethane foam and aerogel in our exercise, are selected based on their thermal properties, especially low thermal conductivity. By incorporating insulation with a low thermal conductivity, the rate of heat loss can be controlled, ensuring the crude oil in the pipe maintains the required temperature for efficient transport.

The type of insulation material and its thickness are pivotal in determining the thermal resistance of the system. For example, replacing a section of the polyurethane with aerogel insulation (with lower thermal conductivity \(k_a = 0.012 \mathrm{~W/m\cdot K}\)) demonstrates this. The aerogel provides a higher thermal resistance despite its thinness, which prevents the inner pipe's heat from being lost to the cold surroundings, allowing for higher crude oil temperatures without jeopardizing the insulation's integrity.