Chapter 7, Problem 7.51P

To determine

Dynamic model for the system.

Temperature $T\left(t\right)$ as function of time.

Answer to Problem 7.51P

Dynamic model for the system is $388.34\frac{dT}{dt}=\left(70°\text{F}-T\right)$.

Temperature $T\left(t\right)$ as function of time is $e^{\left[3.912-\frac{t}{388.34}\right]}+70°\text{F}$.

Explanation of Solution

Given information:

The initial temperature of water is $120°\text{F}$, the length of copper pipe is $6\text{ft}$, inner radius of copper pipe is $1/4\text{in}$, the outer radius of the pipe is $3/8\text{in}$, the temperature of the surrounding air is $70°\text{F}$, thermal conductivity of copper is $50\text{lb}/\text{sec}\cdot \text{°F}$, the convection coefficient at the inner surface between water and pipe is $6\text{lb}/\text{sec}\cdot \text{ft}\cdot \text{°F}$, the convection coefficient at the outer surface is $1.1\text{lb}/\text{sec}\cdot \text{ft}\cdot \text{°F}$, specific heat of water is $25000\text{ft}\cdot \text{lb}/\text{slug}\cdot \text{°F}$, the density of water is $1.94\text{slug}/{\text{ft}}^{\text{3}}$.

Write the expression for area of the inner cylinder.

$A_i=2\pi r_{i}L$        ....... (I)

Here, the area of the inner cylinder is $A_i$, the inner radius is $r_i$, the length of the cylinder is $L$.

Write the expression for area of the outer cylinder.

$A_o=2\pi r_{o}L$        ....... (II)

Here, the area of the outer cylinder is $A_o$, the outer radius is $r_o$.

Write the expression for total contact resistance.

$R=R_i+R_o+R_c$        ....... (III)

Here, total contact resistance is $R$ ,the resistance between the inner cylinder surface and water is $R_i$ the resistance between the outer cylinder surface and air is $R_o$ the resistance offered by the cylinder is $R_c$.

Substitute $\frac{1}{h_i{A}_i}$ for $R_i$, $\frac{1}{h_o{A}_o}$ for $R_o$ and $\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi Lk}$ for $R_c$ in Equation (III)

$R=\frac{1}{h_i{A}_i}+\frac{1}{h_o{A}_o}+\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi Lk}$        ....... (IV)

Here, the convection coefficient at the inner surface between water and pipe is $h_i$, the convection coefficient at the outer surface is $h_o$, thermal conductivity of copper is $k$.

Write the expression for conservation of energy.

$m{c}_p\frac{dT}{dt}=-\frac{1}{R}\left(T-T_o\right)$..... (V)

Here, the mass of the copper pipe is $m$, the specific heat of water at constant volume is $c_p$, the change of temperature with respect to time is $\frac{dT}{dt}$, the temperature of the surrounding air is $T_o$, temperature of cylinder at any instant is $T$.

Write the expression for mass of the water flowing inside the pipe.

$m=\rho V$        ....... (VI)

Here, the density of water is $\rho$, the volume of the cylinder is $V$.

Write the expression for volume of the cylinder.

$V=\pi r_i^{2}L$

Here, the volume of the cylinder is $V$.

Substitute $\pi r_i^{2}L$ for $V$ in Equation (VI) $$

$m=\rho \left(\pi r_i^{2}L\right)$        ....... (VII)

Write the temperature difference between the cylinder and the surrounding.

$\theta =T-70°\text{F}$        ....... (IX)

Here, the temperature difference is $\theta$.

Differentiate Equation (IX) with respect to $t$

$\frac{d\theta }{dt}=\frac{dT}{dt}$.

Calculation:

Substitute $1/4\text{ft}$ for $r_i$, $6\text{ft}$ for $L$ in Equation (I).

$\begin{array}{c}{A}_i=2\pi \left(6\text{ft}\right){\left(\frac{1}{4}\text{in}\right)}_{}\\ =2\pi \left(6\text{ft}\right)\left(\frac{1}{4}\text{in}\left(\frac{0.083\text{ft}}{1\text{in}}\right)\right)\\ =2\pi \left(6\text{ft}\right)\left(0.0207\text{ft}\right)\\ =0.78{\text{ft}}^{\text{2}}\end{array}$

Substitute $3/8\text{ft}$ for $r_o$, $6\text{ft}$ for $L$ in Equation (II).

$\begin{array}{c}{A}_o=2\pi \left(6\text{ft}\right)\left(\frac{3}{8}\text{in}\right)\\ =2\pi \left(6\text{ft}\right)\left(\frac{3}{8}\text{in}\left(\frac{0.083\text{ft}}{1\text{in}}\right)\right)\\ =2\pi \left(6\text{ft}\right)\left(0.0311\text{ft}\right)\\ =1.172{\text{ft}}^{\text{2}}\end{array}$

Substitute $1.172{\text{ft}}^{\text{2}}$ for $A_o$, $0.78{\text{ft}}^{\text{2}}$ for $A_i$, $50\text{lb}/\text{sec}\cdot \text{°F}$ for $k$, $6\text{lb}/\text{sec}\cdot \text{ft}\cdot \text{°F}$ for $h_i$, $1.1\text{lb}/\text{sec}\cdot \text{ft}\cdot \text{°F}$ for $h_o$, $1/4\text{ft}$ for $r_i$, $3/8\text{ft}$ for $r_o$ and $6\text{ft}$ for $L$ in Equation (IV).

