Chapter 7, Problem 7.55P

To determine

(a)

The model for the behavior of temperature $T_1$ and $T_2$.

Answer to Problem 7.55P

The model for the behavior of temperature $T_1$ is $\frac{d{T}_1}{dt}=\frac{1}{C_1}\left[\frac{1}{R_2}\left(T_2-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\right]$.

The model for the behavior of temperature $T_2$ is $\frac{d{T}_2}{dt}=-\frac{1}{C_2}\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]$.

Explanation of Solution

Calculation:

Write the expression for the thermal capacitance of the first mass.

$\begin{array}{l}{C}_1=\frac{d{q}_1}{d{T}_1}\\ d{q}_1=C_{1}d{T}_1\end{array}$

Here, the thermal capacitance is of the first mass is $C_1$, the small quantity of heat added to the first mass is $d{q}_1$, and the small change in the temperature of the first mass is $d{T}_1$.

Divide both side by $dt$ in the above expression.

$\frac{d{q}_1}{dt}=C_1\frac{d{T}_1}{dt}$

Write the expression for the thermal capacitance of the second mass.

$\begin{array}{l}{C}_2=\frac{d{q}_2}{d{T}_2}\\ d{q}_2=C_{2}d{T}_2\end{array}$

Here, the thermal capacitance is of the first mass is $C_2$, the small quantity of heat added to the second mass is $d{q}_2$, and the small change in the temperature of the second mass is $d{T}_2$.

Divide both sides by $dt$.

$\frac{d{q}_2}{dt}=C_2\frac{d{T}_2}{dt}$

Write the expression for the conservation of heat of the first mass.

$\frac{d{q}_1}{dt}=q_2-q_1$.... (I)

Here, the heat inflow rate to the first mass from the second mass is $q_2$ and the heat outflow rate from the first mass is $q_1$.

Since the second mass is insulated from the three sides, so there is only heat transfer takes place between the first mass and second mass.

Write the expression for the conservation of heat of the first mass.

$\frac{d{q}_2}{dt}=-q_2$.... (II)

Substitute $C_1\frac{d{T}_1}{dt}$ for $\frac{d{q}_1}{dt}$ in Equation (I).

$C_1\frac{d{T}_1}{dt}=q_2-q_1$.... (III)

Substitute $C_2\frac{d{T}_2}{dt}$ for $\frac{d{q}_2}{dt}$ in Equation (II).

$C_2\frac{d{T}_2}{dt}=-q_2$.... (IV)

Write the expression for the convective heat transfer rate from the first mass to the air.

$q_1=h_{1}A\left(T_1-T_0\right)$.... (V)

Here, the convective heat transfer coefficient is $h_1$ and the area of the wall boundary is $A$.

Write the expression for the convective heat transfer rate from the second mass to the first mass.

$q_2=h_{2}A\left(T_2-T_1\right)$.... (VI)

Here, the convective heat transfer coefficient is $h_2$.

Write the expression for the thermal resistance of the heating element.

$\begin{array}{c}{R}_1=\frac{1}{h_{1}A}\\ h_{1}A=\frac{1}{R_1}\end{array}$

Write the expression for the thermal resistance for the surrounding heat transfer.

$\begin{array}{c}{R}_2=\frac{1}{h_{2}A}\\ h_{2}A=\frac{1}{R_2}\end{array}$

Substitute $\frac{1}{R_1}$ for $h_{1}A$ in Equation (V).

$q_1=\frac{1}{R_1}\left(T_1-T_0\right)$

Substitute $\frac{1}{R_2}$ for $h_{2}A$ in Equation (VI).

$q_2=\frac{1}{R_2}\left(T_2-T_1\right)$

Substitute $\frac{1}{R_1}\left(T_1-T_0\right)$ for $q_1$ and $\frac{1}{R_2}\left(T_2-T_1\right)$ for $q_2$ in Equation (III).

$\begin{array}{l}{C}_1\frac{d{T}_1}{dt}=\frac{1}{R_2}\left(T_2-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\\ \frac{d{T}_1}{dt}=\frac{1}{C_1}\left[\frac{1}{R_2}\left(T_2-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\right]\end{array}$

Thus, the model for the behavior of temperature $T_1$ is $\frac{d{T}_1}{dt}=\frac{1}{C_1}\left[\frac{1}{R_2}\left(T_2-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\right]$.

Substitute $\frac{1}{R_2}\left(T_2-T_1\right)$ for $q_2$ in Equation (IV).

$\begin{array}{l}{C}_2\frac{d{T}_2}{dt}=-\frac{1}{R_2}\left(T_2-T_1\right)\\ \frac{d{T}_2}{dt}=-\frac{1}{C_2}\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]\end{array}$

Thus, the model for the behavior of temperature $T_2$ is $\frac{d{T}_2}{dt}=-\frac{1}{C_2}\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]$.

To determine

(b)

The effect on the dynamic model of temperature if the thermal capacitance $C_2$ is very small.

Answer to Problem 7.55P

The effect on dynamic model of temperature if the thermal capacitance $C_2$ is very small is $C_1\frac{d{T}_1}{dt}=-\frac{1}{R_1}\left(T_1-T_0\right)$.

Explanation of Solution

Calculation:

Write the expression for the model behavior of temperature $T_2$.

$\begin{array}{l}\frac{d{T}_2}{dt}=-\frac{1}{C_2}\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]\\ C_2\frac{d{T}_2}{dt}=-\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]\end{array}$.... (VII)

Write the expression for the model behavior of temperature $T_1$.

$\begin{array}{l}\frac{d{T}_1}{dt}=\frac{1}{C_1}\left[\frac{1}{R_2}\left(T_2-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\right]\\ C_1\frac{d{T}_1}{dt}=\left[\frac{1}{R_2}\left(T_2-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\right]\end{array}$.... (VIII)

Since the thermal capacitance $C_2$ is very small, so its value is supposed to be zero.

$C_2=0$

Substitute $0$ for $C_2$ in Equation (VII).

$\begin{array}{l}\left(0\right)\frac{d{T}_2}{dt}=-\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]\\ -\left[\frac{1}{R_2}\left(T_2-T_1\right)\right]=0\\ T_2-T_1=0\\ T_2=T_1\end{array}$

Substitute $T_1$ for $T_2$ in Equation (VIII).

$\begin{array}{l}{C}_1\frac{d{T}_1}{dt}=\left[\frac{1}{R_2}\left(T_1-T_1\right)-\frac{1}{R_1}\left(T_1-T_0\right)\right]\\ C_1\frac{d{T}_1}{dt}=\left[0-\frac{1}{R_1}\left(T_1-T_0\right)\right]\\ C_1\frac{d{T}_1}{dt}=-\frac{1}{R_1}\left(T_1-T_0\right)\end{array}$

Thus, the effect on dynamic model of temperature if the thermal capacitance $C_2$ is very small is $C_1\frac{d{T}_1}{dt}=-\frac{1}{R_1}\left(T_1-T_0\right)$.