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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.

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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.

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Most popular questions from this chapter

5.53 Stone mix concrete slabs are used to absorb thermal energy from flowing air that is carried from a large concentrating solar collector. The slabs are heated during the day and release their heat to cooler air at night. If the daytime airflow is characterized by a temperature and convection heat transfer coefficient of \(T_{\infty}=200^{\circ} \mathrm{C}\) and \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the slab thickness \(2 L\) required to transfer a total amount of energy such that \(Q / Q_{o}=0.90\) over a \(t=8\)-h period. The initial concrete temperature is \(T_{i}=40^{\circ} \mathrm{C}\).

A thick steel slab \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, c=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is initially at \(300^{\circ} \mathrm{C}\) and is cooled by water jets impinging on one of its surfaces. The temperature of the water is \(25^{\circ} \mathrm{C}\), and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?

A sphere \(30 \mathrm{~mm}\) in diameter initially at \(800 \mathrm{~K}\) is quenched in a large bath having a constant temperature of \(320 \mathrm{~K}\) with a convection heat transfer coefficient of \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the sphere material are: \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Show, in a qualitative manner on \(T \dashv\) coordinates, the temperatures at the center and at the surface of the sphere as a function of time. (b) Calculate the time required for the surface of the sphere to reach \(415 \mathrm{~K}\). (c) Determine the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the outer surface of the sphere at the time determined in part (b). (d) Determine the energy (J) that has been lost by the sphere during the process of cooling to the surface temperature of \(415 \mathrm{~K}\). (e) At the time determined by part (b), the sphere is quickly removed from the bath and covered with perfect insulation, such that there is no heat loss from the surface of the sphere. What will be the temperature of the sphere after a long period of time has elapsed? (f) Compute and plot the center and surface temperature histories over the period \(0 \leq t \leq 150 \mathrm{~s}\). What effect does an increase in the convection coefficient to \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) have on the foregoing temperature histories? For \(h=75\) and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), compute and plot the surface heat flux as a function of time for \(0 \leq t \leq 150 \mathrm{~s}\).

One end of a stainless steel (AISI 316) rod of diameter \(10 \mathrm{~mm}\) and length \(0.16 \mathrm{~m}\) is inserted into a fixture maintained at \(200^{\circ} \mathrm{C}\). The rod, covered with an insulating sleeve, reaches a uniform temperature throughout its length. When the sleeve is removed, the rod is subjected to ambient air at \(25^{\circ} \mathrm{C}\) such that the convection heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Using the explicit finite-difference technique with a space increment of \(\Delta x=0.016 \mathrm{~m}\), estimate the time required for the midlength of the rod to reach \(100^{\circ} \mathrm{C}\). (b) With \(\Delta x=0.016 \mathrm{~m}\) and \(\Delta t=10 \mathrm{~s}\), compute \(T(x, t)\) for \(0 \leq t \leq t_{1}\), where \(t_{1}\) is the time required for the midlength of the rod to reach \(50^{\circ} \mathrm{C}\). Plot the temperature distribution for \(t=0,200 \mathrm{~s}, 400 \mathrm{~s}\), and \(t_{1}\).

Standards for firewalls may be based on their thermal response to a prescribed radiant heat flux. Consider a \(0.25\)-m-thick concrete wall \(\left(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is at an initial temperature of \(T_{i}=25^{\circ} \mathrm{C}\) and irradiated at one surface by lamps that provide a uniform heat flux of \(q_{s}^{\prime \prime}=10^{4} \mathrm{~W} / \mathrm{m}^{2}\). The absorptivity of the surface to the irradiation is \(\alpha_{s}=1.0\). If building code requirements dictate that the temperatures of the irradiated and back surfaces must not exceed \(325^{\circ} \mathrm{C}\) and \(25^{\circ} \mathrm{C}\), respectively, after \(30 \mathrm{~min}\) of heating, will the requirements be met?
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