/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A \(1 \mathrm{~m}\)-diameter liq... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(0.213 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).

Short Answer

Expert verified
The boil-off rate of the LOX is approximately 0.01974 kg/s.

Step by step solution

01

Calculate the Area of the Tank

The tank can be thought of as a cylinder, and its surface area (excluding the ends) is given by the formula for the lateral surface area of a cylinder, which is \( A = \pi \cdot d \cdot h \). Since the height isn't provided, we only consider the cylinder's diameter for heat transfer rate calculations. Therefore, the area used in this case will be the area of the tank calculated as \( A = \pi \cdot d^2 \). With \( d = 1 \text{ m} \), the area is \( A = \pi \times (1 \text{ m})^2 = \pi \text{ m}^2\).
02

Determine the Heat Transfer Through Insulation

Using Fourier's law for steady heat conduction through the insulation, the heat transfer rate \( Q \) can be calculated as: \[ Q = \frac{(T_\text{ambient} - T_\text{boil})}{\frac{\Delta x}{k} \cdot A} \] where \( T_\text{ambient} = 310 \text{ K} \), \( T_\text{boil} = 90 \text{ K} \), \( \Delta x = 0.1 \text{ m} \), \( k = 0.022 \text{ W/m K} \), and \( A = \pi \text{ m}^2 \). Substitute these values into the equation to get: \[ Q = \frac{(310 - 90) \cdot \pi}{\frac{0.1}{0.022}} \approx 763.82 \text{ W}\]
03

Calculate the Total Heat Transfer from Convection and Radiation

The total heat transfer coefficient \( h_t \) is the sum of the convective \( h_c \) and radiative \( h_r \) heat transfer coefficients: \( h_t = h_c + h_r = 3 + 2 = 5 \text{ W/m}^2 \text{ K} \). The heat transfer from convection and radiation is: \[ Q_\text{total} = h_t \cdot A \cdot (T_\text{ambient} - T_\text{boil}) = 5 \cdot \pi \cdot (310 - 90) \approx 3439.03 \text{ W}\]
04

Determine Total Heat Transfer to LOX

The total heat transfer to the LOX tank is the sum of the heat through conduction (Step 2) and the convection/radiation (Step 3): \[ Q_\text{LOX} = Q + Q_\text{total} = 763.82 + 3439.03 = 4202.85 \text{ W}\]
05

Calculate the Boil-Off Rate

The boil-off rate \( \dot{m} \) of the LOX is calculated using the formula: \[ \dot{m} = \frac{Q_\text{LOX}}{\Delta H_v} \] where \( \Delta H_v = 0.213 \times 10^{6} \text{ J/kg} \). Substitute \( Q_\text{LOX} = 4202.85 \text{ W} \) and \( \Delta H_v = 0.213 \times 10^6 \text{ J/kg} \): \[ \dot{m} = \frac{4202.85}{0.213 \times 10^6} \approx 0.01974 \text{ kg/s}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property that describes how well a material conducts heat. In the context of the liquid oxygen (LOX) tank, thermal conductivity is crucial in determining how much heat passes through the fiberglass insulation. The insulation is specified to have a thermal conductivity value of \(0.022\ \mathrm{W/m\ K}\). This relatively low value indicates that fiberglass is a good insulator, meaning it effectively slows down the rate at which heat can pass through it.

When calculating the heat transfer through the insulation, we use Fourier's law for steady-state heat conduction. This law is represented by the equation:
\[Q = \frac{(T_\text{ambient} - T_\text{boil})}{\frac{\Delta x}{k} \cdot A}\]
Here, \(T_\text{ambient}\) is the ambient temperature, \(T_\text{boil}\) is the boiling temperature of the LOX, \(\Delta x\) is the thickness of the insulation, \(k\) is the thermal conductivity, and \(A\) is the surface area of the tank. By plugging in these values, we can find how much heat is transferred through the insulation.

Understanding thermal conductivity helps in choosing appropriate materials for insulation to minimize unwanted heat transfer.
Convection Heat Transfer
Convection heat transfer involves the movement of heat between a surface and a fluid in motion, such as air flowing over the surface of an object. In our exercise, convection contributes to the heat gain in the liquid oxygen tank from the surrounding air.

The heat transfer due to convection can be calculated using the equation:
\[Q_c = h_c \cdot A \cdot (T_\text{ambient} - T_\text{boil})\]
where \(h_c\) is the convective heat transfer coefficient, \(A\) is the surface area of the object, and \(T_\text{ambient} - T_\text{boil}\) is the temperature difference between the fluid and the surface.

