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Chapter 2: Problem 16

Saturated steam at \(200^{\circ} \mathrm{C}\) flows through an AISI 1010 steel tube with an outer diameter of \(10 \mathrm{~cm}\) and a \(4 \mathrm{~mm}\) wall thickness. It is proposed to add a \(5 \mathrm{~cm}\)-thick layer of \(85 \%\) magnesia insulation. Compare the heat loss from the insulated tube to that from the bare tube when the ambient air temperature is \(20^{\circ} \mathrm{C}\). Take outside heat transfer coefficients of 6 and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) for the bare and insulated tubes, respectively.

Short Answer

Expert verified
The insulated tube loses more heat (566.4 W) than the bare tube (339.84 W) due to increased surface area.

Step by step solution

01

Calculate the surface area

The outer diameter of the tube is given as 10 cm, thus the radius is 5 cm or 0.05 m. The surface area of the tube can be calculated using the formula for the lateral surface area of a cylinder: \[ A = 2\pi r L \] Where \( r \) is the radius and \( L \) is the length of the cylinder. Assume a unit length of 1 m for simplicity. Therefore, the surface area is \[ A = 2 \pi (0.05) (1) = 0.314 \text{ m}^2 \]
02

Calculate heat loss for the bare tube

Use the heat loss formula: \[ Q = hA (T_{s} - T_{a}) \] for the bare tube, where \( h = 6 \text{ W/m}^2\text{K} \), \( A = 0.314 \text{ m}^2 \), \( T_{s} = 200^{\circ}C \), \( T_{a} = 20^{\circ}C \). Therefore, \[ Q_{bare} = 6 \times 0.314 \times (200 - 20) = 339.84 \text{ W} \]
03

Calculate the new diameter with insulation

The thickness of the insulation is 5 cm, so the new outer radius of the insulated tube is \[ 0.05 + 0.05 = 0.1 \text{ m} \] The new surface area is then \[ A_{insulated} = 2\pi (0.1)(1) = 0.628 \text{ m}^2. \]
04

Calculate heat loss for the insulated tube

For the insulated tube, use the heat loss formula with the adjusted area and heat transfer coefficient \( h = 5 \text{ W/m}^2\text{K} \):\[ Q_{insulated} = 5 \times 0.628 \times (200 - 20) = 566.4 \text{ W}. \]
05

Comparison of heat losses

The heat loss from the bare tube is 339.84 W, whereas the heat loss from the insulated tube is 566.4 W. Despite adding insulation, the heat loss increased due to a larger surface area, highlighting the importance of considering both insulation and surface area.

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Key Concepts

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

Saturated Steam
Saturated steam refers to steam that is in equilibrium with water at a certain pressure and temperature. This means that the steam contains as much water vapor as possible without condensing into water. At a specific pressure, saturated steam has a unique temperature at which it exists in its equilibrium state. In the context of our exercise, we are dealing with saturated steam at 200掳C.
  • At this temperature, the steam has distinct properties such as specific enthalpy and specific volume, which are essential for calculations related to heat transfer.
  • Understanding these properties is pivotal in ensuring accurate heat loss calculations, as they define how energy moves from the steam to its surroundings.
  • Saturated steam is often used in industrial applications due to its efficient energy transfer capabilities, as it can condense to release the latent heat.
Grasping the idea of saturated steam helps in designing systems for effective insulation and managing heat distribution.
AISI 1010 Steel
AISI 1010 steel is a type of carbon steel that is commonly used in manufacturing applications. It consists mainly of iron and carbon, where the carbon content is relatively low, making it relatively soft but ductile. When considering AISI 1010 steel pipes as in our problem, several attributes are key:
  • The low carbon content makes it suitable for forming and welding, which is why it is often used in tubing.
  • Its thermal conductivity is an important factor in heat transfer calculations. Lower conductivity compared to other metals means more heat is retained within the pipe, affecting the heat transfer rate.
  • Resistance to wear and moderately good machinability make it ideal for pipes in systems where heat loss management is crucial.
These features of AISI 1010 steel are integral in calculating the efficiency of thermal systems and managing heat transfer effectively.
Magnesia Insulation
Magnesia insulation is a type of thermal insulation known for its exceptional heat resistance properties. Consisting mostly of magnesium oxide, it is often employed in high-temperature insulation applications.
  • One of the key properties is its low thermal conductivity, making it excellent for reducing heat loss in industrial applications.
  • In our example, adding magnesia insulation increases the outer radius of the tube, impacting the surface area involved in heat transfer.
  • This type of insulation is lightweight and often used where minimization of thermal energy waste is required, as it effectively slows down the rate of heat loss due to its low conductivity.
Understanding how magnesia insulation works is crucial for predicting and enhancing the performance of insulated systems.
Heat Loss Calculation
Heat loss calculation is an essential part of thermal analysis in engineering. In this exercise, the focus is on comparing heat loss from a bare steel tube to one with insulation.
  • The basic formula used for heat loss through a surface is given by: \[ Q = hA(T_s - T_a) \]where \( Q \) is the heat transfer rate, \( h \) is the heat transfer coefficient, \( A \) is the surface area, and \( T_s - T_a \) is the temperature difference between the surface and the ambient air.
  • For the bare tube, lesser surface area and a higher heat transfer coefficient result in lower heat loss.
  • When insulation is added, the area increases significantly due to greater thickness. Despite a lower heat transfer coefficient with insulation, the overall heat loss increases because of the expanded area.
    This exercise illustrates the importance of balancing surface area and insulation thickness when optimizing for heat loss reduction.
Mastering heat loss calculations allows engineers to enhance energy efficiency in thermal systems efficiently.

