Chapter 14.7, Problem 113P

To determine

The exergy lost in the cooling tower.

Answer to Problem 113P

The exergy lost in the cooling tower is $\text{1}\text{.93}\text{kJ}/\text{kg}\text{dry}\text{air}$.

Explanation of Solution

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

$\begin{array}{c}{\dot{m}}_{a1}={\dot{m}}_{a2}\\ ={\dot{m}}_a\end{array}$

Here, the mass flow rate of air at inlet is ${\dot{m}}_{a1}$, mass flow rate of dry air at exit is ${\dot{m}}_{a2}$ and mass flow rate of dry air is ${\dot{m}}_a$.

Express the water mass balance:

$\begin{array}{l}{\displaystyle \sum {\dot{m}}_{w,i}}={\displaystyle \sum {\dot{m}}_{w,e}}\\ {\dot{m}}_3+{\dot{m}}_{a1}{\omega }_1={\dot{m}}_4+{\dot{m}}_2{\omega }_2\\ {\dot{m}}_3-{\dot{m}}_4={\dot{m}}_a\left({\omega }_2-{\omega }_1\right)\end{array}$

Here, mass flow rate of water at inlet and exit is ${\dot{m}}_{w,i}\text{and}{\dot{m}}_{w,e}$ respectively, specific humidity at state 1 and 2 is ${\omega }_1\text{and}{\omega }_2$ respectively; mass flow rate at state 2, 3 and 4 is ${\dot{m}}_2,{\dot{m}}_3\text{and}{\dot{m}}_4$ respectively.

Express the energy balance.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}\\ {\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\displaystyle \sum {\dot{m}}_i{h}_i=}{\displaystyle \sum {\dot{m}}_e{h}_e}\end{array}$

$\begin{array}{l}0={\displaystyle \sum {\dot{m}}_e{h}_e}-{\displaystyle \sum {\dot{m}}_i{h}_i}\\ 0={\dot{m}}_{a2}{h}_2+{\dot{m}}_4{h}_4-{\dot{m}}_{a1}{h}_1-{\dot{m}}_3{h}_3\\ {\dot{m}}_a=\frac{{\dot{m}}_3\left(h_3-h_4\right)}{\left(h_2-h_1\right)-\left({\omega }_2-{\omega }_1\right)h_4}\end{array}$ (I)

Here, the rate of total energy entering the system is ${\dot{E}}_{\text{in}}$, the rate of total energy leaving the system is ${\dot{E}}_{\text{out}}$, the rate of change in the total energy of the system is $\Delta {\dot{E}}_{\text{system}}$, initial and exit mass flow rate is ${\dot{m}}_i\text{and}{\dot{m}}_e$ respectively, enthalpy at inlet and exit is $h_i\text{and}{h}_e$ respectively and enthalpy at state 1, 2, 3 and 4 is $h_1,h_2,h_3\text{and}{h}_4$ respectively.

Determine the mass flow rate of steam at state 3 per unit mass of dry air.

$m_3=\frac{{\dot{m}}_3}{{\dot{m}}_a}$ (II)

Determine the mass flow rate of steam at state 4 per unit mass of dry air.

$m_4=m_3-\left({\omega }_2-{\omega }_1\right)$ (III)

Determine the change in entropy of water steam.

$\Delta s_{\text{water}}=m_4{s}_4-m_3{s}_3$ (IV)

Determine the change in entropy of water vapour in the air stream.

$\Delta s_{\text{vapour}}={\omega }_2{s}_{g,2}-{\omega }_1{s}_{g,1}$ (V)

Determine the partial pressure of water vapour at state 1 for air steam.

$\begin{array}{c}{P}_{vap,1}={\varphi }_1\times P_{g1}\\ ={\varphi }_1\times P_{\text{sat@15°C}}\end{array}$ (VI)

Determine the partial pressure of dry air at state 1 for air steam.

$P_{a1}=P_1-P_{vap,1}$ (VII)

Here, the pressure at the state 1 is $P_1$.

Determine the partial pressure of water vapour at state 2 for air steam.

