- 230 MW Burning HFO
- 80% Load Burning Bio Fuel
- 80% Load Burning Coal
- 80% Load Burning Coal with soot built up.

`[latex]\eta_{boiler} = \frac{Energy\ to\ steam}{Energy\ from\ fuel}[/latex]`

h_{2}=specific enthalpy of steam formed, [kJ/kg],

h_{1}=specific enthalpy of feed water, [kJ/kg].

[latex]\eta_{boiler} = \frac{m_{s}(h_{2}-h_{1})}{m_{f}HV}100\%[/latex]

- I10 230 MW Burning HFO
- I15 80% Load Burning Bio Fuel
- I14 80% Load Burning Coal
- I14 80% Load Burning Coal and use MD250 to set up soot variables.

- H00810 HFO heat value
- H00870 Pellet heat value (bio fuel)
- H00830 Coal heat value

**Trend plots:**Supply all plots taken for each of the 4 conditions,**Computation:**Use MATLAB or MS Excel and calculate the boiler thermal efficiency for the 4 conditions specified,**Conclusion:**Write a summary (max. 500 words, in a text box if using Excel) comparing your results and suggestions for further study.

Further Reading:

]]>- Fundamentals of Classical Thermodynamics SI Version by G. J. Van Wylen and R. E. Sonntag: Evaluation of Actual Combustion Processes.
- Thermal Engineering by H.L. Solberg, O.C. Cromer and A.R. Spalding: Capacity and Efficiency of Steam Generating Units.
- Basic Engineering Thermodynamics in SI Units by R. Joel: Boiler calculations.

- 35% Load (I13),
- 80% Load (I14),
- 230 MW (I10).

`[latex]\eta_{Turbine} = \frac{Actual\ change\ in\ enthalpy}{Isentropic\ change\ in\ enthalpy}[/latex]`

[latex]\eta_{Turbine} = \frac{(H_{1}-H_{2\prime})}{(H_{1}-H_{2})}[/latex]

The difference in enthalpy H- 35% Load (I13),
- 80% Load (I14),
- 230 MW (I10).

- Z03020
- E03018

**Trend plots:**Supply all plots taken for each of the 3 conditions,**Computation:**Use MATLAB or MS Excel and calculate the turbine efficiency for the 3 conditions specified,**Conclusion:**Write a summary (max. 500 words, in a text box if using Excel) comparing your results and suggestions for further study.

Further Reading:

]]>- Thermodynamics and Heat Power by I. Granet: Vapor power cycles.

- Under normal conditions,
- With the cooling water temperature very high (lake water temperature: 35°C),
- Without regeneration.

`[latex]\eta_{Rankine} = \frac{Net\ work\ output}{Heat\ supplied\ in\ the\ boiler}[/latex]`

`[latex]\eta_{Rankine} = \frac{W_{Turbine}-W_{Pump}}{Q_{boiler}}[/latex] `

referring to the diagram above and using the enthalpy values in the Rankine cycle, we can write:

`[latex]\eta_{Rankine} = \frac{(h_{1}-h_{2})-(h_{4}-h_{3})}{(h_{1}-h_{4})}[/latex]`

`[latex]\eta_{thermal} = \frac{W_{12}+W_{67}-W_{43}}{Q_{41}+Q_{26}}[/latex] `

`[latex]\eta_{thermal} = \frac{W_{Turbines}-W_{Pumps}}{Q_{boiler}+Q_{Reheat1}+Q_{Reheat2}}[/latex] `

- Draw a T-S diagram of the Rankine cycle (not to scale) including reheat and regeneration,
- Using Trend Group Directory, collect the relevant process values,
- Calculate the overall thermal efficiency of the plant:
- Under normal conditions,
- When the cooling water temperature is very high (Set the Variable List Page 0100, tag#: T00305 to 35°C),
- When all the steam extraction valves are closed (i.e. no regeneration and T00305 set to 10°C).

