Chapter 7.13, Problem 158P

a)

To determine

The mass of the iron block.

Answer to Problem 158P

The mass of the iron block is $11.4\text{lbm}$.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

$E_{in}-E_{out}=\Delta E_{system}$ (I)

Here, energy transfer into the control volume is $E_{in}$, energy transfer exit from the control volume is $E_{out}$ and change in internal energy of system is $\Delta E_{system}$.

Write the expression to calculate the mass of the water.

$m_{water}=\rho \nu$ (II)

Here, density of the water is $\rho$ and volume of the iron is $\nu$.

Write the expression to calculate the total work done by the paddle wheel.

$W_{pw}={\dot{W}}_{pw}\left(\Delta t\right)$ (III)

Here, rate of paddle wheel by fan is ${\dot{W}}_{pw}$ and time is $\Delta t$.

Conclusion:

Substitute $W_{pw,in}$0 for $E_{in}$, $0$ for $E_{out}$ and $\Delta U_{iron}+\Delta U_{water}$ for $\Delta E_{system}$ in Equation (I).

$\begin{array}{l}{W}_{pw,in}-0=\Delta U_{iron}+\Delta U_{water}\\ W_{pw,in}=\left\{\begin{array}{l}{\left[m_{iron}{c}_{p,iron}\left({\left(T_2\right)}_{iron}-{\left(T_1\right)}_{iron}\right)\right]}_{iron}\\ +{\left[m_{water}{c}_{p,water}\left({\left(T_2\right)}_{water}-{\left(T_1\right)}_{water}\right)\right]}_{water}\end{array}\right\}\end{array}$ (IV)

Here, change in total internal energy for iron is $\Delta U_{iron}$, change in total internal energy for iron for water is $\Delta U_{water}$, paddle wheel work input is $W_{pw,in}$, mass of the iron is $m_{iron}$, mass of the water is $m_{water}$, initial temperature of water is ${\left(T_1\right)}_{water}$, final temperature of water is ${\left(T_2\right)}_{water}$, specific heat at constant pressure for iron is $c_{p,iron}$ , specific heat at constant pressure for water is $c_{p,water}$ initial temperature of iron is ${\left(T_1\right)}_{iron}$ and final temperature of iron is ${\left(T_2\right)}_{iron}$.

From the Table A-3E, “the properties of common liquids, solids and foods table”, select the density of water at room temperature as $62.1\text{lbm}/{\text{ft}}^3$.

Substitute $62.1\text{lbm}/{\text{ft}}^3$ for $\rho$ and $\text{8}{\text{ft}}^3$ for $\nu$ in Equation (II).

$\begin{array}{c}{m}_{water}=\left(62.1\text{lbm}/{\text{ft}}^3\right)\left(\text{8}{\text{ft}}^3\right)\\ =49.7\text{lbm}\end{array}$

Substitute $0.2\text{kJ}/\text{s}$ for ${\dot{W}}_{pw}$ and $\text{10}\text{min}$ for $\Delta t$ in Equation (III).

$\begin{array}{c}{W}_{pw,in}=\left(0.2\text{kJ}/\text{s}\right)\left(\text{10}\text{min}\right)\\ =\left(0.2\text{kJ}/\text{s}\right)\left(\frac{1\text{Btu}}{1.055\text{kJ}\text{}}\right)\left(\text{10}\text{min}\right)\left(\frac{60\sec }{1\min }\right)\\ =113.7\text{Btu}\end{array}$

From the Table A-3E, “the properties of common liquids, solids and foods table”, select the specific heat at constant pressure at room temperature for water and iron as $1.00\text{Btu}/\text{lbm}\cdot \text{°F}$ and $0.107\text{Btu}/\text{lbm}\cdot \text{°F}$ respectively.

Substitute $0.107\text{Btu}/\text{lbm}\cdot \text{°F}$ for $c_{p,iron}$, $70 °\text{F}$ for ${\left(T_1\right)}_{water}$, $75° \text{F}$ for ${\left(T_2\right)}_{water}$, $113.7\text{Btu}$ for $W_{pw,in}$, $49.7\text{lbm}$ for $m_{water}$, $1.00\text{Btu}/\text{lbm}\cdot \text{°F}$ for $c_{p,water}$ $75° \text{F}$ for ${\left(T_2\right)}_{iron}$, and $185 °\text{F}$ for ${\left(T_1\right)}_{iron}$ in Equation (IV).