$\begin{array}{c}{P}_{vap,2}={\varphi }_2\times P_{g2}\\ ={\varphi }_2\times P_{\text{sat@18°C}}\end{array}$ (VIII)

Determine the partial pressure of dry air at state 2 for air steam.

$P_{a2}=P_2-P_{vap,2}$ (IX)

Here, the pressure at the state 1 is $P_1$.

Determine the entropy change of dry air.

$\begin{array}{c}\Delta s_a=s_2-s_1\\ =c_p\times \ln \frac{T_2}{T_1}-R\times \ln \frac{P_{a,2}}{P_{a,1}}\end{array}$ (X)

Here, the temperature at the state 1 is $T_1$, the temperature at the state 2 is $T_2$, and the universal gas constant is $R$.

Determine the entropy generation in the cooling tower is the total entropy change.

$s_{\text{gen}}=\Delta s_{\text{water}}+\Delta s_{\text{vapour}}+\Delta s_{\text{a}}$ (XI)

Determine the exergy destruction per unit mass of dry air.

$x_{\text{dest}}=T_0{s}_{\text{gen}}$ (XII)

Conclusion:

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{15°C}$ and relative humidity of $\text{25\%}$.

$\begin{array}{l}{h}_1=21.8\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_1=0.00264\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ {\nu }_1=0.820{\text{m}}^{\text{3}}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Figure A-31, “psychometric chart at $\text{1 atm}$ total pressure”, and write the properties corresponding to dry bulb temperature of $\text{18°C}$ and relative humidity of $\text{95\%}$.

$\begin{array}{l}{h}_2=49.3\text{kJ}/\text{kg}\text{dry}\text{air}\\ {\omega }_2=0.0123\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy and entropy at state 3 at temperature of $\text{30°C}$.

$\begin{array}{c}{h}_3=h_f\\ =125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\end{array}$

$\begin{array}{c}{s}_3=s_f\\ =8.7803\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, enthalpy of saturation liquid is $h_f$.

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy at state 4 at temperature of $\text{22°C}$ using an interpolation method.

$h_4=h_{f@22\text{°C}}$ (XIII)

Here, enthalpy of saturation liquid at temperature of $\text{22°C}$ is $h_{f@22\text{°C}}$.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (XIV)

Here, the variables denote by x and y is temperature and specific enthalpy at state 4 respectively.

Show the specific enthalpy at state 4 corresponding to temperature as in Table (1).

Temperature $T\left(\text{°C}\right)$	Specific enthalpy at state 4 $h_{4@h_f}\left(\text{kJ}/\text{kg}\right)$
20 $\left(x_1\right)$	83.915 $\left(y_1\right)$
22 $\left(x_2\right)$	$\left(y_2=?\right)$
25 $\left(x_3\right)$	104.83 $\left(y_3\right)$

Substitute $\text{20°C,}\text{22°C}\text{and}\text{25°C}$ for $x_1,x_2\text{and}{x}_3$ respectively, $83.915\text{kJ}/\text{kg}$ for $y_1$ and $\text{104}\text{.83}\text{kJ}/\text{kg}$ for $y_3$ in Equation (XIV).

$\begin{array}{c}{y}_2=\frac{\left(\text{22°C}-\text{20°C}\right)\left(\text{104}\text{.83}\text{kJ}/\text{kg}-83.915\text{kJ}/\text{kg}\right)}{\left(\text{25°C}-\text{20°C}\right)}+83.915\text{kJ}/\text{kg}\\ =\text{92}\text{.28}\text{kJ}/\text{kg}\\ =h_{f@22\text{°C}}\end{array}$

Substitute $\text{92}\text{.28}\text{kJ}/\text{kg}$ for $h_{f@22\text{°C}}$ in Equation (XIII).

$h_4=92.28\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$

Repeat the Equation (XIV), obtain the value of entropy for water streams at $22°\text{C}$ of temperature as $0.3249\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $\text{5}\text{kg}/\text{s}$ for ${\dot{m}}_3$, $125.74\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_3$, $92.28\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}$ for $h_4$, $49.3\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_2$, $21.8\text{kJ}/\text{kg}\text{dry}\text{air}$ for $h_1$, $0.00264\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.0123\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (I).

