Chapter 5, Problem 25P

To determine

The missing properties of the table.

Explanation of Solution

Given:

$Q_{in}\left(\text{Btu}\right)$	$W_{out}\left(\text{Btu}\right)$	$E_1\left(\text{Btu}\right)$	$E_2\left(\text{Btu}\right)$	$m\left(\text{lbm}\right)$	$e_2-e_1\left(\text{Btu}/\text{lbm}\right)$
$350$		1020	860	3
$350$	130	550		5
	260	600		2	150
$-500$	0	1400	900	7
$-650$	$-50$	1000		3	$-200$

Calculation:

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 rate is $e_2$ and initial internal energy rate 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 $E_1$.

Substitute $Q_{in}=350\text{Btu}$, $E_2=1020\text{Btu}$, and $E_1=860\text{Btu}$ in Equation (III).

  $\begin{array}{l}350\text{Btu}-W_{b,out}=1020\text{Btu}-860\text{Btu}\\ W_{b,out}=510\text{Btu}\end{array}$

Substitute $Q_{in}=350\text{Btu}$, $m=3\text{lbm}$, and $W_{b,out}=510\text{Btu}$ in Equation (II).

  $\begin{array}{l}350\text{Btu}-510\text{Btu}=3\left(e_2-e_1\right)\\ \left(e_2-e_1\right)=-53.3\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{l}350\text{Btu}-130\text{Btu}=550\text{Btu}-E_2\\ E_2=770\text{Btu}\end{array}$

Substitute $Q_{in}=350\text{Btu}$, $m=5\text{lbm}$, and $W_{b,out}=130\text{Btu}$ in Equation (II).

  $\begin{array}{l}350\text{Btu}-130\text{Btu}=5\left(e_2-e_1\right)\\ \left(e_2-e_1\right)=44\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{l}{Q}_{in}-260\text{Btu}=2\text{lbm}\times 150\text{Btu}/\text{lbm}\\ Q_{in}=560\text{Btu}\end{array}$

Substitute $Q_{in}=560\text{Btu}$, $E_1=600\text{Btu}$, and $W_{b,out}=260\text{Btu}$ in Equation (II).

  $\begin{array}{l}560\text{Btu}-260\text{Btu}=E_2-600\text{Btu}\\ E_2=900\text{Btu}\end{array}$

Substitute $Q_{in}=-500\text{Btu}$, $E_2=1400\text{Btu}$, and $E_1=900\text{Btu}$ in Equation (III).

  $\begin{array}{l}-500\text{Btu}-W_{b,out}=160\text{Btu}-900\text{Btu}\\ W_{b,out}=0\text{Btu}\end{array}$

Substitute $Q_{in}=-500\text{Btu}$, $m=7\text{lbm}$, and $W_{b,out}=0\text{Btu}$ in Equation (II).

  $\begin{array}{l}-500\text{Btu}-0=7\left(e_2-e_1\right)\\ \left(e_2-e_1\right)=-71.4\text{Btu}/\text{lbm}\end{array}$

Substitute $W_{b,out}=-50\text{Btu}$, $m=3\text{lbm}$, and $\left(e_2-e_1\right)=-200\text{Btu}/\text{lbm}$ in Equation (III).

  $\begin{array}{l}{Q}_{in}+50\text{Btu}=3\text{lbm}\times -200\text{Btu}/\text{lbm}\\ Q_{in}=-650\text{Btu}\end{array}$

Substitute $Q_{in}=-650\text{Btu}$, $E_1=1000\text{Btu}$, and $W_{b,out}=-50\text{Btu}$ in Equation (II).

  $\begin{array}{l}-650\text{Btu}+\text{5}0\text{Btu}=E_2-1000\text{Btu}\\ E_2=400\text{Btu}\end{array}$

Tabulate the calculated values as shown in the below table.

$Q_{in}\left(\text{Btu}\right)$	$W_{out}\left(\text{Btu}\right)$	$E_1\left(\text{Btu}\right)$	$E_2\left(\text{Btu}\right)$	$m\left(\text{lbm}\right)$	$e_2-e_1\left(\text{Btu}/\text{lbm}\right)$
$350$	510	1020	860	3	–53.3
$350$	130	550	770	5	44
560	260	600	900	2	150
$-500$	0	1400	900	7	–71.4
–650	$-50$	1000	400	3	$-200$