Chapter 5.5, Problem 176RP

To determine

The final temperature and mass of the air in the ballast tank.

Answer to Problem 176RP

The final temperature and mass of the air in the ballast tank is $386.8\text{K}$ and $9460\text{kg}$ respectively.

Explanation of Solution

Write the general mass balance equation.

$\begin{array}{l}{\dot{m}}_{\text{in}}-{\dot{m}}_{\text{out}}=\frac{d}{dt}\left(m_{\text{system}}\right)\\ {\dot{m}}_{\text{in}}-{\dot{m}}_{\text{out}}=\frac{d{m}_{\text{system}}}{dt}\end{array}$ (I)

Here, the inlet mass flow rate is ${\dot{m}}_{\text{in}}$, the exit mass flow rate is ${\dot{m}}_{\text{out}}$, and the change in mass of the system is $\frac{d{m}_{\text{system}}}{dt}$.

It is given that the mass of air pumped into the tank to get the submarine to the surface. Here, there is no exit mass of air. Thus, the exit mass $\left({\dot{m}}_{\text{out}}\right)$ of air is negligible.

${\dot{m}}_{\text{out}}=0$

Refer Equation (I).

Write the mass balance equation for air.

$\begin{array}{l}{\dot{m}}_{a\text{,in}}-{\dot{m}}_{a\text{,out}}=\frac{d{m}_a}{dt}\\ {\dot{m}}_{a\text{,in}}-0=\frac{d{m}_a}{dt}\\ {\dot{m}}_{a\text{,in}}=\frac{d{m}_a}{dt}\end{array}$ (II)

Here, the subscript $a$ indicate the air.

While air entering into the ballast tank, the seawater present in the tank is released to the sea. Here, there is no inlet mass of water. Thus, the inlet mass $\left({\dot{m}}_{\text{in}}\right)$ of water is negligible.

${\dot{m}}_{\text{in}}=0$

Refer Equation (I).

Write the mass balance equation for water.

$\begin{array}{l}{\dot{m}}_{w\text{,in}}-{\dot{m}}_{w\text{,out}}=\frac{d{m}_w}{dt}\\ 0-{\dot{m}}_{w\text{,out}}=\frac{d{m}_w}{dt}\\ {\dot{m}}_{w\text{,out}}=-\frac{d{m}_w}{dt}\end{array}$ (III)

Here, the subscript $w$ indicates the seawater.

Write the general energy rate balance equation.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$ (IV)

Here, the rate of total energy in is ${\dot{E}}_{\text{in}}$, the rate of total energy out is ${\dot{E}}_{\text{out}}$, and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Refer Equation (IV).

Write the energy balance equation for the system (ballast tank).

$\frac{d{\left(mu\right)}_a}{dt}+\frac{d{\left(mu\right)}_w}{dt}+h_{w,\text{out}}{\dot{m}}_{w,\text{out}}-h_{a,\text{in}}{\dot{m}}_{a\text{,in}}=0$ (V)

Here, the mass is $m$, the internal energy is $u$, the enthalpy is $h$, and the subscript $a$ indicates air, $w$ indicates water.

Substitute $\frac{d{m}_a}{dt}$ for ${\dot{m}}_{a\text{,in}}$ and $-\frac{d{m}_w}{dt}$ for ${\dot{m}}_{w\text{,out}}$ in Equation (V).

$\begin{array}{l}\frac{d{\left(mu\right)}_a}{dt}+\frac{d{\left(mu\right)}_w}{dt}+h_{w,\text{out}}\left(-\frac{d{m}_w}{dt}\right)-h_{a,\text{in}}\left(\frac{d{m}_a}{dt}\right)=0\\ \frac{d{\left(mu\right)}_a}{dt}+\frac{d{\left(mu\right)}_w}{dt}-h_{w,\text{out}}\frac{d{m}_w}{dt}-h_{a,\text{in}}\frac{d{m}_a}{dt}=0\\ \frac{d{\left(mu\right)}_a}{dt}+\frac{d{\left(mu\right)}_w}{dt}=h_{w,\text{out}}\frac{d{m}_w}{dt}+h_{a,\text{in}}\frac{d{m}_a}{dt}\\ d{\left(mu\right)}_a+d{\left(mu\right)}_w=h_{w,\text{out}}d{m}_w+h_{a,\text{in}}d{m}_a\end{array}$ (VI)

Integrate the Equation (VI) at initial (1) and final (2) states.

$\begin{array}{l}{\displaystyle \underset{1}{\overset{2}{\int }}d{\left(mu\right)}_a+{\displaystyle \underset{1}{\overset{2}{\int }}d{\left(mu\right)}_w}}={\displaystyle \underset{1}{\overset{2}{\int }}{h}_{w,\text{out}}d{m}_w}+{\displaystyle \underset{1}{\overset{2}{\int }}{h}_{a,\text{in}}d{m}_a}\\ \left\{\begin{array}{l}{\left[{\left(mu\right)}_2-{\left(mu\right)}_1\right]}_a\\ +{\left[{\left(mu\right)}_2-{\left(mu\right)}_1\right]}_w\end{array}\right\}=h_{w,\text{out}}{\left(m_2-m_1\right)}_w+h_{a,\text{in}}{\left(m_2-m_1\right)}_a\end{array}$ (VII)

The enthalpy and internal energy is expressed as follows.

$\begin{array}{l}h=c_{p}T\\ u=c_{v}T\end{array}$

The mass of air is expressed as follows.

