Chapter 7, Problem 7.53P

To determine

The thermal resistance $R_1$ of the water in the tank.

Answer to Problem 7.53P

The thermal resistance $R_1$ of the water in the tank is $2.11504\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}$.

Explanation of Solution

Calculation:

Write the expression for the rate of heat transfer in terms of total thermal resistance.

$\frac{dQ}{dt}=-\frac{1}{R_1}\left(T_1-T_0\right)$ (I)

Here, the rate of heat transfer is $\frac{dQ}{dt}$, the total thermal resistance is $R_1$, the tank s water temperature is $T_1$, the temperature of the air surrounding the tank is $T_0$ and the time is $t$.

Write the expression for the thermal capacitance.

$C_1=\rho Vc$

Here, the thermal capacitance is $C_1$, the mass density is $\rho$, the volume of the water in the tank is $V$ and the specific heat of the water at the room temperature is $c$.

Write the expression for the rate of heat transfer in terms of thermal resistance.

$\frac{dQ}{dt}=\rho Vc\frac{d{T}_1}{dt}$ (II)

Substitute $\rho Vc$ for $C_1$ in Equation (II).

$\frac{dQ}{dt}=C_1\frac{d{T}_1}{dt}$ (III)

Equate the values from the Equation (I) and Equation (III) for the conservation of heat energy.

$C_1\frac{d{T}_1}{dt}=-\frac{1}{R_1}\left(T_1-T_0\right)$ (IV)

Write the expression for the temperature.

${\theta }_1=T_1-T_0$ (V)

Differentiate the above expression.

$\begin{array}{c}\frac{d{\theta }_1}{dt}=\frac{d}{dt}\left(T_1-T_0\right)\\ \frac{d{\theta }_1}{dt}=\frac{d{T}_1}{dt}\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^3$ for $\rho$, $1000{\text{ft}}^3$ for $V$ and $25000\text{ft-lb}/\text{slug-}°\text{F}$ for $c$ in Equation (I).

$\begin{array}{c}{C}_1=\left(1.94\text{slug}/{\text{ft}}^3\right)\left(1000{\text{ft}}^3\right)\left(25000\text{ft-lb}/\text{slug-}°\text{F}\right)\\ =\left(1940\text{slug}\right)\left(25000\text{ft-lb}/\text{slug-}°\text{F}\right)\\ =4.85\times {10}^7\text{ft-lb}/°\text{F}\end{array}$

Substitute $70°\text{F}$ for $T_1$ in Equation (V).

${\theta }_1=T_1-\left(70°\text{F}\right)$

Substitute $4.85\times {10}^7\text{ft-lb}/°\text{F}$ for $C_1$ and $70°\text{F}$ for $T_0$ in Equation (IV).

$\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\frac{d{T}_1}{dt}=-\frac{1}{R_1}\left[T_1-\left(70°\text{F}\right)\right]$

Substitute $\frac{d{\theta }_1}{dt}$ for $\frac{d{T}_1}{dt}$ and ${\theta }_1$ for $T_1-\left(70°\text{F}\right)$ in above expression.

$\begin{array}{l}\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\frac{d{\theta }_1}{dt}=-\frac{1}{R_1}\left[{\theta }_1\right]\\ \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\times \frac{d{\theta }_1}{dt}=-\frac{{\theta }_1}{R_1}\\ \frac{d{\theta }_1}{{\theta }_1}=\frac{-dt}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}\end{array}$

Integrate the above expression.

$\begin{array}{c}{\displaystyle \int \frac{d{\theta }_1}{{\theta }_1}}=-{\displaystyle \int \frac{dt}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}}\\ \ln \left({\theta }_1\right)=-\frac{t}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}+C\end{array}$

Substitute ${\theta }_1$ for $T_1-\left(70°\text{F}\right)$ in above expression.

$\ln \left[T_1-\left(70°\text{F}\right)\right]=-\frac{t}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}+C$ (VI)

Apply boundary conditions.

At $t=0\text{sec}$ ; $T_1=90°\text{F}$.

