Chapter 5.5, Problem 177RP

To determine

The final mass of the air in the ballast tank and total heat transfer.

Answer to Problem 177RP

The final mass of air in the ballast tank is $12697\text{kg}$.

The total heat transfer is $885534.2\text{kJ}$.

Explanation of Solution

Write the formula for mass of air $\left(m_a\right)$ at initial and final states.

$m_{a,1}={\left(\frac{P_1{\nu }_1}{R{T}_1}\right)}_a$ (I)

$m_{a,2}={\left(\frac{P_2{\nu }_2}{R{T}_2}\right)}_a$ (II)

Here, the volume of air is $\nu$, the pressure is $P$, the gas constant of air is $R$ and the temperature is $T$; the subscript 1 and 2 indicates the initial and final condition of air.

Write the formula for mass of water at initial.

$m_{w,1}=\frac{{\nu }_2-{\nu }_1}{v_{w,1}}$ (III)

Write the general energy balance equation.

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

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

Refer Equation (IV).

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

$Q_{\text{in}}+m_{\text{in}}{h}_{\text{in}}-m_{\text{out}}{h}_{\text{out}}=m_2{u}_2-m_1{u}_1$ (V)

Here, the mass is $m$, the internal energy is $u$, the enthalpy is $h$, the heat in is $Q_{\text{in}}$.

At final state 2:

The mass of water in the tank is zero. Only air is present in the ballast tank. Hence, the final state energy of the tank is expressed as follows.

$\begin{array}{c}{m}_2{u}_2=m_{2,a}{u}_{2,a}\\ =m_{2,a}{c}_{v,a}{T}_{2,a}\end{array}$

At initial state 1:

Both air and water is present in the tank. The initial state energy of the tank is expressed as follows.

$\begin{array}{c}{m}_1{u}_1=m_{a,1}{u}_{a,1}+m_{w,1}{u}_{w,1}\\ =m_{a,1}{c}_v{T}_{a,1}+m_{w,1}{u}_{w,1}\end{array}$

The inlet mass of air is expressed as,

$m_{\text{in}}=m_{a,2}-m_{a,1}$

Here, the inlet mass is mass of air and the exit mass is mass of water.

Rewrite the Equation (V) as follows.

$\begin{array}{l}{Q}_{\text{in}}+\left[\left(m_{a,2}-m_{a,1}\right)c_{p,a}T\right]-m_w{h}_w=\left(m_{2,a}{c}_{v,a}{T}_{a,2}\right)-\left(m_{a,1}{c}_v{T}_{a,1}+m_{w,1}{u}_{w,1}\right)\\ Q_{\text{in}}=m_{a,2}{c}_{v,a}{T}_{a,2}-\left(m_{a,1}{c}_v{T}_{a,1}+m_{w,1}{u}_{w,1}\right)+m_w{h}_w-\left(m_{a,2}-m_{a,1}\right)c_{p,a}T\\ Q_{\text{in}}=m_{2,a}{c}_{v,a}{T}_{2,a}-m_{a,1}{c}_v{T}_{a,1}-m_{w,1}{u}_{w,1}+m_w{h}_w-\left(m_{a,2}-m_{a,1}\right)c_{p,a}T\end{array}$ (VI)

Refer Table A-4, “Saturated water-Temperature table”.

The enthalpy $\left(h_w=h_{f@15°\text{C}}\right)$ and internal energy $\left(u_w=u_{f@15°\text{C}}\right)$ of water (saturated liquid) corresponding to the temperature of $15°\text{C}$ is, $u_w=h_w=62.98\text{kJ/kg}$.

It is given that the pressure and temperature of air is kept constant.

$T_{a,1}=T_{a,2}=T$

The Equation (VI) is reduced as follows.

