Question

Ice slurry is being transported in a pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\), and \(L=\) \(5 \mathrm{~m}\) ) with an inner surface temperature of \(0^{\circ} \mathrm{C}\). The ambient condition surrounding the pipe has a temperature of \(20^{\circ} \mathrm{C}\), a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a dew point of \(10^{\circ} \mathrm{C}\). If the outer surface temperature of the pipe drops below the dew point, condensation can occur on the surface. Since this pipe is located in a vicinity of high voltage devices, water droplets from the condensation can cause electrical hazard. To prevent such incident, the pipe surface needs to be insulated. Determine the insulation thickness for the pipe using a material with \(k=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to prevent the outer surface temperature from dropping below the dew point.

Short answer

Question: Determine the necessary insulation thickness for a pipe carrying ice slurry to prevent the outer surface temperature from dropping below the dew point. The pipe has the following properties: thermal conductivity (k) = 15 W/m·K, inner diameter (Di) = 2.5 cm, outer diameter (Do) = 3 cm, length (L) = 5 m, inner surface temperature (Ti) = 0 °C, ambient temperature (T_ambi) = 20 °C, ambient convection heat transfer coefficient (h_ambi) = 10 W/m²·K, dew point temperature (T_dew) = 10 °C, insulation material's thermal conductivity (k_ins) = 0.95 W/m·K. Answer: The required insulation thickness to prevent the outer surface temperature from dropping below the dew point is 7.8 mm.

Step-by-step solution

Write down the given information

(Write down the given information) We are provided with the following information: - Pipe's thermal conductivity (k) = 15 W/m·K - Pipe's inner diameter (Di) = 2.5 cm - Pipe's outer diameter (Do) = 3 cm - Pipe's length (L) = 5 m - Inner surface temperature (Ti) = 0 °C - Ambient temperature (T_ambi) = 20 °C - Ambient convection heat transfer coefficient (h_ambi) = 10 W/m²·K - Dew point temperature (T_dew) = 10 °C - Insulation material's thermal conductivity (k_ins) = 0.95 W/m·K

Convert units if necessary

(Convert all the units to SI units. In our case, convert the diameters to meters) Di = 0.025 m Do = 0.03 m

Find the critical radius of insulation

(Find the critical insulation radius using the formula \( r_{cri} = \frac{k}{h_{ambi}} \)) \(r_{cri} = \frac{k}{h_{ambi}}\) \(r_{cri} = \frac{15}{10}\) \(r_{cri} = 1.5 \mathrm{~m}\) Since the critical radius is greater than the outer diameter of the pipe, we should insulate the pipe.

Calculate the heat flow through a cylinder without insulation

(Calculate the heat flow through the cylinder without insulation using the formula \( q = 2 \pi k L(T_i - T_{ambi}) \int_{Di/2}^{Do/2} \frac{dr}{r} \)) Using the given values, we can calculate the heat flow: \(q = 2 \pi (15) (5)(-20) \int_{0.025/2}^{0.03/2} \frac{dr}{r}\) The integral part can be calculated as: \(\int_{0.0125}^{0.015} \frac{dr}{r} = ln(1.2)\) Now, multiplying the values together: \(q = 2 \pi (15) (5)(-20) \cdot ln(1.2)\) \(q = - 113.7 \mathrm{~W}\)

Calculate the insulation thickness to prevent dew point temperature

(Determine the insulation thickness to keep the outer surface temperature above the dew point using the formula \( T_{out} = T_{ambi} + \frac{q}{2 \pi (Do/2 + R_{ins}) h_{ambi} L} \), and solve it for \(R_{ins}\)) Rewrite the above formula for R_ins: \(R_{ins} = \frac{q}{2 \pi h_{ambi} L (T_{out} - T_{ambi})} - \frac{Do}{2}\) Now, plugging the values: \(R_{ins} = \frac{-113.7}{2 \pi (10) (5)(10 - 20)} - 0.015\) \(R_{ins} = 0.0078 \mathrm{~m}\)

Calculate the insulation thickness

(Find the insulation thickness using the formula \( t = (R_{ins} + Do/2) - Di/2 \)) \(t = (0.0078 + 0.03/2) - 0.025/2\) \(t = 0.0078 \mathrm{~m}\) So, the required insulation thickness to prevent the outer surface temperature from dropping below the dew point is 7.8 mm.

Key concepts

Insulation Thickness

Understanding insulation thickness is crucial for maintaining the right temperature and preventing issues like condensation. In scenarios where pipes carry cold substances, like the ice slurry mentioned, the surface temperature may drop below the dew point. This can lead to condensation, which is undesirable, especially near sensitive equipment.

The thickness of insulation is a measure of how well it can restrict heat transfer from the surroundings to the pipe. A thicker insulation layer keeps the heat out more effectively.

• Critical Radius: Before determining the insulation thickness, it's essential to understand the concept of critical radius, which is the radius at which adding insulation actually begins to reduce losses rather than increase them.
• Heat Flow Equation: The exercise calculates the heat flow through a cylinder to determine how much insulation is needed. This is often based on Fourier’s law for cylindrical geometry.

To prevent the outer surface from dropping below the dew point, an insulation thickness of 7.8 mm was calculated in the example, ensuring safety and efficiency.

Convection Heat Transfer

Convection heat transfer is one of the key processes impacting the temperature of the pipe's outer surface. It occurs when the fluid surrounding the pipe moves, transferring heat either towards or away from the pipe. The rate of convection depends on the convection heat transfer coefficient, a crucial factor in calculating heat loss.

In the problem, the convection heat transfer coefficient is given as 10 W/m²·K. This number quantifies how effectively heat is transferred between the pipe surface and the ambient air.

• Importance of Convection: Convection can either add heat to or remove heat from the pipe surface, making it essential to monitor when dealing with temperature-sensitive materials like ice slurry.
• Role in Dew Point Avoidance: By understanding the convection properties, we ensure the outer surface temperature stays above the dew point, preventing condensation.

A key takeaway is the necessity to balance all factors influencing the heat transfer process to maintain operational safety and effectiveness.

Thermal Conductivity

Thermal conductivity is a material property that measures its ability to conduct heat. It indicates how quickly heat can travel through a substance. In the problem, the thermal conductivity of both the pipe and the insulation material is given.

For effective insulation, you'll need a material with low thermal conductivity. This ensures that minimal heat passes through it.

• Ice Slurry Pipe: The pipe in the exercise has a thermal conductivity of 15 W/m·K. The high value means that it's relatively easy for heat to pass through the pipe.
• Insulation Material: The insulation material has a lower thermal conductivity of 0.95 W/m·K, making it effective at hindering heat flow.

When selecting materials for insulation, balancing thermal conductivity with costs and material availability is crucial for efficient insulation solutions.

Dew Point Temperature

Dew point temperature is the temperature at which air becomes saturated with moisture and condensation forms. In our scenario, when the pipe's outer surface temperature falls below the dew point, water droplets can form.

This could lead to significant hazards, such as electrical issues in high-voltage environments. To avoid this, it's important to ensure the surface temperature is kept above the dew point.

• Given Context: The ambient dew point is 10°C in this situation. Insulation must ensure the temperature doesn't drop below this to prevent condensation.
• Calculating Insulation: By factoring in dew point temperature, we calculate the necessary insulation to keep pipes safe and dry.

Understanding dew point and how to maintain temperatures above it is fundamental in designing systems that are safe from moisture-related risks.