$\begin{array}{c}R=\left[\begin{array}{l}\frac{1}{\left(6\text{lb}/\text{sec}\cdot \text{ft}\cdot \text{°F}\right)\left(0.78{\text{ft}}^{\text{2}}\right)}+\frac{1}{\left(1.1\text{lb}/\text{sec}\cdot \text{ft}\cdot \text{°F}\right)\left(1.172{\text{ft}}^{\text{2}}\right)}\\ +\frac{\ln \left(\frac{3/8\text{in}}{1/4\text{in}}\right)}{2\pi \left(6\text{ft}\right)\left(50\text{lb}/\text{sec}\cdot \text{°F}\right)}\end{array}\right]\\ =\left[\begin{array}{l}\frac{1}{4.68\text{lb}\cdot \text{ft}/\text{sec}\cdot \text{°F}}+\frac{1}{1.2892\text{lb}\cdot \text{ft}/\text{sec}\cdot \text{°F}}\\ +\frac{\ln \left(\frac{12}{8}\right)}{1884.9\text{ft}\cdot \text{lb}/\text{sec}\cdot \text{°F}}\end{array}\right]\\ =\left[\begin{array}{l}\frac{1}{4.68\text{lb}\cdot \text{ft}/\text{sec}\cdot \text{°F}}+\frac{1}{1.2892\text{lb}\cdot \text{ft}/\text{sec}\cdot \text{°F}}\\ +\frac{0.405}{1884.9\text{ft}\cdot \text{lb}/\text{sec}\cdot \text{°F}}\end{array}\right]\\ =0.2136+0.7756+2.15\times {10}^{-4}\\ =0.9894\sec \cdot °\text{F}/\text{ft}\cdot \text{lb}\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $1/4\text{in}$ for $r_i$ and $6\text{ft}$ for $L$ in Equation (VII).

$\begin{array}{c}m=\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(\pi \left(1/4\text{in}\right)6\text{ft}\right)\\ =\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\pi {\left(\left(\frac{1}{4}\right)\text{in}\left(\frac{0.083\text{ft}}{1\text{in}}\right)\right)}^{2}6\text{ft}\\ =\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(8.11\times {10}^{-3}{\text{ft}}^{\text{3}}\right)\\ =0.0157\text{slug}\end{array}$

Substitute $0.0157\text{slug}$ for $m$, $25000\text{ft}\cdot \text{lb}/\text{slug}\cdot \text{°F}$ for $c_p$, $0.9894\sec \cdot °\text{F}/\text{ft}\cdot \text{lb}$ for $R$ and $70°\text{F}$ for $T_o$ in Equation (V).

$\begin{array}{l}\left(0.0157\text{slug}\right)\left(25000\text{ft}\cdot \text{lb}/\text{slug}\cdot \text{°F}\right)\frac{dT}{dt}=-\frac{1}{\left(0.9894\sec \cdot °\text{F}/\text{ft}\cdot \text{lb}\right)}\left(T-70°\text{F}\right)\\ 392.5\text{ft}\cdot \text{lb}/\text{°F}\frac{dT}{dt}=-1.0107\text{ft}\cdot \text{lb}/\sec \cdot °\text{F}\left(T-70°\text{F}\right)\\ \frac{392.5\text{ft}\cdot \text{lb}/\text{°F}}{1.0107\text{ft}\cdot \text{lb}/\sec \cdot °\text{F}}\frac{dT}{dt}=\left(70°\text{F}-T\right)\end{array}$

$388.34\frac{dT}{dt}=\left(70°\text{F}-T\right)$        ....... (X)

Substitute ${\theta }_i$ for $\theta$, $T_i$ for $T$ in Equation (IX).

${\theta }_i=T_i-70°\text{F}$

Substitute $120°\text{F}$ for $T_i$.

$\begin{array}{c}{\theta }_i=120°\text{F}-70°\text{F}\\ =\text{50°F}\end{array}$        ....... (XI)

Substitute $T-70°\text{F}$ for $\theta$, $\frac{d\theta }{dt}$ for $\frac{dT}{dt}$ in Equation (X).

$\begin{array}{c}388.34\frac{d\theta }{dt}=-\theta \\ \frac{1}{\theta }d\theta =-\frac{1}{388.34}dt\end{array}$

Integrate both sides of the equation.

$\begin{array}{l}{\displaystyle \underset{0}{\overset{t}{\int }}\frac{1}{\theta }d\theta }=-{\displaystyle \underset{0}{\overset{t}{\int }}\frac{1}{388.34}dt}\\ \ln \theta -\ln {\theta }_i=\frac{-t}{388.34}\end{array}$        ....... (XII)

Substitute $T-70$ for $\theta$ and $50°\text{F}$ for ${\theta }_i$ in Equation (XII).

$\begin{array}{l}\ln \left(T-70°\text{F}\right)-\ln 50°\text{F}=\frac{-t}{388.34}\\ \ln \left(T-70°\text{F}\right)=\frac{-t}{388.34}+\ln 50°\text{F}\\ \ln \left(T-70°\text{F}\right)=\frac{-t}{388.34}+3.912\\ T\left(t\right)=e^{\left[3.912-\frac{t}{388.34}\right]}+70°\text{F}\end{array}$.

Conclusion:

Dynamic model for the system is $388.34\frac{dT}{dt}=\left(70°\text{F}-T\right)$.

Temperature $T\left(t\right)$ as function of time is $e^{\left[3.912-\frac{t}{388.34}\right]}+70°\text{F}$.