Your MATLAB symbolic code attempts to solve the coupled ODE system representing heat transfer between two blocks exchanging heat through their contact surface only. The model and equations are:dT1dt=−UA(T2−T1)\frac{dT_1}{dt} = -U A (T_2 - T_1)dtdT1=−UA(T2−T1)dT2dt=UA(T2−T1)\frac{dT_2}{dt} = U A (T_2 - T_1)dtdT2=UA(T2−T1)
where UA=hA(1ρcVccpc+1ρhsVhscphs)U A = h A \left( \frac{1}{\rho_c V_c c_{p_c}} + \frac{1}{\rho_{hs} V_{hs} c_{p_{hs}}} \right)UA=hA(ρcVccpc1+ρhsVhscphs1) but you've defined coefficients separately and built matrix AAA accordingly.
Verification of your ODE system and solution approach:
- Physical equations as given:
dT1dt=−UA(T2−T1)dT2dt=+UA(T2−T1)\frac{dT_1}{dt} = - U A (T_2 - T_1) \\ \frac{dT_2}{dt} = + U A (T_2 - T_1)dtdT1=−UA(T2−T1)dtdT2=+UA(T2−T1)
- Your matrix A is:
A=[a−a−bb]A = \begin{bmatrix} a & -a \\ -b & b \end{bmatrix}A=[a−b−ab]
where
a=hAρcVccpc,b=hAρhsVhscphsa = \frac{hA}{\rho_c V_c c_{p_c}}, \quad b = \frac{hA}{\rho_{hs} V_{hs} c_{p_{hs}}}a=ρcVccpchA,b=ρhsVhscphshA
and
ddt[T1T2]=−U⋅A[T1T2]\frac{d}{dt} \begin{bmatrix} T_1 \\ T_2 \end{bmatrix} = - U \cdot A \begin{bmatrix} T_1 \\ T_2 \end{bmatrix}dtd[T1T2]=−U⋅A[T1T2]
But this does not match your original scalar equations, because those involve the difference (T2−T1)(T_2 - T_1)(T2−T1) and the matrix AAA you constructed doesn’t reflect that form directly.
Re-deriving the matrix system consistent with scalar ODEs:
From the scalar equations:
dT1dt=−UA(T2−T1)=UA(T1−T2)\frac{dT_1}{dt} = - U A (T_2 - T_1) = U A (T_1 - T_2)dtdT1=−UA(T2−T1)=UA(T1−T2)dT2dt=UA(T2−T1)\frac{dT_2}{dt} = U A (T_2 - T_1)dtdT2=UA(T2−T1)
If you write in vector form T=[T1;T2]\mathbf{T} = [T_1; T_2]T=[T1;T2]:
dTdt=UA[−111−1]T\frac{d\mathbf{T}}{dt} = U A \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \mathbf{T}dtdT=UA[−111−1]T
where
A=hA(1ρcVccpc+1ρhsVhscphs)A = h A \left( \frac{1}{\rho_c V_c c_{p_c}} + \frac{1}{\rho_{hs} V_{hs} c_{p_{hs}}} \right)A=hA(ρcVccpc1+ρhsVhscphs1)
Alternatively, as two separate constants:
a=hAρcVccpc,b=hAρhsVhscphsa = \frac{h A}{\rho_c V_c c_{p_c}}, \quad b = \frac{h A}{\rho_{hs} V_{hs} c_{p_{hs}}}a=ρcVccpchA,b=ρhsVhscphshA
then
dT1dt=−a(T2−T1)\frac{dT_1}{dt} = - a (T_2 - T_1)dtdT1=−a(T2−T1)dT2dt=b(T2−T1)\frac{dT_2}{dt} = b (T_2 - T_1)dtdT2=b(T2−T1)
Note the different denominators imply a≠ba \neq ba=b, so the system is:
ddt[T1T2]=[a−a−bb][T1T2]\frac{d}{dt} \begin{bmatrix} T_1 \\ T_2 \end{bmatrix} = \begin{bmatrix} a & -a \\ -b & b \end{bmatrix} \begin{bmatrix} T_1 \\ T_2 \end{bmatrix}dtd[T1T2]=[a−b−ab][T1T2]
No negative sign out front here. So the ODE system is:
dTdt=AT\frac{d\mathbf{T}}{dt} = A \mathbf{T}dtdT=AT
