/*! 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 61 A long cylinder of \(30-\mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 61

A long cylinder of \(30-\mathrm{mm}\) diameter, initially at a uniform temperature of \(1000 \mathrm{~K}\), is suddenly quenched in a large, constant- temperature oil bath at \(350 \mathrm{~K}\). The cylinder properties are \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}\), while the convection coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the time required for the surface of the cylinder to reach \(500 \mathrm{~K}\). (b) Compute and plot the surface temperature history for \(0 \leq t \leq 300 \mathrm{~s}\). If the oil were agitated, providing a convection coefficient of \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how would the temperature history change?

Short Answer

Expert verified
The time required for the surface of the cylinder to reach 500 K is approximately 141.02 s. The surface temperature history with the initial convection coefficient (h = 50 W/m虏鈰匥) and agitated oil (h = 250 W/m虏鈰匥) reveals that the higher convection coefficient leads to faster cooling of the surface as it approaches the ambient temperature of 350 K.

Step by step solution

01

Calculate the Biot number

The Biot number (Bi) is given by: Bi = \( \frac{hL_c}{k} \) where: h = convection coefficient = 50 W/m虏鈰匥 \(L_c\) = characteristic length = \( \frac{D}{4} \) (for cylinder) D = diameter = 0.03 m k = thermal conductivity = 1.7 W/m鈰匥 Bi = \( \frac{50 \cdot (\frac{0.03}{4})}{1.7} \) Calculate Bi value to check if it's less than 0.1.
02

Solve part (a) - Calculate the time required for the cylinder surface to reach 500 K

Using the lumped capacitance method, the temperature history is given by the equation: \(T(t) = T_\infty + (T_i - T_\infty)e^{-\frac{h}{\rho cL_c}t} \) where: \(T(t)\) = temperature at time t \(T_\infty\) = ambient temperature (oil bath) = 350 K \(T_i\) = initial temperature = 1000 K \(\rho\) = density = 400 kg/m鲁 c = heat capacity = 1600 J/kg鈰匥 Plug in the given values and set \(T(t)\) = 500 K to calculate the time t.
03

Solve part (b) - Compute and plot the surface temperature history for 0 鈮� t 鈮� 300 s

Using the equation from Step 2, compute the temperature history for the given time interval (0 鈮� t 鈮� 300 s) with the initial given convection coefficient (h = 50 W/m虏鈰匥). Then, repeat the process with the agitated oil convection coefficient (h = 250 W/m虏鈰匥) and compare the temperature history.
04

Analyze the influence of the convection coefficient on temperature history

With the plotted surface temperature history, observe the differences between both convection coefficients (h = 50 W/m虏鈰匥 and h = 250 W/m虏鈰匥). Describe how the agitated oil increases the convection coefficient and affects the temperature history.

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 91影视!

Key Concepts

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

Biot number
The Biot number (Bi) plays a crucial role in analyzing heat transfer in objects like cylinders. It is a dimensionless number that compares the rate of heat transfer by conduction within an object to the rate of heat transfer by convection on its surface. To calculate the Biot number, we use the formula:

Bi = \( \frac{hL_c}{k} \)
where \( h \) is the convection coefficient, \( L_c \) is the characteristic length of the object, and \( k \) is the thermal conductivity of the material.

For a cylinder, the characteristic length, \( L_c \), can often be approximated by \( \frac{D}{4} \) where \( D \) is the diameter of the cylinder. In our exercise, the diameter is 0.03 m, the convection coefficient is 50 W/m虏鈰匥, and the thermal conductivity is 1.7 W/m鈰匥. When calculating the Biot number, if it is less than 0.1, it indicates that the temperature variation inside the object is minimal during the heat transfer process. This allows us to use the lumped capacitance method for a simplified analysis of the cooling process.
Lumped capacitance method
The lumped capacitance method simplifies the analysis of transient heat transfer in objects with small Biot numbers. It assumes that the object's temperature is uniform throughout its volume, neglecting any temperature gradients within the object. This is based on the premise that heat conduction inside the object is much faster than heat transfer by convection at the surface, allowing the object's entire mass to stay at a relatively consistent temperature.

