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Chapter 2: Problem 26

Liquid oxygen at \(90 \mathrm{~K}\) flows inside a \(3 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 2 \mathrm{~mm}\)-wall-thickness, AISI 303 stainless steel tube. The tube is insulated with \(3 \mathrm{~cm}\)-thick fiberglass insulation of thermal conductivity \(0.021 \mathrm{~W} / \mathrm{m} \mathrm{K}\). Will atmospheric water vapor condense on the outside of the insulation when the ambient air temperature is \(300 \mathrm{~K}\) and the dewpoint temperature is \(285 \mathrm{~K}\) ? Take the outside convective and radiative heat transfer coefficients as \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively.

Short Answer

Expert verified
Yes, condensation will occur as the surface temperature (284.5 K) is below the dewpoint (285 K).

Step by step solution

01

Understand the problem

The goal is to determine if atmospheric water vapor will condense on the outside of the insulation. For condensation to occur, the temperature at the surface of the insulation must be below the dewpoint temperature of the ambient air, which is given as 285 K.
02

Calculate the thermal resistance

The total thermal resistance comprises the conduction resistance through the insulation and the combined convective and radiative resistance on the surface. The conduction resistance through the insulation can be calculated using \[ R_{conduction} = \frac{thickness}{k \, \cdot \, A} \]where thickness is 0.03 m, \(k\) is the thermal conductivity of fiberglass (0.021 W/m K), and \(A\) is the surface area. Assume unit length for simplicity, \[ A = \pi \cdot d \cdot 1 \text{ m} = \pi \cdot (0.03 + 0.002 \cdot 2) \text{ m} = 0.1017 \text{ m}^2 \] \[ R_{conduction} = \frac{0.03}{0.021 \cdot 0.1017} \approx 1.38 \, K/W \].
03

Calculate the convection and radiative resistances

The surface convective and radiative resistances are calculated using the formula: \[ R_{conv + rad} = \frac{1}{h_{conv} + h_{rad}} \]where \(h_{conv}\) is 5 W/m虏K and \(h_{rad}\) is 4 W/m虏K, \[ R_{conv + rad} = \frac{1}{5 + 4} = \frac{1}{9} \approx 0.111 \, K/W \] per square meter of surface area.
04

Calculate the total resistance

Add the conduction resistance and the convection-radiation combined resistance, \[ R_{total} = R_{conduction} + R_{conv + rad} = 1.38 + 0.111 = 1.491 \, K/W \].
05

Calculate the heat flow rate

The temperature difference between the inside and outside of the insulation is \(90 \text{ K} - 300 \text{ K} = -210 \text{ K}\). The heat flow rate is given by,\[ Q = \frac{\Delta T}{R_{total}} = \frac{-210}{1.491} \approx -140.95 \, W/m \]. The negative sign indicates heat loss from inside the tube.
06

Calculate the outside surface temperature

The temperature on the outer surface of the insulation, \(T_{surface}\), can be found from\[ Q = h_{ext} \cdot A \cdot (T_{surface} - T_{ambient}) \], Rearranging gives\[ T_{surface} = T_{ambient} + \frac{Q}{h_{ext} \cdot A} \], where \(h_{ext} = 9 \, W/m^2 \cdot K\) and \(A\) is the surface area calculated earlier,\[ T_{surface} = 300 \text{ K} + \frac{-140.95}{9 \cdot 0.1017} \approx 284.5 \, K \].
07

Compare with dewpoint

Since the outer surface temperature (284.5 K) is less than the dewpoint temperature (285 K), condensation of atmospheric water vapor will occur on the outer surface of the insulation.

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Key Concepts

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

Condensation
Condensation is a common phenomenon where gas transforms into a liquid. This usually happens when water vapor in the air cools and turns into liquid water. For condensation to occur, the surface temperature must be lower than the dewpoint temperature.
Basically, if the air is cooled to its dewpoint, the water vapor will condense into a liquid form.
This concept is crucial in everyday life as well as engineering applications. Understanding condensation helps in managing moisture levels in buildings and preventing issues such as mold or structural damage.
In the context of this exercise, condensation is used to determine if atmospheric water vapor will transform into liquid on the outside of the insulation material surrounding a cold pipe.
  • Condensation must be controlled in various industries to prevent material deterioration and inefficiencies.
  • It can also be utilized for applications like water harvesting in arid regions.
Dewpoint Temperature
Dewpoint temperature is the temperature at which air reaches 100% relative humidity, saturating and causing water vapor to condense into liquid water. In simpler terms, it's the temperature at which the air "gets full" and no longer holds all the water vapor, leading to condensation.

