/*! 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 221 A 1-m-inner-diameter liquid-oxyg... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum ( \(k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=\) \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the exterior surface of the insulation is (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Solution: Calculate the overall conduction resistance - Calculate Aluminum Resistance - Calculate Insulation Resistance - Compute Overall Resistance Perform energy balance with convection resistance - Compute Convection Resistance - Perform Heat Energy Balance Solve for the exterior surface temperature - Rearrange Energy Balance Equation - Calculate Exterior Surface Temperature Result: Verify which option is closest to the obtained value: (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

Step by step solution

01

Calculate Aluminum Resistance

Using the formula for the resistance of a cylinder, calculate the resistance through the aluminum shell: \(R_{al} = \frac{ln(r_2/r_1)}{2 \pi \; l \; k_{al}}\) Where: \(r_1\) is the inner radius of the aluminum shell (0.5 m) \(r_2\) is the outer radius of the aluminum shell (0.5 m + 0.005 m) \(l\) is the length of the aluminum shell (1 m) \(k_{al}\) is the thermal conductivity of the aluminum (170 W/mâ‹…K)
02

Calculate Insulation Resistance

Calculate the resistance through the insulation: \(R_{ins} = \frac{ln(r_3/r_2)}{2 \pi \; l \; k_{ins}}\) Where: \(r_3\) is the outer radius of the insulation (0.5 m + 0.005 m + 0.1 m) \(k_{ins}\) is the thermal conductivity of the insulation (0.02 W/mâ‹…K)
03

Compute Overall Resistance

Calculate the overall conduction resistance: \(R_{cond} = R_{al} + R_{ins}\) Step 2: Perform energy balance with convection resistance
04

Compute Convection Resistance

Calculate the convection resistance on the exterior side of the insulation: \(R_{conv} = \frac{1}{h \; A}\) Where: \(h\) is the heat transfer coefficient (5 W/m²⋅K) \(A\) is the surface area of the exterior side of the insulation (\(2 \pi \; r_3 \; l\))
05

Heat Energy Balance

Perform a heat energy balance by equating the heat transfer rate through conduction and convection: \(Q = \frac{T_{ins} - T_{liq}}{R_{cond}} = \frac{T_{amb} - T_{ins}}{R_{conv}}\) Where: \(T_{liq}\) is the temperature of the liquid oxygen (90 K or -183.15°C) \(T_{ins}\) is the temperature of the exterior surface of the insulation (unknown) \(T_{amb}\) is the ambient temperature (20°C) Step 3: Solve for the exterior surface temperature
06

Rearrange Energy Balance Equation

Rearrange the heat energy balance equation to solve for the exterior surface temperature: \(T_{ins} = \frac{T_{liq} \; R_{conv} + T_{amb} \; R_{cond}}{R_{conv} + R_{cond}}\)
07

Calculate Exterior Surface Temperature

Calculate the exterior surface temperature of the insulation: \(T_{ins} = \frac{(-183.15 \mathrm{~°C}) \; R_{conv} + (20 \mathrm{~°C}) \; R_{cond}}{R_{conv} + R_{cond}}\) Once we calculated \(T_{ins}\), we can verify which one of the given options is closest to the obtained value: (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

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.

Heat Transfer
Heat transfer is a fundamental concept in physics that describes how thermal energy moves from one object or material to another. In the context of a liquid-oxygen storage tank, understanding heat transfer is essential to keep the tank effectively insulated. There are three primary methods of heat transfer: conduction, convection, and radiation. Each plays a role in the way energy moves through materials. In this exercise, we focus on conduction (through the aluminum and insulation) and convection (on the outer surface).
  • Conduction: Transfer through materials, such as metal or insulation.
  • Convection: Heat transfer through fluids, like air or water, which may occur when the interior and exterior surfaces of the tank interact with their surroundings.
  • Radiation: Not a focal point in this exercise, but generally involves energy transfer through electromagnetic waves.
By reducing heat transfer, the system maintains the liquid oxygen at cryogenic temperatures, preventing evaporation and energy loss.
Conduction Resistance
Conduction resistance is a measure of a material's ability to conduct heat. It essentially tells us how much a material resists the flow of heat through it. In our problem, two main layers contribute to conduction resistance: the aluminum shell and the insulation layer surrounding it. The formula for finding conduction resistance in a cylindrical object is:\[ R_{cond} = \frac{\ln(r_{outer}/r_{inner})}{2 \pi \cdot l \cdot k} \]where:- \( r_{outer} \) and \( r_{inner} \) are the external and internal radii of the material,- \( l \) is the length of the cylinder, - \( k \) is the thermal conductivity.An important point to note is that materials with lower thermal conductivity, like the insulation used in the tank, will have higher conduction resistance. This greatly helps in minimizing the heat transfer into the tank, which is crucial for maintaining the contents at low temperatures.
Convection Resistance
Convection resistance involves the resistance faced by heat as it transfers from the surface of a solid to a fluid surrounding it, or vice versa. This resistance is essential in understanding how quickly heat can move away from a surface, such as the outer side of the insulation on our storage tank.The convection resistance formula is:\[ R_{conv} = \frac{1}{h \cdot A} \]where:- \( h \) is the heat transfer coefficient (measured in \( \mathrm{W/m^2 \cdot K} \)), which indicates how effective a fluid is in transferring heat,- \( A \) is the surface area in contact with the fluid.In our scenario, the ambient air with a transfer coefficient of \( 5 \mathrm{~W/m^2 \cdot K} \) is crucial. The value of \( R_{conv} \) helps in determining how much heat remains on the surface or how much is dissipated into the air. High convection resistance suggests lesser heat escape, which can be important for insulation purposes.
Insulation in Cryogenic Systems
Insulation plays a vital role in cryogenic systems, which are designed to operate at extremely low temperatures. The main goal of insulation in these systems is to minimize heat transfer, ensuring that the interior temperatures remain stable with minimal energy loss.For the liquid-oxygen storage tank, the insulation is structured to handle temperature differentials between the internal storage environment (90 K or \(-183.15^\circ \mathrm{C}\)) and the external ambient air (\(20^\circ \mathrm{C}\)).Key features of effective insulation used in such systems:
  • Low thermal conductivity: This makes it resistant to heat flow.
  • Adequate thickness: More thickness increases resistance, helping trap cold air inside and block external heat.
Insulation must also handle environmental conditions like contact with air and moisture. In cryogenic applications, maintaining the performance of insulation over long periods is essential to prevent faults or system inefficiencies.

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

Consider a pipe at a constant temperature whose radius is greater than the critical radius of insulation. Someone claims that the rate of heat loss from the pipe has increased when some insulation is added to the pipe. Is this claim valid?

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

Consider two metal plates pressed against each other. Other things being equal, which of the measures below will cause the thermal contact resistance to increase? (a) Cleaning the surfaces to make them shinier. (b) Pressing the plates against each other with a greater force. (c) Filling the gap with a conducting fluid. (d) Using softer metals. (e) Coating the contact surfaces with a thin layer of soft metal such as tin.

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a \(10-\mathrm{cm}\)-thick layer of insulation \((k=0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

What is the difference between the fin effectiveness and the fin efficiency?

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.