A 2-in thick plate of 1060 alloy is uniformly heated to 300 °C. It is air-quenched in the forced air. Assume that the temperature of the plate surface is immediately set to 25 °C. The materials density ρ= 0.0975437 lb/in³, the conductivity k = 0.002675 Btu/(s·in·°C), and the specific heat

Question

A 2-in thick plate of 1060 alloy is uniformly heated to 300 °C. It is air-quenched in the forced air. Assume that the temperature of the plate surface is immediately set to 25 °C. The materials density ρ= 0.0975437 lb/in³, the conductivity k = 0.002675 Btu/(s · in ·°C), and the specific heat c = 0.214961 Btu/(lb ·°C). Determine the cooling curve of temperature at the neutral plane of the plate with respect to time in 5 s.

Accepted Answer

Figure 3.28 shows a 1D transient heat transfer element described in the
local coordinate system (LCS). It consists of node i and node j. The state variables at two nodes are the temperatures [latex](T_{i},T_{j})[/latex] and temperature gradient [latex](\dot{T} _{i},\dot{T} _{j})[/latex] . The elemental behavior is governed by

[latex]k\frac{∂^{2}T }{∂x^{2} }-c\frac{∂T}{∂t}+\dot{q} =0 [/latex] (3.117)

where k is conductivity, ρ is the density, and c is the specific heat.
The element model is developed based on Eq. (3.117) as follows. Let the shape functions of an element be [latex]S_{i}[/latex] (x)=1-x/L and [latex]S_{j}[/latex] (x)=x/L, using the shape functions as the test function in the Galerkin method leads to the weak-form solution Eq. (3.117) as (Bi 2018),

[latex] [K_{m} ]{\begin{Bmatrix} T_{i} \ T_{j} \end{Bmatrix}}^{(p+ 1)} =\left\{Q_{m} ight\} [/latex] (3.118)

where p is the step index and [[latex]K_{m}[/latex] ] is the modified conductive matrix as

[latex][K_{M}]=\left(\frac{ ho cAL}{6 heta \Delta t}\begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix} +\frac{kA}{L}\begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} ight) [/latex] (3.119)

[[latex]Q_{M}[/latex] ] is the modified load vector determined by the state of the precedent step (i.e., step i) as

[latex]\left\{Q_{m} ight\} =\begin{Bmatrix}Q_{i} \Q_{j} \end{Bmatrix} +\frac{ ho cAL}{6 heta \Delta t} \begin{bmatrix}2 & 1 \ 1 & 2 \end{bmatrix} {\begin{Bmatrix} T_{i} \ T_{j} \end{Bmatrix}}^{(p)}+\left(\frac{1}{ heta }-1 ight) \frac{ ho cAL}{6} \begin{bmatrix}2 & 1 \ 1 & 2 \end{bmatrix} {\begin{Bmatrix} \dot{T} _{i} \ \dot{T} _{j} \end{Bmatrix}}^{(p)}[/latex] (3.120)

After the temperature at (p + 1) have been found, the temperature gradients can be derived by .

[latex]{\begin{Bmatrix} \dot{T} _{i} \ \dot{T} _{j} \end{Bmatrix}}^{(p+1)}=\frac{1}{ heta \Delta t}\begin{Bmatrix} T^{(p+1)}_{i} -T^{(p)}_{i} \T^{(p+1)}_{j} -T^{(i)}_{j} \end{Bmatrix} -\left(\frac{1}{ heta }-1 ight) {\begin{Bmatrix} \dot{T} _{i} \ \dot{T} _{j} \end{Bmatrix}}^{(p)} [/latex] (3.121)

Since the plate is symmetric about its neutral axis, a 1D FEA model is developed for half of the plate thickness. Accordingly, Fig. 3.29 shows a decomposition which leads to an FEA model with 5 elements and 6 nodes. Every element has the length of 0.2-in.
Let the Fourier number be [latex]F_{0}[/latex] =0.5, and the step time Δt is found from Eq. (3.107) as