Substitute $90°\text{F}$ for $T_1$ at $t=0\text{sec}$ in Equation (VI).

$\begin{array}{l}\ln \left[\left(90°\text{F}\right)-\left(70°\text{F}\right)\right]=-\frac{0\text{sec}}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}+C\\ \ln \left(20°\text{F}\right)=0+C\\ C=\ln \left(20°\text{F}\right)\end{array}$

Substitute $\ln \left(20°\text{F}\right)$ for $C$ in Equation (VI).

$\begin{array}{l}\ln \left[T_1-\left(70°\text{F}\right)\right]=-\frac{t}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}+\ln \left(20°\text{F}\right)\\ \ln \left[T_1-\left(70°\text{F}\right)\right]-\ln \left(20°\text{F}\right)=-\frac{t}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}\\ \ln \left[\frac{T_1-\left(70°\text{F}\right)}{\left(20°\text{F}\right)}\right]=-\frac{t}{R_1\times \left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)}\\ R_1=\frac{t}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{T_1-\left(70°\text{F}\right)}\right]}\end{array}$ (VII)

Substitute $90°\text{F}$ for $T_1$ and $0\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(0\sec \right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(90°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\frac{0}{0}\end{array}$

The above results are not defined.

Substitute $82°\text{F}$ for $T_1$ and $500\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(500\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(82°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\left(2.018159\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\\ \approx \left(2.01816\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Substitute $77°\text{F}$ for $T_1$ and $1000\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(1000\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(77°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\left(1.964\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Substitute $75°\text{F}$ for $T_1$ and $1500\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(1500\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(75°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\left(2.23097\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Substitute $73°\text{F}$ for $T_1$ and $2000\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(2000\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(73°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\left(2.17367\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Substitute $72°\text{F}$ for $T_1$ and $2500\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(2500\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(72°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\left(2.23863\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Substitute $71°\text{F}$ for $T_1$ and $3000\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(3000\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(71°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\left(2.06479\times {10}^{-5}\right)\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Substitute $70°\text{F}$ for $T_1$ and $4000\text{sec}$ for $t$ in Equation (VII).

$\begin{array}{c}{R}_1=\frac{\left(4000\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[\frac{\left(20°\text{F}\right)}{\left(70°\text{F}\right)-\left(70°\text{F}\right)}\right]}\\ =\frac{\left(4000\text{sec}\right)}{\left(4.85\times {10}^7\text{ft-lb}/°\text{F}\right)\ln \left[0\right]}\end{array}$

The above results are not defined.

The Table (1) shows the calculated values of thermal resistance $R_1$.

S. No $\left(i\right)$	$t\left(\sec \right)$	$T_1\left(°\text{F}\right)$	$R_1\left(\text{sec}°\text{F}/\text{ft-lb}\right)$
$1$	$0$	$90$	$-$
$2$	$500$	$82$	$2.01816\times {10}^{-5}$
$3$	$1000$	$77$	$1.964\times {10}^{-5}$
$4$	$1500$	$75$	$2.23097\times {10}^{-5}$
$5$	$2000$	$73$	$2.17367\times {10}^{-5}$
$6$	$2500$	$72$	$2.23863\times {10}^{-5}$
$7$	$3000$	$71$	$2.06479\times {10}^{-5}$
$8$	$4000$	$70$	$-$

Table-(1)

Calculate the average of the thermal resistance from the above table.

$\begin{array}{c}{\left(R_1\right)}_{\text{avg}}=\frac{{\displaystyle \sum _{i=2}^7{\left(R_1\right)}_i}}{6}\\ =\left[\frac{\left\{\begin{array}{l}\left(2.01816\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\right)+\\ \left(1.964\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\right)+\left(2.23097\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\right)\\ +\left(2.17367\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\right)\\ +\left(2.23863\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\right)+\left(2.06479\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\right)\end{array}\right\}}{6}\right]\\ =2.115036\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\\ \approx 2.11504\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}\end{array}$

Thus, the thermal resistance $R_1$ of the water in the tank is $2.11504\times {10}^{-5}\text{sec}°\text{F}/\text{ft-lb}$.