Lumped Capacitance Method for a conductive wall

[Jiec]

Homework Statement

Built the physical model (considering lumped capacitance mthod) of the three-layer wall, using one node for each layer placed at the centre of the latter, one node for the inner surface and one node for the outer surface.
The layers have conductivity k_i (for i=1:3) and thickness l_i (for i=1:3) and cross-sectional area A, only layer number 2 has a heat capacitance C2. The temperature of the inner and outer surface are assigned.
Then write the equations of the mathematical model.

Relevant Equations

R=l_i/(k_i*A)
Q=(T_i-T_j)/R
dQ=C_i*(dT/dt)

Hello,

I have some doubt on the representation of the physical model. I'm not sure about the number and value of the capacitance to be used.
I solved the exercise using this model (see figure) and i would like to know if the solution is correct or if there is something to fix.

Regarding the solution:
- R_a= R1/2
- R2_a=R1/2+R2/2
- R2_b=R2/2+R3/2
- R_b=R3/2
where R1, R2, R3 are the thermal resistance of each layer

- C2=l_2*A*c2
where c2 is hte specific heat of the second layer

For the mathematical model the equations are the following:
- (Ti-T1)/R_a-(T1-T2)/R2_a=0
- ( (T1-T2)/R2_a-(T2-T3)/R2_b )=C2*dT2/dt where dT2/dt is the derivative of T2 with respect to the time
- (T2-T3)/R2_b-(T3-Te)/R_b=0

 

[erobz]

Homework Statement: Built the physical model (considering lumped capacitance mthod) of the three-layer wall, using one node for each layer placed at the centre of the latter, one node for the inner surface and one node for the outer surface.
The layers have conductivity k_i (for i=1:3) and thickness l_i (for i=1:3) and cross-sectional area A, only layer number 2 has a heat capacitance C2. The temperature of the inner and outer surface are assigned.
Then write the equations of the mathematical model.
Relevant Equations: R=l_i/(k_i*A)
Q=(T_i-T_j)/R
dQ=C_i*(dT/dt)

Hello,

I have some doubt on the representation of the physical model. I'm not sure about the number and value of the capacitance to be used.
I solved the exercise using this model (see figure) and i would like to know if the solution is correct or if there is something to fix.

Regarding the solution:
- R_a= R1/2
- R2_a=R1/2+R2/2
- R2_b=R2/2+R3/2
- R_b=R3/2
where R1, R2, R3 are the thermal resistance of each layer

- C2=l_2*A*c2
where c2 is hte specific heat of the second layer

For the mathematical model the equations are the following:
- (Ti-T1)/R_a-(T1-T2)/R2_a=0
- ( (T1-T2)/R2_a-(T2-T3)/R2_b )=C2*dT2/dt where dT2/dt is the derivative of T2 with respect to the time
- (T2-T3)/R2_b-(T3-Te)/R_b=0


View attachment 349110

Are you trying to find the temperature of mass 2 as a function of time? Is the system initially at a uniform temperature? Please fully describe the initial conditions of this wall.

Probably more questions necessary to further tease out your intent.

 

[Jiec]

Are you trying to find the temperature of mass 2 as a function of time? Is the system initially at a uniform temperature? Please fully describe the initial conditions of this wall.

Probably more questions necessary to further tease out your intent.

I need to find the temperature T1, T2, T3 as a function of time.
At initial time the system is at a uniform temperature (say 300K); the inner surface temperature rises to 900K in the first seconds (2 seconds) and then remains constant; the outer surface temperature remains constant (300K) all the time.

I am mostly interested in whether the model with resistors and capacitors is correct :)

 

[erobz]

I need to find the temperature T1, T2, T3 as a function of time.
At initial time the system is at a uniform temperature (say 300K); the inner surface temperature rises to 900K in the first seconds (2 seconds) and then remains constant; the outer surface temperature remains constant (300K) all the time.

I am mostly interested in whether the model with resistors and capacitors is correct :)

What do you end with for a solution for $T$ (the temperature of the chunk within the middle that can store heat)?

$$ M c_p \frac{dT}{dt} = \frac{k_1A ( T_i - T)}{\Delta L_1} - \frac{k_2A ( T - T_e)}{\Delta L_2}$$

Once you have $T(t)$, you find the temp of the flanking walls in space and time, by noting that in the walls surrounding $M$ the temperature gradients are linear in space. So over time the temp profile moves from the solid toward the dashed (etc...) in this simplified model.