The convective heat transfer coefficient \(h_c\) represents how effectively heat is transferred between the fluid and the surface. In this problem, it is given as \(3\ \mathrm{W/m^2\ K}\). Understanding this process is essential to evaluate how much heat the insulation can prevent from reaching the LOX, based on the fluid motion around the tank.
Radiation Heat Transfer
Radiation heat transfer occurs through the emission of electromagnetic waves, and it does not require direct contact between objects. It is one of the means by which heat transfer can occur across a vacuum or transparent media like air.

In the LOX tank scenario, radiation constitutes a part of the heat addition to the tank. The radiative heat transfer coefficient \(h_r\) is given as \(2\ \mathrm{W/m^2\ K}\). Even though radiation may seem less significant compared to other forms of heat transfer, it is always present and can become dominant in certain temperature ranges and conditions.

The heat transfer due to radiation can be summarized in the following equation:
\[Q_r = h_r \cdot A \cdot (T_\text{ambient} - T_\text{boil})\]
Again, \(A\) signifies the area, and \(T_\text{ambient} - T_\text{boil}\) is the temperature differential.

Radiation is especially important when high temperatures or low thermal conductivity layers are involved, as its effect can be amplified under such conditions.
Heat Transfer Coefficients
Heat transfer coefficients are crucial figures in calculating how efficiently heat is transferred from one medium to another. They are used in equations for both convection and radiation heat transfer in exercises like the liquid oxygen tank problem.

The total heat transfer coefficient \(h_t\) in this problem combines both the contributions from convection and radiation. It is calculated as the sum of the individual coefficients:
\[h_t = h_c + h_r = 3\ \mathrm{W/m^2\ K} + 2\ \mathrm{W/m^2\ K} = 5\ \mathrm{W/m^2\ K}\]
These coefficients facilitate the calculation of the total heat transfer from the ambient environment to the LOX tank with the formula:
\[Q_\text{total} = h_t \cdot A \cdot (T_\text{ambient} - T_\text{boil})\]

Accurate knowledge of the heat transfer coefficients allows engineers to predict and manage how much heat will be transferred in a system, thus designing better insulation and heat exchange systems. They are key to ensuring that systems like the LOX tank operate safely and efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electrical current is passed through a horizontal copper rod \(2 \mathrm{~mm}\) in diameter and \(30 \mathrm{~cm}\) long, located in an air stream at \(20^{\circ} \mathrm{C}\). If the ends of the rod are also maintained at \(20^{\circ} \mathrm{C}\) and the convective heat transfer coefficient is estimated to be \(30 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), determine the maximum current that can be passed if the midpoint temperature is not to exceed \(50^{\circ} \mathrm{C}\). (i) Ignore thermal radiation. (ii) Include the effect of thermal radiation. For the copper rod, take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}, \varepsilon=0.8\), and electrical resistivity of \(1.72 \times 10^{-8} \Omega \mathrm{m}\).

A \(1 \mathrm{~mm}\)-diameter resistor has a sheath of thermal conductivity \(k=0.12 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and is located in an evacuated enclosure. Determine the radius of the sheath that maximizes the heat loss from the resistor when it is maintained at \(450 \mathrm{~K}\) and the enclosure is at \(300 \mathrm{~K}\). The surface emittance of the sheath is \(0.85\).

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.

Consider a heat barrier consisting of a \(2 \mathrm{~mm}\)-thick brass plate to which \(3 \mathrm{~mm}\) copper tubing is soldered. The tubes are spaced \(10 \mathrm{~cm}\) apart. Cooling water passed through the tubes keeps them at approximately \(315 \mathrm{~K}\). The underside of the brass wall is insulated with a \(1.5 \mathrm{~cm}\)-thick asbestos layer, which in turn contacts a hot wall at \(600 \mathrm{~K}\). Assuming that the heat transfer from the cold side of the brass plate is negligible, estimate the temperature of the hottest spot on the brass wall. Take \(k=0.16 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\) for the asbestos and \(111 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the brass.

Heat is generated at a rate \(\dot{Q}_{v}^{\prime \prime \prime}\) in a long solid cylinder of radius \(R\). The cylinder has a thin metal sheath and is immersed in a liquid at temperature \(T_{e}\). Heat transfer from the cylinder surface to the liquid can be characterized by an overall heat transfer coefficient \(U\). Obtain the steady-state temperature distributions for the following cases: (i) \(\dot{Q}_{v}^{\prime \prime \prime}\) is constant. (ii) \(\dot{Q}_{v}^{\prime \prime \prime}=\dot{Q}_{v 0}^{\prime \prime \prime}\left[1-(r / R)^{2}\right]\). (iii) \(\dot{Q}_{v}^{\prime \prime \prime \prime}=a+b\left(T-T_{e}\right)\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.