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Chapter 2: Problem 70

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of \(70^{\circ} \mathrm{C}\) and flow in subsea surrounding of \(5^{\circ} \mathrm{C}\). As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of \(0.5 \mathrm{~m}\) and wall thickness of \(8 \mathrm{~mm}\) is used for transporting liquid hydrocarbon at an average temperature of \(70^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The subsea surrounding has a temperature of \(5^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipeline is made of material with thermal conductivity of \(60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, \((b)\) determine the inner surface temperature of the pipeline, \((c)\) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and \((d)\) determine the heat flux through the outer pipeline surface.

Short Answer

Expert verified
A1: The formula for inner radius (ri) and outer radius (ro) can be calculated as: \(r_\mathrm{i} = \dfrac{d_\mathrm{i}}{2}\) \(r_\mathrm{o} = r_\mathrm{i} + t_\mathrm{w}\) Q2: What is the formula for the thermal resistance of the pipeline? A2: The thermal resistance for conduction through the pipe wall (Rc), and the convective resistance on the inner (Ri) and outer (Ro) surfaces can be calculated as: \(R_\mathrm{c} = \dfrac{\ln(r_\mathrm{o} / r_\mathrm{i})}{2 \pi L k}\), \(R_\mathrm{i} = \dfrac{1}{h_\mathrm{i} 2 \pi r_\mathrm{i} L}\), \(R_\mathrm{o} = \dfrac{1}{h_\mathrm{o} 2 \pi r_\mathrm{o} L}\) Q3: How can we find the temperature variation in the pipeline wall? A3: To find the temperature variation in the pipeline wall, the total thermal resistance (Rtotal) must be calculated by summing the thermal resistances for conduction and convection. The temperature difference (ΔT) between the inner and outer surfaces of the pipeline wall can be obtained using the formula: \(\Delta T = T_\mathrm{r} - T_\mathrm{s} = Q \cdot R_\mathrm{total}\) Q4: How can we calculate the inner surface temperature of the pipeline? A4: The inner surface temperature (Ti) can be found using the following equation: \(T_\mathrm{i} = T_\mathrm{r} - Q \cdot R_\mathrm{i}\) Q5: What is the formula for the rate of heat loss from the hydrocarbon liquid? A5: The rate of heat loss from the hydrocarbon liquid (dQ/dL) can be obtained by deriving Q with respect to the length L as: \(\dfrac{dQ}{dL} = \dfrac{d (\dfrac{\Delta T}{R_\mathrm{total}})}{dL}\) Q6: How can we calculate the heat flux through the outer pipeline surface? A6: The heat flux through the outer pipeline surface (qo) can be found by multiplying the heat transfer coefficient for the outer surface (ho) by the temperature difference between the outer surface temperature (To) and the subsea surrounding temperature (Ts) as: \(q_\mathrm{o} = h_\mathrm{o} (T_\mathrm{o} - T_\mathrm{s})\)

Step by step solution

01

Find the inner and outer radii of the pipeline

First, we have to determine the inner and outer radii of the pipeline. We are given the inner diameter of 0.5 m and the wall thickness of 8 mm, which can be converted into meters like this: 8 mm * 0.001 m/mm = 0.008 m. The inner and outer radii can be calculated as follows: \(r_\mathrm{i} = \dfrac{d_\mathrm{i}}{2} = \dfrac{0.5}{2} = 0.25 \mathrm{~m}\) and \(r_\mathrm{o} = r_\mathrm{i} + t_\mathrm{w} = 0.25 + 0.008 = 0.258 \mathrm{~m}\).
02

Calculate the thermal resistance

To find the temperature variation in the pipeline wall, we need to calculate the thermal resistance of the pipeline. The thermal resistance for the conduction through the pipe wall \((R_\mathrm{c})\) and the convective resistance on the inner \((R_\mathrm{i})\) and outer \((R_\mathrm{o})\) surfaces can be calculated as: \(R_\mathrm{c} = \dfrac{\ln(r_\mathrm{o} / r_\mathrm{i})}{2 \pi L k}\), \(R_\mathrm{i} = \dfrac{1}{h_\mathrm{i} 2 \pi r_\mathrm{i} L}\), \(R_\mathrm{o} = \dfrac{1}{h_\mathrm{o} 2 \pi r_\mathrm{o} L}\), where L is the length of the pipeline, \(h_\mathrm{i}\) and \(h_\mathrm{o}\) are the inner and outer surface heat transfer coefficients, and k is the thermal conductivity of the pipe material.
03

Find the temperature variation

The total thermal resistance \((R_\mathrm{total})\) can be found by summing the thermal resistances for conduction and convection: \(R_\mathrm{total} = R_\mathrm{c} + R_\mathrm{i} + R_\mathrm{o}\). The temperature difference \((\Delta T)\) between the inner and outer surfaces of the pipeline wall can be obtained using the formula: \(\Delta T = T_\mathrm{r} - T_\mathrm{s} = Q \cdot R_\mathrm{total}\), where \(T_\mathrm{r}\) is the reservoir temperature (70°C), \(T_\mathrm{s}\) is the subsea surrounding temperature (5°C), and Q is the heat transfer rate per unit length.
04

Calculate the inner surface temperature

To find the inner surface temperature \((T_\mathrm{i})\), we need to rearrange the temperature difference equation: \(T_\mathrm{i} = T_\mathrm{r} - Q \cdot R_\mathrm{i}\).
05

Obtain the rate of heat loss from the hydrocarbon liquid

The rate of heat loss from the hydrocarbon liquid \((\dfrac{dQ}{dL})\) can be obtained by deriving Q with respect to the length L. This can be done by expressing the heat transfer rates in terms of the thermal resistance and temperature difference, and then taking the derivative with respect to L: \(\dfrac{dQ}{dL} = \dfrac{d \left(\dfrac{\Delta T}{R_\mathrm{total}}\right)}{dL}\).
06

Calculate the heat flux through the outer pipeline surface

To find the heat flux through the outer pipeline surface \((q_\mathrm{o})\), we need to multiply the heat transfer coefficient for the outer surface \((h_\mathrm{o})\) by the temperature difference between the outer surface temperature \((T_\mathrm{o})\) and the subsea surrounding temperature \((T_\mathrm{s})\): \(q_\mathrm{o} = h_\mathrm{o} (T_\mathrm{o} - T_\mathrm{s})\).

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Most popular questions from this chapter

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

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