Question

Steam at \(250^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The pipe is covered with \(3.5-\mathrm{cm}-\) thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose outer surface has an emissivity of \(0.3\). Heat is lost to the surrounding air and surfaces at \(3^{\circ} \mathrm{C}\) by convection and radiation. 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 when air is flowing across the pipe at \(4 \mathrm{~m} / \mathrm{s}\). Evaluate the air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

#Answer# Based on the outlined steps, first, we calculate the thermal resistance of each layer: For the stainless steel pipe: $$R_{pipe}=\frac{\ln(0.023 / 0.02)}{2 \pi \times 15 \times 1} \approx 0.00399\, \mathrm{K/W}$$ For the glass wool insulation: $$R_{insulation}=\frac{\ln(0.056 / 0.023)}{2 \pi \times 0.038 \times 1} \approx 0.834\, \mathrm{K/W}$$ Next, we find the Nusselt number and the heat transfer coefficient for the air: Suppose we have the following air properties: \(\rho = 1.2\, \mathrm{kg/m^3}\), \(\mu = 1.81 \times 10^{-5}\, \mathrm{kg/m\cdot s}\), \(k_{air} = 0.024\, \mathrm{W/m\cdot K}\), and \(Pr = 0.7\). Then calculate the Reynolds number: $$Re = \frac{(1.2)(4)(0.056)}{1.81 \times 10^{-5}} \approx 15,\!212$$ Calculate the Nusselt number: $$Nu = 0.023 \times Re^{0.8} \times Pr^{0.4} \approx 68.8$$ Determine the air heat transfer coefficient: $$h_{air} = \frac{Nu \times k_{air}}{D} \approx 29.9\, \mathrm{W/m^2K}$$ Then, use the assumption that the overall heat transfer coefficient is approximately equal to \(h_{air}\) (\(U_{overall} = h_{air}\)) to calculate the heat transfer rate per unit length: $$q = \frac{T_{steam} - T_{\infty}}{R_{pipe} + R_{insulation} + \frac{1}{U_{overall}}} = \frac{373 - 273}{0.00399 + 0.834 + \frac{1}{29.9}} \approx 117.8\, \mathrm{W/m}$$ Thus, the rate of heat loss per unit length from the steam flowing inside a stainless steel pipe with glass wool insulation is approximately \(117.8\, \mathrm{W/m}\).

Step-by-step solution

Calculate the Resistance of each layer of the pipe

First, we need to calculate the thermal resistance of each of the layers: the stainless steel pipe and the glass wool insulation. The thermal resistance of a cylindrical layer can be found using the formula: $$R_{cylinder}=\frac{\ln(r_2 / r_1)}{2 \pi k L}$$ where \(r_1\) and \(r_2\) are the inner and outer radii of the cylinder, \(k\) is the thermal conductivity of the material, and \(L\) is the length of the pipe. For the stainless steel pipe, we have \(r_1 = 0.02\) m, \(r_2 = 0.023\) m, and \(k = 15 \,\mathrm{W/mK}\). For the glass wool insulation, we have \(r_1 = 0.023\) m, \(r_2 = 0.056\) m, and \(k = 0.038\, \mathrm{W/mK}\). Calculate the resistances for both the pipe and the insulation.

Calculate the Nusselt number for the airflow over the insulated pipe

To find the heat transfer coefficient for the air flowing over the insulated pipe, we need to determine the Nusselt number. In this case, we should use the Dittus-Boelter correlation, given by: $$Nu = 0.023 \, Re^{0.8} Pr^{0.4}$$ where \(Re\) is the Reynolds number and \(Pr\) is the Prandtl number of the air. These properties should be evaluated at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) condition. First, determine the physical properties of the air, and then calculate the Reynolds number using the formula: $$Re = \frac{\rho V D}{\mu}$$ where \(\rho\) is the air density, \(V = 4 \, \mathrm{m/s}\) is the air velocity, \(D = 0.056 \, \mathrm{m}\) is the diameter of the insulated pipe, and \(\mu\) is the air dynamic viscosity. Once you have the Reynolds and Prandtl numbers, calculate the Nusselt number and then calculate the air heat transfer coefficient \(h_{air}\) using the formula: $$h_{air} = \frac{Nu \, k_{air}}{D}$$ where \(k_{air}\) is the air thermal conductivity.