In our exercise, this method is applied to determine the time it takes for the cylinder's surface to reach 500 K after being quenched in an oil bath. By using the lumped capacitance formula:

\( T(t) = T_{\infty} + (T_i - T_{\infty})e^{-\frac{h}{\rho cL_c}t} \)

we can model the temperature at any time \( t \), where \( T_{\infty} \) is the ambient temperature, \( T_i \) is the initial temperature, \( \rho \) is the density, and \( c \) is the heat capacity. By substituting the appropriate values into the equation, we can solve for the specific time at which the surface temperature reaches the desired level.
Convection coefficient
The convection coefficient, represented by \( h \) in our equations, is a measure of the heat transfer rate per unit area and per unit temperature difference between a solid surface and the surrounding fluid. It is a key parameter in the analysis of convective heat transfer and is expressed in units of W/m虏鈰匥.

In our situation, the convection coefficient varies between two scenarios: the initial quenching in oil with \( h = 50 \) W/m虏鈰匥 and a hypothetical scenario with agitated oil, resulting in \( h = 250 \) W/m虏鈰匥. This increase in the convection coefficient due to agitation means that the oil is moving more rapidly, enhancing the heat transfer from the cylinder to the oil. As a result, if we were to compute and compare the surface temperature history with these different convection coefficients, we would expect the cylinder to cool more quickly in the agitated oil due to the higher rate of convective heat transfer.

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

A thin circular disk is subjected to induction heating from a coil, the effect of which is to provide a uniform heat generation within a ring section as shown. (a) Derive the transient, finite-difference equation for node \(m\), which is within the region subjected to induction heating. (b) On \(T r\) coordinates sketch, in qualitative manner, the steady-state temperature distribution, identifying important features.

The density and specific heat of a particular material are known \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), but its thermal conductivity is unknown. To determine the thermal conductivity, a long cylindrical specimen of diameter \(D=40 \mathrm{~mm}\) is machined, and a thermocouple is inserted through a small hole drilled along the centerline. The thermal conductivity is determined by performing an experiment in which the specimen is heated to a uniform temperature of \(T_{i}=100^{\circ} \mathrm{C}\) and then cooled by passing air at \(T_{\infty}=25^{\circ} \mathrm{C}\) in cross flow over the cylinder. For the prescribed air velocity, the convection coefficient is \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If a centerline temperature of \(T(0, t)=40^{\circ} \mathrm{C}\) is recorded after \(t=1136 \mathrm{~s}\) of cooling, verify that the material has a thermal conductivity of \(k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) For air in cross flow over the cylinder, the prescribed value of \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) corresponds to a velocity of \(V=6.8 \mathrm{~m} / \mathrm{s}\). If \(h=C V^{0.618}\), where the constant \(C\) has units of \(\mathrm{W} \cdot \mathrm{s}^{0.618} / \mathrm{m}^{2.618} \cdot \mathrm{K}\), how does the centerline temperature at \(t=1136 \mathrm{~s}\) vary with velocity for \(3 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\) ? Determine the centerline temperature histories for \(0 \leq t \leq 1500 \mathrm{~s}\) and velocities of 3,10 , and \(20 \mathrm{~m} / \mathrm{s}\).

Steel balls \(12 \mathrm{~mm}\) in diameter are annealed by heating to \(1150 \mathrm{~K}\) and then slowly cooling to \(400 \mathrm{~K}\) in an air environment for which \(T_{\infty}=325 \mathrm{~K}\) and \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the properties of the steel to be \(k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c=600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), estimate the time required for the cooling process. .

As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Waves Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Engineering Physics

Read Explanation

Rotational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation
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.