$\begin{array}{c}{\dot{m}}_a=\frac{\left(\text{5}\text{kg}/\text{s}\right)\left(125.74-92.28\right)\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}}{\left(49.3-21.8\right)\text{kJ}/\text{kg}-\left(0.0123-0.00264\right)\left(92.28\text{kJ}/\text{kg}{\text{H}}_{\text{2}}\text{O}\right)}\\ =6.29\text{kg}/\text{s}\end{array}$

Substitute $5\text{}\text{kg}\text{water}/\text{s}$ for ${\dot{m}}_3$ and $6.29\text{}\text{kg}\text{dry}\text{}\text{air}/\text{s}$ for ${\dot{m}}_a$ in Equation (II).

$\begin{array}{c}{m}_3=\frac{5\text{}\text{kg}\text{water}/\text{s}}{6.29\text{}\text{kg}\text{dry}\text{}\text{air}/\text{s}}\\ =0.7949\text{kg}\text{water}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.7949\text{kg}\text{water}/\text{kg}\text{dry}\text{air}$ for $m_3$, $0.00264\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.0123\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (III).

$\begin{array}{c}{m}_4=\left(0.7949\text{kg}\text{water}/\text{kg}\text{dry}\text{air}\right)-\left(0.0123-0.00264\right)\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\\ =\left(0.7949\text{kg}\text{water}/\text{kg}\text{dry}\text{air}\right)-\left(0.00966\text{}\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\right)\\ =0.7852\text{kg}\text{water}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.7852\text{kg}\text{water}/\text{kg}\text{dry}\text{air}$ for $m_4$, $0.7949\text{kg}\text{water}/\text{kg}\text{dry}\text{air}$ for $m_3$, $0.3249\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$, and $8.7803\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$ in Equation (IV).

$\begin{array}{c}\Delta s_{\text{water}}=\left[\begin{array}{l}\left(0.7852\text{kg}\text{water}/\text{kg}\text{dry}\text{air}\right)\times \left(0.3249\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(0.7949\text{kg}\text{water}/\text{kg}\text{dry}\text{air}\right)\times \left(8.7803\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =\left(0.255111\text{kJ}/\text{kg}\cdot \text{K}\text{dry}\text{air}\right)-\left(0.347212\text{kJ}/\text{kg}\cdot \text{K}\text{dry}\text{air}\right)\\ =-0.0921\text{kJ}/\text{kg}\cdot \text{K}\text{dry}\text{air}\end{array}$

Refer Table A-4, “saturated water-temperature table”, and write the entropy at state 1 at temperature of $\text{15°C}$.

$\begin{array}{c}{s}_{g1}=s_{g@15°\text{C}}\\ =8.7803\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Repeat the Equation (XIV), obtain the value of entropy for water vapour at $18°\text{C}$ of temperature as $8.7112\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $8.7803\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{g1}$, $8.7112\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{g2}$, $0.00264\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_1$ and $0.0123\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}$ for ${\omega }_2$ in Equation (V).

$\begin{array}{c}\Delta s_{\text{vapour}}=\left[\begin{array}{l}\left(0.0123\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\right)\times \left(8.7112\text{kJ}/\text{kg}\cdot \text{K}\right)\\ -\left(0.00264\text{kg}{\text{H}}_{\text{2}}\text{O}/\text{kg}\text{dry}\text{air}\right)\times \left(8.7803\text{kJ}/\text{kg}\cdot \text{K}\right)\end{array}\right]\\ =\left[\left(0.107148\text{kJ}/\text{K}\cdot \text{kg}\text{dry}\text{air}\right)-\left(0.02318\text{kJ}/\text{K}\cdot \text{kg}\text{dry}\text{air}\right)\right]\\ =0.083968\text{kJ}/\text{K}\cdot \text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.25$ for ${\varphi }_1$ and $1.7057\text{kPa}$ for $P_{sat@15°\text{C}}$ in Equation (VI).