- Q02395 Reheater 1 transferred heat
- Q02375 Reheater 2 transferred heat

**T-S diagram:**As per instructions above,**Trend plots:**Supply all plots taken for this lab,**Computation:**Use MATLAB or MS Excel and calculate the overall thermal efficiency of the plant as per Lab Instructions.**Conclusion:**Write a summary (max. 500 words, in a text box if using Excel) comparing your results and suggestions for further study.

Further Reading:

]]>- Applied Thermodynamics for Engineering Technologists by T. D. Eastop and A. McConkey: Steam Plant.
- Fundamentals of Classical Thermodynamics SI Version by G. J. Van Wylen and R. E. Sonntag: Vapor power cycles.
- Thermodynamics and Heat Power by I. Granet: Vapor power cycles.

- Perform combustion analyses for two types of coal,
- Compare results.

O_{2}: 23%
N_{2}: 77%

Element |
Symbol |
Molecular Weight |

Carbon | C | 12 |

Sulphur | S_{2} |
32 |

Hydrogen | H_{2} |
2 |

Oxygen | O_{2} |
32 |

Nitrogen | N_{2} |
28 |

[latex]C_{n}H_{m}+yO_{2}\rightarrow aCO_{2}+bH_{2}O[/latex]

Where the balance should be satisfied following the moles for any mathematcial equation: Carbon balance:[latex]a=n[/latex] kmol CO_{2}/ kmol fuel

Hydrogen balance:

[latex]2b=m[/latex]

[latex]b = \frac{m}{2}[/latex] kmol H_{2}O/ kmol fuel

[latex]2y = 2a+b[/latex]

[latex]y = a+\frac{b}{2}[/latex] kmol O_{2}/ kmol fuel

[latex]C_{n}H_{m}+y(O_{2}+\frac{79}{21}N_{2}\rightarrow aCO_{2}+bH_{2}O+cN_{2}[/latex]

The balance equations are: Carbon balance:[latex]a=n[/latex] kmol CO_{2}/ kmol fuel

Hydrogen balance:

[latex]2b=m[/latex]

[latex]b = \frac{m}{2}[/latex] kmol H_{2}O/ kmol fuel

[latex]2y = 2a+b[/latex]

[latex]y = a+\frac{b}{2}[/latex] kmol O_{2}/ kmol fuel

[latex]y\frac{79}{21}=c[/latex] kmol N_{2}/kmol fuel

[latex]C_{8}H_{18}+y(O_{2}+\frac{79}{21}N_{2}\rightarrow aCO_{2}+bH_{2}O+cN_{2}[/latex]

Carbon balance:[latex]a=n=8[/latex] kmol CO_{2}/ kmol fuel

Hydrogen balance:

[latex]2b=18[/latex]

[latex]b = \frac{18}{2}=9[/latex] kmol H_{2}O/ kmol fuel

[latex]2y = 2(8)+b[/latex]

[latex]y = 8+\frac{9}{2}=12.5[/latex] kmol O_{2}/ kmol fuel

[latex]y\frac{79}{21}=c=12.5\frac{79}{21}=47[/latex] kmol N_{2}/kmol fuel

[latex]C_{8}H_{18}+12.5(O_{2}+\frac{79}{21}N_{2}\rightarrow 8CO_{2}+9H_{2}O+47N_{2}[/latex]

Air/ fuel mass based ratio considering that 1 kmol fuel is 114 kg of fuel ( 8*12 + 18 *(1) = 114):

[latex]12.5\frac{kmol\ O_{2}}{kmol\ fuel}\frac{1\ kmol\ fuel}{114\ kg\ fuel}32\frac{kg\ O_{2}}{kmol\ O_{2}}\frac{100\ kg\ air}{23\ kg\ O_{2}}=15.25\frac{kg\ air}{kg\ fuel}[/latex]

Flue gas composition on molar basis is:

Total number of kmol of flue gasses = kmol CON_{2} = 47/64 = 73.5%

C+O_{2}=CO_{2}

12+32=44

1 kg C + 2⅔ kg O_{2} = 3⅔ kg CO_{2}

2H_{2} +O_{2} =2H_{2}O

4+32=36

1 kg H_{2} + 8 kg O_{2} = 9 kg H_{2}O

Sulphur burns as follows:

S +O_{2} =SO_{2}

32+32=64

1 kg S + 1 kg O_{2} = 2 kg SO_{2}

1 kg S + 4.3 kg air = 2 kg SO_{2}+3.3 kg N_{2}

If the analysis of fuel is given by mass, follow the steps below:

**Total O**Determine the mass of O_{2}required:_{2}required for each constituent and find the total mass of O_{2 }(Subtract any O_{2}which may be in the fuel)**Stoichiometric air:**Stoichiometric mass of air = O_{2}required/0.232**Total****mass of combustion products:**Determine the % mass of each combustion product. For example, given C content of 84.9%, CO_{2}=84.9/100*2⅔=3.11% and find the total mass of combustion products.**Analysis of combustion products by mass:**Suppose the total mass of combustion products in step 3 has been found as 12.09 kg/kg fuel, then CO_{2}=3.11/12.09*100=25.74. This means that 25.74% of the flue gas is CO_{2}. Repeat this calculation for each constituent.

X00820 X00821 X00822 X00823 X00824 E23356

G02196 G02197 X32419 X02419 G00831 H00830

**Burning default coal:**After 10 minutes of running the simulator, freeze simulator and print the two trends. This is the reference point for the next step.**Burning poor quality****coal:**Switch to run mode and access Variable List page 0111 on MD180. Set the new values as shown below. After 10 minutes, freeze simulator and print the two trends.- X00820: 70.40
- X00821: 5.10
- X00822: 1.10
- X00823: 12.50
- X00824: 1.60
- X00825: 9.30

**Computation:**Compute the following values for both types of fuels:- Total O
_{2}required - Stoichiometric air
- Total mass of combustion products
- Analysis of combustion products by mass [%]: CO
_{2}, H_{2}O, SO_{2}, N_{2}

- Total O
**Comparison:**Compare your findings based on the following data:- Furnace outlet SO
_{x}flow - Furnace outlet NO
_{x}flow - CO content in flue gas
- Oxygen content in flue gas
- Theoretical combustion air
- Coal Heat Value

- Furnace outlet SO

**Trend plots:**Supply all plots taken for each of the 2 fuels,**Computation:**As per lab instructions above, perform combustion analyses using MATLAB or MS Excel.**Conclusion:**Write a summary (max. 500 words, in a text box if using Excel) comparing your results and suggestions for further study.

Further Reading:

]]>- Basic Engineering Thermodynamics in SI Units by R. Joel: Combustion.
- Thermal Engineering by H.L. Solberg, O.C. Cromer and A.R. Spalding: Fossil fuels and their combustion.

- Under normal conditions,
- During over fire air damper failure,
- During over burner air control failure,
- Burning poor quality fuel,
- With the DeNO
_{x}plant bypassed.

G02197 X17821 G17107 X17106 D17104

T17103 C08444 G08444 G08443 C08400

**Stable operation:**After 5 minutes of running a stable operation, freeze simulator and print the two trends. This is the reference point for the rest of the lab.**OFA damper failure:**Switch to run mode and activate malfunction 0881 on MD200. After 5 minutes, freeze simulator and print the two trends. Before moving on to the next step deactivate the malfunction.**OBA control failure:**Switch to run mode and activate malfunction 0780 on MD180. After 5 minutes, freeze simulator and print the two trends. Before moving on to the next step deactivate the malfunction.**Burning poor quality****fuel:**Switch to run mode and access Variable List page 0111 on MD180. Set the new values as shown below. After 5 minutes, freeze simulator and print the two trends.- X00820: 70.40
- X00821: 5.10
- X00822: 1.10
- X00823: 12.50
- X00824: 1.60
- X00825: 9.30

**DeNO**Switch to run mode and bypass the SCR 1 and SCR 2 on MD710 and MD720 respectively. After 5 minutes, freeze simulator and print the two trends._{x}plant bypassed:

`[latex]Percentage\ deviation = \frac{Current\ Value-Reference\ Value}{Reference\ Value}100\%[/latex]`

and tabulated as follows:

[caption id="attachment_41" align="aligncenter" width="701"] The deNOx plant deviation data.[/caption]**Trend plots:**Supply all plots taken for this lab (make sure plots are labeled properly),**Computation:**Use MATLAB or MS Excel to process your data. Calculate the percentage deviation for each operation and plot your results,**Conclusion:**Write a summary (max. 500 words in a text box, if using Excel) comparing your results and suggestions for further study.