 

[erobz]

Also: moving forward please see LaTeX Guide for how to properly format mathematics on the site. It doesn't take long to learn.

 

[Jiec]

What do you end with for a solution for $T$ (the temperature of the chunk within the middle that can store heat)?

$T_{1}$, $T_{2}$, $T_{3}$ are the temperatures of the section placed in the middle of layers 1, 2, 3.
My doubt is due to the fact that using the nodes shown in the figure the thermal resistances $R_{2a}$ and $R_{2b}$ consider the resistance of two layers and not just the layer 2. So I'm not sure if using one capacitance on the node 2 is correct or it is better to use two/three capacitances as in the following figures.

 

[erobz]

$T_{1}$, $T_{2}$, $T_{3}$ are the temperatures of the section placed in the middle of layers 1, 2, 3.
My doubt is due to the fact that using the nodes shown in the figure the thermal resistances $R_{2a}$ and $R_{2b}$ consider the resistance of two layers and not just the layer 2. So I'm not sure if using one capacitance on the node 2 is correct or it is better to use two/three capacitances as in the following figures.

View attachment 349160View attachment 349161

I don't know what the big deal is about using the circuit analogy. If the flanking walls have no mass they aren't accumulation their own internal energy, they are just passing it through the system.

Solve the ODE in post #4 to get the solution ( subject to boundary conditions $T_i,T_e = \text{const.}$ and $T_o$ the initial temp of the sandwiched mass $M$ ) The diagram at the bottom of #4 is the temperature profile throughout the system at a particular instant in time. Solving the ODE tells you how it changes as time progresses

 

[Jiec]

I don't know what the big deal is about using the circuit analogy. If the flanking walls have no mass they aren't accumulation their own internal energy, they are just passing it through the system.

Solve the ODE in post #4 to get the solution ( subject to $T_i,T_e = \text{const.}$) The diagram at the bottom of #4 is the temperature profile throughout the system at a particular instant in time.

I used the analogy because i need also to create a model in simscape using elements such as thermal mass, thermal resistance, conductive heat transfer etc...
I'm not sure where to put the thermal masses in that model.

 

[erobz]

I used the analogy because i need also to create a model in simscape using elements such as thermal mass, thermal resistance, conductive heat transfer etc...
I'm not sure where to put the thermal masses in that model.

walls 1 and 3 have no thermal mass. They have no specific heat.

 

[erobz]

If you are trying more advance calculations maybe @Chestermiller will help.

 

[Jiec]

walls 1 and 3 have no thermal mass. They have no specific heat.

Ok, this was what I thought :)
Probably i'm mixing up with a model that uses 3 capacitance when there is convective heat transfer on the external surfaces (but this is not the case)...do you know if there is a more refined model using always the same nodes?

 

[Chestermiller]

If the capacitances on the left and right are zero, then this is the same as convective heat transfer on the left and right surfaces of region 2. So, at the left surface, $$-k_2\frac{\partial T}{\partial x}= h_L(900-T)$$and on the right surface, $$-k_2\frac{\partial T}{\partial x}=h_R(T-300)$$

 

[Jiec]

If the capacitances on the left and right are zero, then this is the same as convective heat transfer on the left and right surfaces of region 2. So, at the left surface, $$-k_2\frac{\partial T}{\partial x}= h_L(900-T)$$and on the right surface, $$-k_2\frac{\partial T}{\partial x}=h_R(T-300)$$

With $h_R$ and $h_L$ do you mean convective heat coefficients? Because I don't have them; I think that the convective coefficients are high and so the inner and outer surface act as an ideal temperature source

 

[Chestermiller]

With $h_R$ and $h_L$ do you mean convective heat coefficients? Because I don't have them; I think that the convective coefficients are high and so the inner and outer surface act as an ideal temperature source

Convective is mathematically the same as resistive without capacitance.

 

[Jiec]

Sorry for not replying, i was a little busy and I forgot
Thanks @Chestermiller and @erobz for the help