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Most popular questions from this chapter

An electrical current is passed through a \(1 \mathrm{~mm}\)-diameter, \(20 \mathrm{~cm}\)-long copper wire located in an air flow at \(290 \mathrm{~K}\). If the ends of the wire are maintained at \(300 \mathrm{~K}\), determine the maximum current that can be passed if the midpoint temperature is not to exceed \(400 \mathrm{~K}\). The convective heat transfer coefficient is estimated to be 20 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\). For the copper wire, take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}, \varepsilon=0.8\). and an electrical resistance of \(2.2 \times 10^{-2} \Omega / \mathrm{m}\).

A \(6 \mathrm{~mm}\)-O.D. tube is to be insulated with an insulation of thermal conductivity \(0.08 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and a very low surface emittance. Heat loss is by natural convection, for which the heat transfer coefficient can be taken as \(\bar{h}_{c}=1.3(\Delta T / D)^{1 / 4} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) for \(\Delta T=T_{s}-T_{e}\) in kelvins and the diameter \(D\) in meters. Determine the critical radius of the insulation and the corresponding heat loss for a tube surface temperature of \(350 \mathrm{~K}\), and an ambient temperature of \(300 \mathrm{~K}\).

So-called "compact" heat exchanger cores often consist of finned passages between parallel plates. A particularly simple configuration has square passages with the effective fin length equal to half the plate spacing \(L\). In a particular application with \(L=5 \mathrm{~mm}\), a convective heat transfer coefficient of \(160 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) is expected. If \(95 \%\) efficient fins are desired, how thick should they be if the core is constructed from (i) an aluminum alloy with \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) ? (ii) mild steel with \(k=64 \mathrm{~W} / \mathrm{m} \mathrm{K}\) ? (iii) a plastic with \(k=0.33 \mathrm{~W} / \mathrm{m} \mathrm{K}\) ? Discuss the significance of your results to the design of such cores.

A \(60 \mathrm{~cm}\)-long, \(3 \mathrm{~cm}\)-diameter AISI 1010 steel rod is welded to a furnace wall and passes through \(20 \mathrm{~cm}\) of insulation before emerging into the surrounding air. The furnace wall is at \(300^{\circ} \mathrm{C}\), and the air temperature is \(20^{\circ} \mathrm{C}\). Estimate the temperature of the bar tip if the heat transfer coefficient between the rod and air is taken to be \(13 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A \(4 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness stainless steel tube is insulated with a 5 \(\mathrm{cm}\)-thick layer of cork. Chilled milk flows through the tube. At a given location the milk temperature is \(5^{\circ} \mathrm{C}\) when the ambient temperature is \(25^{\circ} \mathrm{C}\). If the inside and outside heat transfer coefficients are estimated to be 50 and \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively, calculate the rate of heat gain per meter length of tube.
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