Calculate the radiation heat transfer coefficient

Next, calculate the radiation heat transfer coefficient \(h_{rad}\) using the following formula: $$h_{rad} = \sigma \epsilon(T_s^3 + T_{\infty}^3)(T_s + T_{\infty})$$ where \(\sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4}\) is the Stefan-Boltzmann constant, \(\epsilon = 0.3\) is the outer surface emissivity of the insulation, \(T_s\) and \(T_{\infty}\) are the temperatures of the outer surface and surrounding air, respectively. Since we have not yet determined the outer surface temperature, we cannot calculate \(h_{rad}\) directly. Instead, we shall combine it with the overall heat transfer coefficient, as described in the next step.

Calculate the overall heat transfer coefficient and heat transfer rate

Now, the total heat transfer rate can be calculated using the following formula: $$Q = \frac{T_{steam} - T_{\infty}}{R_{pipe} + R_{insulation} + R_{conv} + R_{rad}}$$ where \(R_{conv}\) and \(R_{rad}\) are the convection and radiation thermal resistances, respectively. Assuming that the air heat transfer coefficient is much larger than the radiation heat transfer coefficient, we can express the overall heat transfer coefficient \(U_{overall}\) as: $$U_{overall} = h_{air} + h_{rad} \approx h_{air} $$ Thus, the heat transfer rate for unit length \(L = 1\, \mathrm{m}\) of the pipe can be written as: $$q = \frac{T_{steam} - T_{\infty}}{R_{pipe/1\mathrm{m}} + R_{insulation/1\mathrm{m}} + \frac{1}{U_{overall}}}$$ Calculate the total heat transfer rate per unit length \(q\) using the previously determined values in the equation above. Ensure that the temperature units are in Kelvin for the calculation.

Key concepts

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist heat flow. It is an essential concept in understanding how heat transfers through different layers of a material. Thermal resistance can be particularly significant when examining pipes covered with insulation, like in the given exercise.

In a cylindrical layer, thermal resistance is calculated using the formula:

• \( R_{cylinder} = \frac{\ln(r_2 / r_1)}{2 \pi k L} \)

where:

• \( r_1 \) and \( r_2 \) are the inner and outer radii, respectively,
• \( k \) is the thermal conductivity of the material,
• \( L \) is the pipe length.

The thermal resistance of both the steel pipe and the glass wool insulation needs to be calculated by this formula. The concept here is similar to electrical resistance, but instead of resisting electric flow, it resists heat flow. Just like in electrical circuits, a higher thermal resistance means less heat loss, which is crucial in maintaining the desired temperature of the steam inside the pipe.

Understanding thermal resistance helps to design more efficient heating systems, as minimizing heat loss is both energy-effective and economical.

Nusselt Number

The Nusselt number (Nu) is a dimensionless number that describes the ratio of convective to conductive heat transfer. It is used to determine the effectiveness of convection compared to conduction across a boundary, such as the surface of the insulated pipe in our problem.

The formula for the Nusselt number using the Dittus-Boelter correlation is:

• \( Nu = 0.023 \times Re^{0.8} \times Pr^{0.4} \)

where:

• \( Re \) is the Reynolds number, which describes fluid flow characteristics,
• \( Pr \) is the Prandtl number, a measure of the fluid's momentum diffusivity against its thermal diffusivity.

A higher Nusselt number indicates more effective convective heat transfer compared to conduction. In practical terms, it helps engineers determine if their system efficiently transfers heat away from the pipe under given conditions. Evaluating the Nusselt number involves determining fluid properties such as density and viscosity, typically at a specific film temperature, to calculate Reynolds and Prandtl numbers.

This is critical in optimizing systems where air flows over heated surfaces, ensuring that heat is transferred as intended and helps prevent potential performance shortcomings.

Convection and Radiation

Convection and radiation are two primary modes of heat transfer that occur across boundaries. In the context of a pipe, convection refers to heat transfer due to fluid motion across the pipe's surface, while radiation involves heat transfer through electromagnetic waves.

• Convection: The convective heat transfer coefficient \( h_{air} \) can be found using the Nusselt number obtained earlier: \( h_{air} = \frac{Nu \times k_{air}}{D} \), where \( D \) is the diameter of the pipe. This formula quantifies how efficiently the air flowing over the pipe carries away heat.
• Radiation: Radiation heat transfer depends heavily on the emissivity of the pipe's surface, \( \epsilon \), which is a material property. The radiation heat transfer coefficient \( h_{rad} \) is calculated using the formula: \( h_{rad} = \sigma \epsilon(T_s^3 + T_{\infty}^3)(T_s + T_{\infty}) \), where \( \sigma \) is the Stefan-Boltzmann constant, and \( T_s \) and \( T_{\infty} \) are outer surface and surrounding air temperatures, respectively.

Both convection and radiation play vital roles in determining the overall heat loss from the pipe. They influence how well the system controls temperature drops, impacting both efficiency and energy use in applications from pipelines to HVAC systems.