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Chapter 7: Problem 126

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

Expert verified
#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

01

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.
02

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.
03

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.
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Hydrogen gas at \(1 \mathrm{~atm}\) is flowing in parallel over the upper and lower surfaces of a 3-m-long flat plate at a velocity of \(2.5 \mathrm{~m} / \mathrm{s}\). The gas temperature is \(120^{\circ} \mathrm{C}\) and the surface temperature of the plate is maintained at \(30^{\circ} \mathrm{C}\). Using the EES (or other) software, investigate the local convection heat transfer coefficient and the local total convection heat flux along the plate. By varying the location along the plate for \(0.2 \leq x \leq 3 \mathrm{~m}\), plot the local convection heat transfer coefficient and the local total convection heat flux as functions of \(x\). Assume flow is laminar but make sure to verify this assumption. 7-31 Carbon dioxide and hydrogen as ideal gases at \(1 \mathrm{~atm}\) and \(-20^{\circ} \mathrm{C}\) flow in parallel over a flat plate. The flow velocity of each gas is \(1 \mathrm{~m} / \mathrm{s}\) and the surface temperature of the 3 -m-long plate is maintained at \(20^{\circ} \mathrm{C}\). Using the EES (or other) software, evaluate the local Reynolds number, the local Nusselt number, and the local convection heat transfer coefficient along the plate for each gas. By varying the location along the plate for \(0.2 \leq x \leq 3 \mathrm{~m}\), plot the local Reynolds number, the local Nusselt number, and the local convection heat transfer coefficient for each gas as functions of \(x\). Discuss which gas has higher local Nusselt number and which gas has higher convection heat transfer coefficient along the plate. Assume flow is laminar but make sure to verify this assumption.

Jakob (1949) suggests the following correlation be used for square tubes in a liquid cross-flow situation: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.675} \mathrm{Pr}^{1 / 3} $$ Water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6)\) at \(50^{\circ} \mathrm{C}\) flows across a \(1-\mathrm{cm}\) square tube with a Reynolds number of 10,000 and surface temperature of \(75^{\circ} \mathrm{C}\). If the tube is \(2 \mathrm{~m}\) long, the rate of heat transfer between the tube and water is (a) \(6.0 \mathrm{~kW}\) (b) \(8.2 \mathrm{~kW}\) (c) \(11.3 \mathrm{~kW}\) (d) \(15.7 \mathrm{~kW}\) (e) \(18.1 \mathrm{~kW}\)

Air at \(25^{\circ} \mathrm{C}\) flows over a 5 -cm-diameter, \(1.7\)-m-long pipe with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). A refrigerant at \(-15^{\circ} \mathrm{C}\) flows inside the pipe and the surface temperature of the pipe is essentially the same as the refrigerant temperature inside. Air properties at the average temperature are \(k=0.0240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735\), \(\nu=1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The rate of heat transfer to the pipe is (a) \(343 \mathrm{~W}\) (b) \(419 \mathrm{~W}\) (c) \(485 \mathrm{~W}\) (d) \(547 \mathrm{~W}\) (e) \(610 \mathrm{~W}\)

A 12 -ft-long, \(1.5-\mathrm{kW}\) electrical resistance wire is made of \(0.1\)-in-diameter stainless steel \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). The resistance wire operates in an environment at \(85^{\circ} \mathrm{F}\). Determine the surface temperature of the wire if it is cooled by a fan blowing air at a velocity of \(20 \mathrm{ft} / \mathrm{s}\). For evaluations of the air properties, the film temperature has to be found iteratively. As an initial guess, assume the film temperature to be \(200^{\circ} \mathrm{F}\).

Consider a person who is trying to keep cool on a hot summer day by turning a fan on and exposing his entire body to air flow. The air temperature is \(85^{\circ} \mathrm{F}\) and the fan is blowing air at a velocity of \(6 \mathrm{ft} / \mathrm{s}\). If the person is doing light work and generating sensible heat at a rate of \(300 \mathrm{Btu} / \mathrm{h}\), determine the average temperature of the outer surface (skin or clothing) of the person. The average human body can be treated as a 1-ft-diameter cylinder with an exposed surface area of \(18 \mathrm{ft}^{2}\). Disregard any heat transfer by radiation. What would your answer be if the air velocity were doubled? Evaluate the air properties at \(100^{\circ} \mathrm{F}\).
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