$\begin{array}{l}{Q}_{\text{in}}=m_{2,a}{c}_{v,a}{T}_{2,a}-m_{a,1}{c}_v{T}_{a,1}-m_{w,1}{u}_{w,1}+m_w{h}_w-\left(m_{a,2}-m_{a,1}\right)c_{p,a}T\\ Q_{\text{in}}=\left(m_{2,a}-m_{a,1}\right)c_{v,a}T-0-\left(m_{a,2}-m_{a,1}\right)c_{p,a}T\\ Q_{\text{in}}=\left(m_{2,a}-m_{a,1}\right)c_{v,a}T-\left(m_{a,2}-m_{a,1}\right)c_{p,a}T\end{array}$ (VII)

The specific volume $\left(v_w=v_{f@15°\text{C}}\right)$ of water corresponding to the temperature of $15°\text{C}$ is $0.001001{\text{m}}^3\text{/kg}$.

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-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ of air is $1.005\text{kJ/kg}\cdot \text{K}$.

The specific heat at constant volume $\left(c_v\right)$ of air is $0.718\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Here, $P_{a,2}=P_{a,1}=1500\text{kPa}$.

Substitute $1500\text{kPa}$ for $P_1$, $100{\text{m}}^3$ for $v_1$, $0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $\left(15+273.15\right)\text{K}$ for $T_1$ in Equation (I).

$\begin{array}{c}{\dot{m}}_{a,1}=\frac{1500\text{kPa}\left(100{\text{m}}^3\right)}{0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\times \left(15+273.15\right)\text{K}}\\ =\frac{150000\text{kPa}\cdot {\text{m}}^3}{82.69905\text{kPa}{\text{.m}}^3\text{/kg}}\\ =1813.8056\text{kg}\\ \simeq 1814\text{kg}\end{array}$

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 $\left(15+273.15\right)\text{K}$ for $T_2$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{2,a}=\frac{1500\text{kPa}\left(700{\text{m}}^3\right)}{0.287\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\times \left(15+273.15\right)\text{K}}\\ =\frac{1050000\text{kPa}\cdot {\text{m}}^3}{86.69905\text{kPa}{\text{.m}}^3\text{/kg}}\\ =12696.6362\text{kg}\\ \simeq 12697\text{kg}\end{array}$

Thus, the final mass of air in the ballast tank is $12697\text{kg}$.

Substitute $700{\text{m}}^3$ for $v_2$, $100{\text{m}}^3$ for $v_1$, and $0.001001{\text{m}}^3\text{/kg}$ for $v_{w,1}$ in Equation (III).

$\begin{array}{c}{m}_{w,1}=\frac{700{\text{m}}^3-100{\text{m}}^3}{0.001001{\text{m}}^3\text{/kg}}\\ =599400.5994\text{kg}\\ \simeq 600000\text{kg}\end{array}$

Substitute $12697\text{kg}$ for $m_{2,a}$, $1814\text{kg}$ for $m_{a,1}$, $0.718\text{kJ/kg}\cdot \text{K}$ for $c_{v,a}$, $288.15\text{K}$ for $T_{2,a}$, , $288.15\text{K}$ for $T_{a,1}$, $600000\text{kg}$ for $m_w$, $62.98\text{kJ/kg}$ for $h_w$,

$\begin{array}{c}{Q}_{\text{in}}=\left[\begin{array}{l}\left(12697\text{kg}-1814\text{kg}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)\left(288.15\text{K}\right)\\ -\left(12697\text{kg}-1814\text{kg}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(288.15\text{K}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(10883\text{kg}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)\left(288.15\text{K}\right)\\ -\left(10883\text{kg}\right)\left(1.005\text{kJ/kg}\cdot \text{K}\right)\left(288.15\text{K}\right)\end{array}\right]\\ =2251602.371\text{kJ}-3137136.595\text{kJ}\\ =-885534.2238\text{kJ}\end{array}$

Here, the negative sign indicates that the heat is removed.

$\begin{array}{c}{Q}_{\text{out}}=885534.2238\text{kJ}\\ \simeq 885534.2\text{kJ}\end{array}$

Thus, the total heat transfer is $885534.2\text{kJ}$.