Chapter 4.5, Problem 26P

To determine

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

Answer to Problem 26P

$Q_{in}\left(\text{Btu}\right)$	$W_{out}\left(\text{Btu}\right)$	$E_1\left(\text{Btu}\right)$	$E_2\left(\text{Btu}\right)$	$m\left(\text{lbm}\right)$	$e_2-e_1\left(\text{Btu}/\text{lbm}\right)$
$350$	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	$-200$

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:

Substitute $350\text{Btu}$ for $Q_{in}$, $1020\text{Btu}$ for $E_2$, and $860\text{Btu}$ for $E_1$ 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 $350\text{Btu}$ for $Q_{in}$, $3\text{lbm}$ for $m$, and $510\text{Btu}$ for $W_{b,out}$ 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 $350\text{Btu}$ for $Q_{in}$, $130\text{Btu}$ for $W_{b,out}$, and $550\text{Btu}$ for $E_1$ 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 $350\text{Btu}$ for $Q_{in}$, $5\text{lbm}$ for $m$, and $130\text{Btu}$ for $W_{b,out}$ 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 $260\text{Btu}$ for $W_{b,out}$, $2\text{lbm}$ for $m$, and $150\text{Btu}/\text{lbm}$ for $\left(e_2-e_1\right)$ 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 $560\text{Btu}$ for $Q_{in}$, $600\text{Btu}$ for $E_1$, and $260\text{Btu}$ for $W_{b,out}$ 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 $-500\text{Btu}$ for $Q_{in}$, $1400\text{Btu}$ for $E_2$, and $900\text{Btu}$ for $E_1$ 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 $-500\text{Btu}$ for $Q_{in}$, $7\text{lbm}$ for $m$, and $0\text{Btu}$ for $W_{b,out}$ 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 $-50\text{Btu}$ for $W_{b,out}$, $3\text{lbm}$ for $m$, and $-200\text{Btu}/\text{lbm}$ for $\left(e_2-e_1\right)$ 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 $-650\text{Btu}$ for $Q_{in}$, $1000\text{Btu}$ for $E_1$, and $-50\text{Btu}$ for $W_{b,out}$ in Equation (II).

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