non-numerical answer for solving complex non-linear equation

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uzzi 2023-5-19

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评论： Star Strider 2023-5-19

Hello, I am trying to sollve for T_TP according to the following equation. I used "solve" but as for the answer, I am not getting numerical value and I am getting the answer with the variables. Can someone help me?

d_T = 0.0001; % diameter of thermocouple junction (m)

L = 4; % length of thermocouple (m)

d_TP = 0.0005; % outer diameter of thermocouple sheath (m)

emi = 0.20; % emissivity of thermocouple

k = 30; % heat conduction of thermocouple (W/mK)

T_env = 25+273.15; % environment temperature (°C)

T_fg = 1190+273.15; % flue gas temperature (°C)

T_wall = 1350+273.15; % wall temperature (°C)

alpha = 8; % heat transfer coefficient of flue gas (W/m^2K)

Sigma=5.67e-8;

syms d_T L d_TP emi k T_env T_fg T_wall alpha Sigma T_TP

eq1 = ((pi*d_TP^2)/4) * emi* Sigma* (T_fg^4 - T_TP^4);

eq2 = ((pi*d_TP^2)/4) * alpha* (T_fg - T_TP);

eq3 = ((pi*d_TP^2)/4) * emi* Sigma* (T_TP^4 - T_wall^4);

eq4 = ((pi*d_T^2)/4) * (k/L) * (T_TP - T_env);

sol = solve([eq1+eq2 == eq3+eq4],[T_TP])

sol =

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Answer 1

Star Strider 2023-5-19

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编辑：Star Strider 2023-5-19

在 MATLAB Online 中打开

You define all the variables numerically, however your code then declares them as symbolic.

The easy solution is simply to put the syms call before the numerical assignments —

syms d_T L d_TP emi k T_env T_fg T_wall alpha Sigma T_TP

d_T = 0.0001; % diameter of thermocouple junction (m)

L = 4; % length of thermocouple (m)

d_TP = 0.0005; % outer diameter of thermocouple sheath (m)

emi = 0.20; % emissivity of thermocouple

k = 30; % heat conduction of thermocouple (W/mK)

T_env = 25+273.15; % environment temperature (°C)

T_fg = 1190+273.15; % flue gas temperature (°C)

T_wall = 1350+273.15; % wall temperature (°C)

alpha = 8; % heat transfer coefficient of flue gas (W/m^2K)

Sigma=5.67e-8;

% syms d_T L d_TP emi k T_env T_fg T_wall alpha Sigma T_TP

eq1 = ((pi*d_TP^2)/4) * emi* Sigma* (T_fg^4 - T_TP^4);

eq2 = ((pi*d_TP^2)/4) * alpha* (T_fg - T_TP);

eq3 = ((pi*d_TP^2)/4) * emi* Sigma* (T_TP^4 - T_wall^4);

eq4 = ((pi*d_T^2)/4) * (k/L) * (T_TP - T_env);

sol = solve([eq1+eq2 == eq3+eq4],[T_TP])

sol =

vpasol = vpa(sol)

vpasol =

format longE

doublesol = double(sol)

doublesol =

1.546247219949594e+03 + 0.000000000000000e+00i 3.650226045888263e+01 - 1.583591092350079e+03i 3.650226045888263e+01 + 1.583591092350079e+03i -1.619251740867359e+03 + 0.000000000000000e+00i

EDIT — Corrected typographical errors.

.

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John D'Errico 2023-5-19

编辑：John D'Errico 2023-5-19

@uzzi

Sadly, what you don't understand is that while it did give you numerical ansers, you do not understand what you did, or what was happening.

The pre-definition of those variables as syms had ABSOLUTELY NO IMPACT on the result. It did not create symbolic variables that were then used. In fact, it just wasted CPU cycles.

Star Strider 2023-5-19

@uzzi — As always, my pleasure!

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Answer 2

John D'Errico 2023-5-19

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编辑：John D'Errico 2023-5-19

在 MATLAB Online 中打开

I think you do not understand how variables work in MATLAB. Variables are dynamically assigned. When you assign them as you have done, the last assignment takes precendent, completely overwriting the previous variable.

For example,

a = 1;
whos a
  Name      Size            Bytes  Class     Attributes

  a         1x1                 8  double              

So a is a double. It has the value 1.

But now, if I use syms to define a, what happens?

syms a
whos a
  Name      Size            Bytes  Class    Attributes

  a         1x1                 8  sym                

a is now a symbolic variable. It no longer has the value 1 anymore. a is just a, an unknown variable. We overwrote the original double precision variable with a new symbolic variable. What we did not do was convert the value a into a symbolic variable that contains the number 2.

Similarly, if you create a variable b, using syms

syms b
whos b
  Name      Size            Bytes  Class    Attributes

  b         1x1                 8  sym                

But now assign b to have a value like this:

b = 2;
whos b
  Name      Size            Bytes  Class     Attributes

  b         1x1                 8  double              

So b is no longer symbolic! I have replaced it with the number 2, and b is now a double.

Let me try one more thing.

c = sym(2)
c = 2
whos c
  Name      Size            Bytes  Class    Attributes

  c         1x1                 8  sym                

Do you see that sym allows me to explicitly assign the value 2 to a symbolic variable, named c? We can work with it, and MATLAB has stored 2 in the variable c, but in symbolic form.

c + 1
ans = 3

Now, what did you do? You made the first mistake I show above.

Can you take a double precision variable, and convert it into a symbolic one, that contains that number?

d = 4 % d is a double
d = 4

There would be multiple ways to do this operation. You might use cast.

d = cast(d,'sym')
d = 4
whos d
  Name      Size            Bytes  Class    Attributes

  d         1x1                 8  sym                

Another way to do it is to assign using an index. For example:

syms e
whos e
  Name      Size            Bytes  Class    Attributes

  e         1x1                 8  sym                
e(1) = 5
e = 5
whos e
  Name      Size            Bytes  Class    Attributes

  e         1x1                 8  sym                

Do you see the difference? By assignming one element of e to be the number 5, MATLAB automatically performs the cast of that value into the variable e. Even though e is a scalar, using the index makes it work, as if I had used cast or sym directly.