$\begin{array}{l}113.7\text{Btu}=\left\{\begin{array}{l}{\left[m_{iron}\left(0.107\text{Btu}/\text{lbm}\cdot \text{°F}\right)\left(75° \text{F}-185 °\text{F}\right)\right]}_{iron}\\ +{\left[\left(49.7\text{lbm}\right)\left(1.00\text{Btu}/\text{lbm}\cdot \text{°F}\right)\left(75° \text{F}-70 °\text{F}\right)\right]}_{water}\end{array}\right\}\\ m_{iron}=11.4\text{lbm}\end{array}$

Thus, the mass of the iron block is $11.4\text{lbm}$.

b)

To determine

The entropy generation during the process.

Answer to Problem 158P

The entropy generation during the process is $0.238\text{Btu}/\text{R}$.

Explanation of Solution

Write the expression for the entropy balance equation of the system.

$S_{in}-S_{out}+S_{gen}=\Delta S_{system}$ (V)

Here, rate of net entropy in is $S_{in}$, rate of net entropy out is $S_{out}$, rate of entropy generation is $S_{gen}$ and change of entropy of the system is $\Delta S_{system}$

Conclusion:

Substitute 0 for $S_{in}$, $0$ for $S_{out}$ and $\Delta S_{iron}+\Delta S_{water}$ for $\Delta S_{system}$ in Equation (V).

$0-0+S_{gen}=\Delta S_{iron}+\Delta S_{water}$

$S_{gen}=\left\{\begin{array}{l}{\left[m_{iron}{c}_{p,iron}\ln \left(\frac{{\left(T_2\right)}_{iron}}{{\left(T_1\right)}_{iron}}\right)\right]}_{iron}\\ +{\left[m_{water}{c}_{p,water}\ln \left(\frac{{\left(T_2\right)}_{water}}{{\left(T_1\right)}_{water}}\right)\right]}_{water}\end{array}\right\}$ (VI)

Substitute $0.107\text{Btu}/\text{lbm}\cdot \text{°F}$ for $c_{p,iron}$, $11.4\text{lbm}$ for $m_{iron}$, $70 °\text{F}$ for ${\left(T_1\right)}_{water}$, $75° \text{F}$ for ${\left(T_2\right)}_{water}$, $49.7\text{lbm}$ for $m_{water}$, $1.00\text{Btu}/\text{lbm}\cdot \text{°F}$ for $c_{p,water}$ $75° \text{F}$ for ${\left(T_2\right)}_{iron}$, and $185 °\text{F}$ for ${\left(T_1\right)}_{iron}$ in Equation (VI).

$S_{gen}=\left\{\begin{array}{l}{\left[\left(11.4\text{lbm}\right)\left(0.107\text{Btu}/\text{lbm}\cdot \text{°F}\right)\ln \left(\frac{75° \text{F}}{185 °\text{F}}\right)\right]}_{iron}\\ +{\left[\left(49.7\text{lbm}\right)\left(1.00\text{Btu}/\text{lbm}\cdot \text{°F}\right)\ln \left(\frac{75° \text{F}}{70 °\text{F}}\right)\right]}_{water}\end{array}\right\}$

$\begin{array}{c}{S}_{gen}=\left\{\begin{array}{l}{\left[\left(11.4\text{lbm}\right)\left(0.107\text{Btu}/\text{lbm}\cdot \text{°F}\right)\ln \left(\frac{\left(75+460\right)\text{R}}{\left(185+460\right)\text{R}}\right)\right]}_{iron}\\ +{\left[\left(49.7\text{lbm}\right)\left(1.00\text{Btu}/\text{lbm}\cdot \text{°F}\right)\ln \left(\frac{\left(75+460\right)\text{R}}{\left(70+460\right)\text{R}}\right)\right]}_{water}\end{array}\right\}\\ =-0.228\text{Btu}/\text{R}+0.466\text{Btu}/\text{R}\\ =0.238\text{Btu}/\text{R}\end{array}$

Thus, the entropy generation during the process is $0.238\text{Btu}/\text{R}$.