Question

Steam at \(320^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Heat is lost to the surroundings at \(5^{\circ} \mathrm{C}\) by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

Short answer

Question: Calculate the rate of heat loss from the steam per unit length of the pipe and the temperature drops across the pipe shell and the insulation. Answer: To calculate the rate of heat loss (q) and the temperature drops across the pipe shell (∆T_s) and insulation (∆T_i), follow these steps: 1. Calculate the thermal resistances of each layer (inside pipe, pipe wall, insulation, and convection at the outer surface) using the given dimensions, materials, and heat transfer coefficients. 2. Calculate the overall resistance (R_total) by adding up the individual thermal resistances. 3. Calculate the heat loss rate (q) based on the temperature difference between the steam and the ambient, and the overall resistance. 4. Calculate the temperature drops across the pipe shell (∆T_s) and insulation (∆T_i) using the heat loss rate and the individual thermal resistances. Note that this requires the input values for dimensions, materials, and heat transfer coefficients to perform the calculations.

Step-by-step solution

Calculate the resistances of each layer of the pipe and insulation

We can model the heat transfer process as a series of thermal resistances. We are given the inner and outer diameters of the stainless steel pipe, the thickness of the glass wool insulation, the thermal conductivities of the materials, as well as the heat transfer coefficients for the inside of the pipe and the natural convection. Using this information, we can calculate the thermal resistance for each layer (inside pipe, pipe wall, insulation, and convection at the outer surface). Let \(R_i\) be the inside heat transfer resistance, \(R_s\) be the steel pipe resistance, \(R_i\) be the insulation resistance, and \(R_o\) be the outside heat transfer resistance. For the inside heat transfer resistance: $$R_i = \frac{1}{h_i A_i}$$ For the steel pipe resistance: $$R_s = \frac{\ln(\frac{D_o}{D_i})}{2 \pi k_s L}$$ For the insulation resistance: $$R_i = \frac{\ln(\frac{D_o + 2 t_i}{D_o})}{2 \pi k_i L}$$ For the outside heat transfer resistance: $$R_o = \frac{1}{h_o A_o}$$ Where \(h_i\) is the inside heat transfer coefficient, \(h_o\) is the outside heat transfer coefficient, \(A_i\) and \(A_o\) are the inside and outside surface areas of the pipe, \(D_i\) and \(D_o\) are the inner and outer diameters of the pipe, \(t_i\) is the insulation thickness, \(k_s\) is the thermal conductivity of the stainless steel pipe, \(k_i\) is the thermal conductivity of the insulation, and \(L\) is the length of the pipe.

Calculate the overall resistance and the heat loss rate

The overall resistance of the system can be obtained by adding the thermal resistances of each layer: $$R_{total} = R_i + R_s + R_i + R_o$$ Using the given values for heat transfer coefficients and thermal conductivities, as well as the inside and outside surface areas of the pipe, we can calculate the overall resistance. Now, the heat loss rate per unit length of the pipe can be calculated using the temperature difference between the steam and the ambient and the overall resistance: $$q = \frac{T_{steam} - T_{ambient}}{R_{total}}$$ Where \(q\) is the heat loss rate per unit length, \(T_{steam}\) is the steam temperature, and \(T_{ambient}\) is the ambient temperature.

Calculate the temperature drops across the pipe shell and insulation

We can now determine the temperature drops across the pipe shell and the insulation using the heat loss rate and the individual thermal resistances. Temperature drop across the pipe shell (\(\Delta T_s\)): $$\Delta T_s = q \cdot R_s$$ Temperature drop across the insulation (\(\Delta T_i\)): $$\Delta T_i = q \cdot R_i$$

Key concepts

Thermal Resistance

Thermal resistance is a crucial concept in understanding heat transfer. In simple terms, it's like an obstacle that heat energy has to overcome to flow through a material. In heat transfer, we visualize this process similar to how electrical resistance works in circuit theory. Just as an electric resistor impedes the flow of electrical current, thermal resistance impedes the flow of heat.

It is calculated using the formula \[ R = \frac{L}{kA} \]where:

• L is the thickness of the material.
• k is the thermal conductivity.
• A is the area through which heat is transferred.

In the given problem, each layer (the pipe wall, the insulation, and the boundaries inside and outside of the pipe) has its own thermal resistance. By adding the resistances together, we can determine the total resistance faced by the heat as it travels from the steam to the surrounding air.

Convection

Convection is a type of heat transfer that occurs in fluids such as liquids and gases. During convection, heat is moved through the movement of fluid particles. This is how heat is transferred from the surface of the steel pipe to the surrounding air in our problem.

There are two types of convection: natural and forced. Natural convection occurs due to the natural movement of the fluid particles because of density differences, whereas forced convection is caused by external forces, such as fans or pumps. In our example, the convection considered is natural.
Convection heat transfer is calculated using the formula:\[ q = hA(T_{surface} - T_{fluid}) \]where:

• h is the convection heat transfer coefficient.
• A is the surface area.
• T_{surface} is the surface temperature.
• T_{fluid} is the fluid temperature.

In the solution, the surface temperature is the pipe's outer surface, and the fluid temperature is the ambient air temperature.

Insulation

Insulation is a material or layer that reduces the rate of heat transfer. Using materials with low thermal conductivity achieves his and in this example, glass wool is used.

Insulation works by increasing the thermal resistance between two surfaces. Think of it as a thick sweater keeping the heat from leaving quickly. Because the glass wool has a very low thermal conductivity, it prevents the heat from escaping as fast as it would through the steel pipe alone.
The importance of insulation lies in its capacity to effectively manage and reduce energy loss. This helps in maintaining temperature inside pipes or buildings and saving energy costs. The effectiveness of insulation can be even more understood by calculating its resistance with:\[ R = \frac{\ln\left(\frac{D_o + 2 t_i}{D_o}\right)}{2 \pi k L} \]This formula gives the ease of identifying how much resistance the insulation adds to the system.

Conduction

Conduction is the simplest form of heat transfer. It occurs when heat flows through a material from a high-temperature area to a lower-temperature one, typically through physical contact.

In the context of the exercise, conduction is the mode of heat transfer through the steel pipe wall and the glass wool insulation. It depends on the thermal conductivity of the materials. Good conductors have high thermal conductivity, while insulators have low thermal conductivity.
The rate of conduction heat transfer can be expressed as:\[ q = -kA \frac{dT}{dx} \]where:

• k is the thermal conductivity.
• A is the area through which heat is conducted.
• \frac{dT}{dx} is the temperature gradient in the direction of heat flow.

Conduction is essential in the overall heat loss calculation in our problem, as it represents the first phase of energy movement after the steam, before reaching the insulation.