A pressure vessel operating at 300°F is subjected to a short excursion temperature of 600°F. At a given time, the temperature distribution in the wall is shown in Fig. 5.27. Find the maximum thermal stress at that instance. Let μ = 0.3, E = 30 × 10^6 psi, and α = 6.0 × 10^-6 in./in.°F

Question

A pressure vessel operating at 300°F is subjected to a short excursion temperature of 600°F. At a given time, the temperature distribution in the wall is shown in Fig. 5.27. Find the maximum thermal stress at that instance. Let μ = 0.3, E = 30 × 10[latex]^6[/latex] psi, and α = 6.0 × 10[latex]^{-6}[/latex] in./in.°F

Accepted Answer

This problem can be visualized as a biaxial case where the inner surface heats quickly while the rest of the wall remains at 300°F. Using Eq. 5.29 b results in

[latex]\sigma_{x}=\sigma_{y}=\frac{\alpha \Delta T E}{1-\mu}[/latex] for a biaxial case (5.29b)

[latex]\begin{aligned}\sigma &=\frac{\left(6 imes 10^{-6} ight)(600-300)\left(30 imes 10^{6} ight)}{1-0.3} \&=-77,100 psi\end{aligned}[/latex]

which is extremely high and is based on very limiting assumptions. A more realistic approach is that based on Eq. 5.46. The mean temperature is obtained from Fig. 5.27 and tabulated as follows:

[latex]\sigma_{ heta}=\frac{E \alpha}{1-\mu}\left(T_{m}-T ight)[/latex] (5.46)

where [latex]T_{m}[/latex] = mean value of temperature distribution through the wall

T = temperature at desired location

Locations as Ratios of Thickness	Temperature °F	Area
0	600
0.1	460	53.0
0.2	400	43.0
0.3	370	38.5
0.4	340	35.5
0.5	320	33.0
0.6	310	31.5
0.7	305	30.8
0.8	300	30.3
0.9	300	30.0
1.0	300	30.0
		∑355.6

And [latex]T_{m} \approx 356^{\circ}F[/latex].

From Eq. 5.46, at inner surface,

[latex]\begin{aligned}\sigma &=\frac{\left(30 imes 10^{6} ight)\left(6.0 imes 10^{-6} ight)}{1-0.3}(356-600) \&=-62,700 psi\end{aligned}[/latex]

and at outer surface

[latex]\begin{aligned}\sigma &=\frac{\left(30 imes 10^{6} ight)\left(6.0 imes 10^{-6} ight)}{1-0.3}(356-300) \&=14,400 psi\end{aligned}[/latex]

It is of interest to note that the high stress occurs at the surface only. Thus at one-tenth of the thickness inside the surface, the stress is

[latex]\begin{aligned}\sigma &=\frac{\left(30 imes 10^{6} ight)\left(6 imes 10^{-6} ight)}{1-0.3}(356-460) \&=-26,700 psi\end{aligned}[/latex]

The high stress at the inner surface indicates that local yielding will occur.

Question 5.16: A pressure vessel operating at 300°F is subjected to a short...

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