Further Reading:

]]>- Thermal Power Plant Simulator Course Manual by BCIT: The DeNO
_{x}plant

[latex]Q_{hot} = m_{hot}c_{p hot}\Delta T_{hot}[/latex] [latex]Q_{cold} = m_{cold}c_{p cold}\Delta T_{cold}[/latex]

Considering the surface area involved in heat transfer, Newton’s Law of cooling states that the rate of heat loss is proportional to the difference in temperatures between the body and its surroundings and given by,[latex]Q = \alpha Area \Delta T[/latex]

where α is called the heat transfer coefficient [W/m[latex]Q = \frac{F(LMTD)}{R_{T}}[/latex]

Or[latex]Q = F(UA)(LMTD)[/latex]

where F is the correction factor which equals 1 for this SIMLAB (it takes values between 0.5 and 1). R[latex]R_{T}=R_{hf}+R_{w}+R_{cf}[/latex]

[latex]R_{hf} = \frac{1}{A_{1}\alpha_{h}}[/latex]

[latex]R_{w} = \frac{ln\frac{D_{2}}{D_{1}}}{2\pi L\lambda_{w}}[/latex]

[latex]R_{cf} = \frac{1}{A_{2}\alpha_{c}}[/latex]

Heat transfer coefficients ah and ac can be calculated using the following expression for Nusselt number for hot and cold water: For cooling[latex]\alpha_{h} = \frac{Nu_{h}\lambda_{h}}{D_{h}}[/latex]

[latex]Nu_{h} = 0.3Re_{h}^{0.8}Pr_{h}^{0.3}[/latex]

For heating[latex]\alpha_{c} = \frac{Nu_{c}\lambda_{c}}{D_{c}}[/latex]

[latex]Nu_{c} = 0.3Re_{c}^{0.8}Pr_{c}^{0.3}[/latex]

And the LMTD is given by the following correlation where 1 and 2 presents the ends of the heat exchanger:[latex]\Delta T_{LMTD} = \frac{\Delta T_{1}-\Delta T_{2}}{ln\frac{\Delta T_{1}}{\Delta T_{2}}}[/latex]

[latex]\epsilon = \frac{Q}{Q_{max}}[/latex]

[latex]Q_{max} = C_{min}(T_{hi}-T_{ci})[/latex]

Where C[latex]C_{h} = m_{h}c_{ph}[/latex]

[latex]C_{c} = m_{c}c_{pc}[/latex]

Heat Area Factor | C34201 |

Q | Q34228 |

T_{cold1} |
T24214 |

T_{cold2} |
T34214 |

m_{cold} |
G34213 |

T_{hot1} |
T34204 |

T_{hot2} |
T34227 |

m_{hot} |
G34203 |

**Heat Area Factor set to 0.5:**Using MD420 and Variable List 4210 set C34201 to 0.5. Run the simulator with this setting for 15 minutes. Freeze and print the two trends.**Heat Area Factor set to 1:**As in step 1, set C34201 to 1. This is the default setting for the Heat Area Factor (heat transfer coefficient x area). After 15 minutes of running the simulator, freeze simulator and print the two trends.**Heat Area Factor set to 1.5:**This time, set C34201 to 1.5. Run the simulator with this setting for 15 minutes. Freeze and print the two trends.

**Trend plots:**Supply all plots taken for this lab, make sure they are labelled properly.**Computation:**Use MATLAB or MS Excel and calculate the LMTD and Heat Exchanger effectiveness values for the 3 tests.**Conclusion:**Write a summary (max. 500 words, in a text box if using Excel) comparing your results and suggestions for further study.

Further Reading:

]]>- Applied Thermodynamics for Engineering Technologists by T. D. Eastop and A. McConkey: Heat Transfer.