Chapter 4.5, Problem 29P

Complete each line of the following table on the basis of the conservation of energy principle for a closed system.

Qin kJ	Wout kJ	E1 kJ	E2 kJ	m kg	e2 – e1 kJ/kg
280	—	1020	860	3	—
−350	130	550	—	5	—
—	260	300	—	2	−150
300	—	750	500	1	—
—	−200	—	300	2	−100

FIGURE P4–29

To determine

Fill the blanks from the below table on the basis of the conservation of energy principle for a closed system.

$Q_{\text{in}}\left(\text{kJ}\right)$	$W_{\text{out}}\left(\text{kJ}\right)$	$E_1\left(\text{kJ}\right)$	$E_2\left(\text{kJ}\right)$	$m\left(\text{kg}\right)$	$e_2-e_1\left(\text{kJ}/\text{kg}\right)$
280		1020	860	3
$-350$	130	550		5
	260	300		2	–150
$300$		750	500	1
	$-200$		300	2	$-100$

Explanation of Solution

Write the energy balance equation for closed system.

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

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

Substitute $Q_{in}$ for $E_{in}$, $W_{b,out}$ for $E_{out}$ and $m\left(e_2-e_1\right)$ for $\Delta E_{system}$ in Equation (I).

$Q_{in}-W_{b,out}=m\left(e_2-e_1\right)$ . (II)

Here, amount of heat transfer into the system is $Q_{in}$, mass in the system is m, work done output is $W_{b,out}$, final internal energy is $e_2$ and initial internal energy is $e_1$.

Substitute $Q_{in}$ for $E_{in}$, $W_{b,out}$ for $E_{out}$ and $E_2-E_1$ for $\Delta E_{system}$ in Equation (I).

$Q_{in}-W_{b,out}=E_2-E_1$ (III)

Here, initial and final internal energy are $E_2$ and $E_1$.

Conclusion:

For row I

Substitute $280\text{kJ}$ for $Q_{in}$, $860\text{kJ}$ for $E_2$, and $1020\text{kJ}$ for $E_1$ in Equation (III).

$\begin{array}{l}280\text{kJ}-W_{b,\text{out}}=860\text{kJ}-1020\text{kJ}\\ 280\text{kJ}-W_{b,\text{out}}=-160\text{kJ}\\ W_{b,\text{out}}=280\text{kJ}+160\text{kJ}\\ W_{b,\text{out}}=\underset{\_}{440\text{kJ}}\end{array}$

Substitute $280\text{kJ}$ for $Q_{in}$, $3\text{kg}$ for $m$, and $440\text{kJ}$ for $W_{b,out}$ in Equation (II).

$\begin{array}{l}280\text{kJ}-440\text{kJ}=3\text{kg}\left(e_2-e_1\right)\\ \left(e_2-e_1\right)=\underset{\_}{-53.3\text{kJ}/\text{kg}}\end{array}$

For row II

Substitute $-350\text{kJ}$ for $Q_{in}$, $130\text{kJ}$ for $W_{b,out}$, and $550\text{kJ}$ for $E_1$ in Equation (III).

$\begin{array}{l}-350\text{kJ}-130\text{kJ}=E_2-550\text{kJ}\\ E_2=-350\text{kJ}-130\text{kJ}+550\text{kJ}\\ E_2=\underset{\_}{70\text{kJ}}\end{array}$

Substitute $-350\text{kJ}$ for $Q_{in}$, $5\text{kg}$ for $m$, and $130\text{kJ}$ for $W_{b,out}$ in Equation (II).

$\begin{array}{l}-350\text{kJ}-130\text{kJ}=5\text{kg}\left(e_2-e_1\right)\\ \left(e_2-e_1\right)=\frac{-350\text{kJ}-130\text{kJ}}{5\text{kg}}\\ \left(e_2-e_1\right)=\underset{\_}{-96\text{kJ/kg}}\end{array}$

For row III

Substitute $260\text{kJ}$ for $W_{b,out}$, $2\text{kg}$ for $m$, and $-150\text{kJ}/\text{kg}$ for $\left(e_2-e_1\right)$ in Equation (III).

$\begin{array}{l}{Q}_{\text{in}}-260\text{kJ}=2\text{kg}\times \left(-150\text{kJ}/\text{kg}\right)\\ Q_{\text{in}}=\underset{\_}{-40\text{kJ}}\end{array}$

Substitute $-40\text{kJ}$ for $Q_{\text{in}}$, $300\text{kJ}$ for $E_1$, and $260\text{kJ}$ for $W_{b,out}$ in Equation (II).

$\begin{array}{l}-40\text{kJ}-260\text{kJ}=E_2-300\text{kJ}\\ E_2=\underset{\_}{0}\end{array}$

For row IV

Substitute $300\text{kJ}$ for $Q_{in}$, $500\text{kJ}$ for $E_2$, and $750\text{kJ}$ for $E_1$ in Equation (III).

$\begin{array}{l}300\text{kJ}-W_{b,out}=500\text{kJ}-750\text{kJ}\\ W_{b,out}=\underset{\_}{550\text{kJ}}\end{array}$

Substitute $300\text{kJ}$ for $Q_{in}$, $1\text{kg}$ for $m$, and $550\text{kJ}$ for $W_{b,out}$ in Equation (II).

$\begin{array}{l}300\text{kJ}-550\text{kJ}=1\text{kg}\left(e_2-e_1\right)\\ \left(e_2-e_1\right)=\underset{\_}{-250\text{kJ/kg}}\end{array}$

For row V

Substitute $-200\text{kJ}$ for $W_{b,out}$, $2\text{kg}$ for $m$, and $-100\text{kJ/kg}$ for $\left(e_2-e_1\right)$ in Equation (III).

$\begin{array}{l}{Q}_{\text{in}}-\left(-200\text{kJ}\right)=2\text{kg}\times \left(-100\text{kJ/kg}\right)\\ Q_{\text{in}}=-200\text{kJ}-200\text{kJ}\\ Q_{\text{in}}=\underset{\_}{-400\text{kJ}}\end{array}$

Substitute $-400\text{kJ}$ for $Q_{in}$, $300\text{kJ}$ for $E_2$, and $-200\text{kJ}$ for $W_{b,out}$ in Equation (II).

$\begin{array}{l}-400\text{kJ}-\left(-200\text{kJ}\right)=300\text{kJ}-E_1\\ E_1=\underset{\_}{500\text{kJ}}\end{array}$

The following table blanks are filled and are shown below as summarized.

$Q_{\text{in}}\left(\text{kJ}\right)$	$W_{\text{out}}\left(\text{kJ}\right)$	$E_1\left(\text{kJ}\right)$	$E_2\left(\text{kJ}\right)$	$m\left(\text{kg}\right)$	$e_2-e_1\left(\text{kJ}/\text{kg}\right)$
280	$\underset{\_}{440\text{kJ}}$	1020	860	3	$\underset{\_}{-53.3\text{kJ}/\text{kg}}$
$-350$	130	550	$\underset{\_}{70\text{kJ}}$	5	$\underset{\_}{-96\text{kJ/kg}}$
$\underset{\_}{-40\text{kJ}}$	260	300	$\underset{\_}{0}$	2	–150
$300$	$\underset{\_}{550\text{kJ}}$	750	500	1	$\underset{\_}{-250\text{kJ/kg}}$
$\underset{\_}{-400\text{kJ}}$	$-200$	$\underset{\_}{500\text{kJ}}$	300	2	$-100$