Chapter 7, Problem 7.50P

To determine

(a)

The heat loss rate from the water to the air in the radial direction.

Answer to Problem 7.50P

The heat loss rate from the water to the air in the radial direction is $195.3\text{lb}\cdot \text{ft}/\text{sec}$.

Explanation of Solution

Calculation:

Write the radial resistance of the pipe.

$R_p=\frac{\ln \left(\frac{r_o}{r_i}\right)}{2\pi kL}$...... (I)

Here, the inner radius of the pipe is $r_i$, the outer radius of the pipe is $r_o$, the thermal conductivity of iron is $k$ and the length of the pipe $L$.

Write the inner suface area of the pipe.

$A_i=2\pi r_{i}L$...... (II)

Write the inner convective resistance of the pipe.

$R_i=\frac{1}{h_i{A}_i}$...... (III)

Here, the inner surface convective coefficient is $h_i$.

Write the outer suface area of the pipe.

$A_o=2\pi r_{o}L$...... (IV)

Write the outer convective resistance of the pipe.

$R_o=\frac{1}{h_o{A}_o}$...... (V)

Here, the outer surface convective coefficient is $h_o$.

Write the total resistance of the pipe.

$R=R_i+R_p+R_o$...... (VI)

Calculate the heat loss rate from the water to the air.

$q=\frac{\left(T_w-T_o\right)}{R}$...... (VII)

Here, the temperature of the water is $T_w$ and the temperature of the surrounding air is $T_o$.

Substitute $10\text{ft}$ for $L$, $\frac{1}{2}\text{in}$ for $r_i$, $\frac{3}{4}\text{in}$ for $r_o$, and $10.1\text{lb}/\text{sec}\cdot °\text{F}$ for $k$ in Equation (I).

$\begin{array}{c}{R}_p=\frac{\ln \left(\frac{\left(\frac{3}{4}\text{in}\right)}{\left(\frac{1}{2}\text{in}\right)}\right)}{2\pi \left(10.1\text{lb}/\text{sec}\cdot °\text{F}\right)\left(10\text{ft}\right)}\\ =\frac{0.405}{2\pi \left(10.1\text{lb}/\text{sec}\cdot °\text{F}\right)\left(10\text{ft}\right)}\\ =6.389\times {10}^{-4}\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\\ \approx 6.4\times {10}^{-4}\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\end{array}$

Substitute $10\text{ft}$ for $L$, and $\frac{1}{2}\text{in}$ for $r_i$ in Equation (II).

$\begin{array}{c}{A}_i=2\pi \left(\frac{1}{2}\text{in}\right)\left(10\text{ft}\right)\\ =2\pi \left(\frac{1}{2}\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\left(10\text{ft}\right)\\ =\left(0.2617\text{ft}\right)\left(10\text{ft}\right)\\ \approx 2.62{\text{ft}}^2\end{array}$

Substitute $2.62{\text{ft}}^2$ for $A_i$ and $16\text{lb}/\text{ft-sec-}°\text{F}$ for $h_i$ in Equation (III).

$\begin{array}{c}{R}_i=\frac{1}{\left(16\text{lb}/\text{ft}\cdot \text{sec}\cdot °\text{F}\right)\left(2.62{\text{ft}}^2\right)}\\ =\frac{1}{\left(41.92\text{lb}\cdot \text{ft}/\text{sec}\cdot °\text{F}\right)}\\ =0.0238\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\\ \approx 0.024\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\end{array}$

Substitute $10\text{ft}$ for $L$, and $\frac{3}{4}\text{in}$ for $r_o$ in Equation (IV).

$\begin{array}{c}{A}_o=2\pi \left(\frac{3}{4}\text{in}\right)\left(10\text{ft}\right)\\ =2\pi \left(\frac{3}{4}\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\left(10\text{ft}\right)\\ =3.926{\text{ft}}^2\\ \approx 3.93{\text{ft}}^2\end{array}$

Substitute $3.93{\text{ft}}^2$ for $A_o$ and $1.1\text{lb}/\text{ft}\cdot \text{sec}\cdot °\text{F}$ for $h_o$ in Equation (V).

$\begin{array}{c}{R}_o=\frac{1}{\left(1.1\text{lb}/\text{ft}\cdot \text{sec}\cdot °\text{F}\right)\left(3.93{\text{ft}}^2\right)}\\ =\frac{1}{\left(4.323\text{lb}\cdot \text{ft}/\text{sec}\cdot °\text{F}\right)}\\ =0.231\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\end{array}$

Substitute $0.231\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}$ for $R_o$, $6.4\times {10}^{-4}\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}$ for $R_p$ and $0.024\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}$ for $R_i$ in Equation (VI).

$\begin{array}{c}R=\left(0.024\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)+\left(6.4\times {10}^{-4}\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)+\left(0.231\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)\\ =\left(0.024\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)+\left(0.23164\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)\\ =0.25564\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\\ \approx 0.256\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\end{array}$

Substitute $0.256\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}$ for $R$

$120°\text{F}$ for $T_w$ and $70°\text{F}$ for $T_o$ in Equation (VII).

$\begin{array}{c}q=\frac{\left(120°\text{F}-70°\text{F}\right)}{\left(0.256\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)}\\ =\frac{\left(50°\text{F}\right)}{\left(0.256\text{sec}\cdot °\text{F}/\text{lb}\cdot \text{ft}\right)}\\ \approx 195.3\text{lb}\cdot \text{ft}/\text{sec}\end{array}$.

Conclusion:

Thus, the heat loss rate from the water to the air in the radial direction is $195.3\text{lb}\cdot \text{ft}/\text{sec}$.

To determine

(b)

The validity of the constant-temperature assumption.

Answer to Problem 7.50P

The constant-temperature assumption is valid.

Explanation of Solution

Calculation:

Write the mass flow of the water.

$\dot{m}=\rho A_{i}V$...... (VIII)

Here, the densityof the water is $\rho$ and the flow velocity of the water is $V$.

Write the thermal energy of the water in the pipe.

$E=\dot{m}{c}_p{T}_w$...... (IX)

Here, the specific heat at constant pressure of the water is $c_p$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $2.62{\text{ft}}^2$ for $A_i$, and $0.5\text{ft}/\text{s}$ for $V$ in Equation (VIII).

$\begin{array}{c}\dot{m}=\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(2.62{\text{ft}}^2\right)\left(0.5\text{ft}/\text{s}\right)\\ =\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\left(1.31{\text{ft}}^3/\text{s}\right)\\ \approx 2.54\text{slug}/\text{s}\end{array}$

Substitute $2.54\text{slug}/\text{s}$ for $\dot{m}$, $25000\text{lb}\cdot \text{ft}/\text{slug}\cdot °\text{F}$ for $c_p$ and $120°\text{F}$ for $T_w$ in Equation (IX).

$\begin{array}{c}E=\left(2.54\text{slug}/\text{s}\right)\left(25000\text{lb}\cdot \text{ft}/\text{slug}\cdot °\text{F}\right)\left(120°\text{F}\right)\\ =\left(2.54\text{slug}/\text{s}\right)\left(3000000\text{lb}\cdot \text{ft}/\text{slug}\right)\\ =7.62\times {10}^6\text{lb}\cdot \text{ft}/\text{s}\end{array}$

The thermal energy of the water in the pipe is $7.62\times {10}^6\text{lb}\cdot \text{ft}/\text{s}$.

Since the thermal energy of the water in the pipe is greater than the the heat loss rate from the water to the air in the radial direction, so the water will loose very less amount of heat.

Conclusion:

Thus, the constant-temperature assumption is valid.