$m_a=\frac{P\nu }{RT}$

Here, the specific volume is $v$.

For water the specific heat is $c_p=c_v=c_w$.

At final state, the mass sea water is zero and the ballast is completely filled with air.

${\left(m_2\right)}_w=0$

Rewrite the Equation (VII) as follows.

$\begin{array}{l}{\left[\frac{P_2{\nu }_2}{R{T}_2}{c}_v{T}_2-\frac{P_1{\nu }_1}{R{T}_1}{c}_v{T}_1\right]}_a+{\left[0-m_1{c}_w{T}_1\right]}_w=c_w{T}_{1,w}{\left(0-m_1\right)}_w+c_{p,a}{T}_{\text{in},a}{\left(\frac{P_2{\nu }_2}{R{T}_2}-\frac{P_1{\nu }_1}{R{T}_1}\right)}_a\\ {\left[\frac{P_2{\nu }_2}{R{T}_2}{c}_v{T}_2-\frac{P_1{\nu }_1}{R{T}_1}{c}_v{T}_1\right]}_a-m_{1,w}{c}_w{T}_{1,w}=-m_{1,w}{c}_w{T}_{1,w}+c_{p,a}{T}_{\text{in},a}{\left(\frac{P_2{\nu }_2}{R{T}_2}-\frac{P_1{\nu }_1}{R{T}_1}\right)}_a\\ {\left[\frac{P_2{\nu }_2}{R{T}_2}{c}_v{T}_2-\frac{P_1{\nu }_1}{R{T}_1}{c}_v{T}_1\right]}_a=c_{p,a}{T}_{\text{in},a}{\left(\frac{P_2{\nu }_2}{R{T}_2}-\frac{P_1{\nu }_1}{R{T}_1}\right)}_a\\ c_{v,a}{\left(\frac{P_2{\nu }_2}{R}-\frac{P_1{\nu }_1}{R}\right)}_a=c_{p,a}{T}_{\text{in},a}{\left(\frac{P_2{\nu }_2}{R{T}_2}-\frac{P_1{\nu }_1}{R{T}_1}\right)}_a\end{array}$

$\frac{c_{v,a}}{c_{p,a}}{\left(\frac{P_2{\nu }_2}{R}-\frac{P_1{\nu }_1}{R}\right)}_a=T_{\text{in},a}{\left(\frac{P_2{\nu }_2}{R{T}_2}-\frac{P_1{\nu }_1}{R{T}_1}\right)}_a$ (VIII)

Here,

The specific heat ratio is, $k=\frac{c_p}{c_v}$.

The initial and final pressure of air is, $P_2=P_1$.

Rewrite the Equation (VIII) as follows.

$\begin{array}{l}\frac{1}{k_a}{\left({\nu }_2-{\nu }_1\right)}_a=T_{\text{in},a}{\left(\frac{{\nu }_2}{T_2}-\frac{{\nu }_1}{T_1}\right)}_a\\ \frac{1}{k_a{T}_{\text{in},a}}{\left({\nu }_2-{\nu }_1\right)}_a+{\left(\frac{{\nu }_1}{T_1}\right)}_a={\left(\frac{{\nu }_2}{T_2}\right)}_a\\ T_{2,a}=\frac{{\nu }_{2,a}}{\frac{1}{k_a{T}_{\text{in},a}}{\left({\nu }_2-{\nu }_1\right)}_a+{\left(\frac{{\nu }_1}{T_1}\right)}_a}\end{array}$ (IX)

Write the formula for mass flow rate of air at final state.

${\dot{m}}_{2,a}={\left(\frac{P_2{\nu }_2}{R{T}_2}\right)}_a$ (X)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant $\left(R\right)$ of air is $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$.

Refer Table A-2a, “Ideal-gas specific heats of various common gases”.

The specific heat ratio $\left(k_a\right)$ of air at room temperature is $1.4$.

Conclusion:

Substitute $700{\text{m}}^3$ for $v_{2,a}$, $1.4$ for $k_a$, $\left(20+273\right)\text{K}$ for $T_{\text{in,}a}$, $100{\text{m}}^3$ for ${\nu }_1$, and $\left(15+273\right)\text{K}$ for $T_1$ in Equation (IX).

$\begin{array}{c}{T}_{2,a}=\frac{700{\text{m}}^3}{\frac{1}{1.4\left[\left(20+273\right)\text{K}\right]}\left(700{\text{m}}^3-100{\text{m}}^3\right)+\frac{100{\text{m}}^3}{\left(15+273\right)\text{K}}}\\ =\frac{700{\text{m}}^3}{1.4627{\text{m}}^3\text{/K}+0.3472{\text{m}}^3\text{/K}}\\ =386.7828\text{K}\\ \simeq 386.8\text{K}\end{array}$

Substitute $1500\text{kPa}$ for $P_2$, $700{\text{m}}^3$ for $v_2$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $386.8\text{K}$ for $T_2$ in Equation (X).

$\begin{array}{c}{\dot{m}}_{2,a}=\frac{1500\text{kPa}\left(700{\text{m}}^3\right)}{0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\times \left(386.8\text{K}\right)}\\ =\frac{1050000\text{kPa}\cdot {\text{m}}^3}{111.0116\text{kPa}{\text{.m}}^3\text{/kg}}\\ =9458.47\text{kg}\\ \simeq 9460\text{kg}\end{array}$

Thus, the final temperature and mass of the air in the ballast tank is $386.8\text{K}$ and $9460\text{kg}$ respectively.