[latex]\Delta t=\frac{(\Delta x)^{2}F_{0} }{\alpha } \quad \quad (3.107) \ \ =\frac{(0.2)^{2}(0.5) }{({0.002675}/{(0.0975437\ast 0.214961)})} =0.1568(s)[/latex]

If the history of the temperature change in 5 s is concerned, the total number N of the time steps is

[latex]N=\frac{(5)}{\Delta t} =32(steps) [/latex]

Let the Euler parameter θ = 0.5 in Eq. (3.121) where the Crank-Nicolson difference is used in the approximation. At each time step p from 1 to N, the system model can be assembled from 5 element models expressed in Eq. (3.118) as

[latex][K]^{(G)}\begin{Bmatrix} T_{1} \ T_{2} \ T_{3} \ T_{4} \ T_{5} \ T_{6} \end{Bmatrix} ^{(p+1)} =\left\{Q ight\} ^{(p)} =\begin{Bmatrix} 2c_{1}T_{1}^{(p)} +c_{1}T_{2}^{(p)}+2c_{2}\dot{T} _{1}^{(p)} +c_{2}\dot{T} _{2}^{(p)} \ c_{1}T_{1}^{(p)} +4c_{1}T_{2}^{(p)}+ c_{2}\dot{T} _{1}^{(p)}+4c_{2}\dot{T} _{2}^{(p)}+c_{1}T_{3}^{(p)}+c_{2}\dot{T} _{3}^{(p)} \ c_{1}T_{2}^{(p)}+4c_{1}T_{3}^{(p)}+c_{2}\dot{T} _{2}^{(p)}+4c_{2}\dot{T} _{3}^{(p)}+c_{1}T_{4}^{(p)} +c_{2}\dot{T} _{4}^{(p)} \ c_{1}T_{3}^{(p)} +4c_{1}T_{4}^{(p)}+c_{2}\dot{T} _{3}^{(p)} +4c_{2}\dot{T} _{4}^{(p)} +c_{1}T_{5}^{(p)}+c_{2}\dot{T} _{5}^{(p)} \ c_{1}T_{4}^{(p)}+4c_{1}T_{5}^{(p)} +c_{2}\dot{T} _{4}^{(p)} +4c_{2}\dot{T} _{5}^{(p)}+c_{1}T_{6}^{(p)} +c_{2}\dot{T} _{6}^{(p)} \ c_{1}T_{5}^{(p)} +2c_{1}T_{6}^{(p)} +c_{2}\dot{T} _{5}^{(p)} +2c_{2}\dot{T} _{6}^{(p)} \end{Bmatrix} [/latex] (3.122)

where

[latex][K]^{(G)}=\begin{bmatrix} 0.0312 & -0.0045 & 0 & 0 & 0 & 0 \ -0.0045 & 0.0624 & -0.0045 & 0 & 0 & 0 \ 0 & -0.0045 & 0.0624 & -0.0045 & 0 & 0 \0 & 0 & -0.0045 & 0.0624 &-0.0045 & 0 \ 0 & 0 & 0 & -0.0045 & 0.0624 & -0.0045 \ 0 & 0 & 0 & 0 & -0.0045 & 0.0312 \end{bmatrix} [/latex] (3.123)

[latex]c_{1}[/latex] and [latex]c_{2}[/latex] are two constants as below

[latex]\left.\begin{matrix}c_{1}=\frac{ ho cAL}{6 heta \Delta t}=0.0089 \\\ c_{2}=\left(1-\frac{1}{ heta } ight) \frac{ ho cAL}{6}= -0.00069893 \end{matrix} ight\} (3.124)[/latex]

At each step p, after [latex]\left\{T ight\} ^{(p+1)} [/latex] is found from Eq. (3.118), the temperature gradients on nodes can be updated based on Eq. (3.121) as

