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Calculation of the temperature field of a flat radiation heat exchanger

Determination of heat exchanger parameters, calculation scheme, initial and boundary conditions

This example will demonstrate the solution of the heat conduction equation using the Euler method.

Task: Calculate the temperature field for a plane radiation heat exchanger.

Given:

a flat radiation heat exchanger of length 20 cm, thickness 1.25 cm.

Starting conditions:

vacuum,

integral absorption capacity of the body $\xi=0,9$,

spectral absorptivity of a body $a=0,14$,

temperature at each point of the heat exchanger at the initial moment of time ${T_{0,j}} = 293K$.

Boundary conditions:

temperature of the heat transfer fluid ${T_ж} = 293K$.

Assumptions:

heat transfer along the fluid flow is not taken into account, therefore the width of the heat exchanger is taken as 1 cm,

radiation of only one side of the heat exchanger is taken into account, as the second side absorbs the heat re-reflected from the cooled spacecraft.

The general view of the heat exchanger is shown in Fig:

heatex1.png

The classical heat conduction equation was modified to take into account the radiant heat transfer with the environment, which is described by the Stefan-Boltzmann law.

Applying the Euler method to this equation, we obtained:

$${T_{i+1,j}}={T_{i,j}}+h\frac{1}{cm}(\lambda F \frac{{T_{i,j-1}}-{T_{i,j}}}{d}+\frac{{T_{i,j+1}}-{T_{i,j}}}{d}-\xi \sigma{F_{izl}}{T^4_{i,j}}+{a_s}{q_s}{F_m})$$

where:

$c=903.7$ (J/kg*K) is the heat capacity of aluminium,

$m=0.003375$ kg - mass of the heat exchanger section to be calculated,

$\lambda=235.9$ W/(m*K) - heat conductivity coefficient,

$F=0.00025$ m2 - area of the calculated section,

$d=0.01$ m - length of the heat exchanger section to be calculated,

$\xi=0.90$ - integral absorption capacity of the body,

$\sigma=0.000000056704$ W/(m2*K4);

${F_m}=Fsin(67\pi/180)$ - midsection area of the calculated area (projection of the area of the section on the perpendicular to the direction of solar radiation)

${a_s}=0.14$ - spectral absorption capacity of the body,

${q_s}=1400$ W/m2 - intensity of solar radiation,

$h=0.001$ s - calculation step.

The location of the calculation segments is shown in the figure:

heatex2.png

Determination of initial conditions, heat exchanger and environmental parameters

Determination of heat exchanger and ambient parameters:

In [ ]: 
# @markdown #### Выбор материала теплообменника с определённой теплоёмкостью (по порядку: алюминий, железо, золото)
c = "903.7" # @param ["903.7", "450.0", "130.0"]
c = parse(Float64, c)
m = 0.003375
A = 235.9
F = 0.00025
d = 0.01
E = 0.90
S = 0.000000056704
Fm = F*sin(67*pi/180)
L = 0.14
# @markdown #### Определение теплового потока от солнца (солнечная сторона/тень)
q = "1400.0" # @param ["1400.0", "0.0"]
q = parse(Float64, q)
# @markdown #### Определение шага расчёта по времени
h = 0.1 # @param {type:"number"}
Out[0]:
0.1

Calculation and data visualisation

Plotting the time solution graph for segments 1 to 10 (from the left edge to the middle of the heat exchanger):

In [ ]: 
using Plots
# Определение начальных условий
T0 = 293.15
T = fill(293.0, (1080,20)); #определение массива для температуры
Q3 = fill(0.094, (1080,20)); #определение массива для теплового потока, излучаемого теплообменником
Q4 = fill(0.0451047, (1080,20)); #определение массива для теплового потока, излучаемого Солнцем

# Расчёт температур и тепловых потоков в каждый момент времени 
# для всех отрезков теплообменника:
for i in 1:1079 #цикл для расчёта по времени  
    # @markdown #### Температура теплоносителя слева:
    T[i,1] = 290.06 # @param {type:"slider", min:250, max:350, step:0.01}
    # @markdown #### Температура теплоносителя справа:
    T[i,20] = 293.13  # @param {type:"slider", min:250, max:350, step:0.01}
        for j in 2:19 #цикл для расчёта по координате
            T[1,j] = T0 #определение начального условия
            T[i+1,j] = T[i,j] + h*(
                (1/(c*m))*
                (A*F*(T[i,j-1]-T[i,j])/d #тепловой поток, передаваемый от левого отрезка к текущему с помощью теплопроводности
                 + A*F*(T[i,j+1]-T[i,j])/d #тепловой поток, передаваемый от правого отрезка к текущему с помощью теплопроводности
                 - E*S*T[i,j]*T[i,j]*T[i,j]*T[i,j]*F #тепловой поток, излучаемый текущим отрезком теплообменника
                 + L*q*Fm) #тепловой поток от Солнца, поглощаемый текущим отрезком теплообменника
                 )
            Q3[i,j] = E*S*T[i,j]*T[i,j]*T[i,j]*T[i,j]*F #заполнение массива для излучаемого теплового потока
            Q4[i,j] = L*q*Fm #заполнение массива для поглощаемого теплового потока
        end
end
T[end,1] = T[end-1,1]
Legend = ["1 отрезок" "2 отрезок" "3 отрезок" "4 отрезок" "5 отрезок" "6 отрезок" "7 отрезок" "8 отрезок" "9 отрезок" "10 отрезок"]
# построение графиков для отрезков с 11 по 20 не имеет смысла, так как они геометрически симметричны с отрезками с 1 по 10
Plots.plot(T[:,1:10], xlabel="Время", ylabel="Температура", label=Legend) # , ylims=(lim_low,lim_high)
Out[0]:

Conclusions:

In this example, the calculation of the temperature field of a planar heat exchanger has been demonstrated. From the calculation of the ratio of radiated and absorbed heat fluxes, it can be seen that the heat exchanger actually absorbs heat brought by the heat transfer medium and radiates it to the environment. In turn, depending on the coordinate, the temperature values at each moment of time were obtained. Thus, the problem of calculating the temperature field in a flat heat exchanger was solved.