Chapter 10.9, Problem 69P

The schematic of a single-flash geothermal power plant with state numbers is given in Fig. P10–69. Geothermal resource exists as saturated liquid at 230°C. The geothermal liquid is withdrawn from the production well at a rate of 230 kg/s and is flashed to a pressure of 500 kPa by an essentially isenthalpic flashing process where the resulting vapor is separated from the liquid in a separator and is directed to the turbine. The steam leaves the turbine at 10 kPa with a moisture content of 5 percent and enters the condenser where it is condensed; it is routed to a reinjection well along with the liquid coming off the separator. Determine (a) the power output of the turbine and the thermal efficiency of the plant, (b) the exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destructions and the second-law efficiencies for (c) the turbine and (d) the entire plant.

FIGURE P10–69

To determine

The temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam at the exit of flash chamber is $1\text{MPa}$.

Answer to Problem 69P

The power output turbine is $10.8\text{kW}$.

The thermal efficiency of the plant is $5.3\%$.

Explanation of Solution

Draw schematic diagram of single flash geothermal power plant as shown in Figure 1.

Write the general energy rate balance equation.

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

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 $\Delta {\dot{E}}_{\text{system}}$.

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

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

The Equation (I) is reduced as follows.

$\begin{array}{l}{\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=0\\ {\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\end{array}$

Refer Figure 1.

The flash chamber is nothing but the expansion valve. At expansion valve, the enthalpy kept constant.

Express the energy balance equation for the flash chamber.

$h_1=h_2$

Express the energy balance equation for the separator.

${\dot{m}}_2{h}_2={\dot{m}}_3{h}_3+{\dot{m}}_{\text{liquid}}{h}_{\text{liquid}}$ (II)

Express the energy balance equation for the turbine.

${\dot{W}}_{\text{T,out}}={\dot{m}}_3\left(h_3-h_{\text{4}}\right)$ (III)

At state 1:

The geothermal water is extracted at the state of saturated liquid at the temperature of $230°\text{C}$.

The enthalpy and entropy at state 1 is as follows.

$\begin{array}{l}{h}_1=h_{f@230°\text{C}}\\ s_1=s_{f@230°\text{C}}\end{array}$

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

The enthalpy $\left(h_1\right)$ and entropy $\left(s_1\right)$ corresponding to the temperature of $230°\text{C}$ is $990.14\text{kJ/kg}$ and $2.6100\text{kJ/kg}\cdot \text{K}$ respectively.

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

At state 2:

The exit pressure of the flash chamber is $500\text{kPa}$.

The geothermal steam is flashed at constant enthalpy. The exit steam of the flash chamber is at the quality of $x_2$.

$x_2=\frac{h_2-h_{f,2}}{h_{fg,2}}$ (IV)

Here, the fluid enthalpy is $h_f$, the vaporization enthalpy is $h_{fg}$ and subscript 2 indicates state 2.

Refer Table A-5, “Saturated water-Pressure table”.

Obtain the following corresponding to the pressure of $500\text{kPa}$.

$\begin{array}{l}{h}_{f,2}=640.09\text{kJ/kg}\\ h_{fg,2}=2108.0\text{kJ/kg}\\ s_{f,2}=1.8604\text{kJ/kg}\cdot \text{K}\\ s_{fg,2}=4.9603\text{kJ/kg}\cdot \text{K}\end{array}$

The entropy $\left(s_2\right)$ at state 2 is expressed as follows.

$s_2=s_{f,2}+x_2{s}_{fg,2}$ (V)

Write the formula for mass flow rate of vapor at entering the turbine.

${\dot{m}}_3=x_2{\dot{m}}_2$ (VI)

Here, the mass flow rate is $\dot{m}$ and the subscripts 2 and 3 indicates the process states.

At state 3:

There is no pressure drop in the separator. The separator separates vapor and liquid form the flashed steam, and the separated vapor alone sent to the turbine.

The enthalpy $\left(h_3\right)$ and entropy $\left(s_3\right)$ at state 3 is expressed as follows.

$\begin{array}{l}{h}_3=h_{g@500\text{kPa}}\\ s_3=s_{g@500\text{kPa}}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”.

The enthalpy $\left(h_3\right)$ and entropy $\left(s_3\right)$ at state 3 corresponding to the pressure of $500\text{kPa}$ is $2748.1\text{kJ/kg}$ and $6.8207\text{kJ/kg}\cdot \text{K}$ respectively.

At state 4:

The steam is at the state of saturated mixture at the pressure of $10\text{kPa}$ with $5\%$ of moisture content.

