Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 6, Problem 168RQ

(a)

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 1MPa.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The inlet mass (min) is 1.2 kg.

The inlet pressure (P1) is 700 kPa.

The inlet temperature (T1) is 230°C.

The mass flow rate (m˙) is 50 kg/s

The pressure at exit (P2) is 20 kPa.

The moisture content (xe) is 5%

Calculation:

Initially find out the properties of the geothermal water.

At state 1:

The geothermal water is extracted at the state of saturated liquid at the temperature of 230°C.

The enthalpy at state 1 is as follows.

  h1=hf@230°C

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

The enthalpy (h1) corresponding to the temperature of 230°C is 990.14kJ/kg.

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

At state 2:

The exit pressure of the flash chamber is 1MPa(1000kPa).

The geothermal steam is flashed at constant enthalpy. The exit steam of the flash chamber is at the quality of x2.

  x2=h2hf,2hfg,2        (IV)

Here, the fluid enthalpy is hf, the vaporization enthalpy is hfg and subscript 2 indicates state 2.

Substitute h2=990.14kJ/kg, hf,2=762.51kJ/kg, and hfg,2=2014.6kJ/kg in Equation (IV).

  x2=990.14kJ/kg762.51kJ/kg2014.6kJ/kg=0.113

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

Obtain the following corresponding to the pressure of 1MPa(1000kPa).

  hf,2=762.51kJ/kghfg,2=2014.6kJ/kg

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 (h3) at state 3 is expressed as follows.

  h3=hg@1000kPa

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

The enthalpy (h3) at state 3 corresponding to the pressure of 1MPa(1000kPa) is 2777.1kJ/kg.

At state 4:

The steam is at the state of saturated mixture at the pressure of 20kPa with 5% of moisture content.

The quality at state 4 is as follows.

  x4=100%5%=95%×1100=0.95

The enthalpy (h4) at state 4 is expressed as follows.

  h4=hf,4+x4hfg,4        (V)

  h4=251.42kJ/kg+0.95(2357.5kJ/kg)=251.42kJ/kg+2239.625kJ/kg=2491.045kJ/kg2491.1kJ/kg

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

Obtain the following corresponding to the pressure of 20kPa.

  hf,4=251.42kJ/kghfg,4=2357.5kJ/kg

The temperature of the steam after flashing process is equal to the saturation temperature at the exit pressure of flash chamber i.e. 1MPa(1000kPa).

  T2=Tsat @ 1000kPa

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

The temperature (T2) at state 2 corresponding to the pressure of 1MPa(1000kPa) is 179.9°C.

Thus, the exit temperature of flash chamber is 179.9°C.

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

  m˙3=x2m˙2        (VI)

Substitute x2=0.113 and m˙2=50kg/s in Equation (VI).

  m˙3=0.113(50kg/s)=5.65kg/s

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

Fundamentals of Thermal-Fluid Sciences, Chapter 6, Problem 168RQ

Write the general energy rate balance equation.

  E˙inE˙out=ΔE˙system        (I)

Here, the rate of total energy in is E˙in, the rate of total energy out is E˙out, and the rate of change in net energy of the system is ΔE˙system.

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

  ΔE˙system=0

The Equation (I) is reduced as follows.

  E˙inE˙out=0E˙in=E˙out

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.

  h1=h2

Express the energy balance equation for the separator.

  m˙2h2=m˙3h3+m˙liquidhliquid        (II)

Express the energy balance equation for the turbine.

  W˙T=m˙3(h3h4)        (III)

Substitute m˙3=5.65kg/s, h3=2777.1kJ/kg, and h4=2491.1kJ/kg in

Equation (III).

  W˙T=5.65kg/s(2777.1kJ/kg2491.1kJ/kg)=5.65kg/s(286kJ/kg)=1615.9kJ/s×1kW1kJ/s=1616kW

Thus, the power output turbine is 1616kW.

(b)

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 500kPa.

(b)

Expert Solution
Check Mark

Explanation of Solution

Similarly for the pressure (P2=500 kPa).

At state 2:

The exit pressure of the flash chamber is 500kPa.

