Chapter 19, Problem 135RQ

(a)

To determine

The rate of heat loss from the water.

Explanation of Solution

Given:

The Length of pipe $\left(l\right)$ is $15\text{m}$.

The thermal conductivity $\left(k\right)$ is $52\text{W}/\text{m}\cdot \text{K}$.

The inner diameter $\left(d_1\right)$ is $4\text{cm}$.

The inner diameter $\left(d_2\right)$ is $4.6\text{cm}$.

The temperature of air $\left(T_{\infty }\right)$ is $10°\text{C}$.

The temperature of hot water $\left(T\right)$ is $90°\text{C}$.

The speed of water $\left(V\right)$ is $1.2\text{m}/\text{s}$.

The emissivity $\left(\epsilon \right)$ is $0.7$.

The convection heat transfer coefficient $\left(h\right)$ is $12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Calculation:

Refer Table A-22 “Properties of saturated air at $1\text{atm}$ pressure”.

Obtain the following properties of air corresponding to the temperature of $90°\text{C}$ as follows:

$\begin{array}{c}k=0.675\text{W}/\text{m}\cdot \text{K}\\ \mathrm{Pr}=1.96\\ \rho =965.3\text{kg}/{\text{m}}^{\text{3}}\\ c_p=4206\text{J}/\text{kg}\cdot \text{K}\\ v=0.326\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\end{array}$

Calculate the mass flow rate of water $\left(m\right)$ using the relation.

    $\begin{array}{c}m=\rho AV\\ =\left(965.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{\pi }{4}{\left(0.04\text{m}\right)}^2\right)\left(1.2\text{m}/\text{s}\right)\\ =1.456\text{kg}/\text{s}\end{array}$

Calculate the Reynolds number $\left(\mathrm{Re}\right)$ using the relation.

  $\begin{array}{c}\mathrm{Re}=\frac{Vd}{v}\\ =\frac{\left(1.2\text{m}/\text{s}\right)\left(4\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}{0.326\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}}\\ =147240\end{array}$

Calculate the Nusselt number $\left(Nu\right)$ using the relation.

    $\begin{array}{c}Nu=\frac{hD}{k}\\ =0.125f\mathrm{Re}{\mathrm{Pr}}^{1/3}\\ =0.125\times 0.032\times 147240\times {1.96}^{1/3}\\ =737.1\end{array}$

Calculate the heat transfer coefficient $\left(h\right)$ using the relation.

    $\begin{array}{c}h=\frac{k}{D}Nu\\ =\frac{0.675\text{W}/\text{m}\cdot \text{K}}{\left(4\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\left(737.1\right)\\ =12440\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\end{array}$

Calculate the rate of heat transfer by convection $\left(Q_c\right)$ using the relation.

    $\begin{array}{c}Q=hA\left(T_s-T_{\infty }\right)\\ =h\pi d_{2}l\left(T-T_{\infty }\right)\\ =\left(12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}\right)\left(\pi \left(4.6\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right)\left(15\text{m}\right)\left(\begin{array}{l}\left(90°\text{C}+273\right)\text{K}-\\ \left(10°\text{C}+273\right)\text{K}\end{array}\right)\\ =\text{2081}\text{W}\end{array}$

Calculate the rate of heat transfer by radiation $\left(Q_r\right)$ using the relation.

    $\begin{array}{c}{Q}_r=\epsilon A\sigma \left(T^4-T_{\infty }^4\right)\\ =\left[\begin{array}{l}\left(0.7\right)\left(\pi \left(4.6\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right)\left(15\text{m}\right)\left(5.67\times {10}^{-8}\text{W}/{\text{m}}^2\cdot {\text{K}}^4\right)\times \\ \left({\left(90°\text{C}+273\text{K}\right)}^4-{\left(10°\text{C}+273\text{K}\right)}^4\right)\end{array}\right]\\ =\text{942}\text{W}\end{array}$

Calculate the total heat transfer $\left(Q\right)$ using the relation.

    $\begin{array}{c}Q=Q_c+Q_r\\ =2081\text{W}+942\text{W}\\ =3023\text{W}\end{array}$

Thus, the rate of heat loss from the water is $3023\text{W}$.

(b)

To determine

The temperature at which the water leaves the basement.

Explanation of Solution

Given:

The Length of pipe is $\left(l\right)$ is $15\text{m}$.

The thermal conductivity $\left(k\right)$ is $52\text{W}/\text{m}\cdot \text{K}$.

The inner diameter $\left(d_1\right)$ is $4\text{cm}$.

The inner diameter $\left(d_2\right)$ is $4.6\text{cm}$.

The temperature of air $\left(T_{\infty }\right)$ is $10°\text{C}$.

The temperature of hot water $\left(T\right)$ is $90°\text{C}$.

The speed of water $\left(V\right)$ is $1.2\text{m}/\text{s}$.

The emissivity $\left(\epsilon \right)$ is $0.7$.

The convection heat transfer coefficient $\left(h\right)$ is $12\text{W}/{\text{m}}^{\text{2}}\cdot \text{K}$.

Calculation:

Calculate the temperature at which water leaves the basement $\left(T_e\right)$ using the relation

  $\begin{array}{c}Q=m{c}_p\left(T-T_e\right)\\ T_e=T-\frac{Q}{m{c}_p}\\ =\left(90°\text{C}+273\right)\text{K}-\frac{3023\text{W}}{\left(1.456\text{kg}/\text{s}\right)\left(4206\text{J}/\text{kg}\cdot \text{K}\right)}\\ =\left(362.5\text{K}-273\right)°\text{C}\\ =89.5°\text{C}\end{array}$

Calculate the temperature at which the water $\left(\Delta T_p\right)$ leaves the basement.

    $\begin{array}{c}\Delta T_P=\frac{Q\left(\ln \left(d_2/d_1\right)\right)}{2\pi kL}\\ =3023\text{W}\times \left(\frac{\left(\ln \left(4.6\text{cm}/4\text{cm}\right)\right)}{2\pi \left(52\text{W}/\text{m}\cdot \text{K}\right)\left(15\text{m}\right)}\right)\\ =0.09\text{K}\end{array}$

The temperature difference is same in $\text{K}$ as well as $°\text{C}$ so the value of $\Delta T_p$ can be taken as $0.09°\text{C}$.

Thus, the temperature at which the water leaves the basement is $0.09°\text{C}$.