The quality at state 4 is as follows.

$\begin{array}{c}{x}_4=100\%-5\%\\ =95\%\times \frac{1}{100}\\ =0.95\end{array}$

The enthalpy $\left(h_4\right)$ and entropy $\left(s_4\right)$ at state 4 is expressed as follows.

$h_4=h_{f,4}+x_4{h}_{fg,4}$ (VII)

$s_4=s_{f,4}+x_4{s}_{fg,4}$ (VIII)

Refer Table A-5, “Saturated water-Pressure table”.

Obtain the following corresponding to the pressure of $10\text{kPa}$.

$\begin{array}{l}{h}_{f,4}=191.81\text{kJ/kg}\\ h_{fg,4}=2392.1\text{kJ/kg}\\ s_{f,4}=0.6492\text{kJ/kg}\cdot \text{K}\\ s_{fg,4}=7.4996\text{kJ/kg}\cdot \text{K}\end{array}$

At state 6:

The saturated water only exits at the bottom of the separator. The enthalpy $\left(h_6\right)$ and entropy $\left(s_6\right)$ at state 6 is expressed as follows.

$\begin{array}{l}{h}_6=h_{f@10\text{kPa}}\\ s_6=s_{f@10\text{kPa}}\end{array}$

Refer Table A-5, “Saturated water-Pressure table”.

The enthalpy $\left(h_6\right)$ and entropy $\left(s_6\right)$ at state 6 corresponding to the pressure of $500\text{kPa}$ is $640.09\text{kJ/kg}$ and $1.8604\text{kJ/kg}\cdot \text{K}$ respectively.

Write the formula for net energy input of the plant.

${\dot{E}}_{\text{in}}={\dot{m}}_1\left(h_1-h_0\right)$ (IX)

Write the formula for thermal efficiency.

${\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{T,out}}}{{\dot{E}}_{\text{in}}}$ (X)

Consider, the surrounding temperature is $25°\text{C}$ and the water re injected in to the well is at saturated liquid state.

$T_0=25°\text{C}=298\text{K}$

The surrounding enthalpy $\left(h_0\right)$ and entropy $\left(s_0\right)$ is expressed as follows.

$\begin{array}{l}{h}_0=h_{f@298\text{K}}\\ s_0=s_{f@298\text{K}}\end{array}$

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

The enthalpy $\left(h_0\right)$ and entropy $\left(s_0\right)$ corresponding to the temperature of $298\text{K}$ is $104.83\text{kJ/kg}$ and $0.3672\text{kJ/kg}\cdot \text{K}$ respectively.

Conclusion:

Substitute $990.14\text{kJ/kg}$ for $h_2$, $640.09\text{kJ/kg}$ for $h_{f,2}$, and $2108.0\text{kJ/kg}$ for $h_{fg,2}$ in Equation (IV).

$\begin{array}{c}{x}_2=\frac{990.14\text{kJ/kg}-640.09\text{kJ/kg}}{2108.0\text{kJ/kg}}\\ =0.1661\end{array}$

Substitute $1.8604\text{kJ/kg}\cdot \text{K}$ for $s_{f,2}$, $0.1661$ for $x_2$, and $4.9603\text{kJ/kg}\cdot \text{K}$ for $s_{fg,2}$ in Equation (V).

$\begin{array}{c}{s}_2=1.8604\text{kJ/kg}\cdot \text{K}+0.1661\left(4.9603\text{kJ/kg}\cdot \text{K}\right)\\ =1.8604\text{kJ/kg}\cdot \text{K}+0.8239\text{kJ/kg}\cdot \text{K}\\ =2.6843\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $0.166$ for $x_2$ and $230\text{kg/s}$ for ${\dot{m}}_2$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_3=0.1661\left(230\text{kg/s}\right)\\ =38.203\text{kg/s}\end{array}$

Substitute $191.81\text{kJ/kg}$ for $h_{f,4}$, $0.95$ for $x_4$, and $2392.1\text{kJ/kg}$ for $h_{fg,4}$ in

Equation (VII).

$\begin{array}{c}{h}_4=191.81\text{kJ/kg}+0.95\left(2392.1\text{kJ/kg}\right)\\ =191.81\text{kJ/kg}+2272.495\text{kJ/kg}\\ =2464.305\text{kJ/kg}\\ \simeq 2464.3\text{kJ/kg}\end{array}$

Substitute $0.6492\text{kJ/kg}\cdot \text{K}$ for $s_{f,4}$, $0.95$ for $x_4$, and $7.4996\text{kJ/kg}\cdot \text{K}$ for $s_{fg,4}$ in Equation (VIII).

$\begin{array}{c}{s}_4=0.6492\text{kJ/kg}\cdot \text{K}+0.95\left(7.4996\text{kJ/kg}\cdot \text{K}\right)\\ =0.6492\text{kJ/kg}\cdot \text{K}+7.1246\text{kJ/kg}\cdot \text{K}\\ =7.7738\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $38.203\text{kg/s}$ for ${\dot{m}}_3$, $2748.1\text{kJ/kg}$ for $h_3$, and $2464.3\text{kJ/kg}$ for $h_4$ in

Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{T,out}}=38.203\text{kg/s}\left(2748.1\text{kJ/kg}-2464.3\text{kJ/kg}\right)\\ =38.203\text{kg/s}\left(283.8\text{kJ/kg}\right)\\ =10842.0114\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =10842\text{kW}\times \frac{1\text{MW}}{{10}^3\text{kW}}\end{array}$

$\begin{array}{c}=10.843\text{MW}\\ \simeq 10.8\text{MW}\end{array}$

Thus, the power output turbine is $10.8\text{kW}$.

Substitute $230\text{kg/s}$ for ${\dot{m}}_1$, $990.14\text{kJ/kg}$ for $h_1$, and $104.83\text{kJ/kg}$ for $h_0$ in

Equation (IX).

$\begin{array}{c}{\dot{E}}_{\text{in}}=230\text{kg/s}\left(990.14\text{kJ/kg}-104.83\text{kJ/kg}\right)\\ =230\text{kg/s}\left(885.31\text{kJ/kg}\right)\\ =203621.3\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =230621.3\text{kW}\end{array}$

Substitute $10842\text{kW}$ for ${\dot{W}}_{\text{T,out}}$ and $230621.3\text{kW}$ for ${\dot{E}}_{\text{in}}$ in Equation (X).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{10842\text{kW}}{230621.3\text{kW}}\\ =0.05324\times 100\\ =5.324\%\\ \simeq 5.3\%\end{array}$

Thus, the thermal efficiency of the plant is $5.3\%$.

To determine

The exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destructions.

Answer to Problem 69P

The exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destruction is $90.2864\text{kJ/kg}$ and $17.3\text{MW}$ respectively.

Explanation of Solution

Write the formula for exergy of the steam at their respective process state.

${\psi }_i=\left(h_i-h_0\right)-T_0\left(s_i-s_0\right)$ (XI)

Here, the enthalpy is $h$, the temperature is $T$, the entropy is $s$, subscripts $i$ indicates the $i^{\text{th}}$ process state, and the subscript $0$ indicates the surrounding state.

Write the formula for rate of exergy destruction at the exit of flash chamber (state 6).

${\dot{X}}_6={\dot{m}}_6{\psi }_6$ (XII)

Here, the rate of exergy destruction at state 6 is ${\dot{X}}_{\text{6}}$.

Conclusion:

For process state 1:

$\begin{array}{l}{h}_i=h_1=990.14\text{kJ/kg}\\ s_i=s_1=2.6100\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $990.14\text{kJ/kg}$ for $h_i$, $104.83\text{kJ/kg}$ for $h_0$, $298\text{}\text{K}$ for $T_0$, $2.6100\text{kJ/kg}\cdot \text{K}$ for $s_i$, and $0.3672\text{kJ/kg}\cdot \text{K}$ for $s_0$ in Equation (XI).

$\begin{array}{c}{\psi }_1=\left(990.14\text{kJ/kg}-104.83\text{kJ/kg}\right)-298\text{}\text{K}\left(2.6100\text{kJ/kg}\cdot \text{K}-0.3672\text{kJ/kg}\cdot \text{K}\right)\\ =885.31\text{kJ/kg}-668.3544\text{kJ/kg}\\ =216.9556\text{kJ/kg}\end{array}$

For process state 2:

$\begin{array}{l}{h}_i=h_2=990.14\text{kJ/kg}\\ s_i=s_2=2.6843\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $990.14\text{kJ/kg}$ for $h_i$, $104.83\text{kJ/kg}$ for $h_0$, $298\text{}\text{K}$ for $T_0$, $2.6843\text{kJ/kg}\cdot \text{K}$ for $s_i$, and $0.3672\text{kJ/kg}\cdot \text{K}$ for $s_0$ in Equation (XI).

$\begin{array}{c}{\psi }_2=\left(990.14\text{kJ/kg}-104.83\text{kJ/kg}\right)-298\text{}\text{K}\left(2.6843\text{kJ/kg}\cdot \text{K}-0.3672\text{kJ/kg}\cdot \text{K}\right)\\ =885.31\text{kJ/kg}-690.4958\text{kJ/kg}\\ =194.814\text{kJ/kg}\end{array}$

For process state 3:

$\begin{array}{l}{h}_i=h_3=2748.1\text{kJ/kg}\\ s_i=s_3=6.8207\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $2748.1\text{kJ/kg}$ for $h_i$, $104.83\text{kJ/kg}$ for $h_0$, $298\text{}\text{K}$ for $T_0$, $6.8207\text{kJ/kg}\cdot \text{K}$ for $s_i$, and $0.3672\text{kJ/kg}\cdot \text{K}$ for $s_0$ in Equation (XI).

$\begin{array}{c}{\psi }_3=\left(2748.1\text{kJ/kg}-104.83\text{kJ/kg}\right)-298\text{}\text{K}\left(6.8207\text{kJ/kg}\cdot \text{K}-0.3672\text{kJ/kg}\cdot \text{K}\right)\\ =2643.27\text{kJ/kg}-1923.143\text{kJ/kg}\\ =720.127\text{kJ/kg}\end{array}$

For process state 4:

$\begin{array}{l}{h}_i=h_4=2464.3\text{kJ/kg}\\ s_i=s_4=7.7738\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $2464.3\text{kJ/kg}$ for $h_i$, $104.83\text{kJ/kg}$ for $h_0$, $298\text{}\text{K}$ for $T_0$, $7.7738\text{kJ/kg}\cdot \text{K}$ for $s_i$, and $0.3672\text{kJ/kg}\cdot \text{K}$ for $s_0$ in Equation (XI).

$\begin{array}{c}{\psi }_4=\left(2464.3\text{kJ/kg}-104.83\text{kJ/kg}\right)-298\text{}\text{K}\left(7.7738\text{kJ/kg}\cdot \text{K}-0.3672\text{kJ/kg}\cdot \text{K}\right)\\ =2359.47\text{kJ/kg}-2207.1668\text{kJ/kg}\\ =152.3032\text{kJ/kg}\end{array}$

For process state 6:

$\begin{array}{l}{h}_i=h_6=640.09\text{kJ/kg}\\ s_i=s_6=1.8604\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $640.09\text{kJ/kg}$ for $h_i$, $104.83\text{kJ/kg}$ for $h_0$, $298\text{}\text{K}$ for $T_0$, $1.8604\text{kJ/kg}\cdot \text{K}$ for $s_i$, and $0.3672\text{kJ/kg}\cdot \text{K}$ for $s_0$ in Equation (XI).

$\begin{array}{c}{\psi }_6=\left(640.09\text{kJ/kg}-104.83\text{kJ/kg}\right)-298\text{}\text{K}\left(1.8604\text{kJ/kg}\cdot \text{K}-0.3672\text{kJ/kg}\cdot \text{K}\right)\\ =535.26\text{kJ/kg}-444.9736\text{kJ/kg}\\ =90.2864\text{kJ/kg}\end{array}$

The mass flow rate of water at the bottom exit of separator (state 6) is expressed as follows.

$\begin{array}{c}{\dot{m}}_6={\dot{m}}_1-{\dot{m}}_3\\ =230\text{kg/s}-38.203\text{kg/s}\\ =191.797\text{kg/s}\end{array}$

Substitute $191.797\text{kg/s}$ for ${\dot{m}}_{\text{6}}$ and $90.2864\text{kJ/kg}$ for ${\psi }_6$ in Equation (XII).

$\begin{array}{c}{\dot{X}}_6=191.797\text{kg/s}\left(90.2864\text{kJ/kg}\right)\\ =17316.66066\text{kJ/s}\times \frac{1\text{MW}}{{10}^3\text{kJ/s}}\\ =17.3166\text{MW}\\ \simeq 17.3\text{MW}\end{array}$

Thus, the exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destruction is $90.2864\text{kJ/kg}$ and $17.3\text{MW}$ respectively.

To determine

The exergy destruction and second law of efficiency for the turbine.

Answer to Problem 69P

The exergy destruction and second law of efficiency for the turbine is $10.9\text{MW}$ and $50\%$ respectively.

Explanation of Solution

Write the formula for rate of exergy destruction of the turbine.

${\dot{X}}_{\text{dest,T}}={\dot{m}}_3\left({\psi }_3-{\psi }_4\right)-{\dot{W}}_{\text{T,out}}$ (XIII)

Write the formula for second law of efficiency of the turbine.

${\eta }_{\text{II,T}}=\frac{{\dot{W}}_{\text{T,out}}}{{\dot{m}}_3\left({\psi }_3-{\psi }_4\right)}$ (XIV)

Conclusion:

Substitute $38.203\text{kg/s}$ for ${\dot{m}}_3$, $720.127\text{kJ/kg}$ for ${\psi }_3$, $152.3032\text{kJ/kg}$ for ${\psi }_4$, and $10842\text{kW}$ for ${\dot{W}}_{\text{T,out}}$ in Equation (XIII).

$\begin{array}{c}{\dot{X}}_{\text{dest,T}}=38.203\text{kg/s}\left(720.127\text{kJ/kg}-152.3032\text{kJ/kg}\right)-10842\text{kW}\\ =\text{21692}\text{.5726}\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}-10842\text{kW}\\ =21692.5726\text{kW}-10842\text{kW}\\ =10850.5726\text{kW}\times \frac{1\text{}\text{MW}}{{10}^3\text{kW}}\end{array}$

$\begin{array}{c}=10.85\text{MW}\\ \simeq 10.9\text{MW}\end{array}$

Substitute $38.203\text{kg/s}$ for ${\dot{m}}_3$, $720.127\text{kJ/kg}$ for ${\psi }_3$, $152.3032\text{kJ/kg}$ for ${\psi }_4$, and $10842\text{kW}$ for ${\dot{W}}_{\text{T,out}}$ in Equation (XIV).

$\begin{array}{c}{\eta }_{\text{II,T}}=\frac{10842\text{kW}}{38.203\text{kg/s}\left(720.127\text{kJ/kg}-152.3032\text{kJ/kg}\right)}\\ =\frac{10842\text{kW}}{\text{21692}\text{.5726}\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}}\\ =\frac{10842\text{kW}}{21692.5726\text{kW}}\\ =0.4998\times 100\end{array}$

$\begin{array}{c}=49.98\%\\ \simeq 50\%\end{array}$

Thus, the exergy destruction and second law of efficiency for the turbine is $10.9\text{MW}$ and $50\%$ respectively.

To determine

The exergy destruction and second law of efficiency for the entire plant.

Answer to Problem 69P

The exergy destruction and second law of efficiency for the plant is $\text{39}\text{MW}$ and $21.7\%$ respectively.

Explanation of Solution

Write the formula for rate of exergy input of the plant.

${\dot{X}}_{\text{in, Plant}}={\dot{m}}_1{\psi }_1$ (XV)

Write the formula for rate of exergy destruction of the plant.

${\dot{X}}_{\text{dest, Plant}}={\dot{X}}_{\text{in,Plant}}-{\dot{W}}_{\text{T,out}}$ (XVI)

Write the formula for second law of efficiency of the plant.

${\eta }_{\text{II,Plant}}=\frac{{\dot{W}}_{\text{T,out}}}{{\dot{X}}_{\text{dest, Plant}}}$ (XVII)

Conclusion:

Substitute $230\text{kg/s}$ for ${\dot{m}}_1$ and $216.9556\text{kJ/kg}$ for ${\psi }_1$ in Equation (XV).

$\begin{array}{c}{\dot{X}}_{\text{in, Plant}}=230\text{kg/s}\left(216.9556\text{kJ/kg}\right)\\ =49899.788\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =49899.788\text{kW}\end{array}$

Substitute $49899.788\text{kW}$ for ${\dot{X}}_{\text{in, Plant}}$ and $10842\text{kW}$ for ${\dot{W}}_{\text{T,out}}$ in Equation (XVI).

$\begin{array}{c}{\dot{X}}_{\text{dest,Plant}}=49899.788\text{kW}-10842\text{kW}\\ =39057.788\text{kW}\times \frac{1\text{}\text{MW}}{{10}^3\text{kW}}\\ =39.057\text{MW}\\ \simeq \text{39}\text{MW}\end{array}$

Substitute $10842\text{kW}$ for ${\dot{W}}_{\text{T,out}}$ and $49899.788\text{kW}$ for ${\dot{X}}_{\text{in, Plant}}$ in Equation (XVII).

$\begin{array}{c}{\eta }_{\text{II,T}}=\frac{10842\text{kW}}{49899.788\text{kW}}\\ =0.217275\times 100\\ =21.7275\%\\ \simeq 21.7\%\end{array}$

Thus, the exergy destruction and second law of efficiency for the plant is $\text{39}\text{MW}$ and $21.7\%$ respectively.