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

Obtain the following corresponding to the pressure of 500kPa.

  hf,2=640.09kJ/kghfg,2=2108.0kJ/kg

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 (h3) at state 3 is expressed as follows.

  h3=hg@500kPa

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

The enthalpy (h3) at state 3 corresponding to the pressure of 500kPa is 2748.1kJ/kg.

Calculation:

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

  T2=Tsat @ 500kPa

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

The temperature (T2) at state 2 corresponding to the pressure of 500kPa is 179.9°C.

Thus, the exit temperature of flash chamber is 151.83°C.

Substitute h2=990.14kJ/kg, hf,2=640.09kJ/kg, and hfg,2=2108.0kJ/kg in Equation (IV).

  x2=990.14kJ/kg640.09kJ/kg2108.0kJ/kg=0.1660

Substitute x2=0.1660 and 50kg/s for m˙2 in Equation (VI).

  m˙3=0.1660(50kg/s)=8.3029kg/s

Substitute m˙3=8.3029kg/s, h3=2748.1kJ/kg, and h4=2491.1kJ/kg in

Equation (III).

  W˙T=8.3029kg/s(2748.1kJ/kg2491.1kJ/kg)=8.3029kg/s(257kJ/kg)=2133.8437kJ/s×1kW1kJ/s=2134kW

Thus, the power output turbine is 2134kW.

(c)

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 100kPa.

(c)

Expert Solution
Check Mark

Explanation of Solution

At state 2:

The exit pressure of the flash chamber is 100kPa.

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

Obtain the following corresponding to the pressure of 100kPa.

  hf,2=417.51kJ/kghfg,2=2257.5kJ/kg

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 (h3) at state 3 is expressed as follows.

  h3=hg@100kPa

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

The enthalpy (h3) at state 3 corresponding to the pressure of 100kPa is 2675.0kJ/kg.

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

  T2=Tsat @ 100kPa

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

The temperature (T2) at state 2 corresponding to the pressure of 100kPa is 99.61°C.

Thus, the exit temperature of flash chamber is 99.61°C.

Substitute h2=990.14kJ/kg, hf,2=417.51kJ/kg, and hfg,2=2257.5kJ/kg in Equation (IV).

  x2=990.14kJ/kg417.51kJ/kg2257.5kJ/kg=0.25366

Substitute x2=0.25366 and m˙2=50kg/s in Equation (VI).

  m˙3=0.25366(50kg/s)=12.683kg/s

Substitute  m˙3=12.683kg/s, h3=2675.0kJ/kg, and h4=2491.1kJ/kg in

Equation (III).

  W˙T=12.683kg/s(2675.0kJ/kg2491.1kJ/kg)=12.683kg/s(183.9kJ/kg)=2332.4037kJ/s×1kW1kJ/s=2332.4kW

Thus, the power output turbine is 2332.4kW.

(d)

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 50kPa.

(d)

Expert Solution
Check Mark

Explanation of Solution

At state 2:

The exit pressure of the flash chamber is 50kPa.

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

Obtain the following corresponding to the pressure of 50kPa.

  hf,2=340.54kJ/kghfg,2=2304.7kJ/kg

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 (h3) at state 3 is expressed as follows.

  h3=hg@50kPa

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

The enthalpy (h3) at state 3 corresponding to the pressure of 50kPa is 2645.2kJ/kg.

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

  T2=Tsat @ 50kPa

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

The temperature (T2) at state 2 corresponding to the pressure of 50kPa is 81.32°C.

Thus, the exit temperature of flash chamber is 81.32°C.

Substitute h2=990.14kJ/kg, hf,2=340.54kJ/kg, and hfg,2=2304.7kJ/kg in Equation (IV).

  x2=990.14kJ/kg340.54kJ/kg2304.7kJ/kg=0.28186

Substitute x2=0.28186 and m˙2=50kg/s in Equation (VI).

  m˙3=0.28186(50kg/s)=14.093kg/s14.1kg/s

Substitute m˙3=14.1kg/s, h3=2645.2kJ/kg, and h4=2491.1kJ/kg in

Equation (III).

  W˙T=14.1kg/s(2645.2kJ/kg2491.1kJ/kg)=14.1kg/s(154.1kJ/kg)=2172.81kJ/s×1kW1kJ/s=2173kW

Thus, the power output turbine is 2173kW.

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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