$\begin{array}{c}{P}_{vap,1}=\left(0.25\right)\times \left(1.7057\text{kPa}\right)\\ =0.4264\text{kPa}\end{array}$

Substitute 1 atm for $P_1$ and $0.4264\text{kPa}$ for $P_{vap,1}$ in Equation (VII).

$\begin{array}{c}{P}_{a1}=1\text{atm}-\left(0.4264\text{kPa}\right)\\ =1\text{atm}\times \left(\frac{\left(101.325\text{kPa}\right)}{1\text{atm}}\right)-\left(0.4264\text{kPa}\right)\\ =101.325\text{kPa}-0.4264\text{kPa}\\ =100.8986\text{kPa}\end{array}$

Substitute $0.95$ for ${\varphi }_2$ and $2.065\text{kPa}$ for $P_{sat@18°\text{C}}$ in Equation (VIII).

$\begin{array}{c}{P}_{vap,2}=\left(0.95\right)\times \left(2.065\text{kPa}\right)\\ =1.962\text{kPa}\end{array}$

Substitute 1 atm for $P_2$ and $1.962\text{kPa}$ for $P_{vap,2}$ in Equation (IX).

$\begin{array}{c}{P}_{a2}=1\text{atm}-\left(1.962\text{kPa}\right)\\ =1\text{atm}\times \left(\frac{\left(101.325\text{kPa}\right)}{1\text{atm}}\right)-\left(1.962\text{kPa}\right)\\ =101.325\text{kPa}-1.962\text{kPa}\\ =99.36\text{kPa}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $15°\text{C}$ for $T_1$, $18°\text{C}$ for $T_2$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, 99.36 kPa for $P_{a2}$, and $100.90\text{kPa}$ for $P_{a1}$ in Equation (X).

$\begin{array}{c}\Delta s_a=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \ln \frac{\left(18°\text{C}\right)}{\left(15°\text{C}\right)}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \ln \frac{\left(99.36\text{kPa}\right)}{\left(100.90\text{kPa}\right)}\\ =\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \ln \frac{\left(18°\text{C+273}\right)}{\left(15°\text{C+273}\right)}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \ln \frac{\left(99.36\text{kPa}\right)}{\left(100.90\text{kPa}\right)}\\ =\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\times \ln \left(1.010417\text{K}\right)-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\times \ln \left(0.984737\text{kPa}\right)\\ =0.01483\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $-0.09210\text{kJ}/\text{kg}\text{dry}\text{air}$ for $\Delta s_{\text{water}}$, $0.08397\text{kJ}/\text{kg}\cdot \text{K}\text{dry}\text{air}$ for $\Delta s_{\text{vapour}}$ and $0.01483\text{kJ}/\text{kg}\text{dry}\text{air}$ for $\Delta s_a$ in Equation (XI)

$\begin{array}{c}{s}_{\text{gen}}=\left[\begin{array}{l}\left(-0.09210\text{kJ}/\text{kg}\text{dry}\text{air}\right)\\ +\left(0.08397\text{kJ}/\text{kg}\cdot \text{K}\text{dry}\text{air}\right)\\ +\left(0.01483\text{kJ}/\text{kg}\text{dry}\text{air}\right)\end{array}\right]\\ =0.00670\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

Substitute $0.00670\text{kJ}/\text{kg}\text{dry}\text{air}$ for $s_{\text{gen}}$ and $15°\text{C}$ for $T_0$ in Equation (XII).

$\begin{array}{c}{x}_{\text{dest}}=\left(15°\text{C}\right)\times \left(0.00670\text{kJ}/\text{kg}\text{dry}\text{air}\right)\\ =\left(15°\text{C}+\text{273}\right)\times \left(0.00670\text{kJ}/\text{kg}\text{dry}\text{air}\right)\\ =\left(288\text{K}\right)\times \left(0.00670\text{kJ}/\text{kg}\text{dry}\text{air}\right)\\ =1.9296\text{kJ}/\text{kg}\text{dry}\text{air}\end{array}$

      $\cong 1.93\text{kJ}/\text{kg}\text{dry}\text{air}$

Thus, the exergy lost in the cooling tower is $\text{1}\text{.93}\text{kJ}/\text{kg}\text{dry}\text{air}$.