[latex]\begin{Bmatrix} \dot{T} _{1} \ \dot{T} _{2} \ \dot{T} _{3} \\dot{T} _{4} \\dot{T} _{5}\\dot{T} _{6} \end{Bmatrix} ^{(p+1)}=\begin{Bmatrix} \frac{1}{ heta \Delta t}\left(T^{(p+1)}_{1}-T^{(p)}_{1} ight)-\left(\frac{1}{ heta }-1 ight)\dot{T} ^{(p)}_{1} \ \frac{1}{ heta \Delta t}\left(T^{(p+1)}_{2}-T^{(p)}_{2} ight)-\left(\frac{1}{ heta }-1 ight)\dot{T} ^{(p)}_{2} \ \frac{1}{ heta \Delta t}\left(T^{(p+1)}_{3}-T^{(p)}_{3} ight)-\left(\frac{1}{ heta }-1 ight)\dot{T} ^{(p)}_{3}\ \frac{1}{ heta \Delta t}\left(T^{(p+1)}_{4}-T^{(p)}_{4} ight)-\left(\frac{1}{ heta }-1 ight)\dot{T} ^{(p)}_{4} \ \frac{1}{ heta \Delta t}\left(T^{(p+1)}_{5}-T^{(p)}_{5} ight)-\left(\frac{1}{ heta }-1 ight)\dot{T} ^{(p)}_{5} \ \frac{1}{ heta \Delta t}\left(25-T^{(p+1)}_{6} ight)-\left(\frac{1}{ heta }-1 ight)\dot{T} ^{(p)}_{6} \end{Bmatrix} [/latex] (3.125)

Note that the calculation for the temperature gradient of the boundary node 6 is different since the temperature at this node is fixed as 25 °C. Accordingly, when the calculation result at step p is moved to next step p + 1, the temperature at node 6 must be set back to 25 °C as the boundary constraint. The temperature changes on nodes with respect to time are obtained in Table. 3.9.

The plots of temperature on node 1 (i.e., the center of plate) are illustrated in Figs. 3.30 and 3.31 respectively.

Table 3.9 The History of nodal temperatures in 5 s

Question 3.8: A 2-in thick plate of 1060 alloy is uniformly heated to 300 ...

Related Answered Questions

A manufacturer produces an integrated ventilating system for customers, and Fig. 3.44 shows one of its products with two blowers. In mounting the blowers, a misalignment of the transmission plane of a blower will cause an excitation to the system. Conduct a modal analysis to confirm that two ...

Verified Answer:

Verified Answer:

Verified Answer:

Use the iterative method to solve [K]{U} = {F} where. [K]=[2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2] and {F}={5 2 2 2 5} ...

Verified Answer:

Figure 3.27a shows a FEA model for a 2D heat transfer problem The model is to find the temperature distribution, and the FEA model consists of 10 nodes for three triangular elements and three rectangle elements. Build a system model based on the given element models as follows: The stiffness ...

Verified Answer:

Analyze the fluid flow around an example racecar model shown in Fig. 3.56a and estimate the drag force when the racecar run at the speed of 90 m/s (~200 miles per hour). ...

Verified Answer:

The par geometry is shown in Fig. 3.54a; it is made of 6061-T6 aluminum alloy, the initial temperature of the heat treatment process is T0 = 532 °C. The ambient temperature is Ta = 30 °C and the coefficient of heat convection is h = 60 W/(m²·K), find out the cycle time of the heat treatment ...

Verified Answer:

Find the first two natural frequencies of the following system model with the given accuracy ε of 0.001: [M]{ü} + [K]{u} = 0 (3.105) where [M] =[20 5 0 5 10 5 0 5 10] , [K]=[2 -1 0 -1 2 -1 0 -1 1] ...

Verified Answer:

Find a weak solution to the following boundary value problem: d²u/dx²=6x-sin(x), 0≤x≤1 subjected to: u(x)|x=1=sin(1) du/dx|x=0 }(3.28) ...

Verified Answer:

Use the principle of minimum potential energy to find the displacement at the free end in Fig. 3.17. ...

Verified Answer: