Chapter 5.5, Problem 186RP

In a single-flash geothermal power plant, geothermal water enters the flash chamber (a throttling valve) at 230°C as a saturated liquid at a rate of 50 kg/s. The steam resulting from the flashing process enters a turbine and leaves at 20 kPa with a moisture content of 5 percent. 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 the flash chamber is (a) 1 MPa, (b) 500 kPa, (c) 100 kPa, (d) 50 kPa.

FIGURE P5–186

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 186RP

The exit temperature of flash chamber is $179.9°\text{C}$.

The power output turbine is $1616\text{kW}$.

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.

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

At state 1:

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

The enthalpy at state 1 is as follows.

$h_1=h_{f@230°\text{C}}$

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

The enthalpy $\left(h_1\right)$ corresponding to the temperature of $230°\text{C}$ is $990.14\text{kJ/kg}$.

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

At state 2:

The exit pressure of the flash chamber is $1\text{MPa}\left(1000\text{kPa}\right)$.

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 $1\text{MPa}\left(1000\text{kPa}\right)$.

$\begin{array}{l}{h}_{f,2}=762.51\text{kJ/kg}\\ h_{fg,2}=2014.6\text{kJ/kg}\end{array}$

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 $\left(h_3\right)$ at state 3 is expressed as follows.

$h_3=h_{g@1000\text{kPa}}$

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

The enthalpy $\left(h_3\right)$ at state 3 corresponding to the pressure of $1\text{MPa}\left(1000\text{kPa}\right)$ is $2777.1\text{kJ/kg}$.

At state 4:

The steam is at the state of saturated mixture at the pressure of $20\text{kPa}$ 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 $\left(h_4\right)$ at state 4 is expressed as follows.

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

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

Obtain the following corresponding to the pressure of $20\text{kPa}$.

$\begin{array}{l}{h}_{f,4}=251.42\text{kJ/kg}\\ h_{fg,4}=2357.5\text{kJ/kg}\end{array}$

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

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

Conclusion:

The temperature of the steam after flashing process is equal to the saturation temperature at the exit pressure of flash chamber i.e. $1\text{MPa}\left(1000\text{kPa}\right)$.

$T_2=T_{\text{sat @ 1000}\text{kPa}}$

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

The temperature $\left(T_2\right)$ at state 2 corresponding to the pressure of $1\text{MPa}\left(1000\text{kPa}\right)$ is $179.9°\text{C}$.

Thus, the exit temperature of flash chamber is $179.9°\text{C}$.

Substitute $990.14\text{kJ/kg}$ for $h_2$, $762.51\text{kJ/kg}$ for $h_{f,2}$, and $2014.6\text{kJ/kg}$ for $h_{fg,2}$ in Equation (IV).

$\begin{array}{c}{x}_2=\frac{990.14\text{kJ/kg}-762.51\text{kJ/kg}}{2014.6\text{kJ/kg}}\\ =0.113\end{array}$

Substitute $0.113$ for $x_2$ and $50\text{kg/s}$ for ${\dot{m}}_2$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_3=0.113\left(50\text{kg/s}\right)\\ =5.65\text{kg/s}\end{array}$

Substitute $251.42\text{kJ/kg}$ for $h_{f,4}$, $0.95$ for $x_4$, and $2357.5\text{kJ/kg}$ for $h_{fg,4}$ in

Equation (V).

$\begin{array}{c}{h}_4=251.42\text{kJ/kg}+0.95\left(2357.5\text{kJ/kg}\right)\\ =251.42\text{kJ/kg}+2239.625\text{kJ/kg}\\ =2491.045\text{kJ/kg}\\ \simeq 2491.1\text{kJ/kg}\end{array}$

Substitute $5.65\text{kg/s}$ for ${\dot{m}}_3$, $2777.1\text{kJ/kg}$ for $h_3$, and $2491.1\text{kJ/kg}$ for $h_4$ in

Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{T}}=5.65\text{kg/s}\left(2777.1\text{kJ/kg}-2491.1\text{kJ/kg}\right)\\ =5.65\text{kg/s}\left(286\text{kJ/kg}\right)\\ =1615.9\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =1616\text{kW}\end{array}$

Thus, the power output turbine is $1616\text{kW}$.

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 $500\text{kPa}$.

Answer to Problem 186RP

The exit temperature of flash chamber is $151.83°\text{C}$.

The power output turbine is $2134\text{kW}$.

Explanation of Solution

At state 2:

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

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

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

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

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 $\left(h_3\right)$ at state 3 is expressed as follows.

$h_3=h_{g@500\text{kPa}}$

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

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

Conclusion:

The temperature of the steam after flashing process is equal to the saturation temperature at the exit pressure of flash chamber i.e. $500\text{kPa}$.

$T_2=T_{\text{sat @ 500}\text{kPa}}$

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

The temperature $\left(T_2\right)$ at state 2 corresponding to the pressure of $500\text{kPa}$ is $179.9°\text{C}$.

Thus, the exit temperature of flash chamber is $151.83°\text{C}$.

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).

$\begin{array}{c}{x}_2=\frac{990.14\text{kJ/kg}-640.09\text{kJ/kg}}{2108.0\text{kJ/kg}}\\ =0.1660\end{array}$

Substitute $0.1660$ for $x_2$ and $50\text{kg/s}$ for ${\dot{m}}_2$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_3=0.1660\left(50\text{kg/s}\right)\\ =8.3029\text{kg/s}\end{array}$

Substitute $8.3029\text{kg/s}$ for ${\dot{m}}_3$, $2748.1\text{kJ/kg}$ for $h_3$, and $2491.1\text{kJ/kg}$ for $h_4$ in

Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{T}}=8.3029\text{kg/s}\left(2748.1\text{kJ/kg}-2491.1\text{kJ/kg}\right)\\ =8.3029\text{kg/s}\left(257\text{kJ/kg}\right)\\ =2133.8437\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =2134\text{kW}\end{array}$

Thus, the power output turbine is $2134\text{kW}$.

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 $100\text{kPa}$.

Answer to Problem 186RP

The exit temperature of flash chamber is $99.61°\text{C}$.

The power output turbine is $2332.4\text{kW}$.

Explanation of Solution

At state 2:

The exit pressure of the flash chamber is $100\text{kPa}$.

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

Obtain the following corresponding to the pressure of $100\text{kPa}$.

$\begin{array}{l}{h}_{f,2}=417.51\text{kJ/kg}\\ h_{fg,2}=2257.5\text{kJ/kg}\end{array}$

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 $\left(h_3\right)$ at state 3 is expressed as follows.

$h_3=h_{g@100\text{kPa}}$

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

The enthalpy $\left(h_3\right)$ at state 3 corresponding to the pressure of $100\text{kPa}$ is $2675.0\text{kJ/kg}$.

Conclusion:

The temperature of the steam after flashing process is equal to the saturation temperature at the exit pressure of flash chamber i.e. $100\text{kPa}$.

$T_2=T_{\text{sat @ 100}\text{kPa}}$

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

The temperature $\left(T_2\right)$ at state 2 corresponding to the pressure of $100\text{kPa}$ is $99.61°\text{C}$.

Thus, the exit temperature of flash chamber is $99.61°\text{C}$.

Substitute $990.14\text{kJ/kg}$ for $h_2$, $417.51\text{kJ/kg}$ for $h_{f,2}$, and $2257.5\text{kJ/kg}$ for $h_{fg,2}$ in Equation (IV).

$\begin{array}{c}{x}_2=\frac{990.14\text{kJ/kg}-417.51\text{kJ/kg}}{2257.5\text{kJ/kg}}\\ =0.25366\end{array}$

Substitute $0.25366$ for $x_2$ and $50\text{kg/s}$ for ${\dot{m}}_2$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_3=0.25366\left(50\text{kg/s}\right)\\ =12.683\text{kg/s}\end{array}$

Substitute $12.683\text{kg/s}$ for ${\dot{m}}_3$, $2675.0\text{kJ/kg}$ for $h_3$, and $2491.1\text{kJ/kg}$ for $h_4$ in

Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{T}}=12.683\text{kg/s}\left(2675.0\text{kJ/kg}-2491.1\text{kJ/kg}\right)\\ =12.683\text{kg/s}\left(183.9\text{kJ/kg}\right)\\ =2332.4037\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =2332.4\text{kW}\end{array}$

Thus, the power output turbine is $2332.4\text{kW}$.

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 $50\text{kPa}$.

Answer to Problem 186RP

The exit temperature of flash chamber is $81.32°\text{C}$.

The power output turbine is $2173\text{kW}$.

Explanation of Solution

At state 2:

The exit pressure of the flash chamber is $50\text{kPa}$.

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

Obtain the following corresponding to the pressure of $50\text{kPa}$.

$\begin{array}{l}{h}_{f,2}=340.54\text{kJ/kg}\\ h_{fg,2}=2304.7\text{kJ/kg}\end{array}$

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 $\left(h_3\right)$ at state 3 is expressed as follows.

$h_3=h_{g@50\text{kPa}}$

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

The enthalpy $\left(h_3\right)$ at state 3 corresponding to the pressure of $50\text{kPa}$ is $2645.2\text{kJ/kg}$.

Conclusion:

The temperature of the steam after flashing process is equal to the saturation temperature at the exit pressure of flash chamber i.e. $50\text{kPa}$.

$T_2=T_{\text{sat @ 50}\text{kPa}}$

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

The temperature $\left(T_2\right)$ at state 2 corresponding to the pressure of $50\text{kPa}$ is $81.32°\text{C}$.

Thus, the exit temperature of flash chamber is $81.32°\text{C}$.

Substitute $990.14\text{kJ/kg}$ for $h_2$, $340.54\text{kJ/kg}$ for $h_{f,2}$, and $2304.7\text{kJ/kg}$ for $h_{fg,2}$ in Equation (IV).

$\begin{array}{c}{x}_2=\frac{990.14\text{kJ/kg}-340.54\text{kJ/kg}}{2304.7\text{kJ/kg}}\\ =0.28186\end{array}$

Substitute $0.28186$ for $x_2$ and $50\text{kg/s}$ for ${\dot{m}}_2$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_3=0.28186\left(50\text{kg/s}\right)\\ =14.093\text{kg/s}\\ \simeq 14.1\text{kg/s}\end{array}$

Substitute $14.1\text{kg/s}$ for ${\dot{m}}_3$, $2645.2\text{kJ/kg}$ for $h_3$, and $2491.1\text{kJ/kg}$ for $h_4$ in

Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{T}}=14.1\text{kg/s}\left(2645.2\text{kJ/kg}-2491.1\text{kJ/kg}\right)\\ =14.1\text{kg/s}\left(154.1\text{kJ/kg}\right)\\ =2172.81\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ =2173\text{kW}\end{array}$

Thus, the power output turbine is $2173\text{kW}$.