Dewpoint is a useful measure to predict weather conditions and for indoor climate control. For instance, a higher dewpoint indicates more moisture in the air, which might cause uncomfortable humidity levels.
  • Dewpoint gives us insight into how damp or dry the air feels.
  • It's an important factor in air conditioning, refrigeration, and even in atmospheric studies.
For the problem at hand, the dewpoint temperature is given as 285 K. If the surface temperature of the insulation is below this level, condensation will occur, as was confirmed in the problem's calculations.
Heat Transfer Resistance
Heat transfer resistance refers to how well a material resists the flow of heat. It's an essential concept in evaluating thermal insulation performance. When you want to prevent heat from escaping or entering a system, understanding and calculating the heat transfer resistance is crucial.
In the exercise, the overall heat transfer resistance was calculated using two major components: conduction through the insulation and combined convection and radiation at the surface.
  • Conduction resistance depends on the material's thermal conductivity and its thickness.
  • Convective and radiative resistance depends on the heat transfer coefficients and surface area.
The lower the heat transfer resistance, the more efficient the heat flow through materials. In this context, calculating the total resistance helps understand how much heat is lost from the cold pipe to the surroundings.

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Most popular questions from this chapter

The convective heat transfer coefficient around a cylinder held perpendicular to a flow varies in a complicated manner. A test cylinder to investigate this behavior consists of a \(0.001\) in-thick, \(12.7 \mathrm{~mm}\)-wide stainless steel heater ribbon (cut from shim stock) wound around a \(2 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness Teflon tube. A single thermocouple is located just underneath the ribbon and measures the local ribbon temperature \(T_{s}(\theta)\). The cylinder is installed in a wind tunnel, and a second thermocouple is used to measure the ambient air temperature \(T_{e}\). The power input to the heater is metered, from which the electrical heat generation per unit area \(\dot{Q} / A\) can be calculated ( \(A\) is the surface area of one side of the ribbon). As a first approximation, the local heat transfer coefficient \(h_{c}(\theta)\) can be obtained from $$ h_{c}(\theta)=\frac{\dot{Q} / A}{T_{s}(\theta)-T_{e}} $$ Hence, by rotating the cylinder with the power held constant, the variation of \(h_{c}\) can be obtained from the variation of \(T_{s}\) : where \(T_{s}(\theta)\) is low, \(h_{c}(\theta)\) is high, and vice versa. A typical variation of \(T_{s}(\theta)\) is shown in the graph. A problem with this technique is that conduction around the circumference of the tube causes the local heat flux \(q_{s}(\theta)\) to not exactly equal \(\dot{Q} / A\). (i) Derive a formula for \(h_{c}(\theta)\) that approximately accounts for circumferential conduction. (ii) The following table gives values of \(T_{s}(\theta)\) in a sector where circumferential conduction effects are expected to be large. Use these values together with \(\dot{Q} / A=5900 \mathrm{~W} / \mathrm{m}^{2}\) and \(T_{e}=25^{\circ} \mathrm{C}\) to estimate the conduction effect at \(\theta=110^{\circ}\). \begin{tabular}{cc} Angle (degrees) & \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline 100 & \(65.9\) \\ 110 & \(65.7\) \\ 120 & \(64.4\) \end{tabular} (iii) Comment on the design of the cylinder. Would a \(3 \mathrm{~mm}\)-thick brass tube, directly heated by an electric current, be a suitable alternative? Use \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel and \(0.38 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for Teflon.

An electrical current of 15 A flows in an 18 gage copper wire ( \(1.02 \mathrm{~mm}\) diameter). If the wire has an electrical resistance of \(0.0209 \Omega / \mathrm{m}\), calculate (i) the rate of heat generation per meter length of wire. (ii) the rate of heat generation per unit volume of copper. (iii) the heat flux across the wire surface at steady state.

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.

A \(2 \mathrm{~cm}\)-thick composite plate has electric heating wires arranged in a grid in its centerplane. On one side there is air at \(20^{\circ} \mathrm{C}\), and on the other side there is air at \(100^{\circ} \mathrm{C}\). If the heat transfer coefficient on both sides is \(40 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), what is the maximum allowable rate of heat generation per unit area if the composite temperature should not exceed \(300^{\circ} \mathrm{C}\) ? Take \(k=0.45 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the composite material.

A \(1 \mathrm{~m}\)-diameter liquid oxygen (LOX) tank is insulated with a \(10 \mathrm{~cm}\)-thick blanket of fiberglass insulation having a thermal conductivity of \(0.022 \mathrm{~W} / \mathrm{m} \mathrm{K}\). The tank is vented to the atmosphere. Determine the boil-off rate if the ambient air is at \(310 \mathrm{~K}\) and the outside convective and radiative heat transfer coefficients are \(3 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(2 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. The boiling point of oxygen is \(90 \mathrm{~K}\), and its enthalpy of vaporization is \(0.213 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).
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