11-19-2016, 11:49 AM

Transients in wide plate with step temperature change [ TXT ]

From the author’s Engineering Collection, included in the ETSII4 module.

This program calculates the temperature T(x) in points x of an infinite flat plate with finite thickness (2L), after experiencing a thermal shock - or sudden change of ambient temperature, from T0 initial to Tf final.

The Biot number is provided indirectly, by means of the heat transfer (or film) coefficient: h= Bi.K/L. The thermal conductivity (K) and thermal diffusivity (alpha) must also be known, where: alpha = K / ro.Cp - i.e. thermal conductivity over the density and specific heat capacity.

The resulting temperature is expressed as an infinite sum as follows:

T(x,t) = Tf + 2 (T0-Tf) SUM { f(x, n) exp[-alpha.t.(Mn/L)^2 ]} ; n = 1,2,...

With: f(x,n) = sin(Mn) cos (Mn.x/L)] / [ Mn + sin(Mn) cos(Mn)

And (Mn) are the n roots of the equation defined by: tan(Mn) = Bi /(Mn)

Which in the program has been replaced by its equivalent form:

(-1)^n cos(Mn) + Mn/[sqrt(Mn)^2+Bi^2] = 0

Solved using the SLV routine, using the truncation of the tangent to its first two terms and using as initial guess:

(Mn)init = pi(n-1) + sqr{ (3/2) [ sqr (1+4h/3) – 1] }

Example.

A 20 cm thick wide plate has a uniform temperature of 1,000 dec C. It is suddenly immersed into a cooling fluid stream at 50 deg C. Calculate the temperature in its center and outer boundary one. two and three hours after the sudden step temperature change. The physical properties of the material are given below:

alpha = 1.66 E-6 m^2/s

h = 20,000 kcal/H.m^2.C = 23,260.0 W/m^2 K

K = 100 kcal/h.m.C = 116.30 W/m.K

The results are shown in the table below: (warning: very slow convergence!)

From the author’s Engineering Collection, included in the ETSII4 module.

This program calculates the temperature T(x) in points x of an infinite flat plate with finite thickness (2L), after experiencing a thermal shock - or sudden change of ambient temperature, from T0 initial to Tf final.

The Biot number is provided indirectly, by means of the heat transfer (or film) coefficient: h= Bi.K/L. The thermal conductivity (K) and thermal diffusivity (alpha) must also be known, where: alpha = K / ro.Cp - i.e. thermal conductivity over the density and specific heat capacity.

The resulting temperature is expressed as an infinite sum as follows:

T(x,t) = Tf + 2 (T0-Tf) SUM { f(x, n) exp[-alpha.t.(Mn/L)^2 ]} ; n = 1,2,...

With: f(x,n) = sin(Mn) cos (Mn.x/L)] / [ Mn + sin(Mn) cos(Mn)

And (Mn) are the n roots of the equation defined by: tan(Mn) = Bi /(Mn)

Which in the program has been replaced by its equivalent form:

(-1)^n cos(Mn) + Mn/[sqrt(Mn)^2+Bi^2] = 0

Solved using the SLV routine, using the truncation of the tangent to its first two terms and using as initial guess:

(Mn)init = pi(n-1) + sqr{ (3/2) [ sqr (1+4h/3) – 1] }

Example.

A 20 cm thick wide plate has a uniform temperature of 1,000 dec C. It is suddenly immersed into a cooling fluid stream at 50 deg C. Calculate the temperature in its center and outer boundary one. two and three hours after the sudden step temperature change. The physical properties of the material are given below:

alpha = 1.66 E-6 m^2/s

h = 20,000 kcal/H.m^2.C = 23,260.0 W/m^2 K

K = 100 kcal/h.m.C = 116.30 W/m.K

The results are shown in the table below: (warning: very slow convergence!)

Code:

`Point t = 1 hour t = 2 hours t = 3 hours`

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center (x=0 cm) 366.5400 133.0300 71.8500

Outer edge (x=0.1 m) 73.5600 56.2700 51.7100