Andreza, Lemuel V. A04
Andreza, Lemuel V. A04
Andreza, Lemuel V. A04
Problem
Tasks
Prepare a brief solution report and develop a MS Excel Model using the Add-ins in ExcelInME website to a
1) Worksheet 1. Input Data and Assumptions. Include the cycle layout and the corresponding T-s diagram
2) Worksheet 2. State Analysis of each point.
3) Worksheets for each Process Analysis. Prepare a worksheet for each process involved in the cycle. The
tabulation of properties, the mass balance, and energy balance.
4) Worksheet on Overall Cycle and Plant Analysis. Specify a particular type of coal as fuel using available r
Note: Use your own nomenclature of variables and parameters. Justify all values used and derive all equa
Reference:
https://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/SteamPlant/steam8.1.html
IDEAL PROCESS IN EVERY POINTS (USE IDEAL PROCESS ANALYSIS FOR POINTS THAT LACKS THE GIVEN A
POINT 1-2: PARTIAL ISENTROPIC EXPANSION IN THE TURBINE (S=C)
POINT 2-3: CONSTANT PRESSURE RE-SUPERHEATING IN THE REHATER (P=C)
POINT 3-4: COMPLETE ISENTROPIC PROCESS IN THE TURBINE (S=C)
POINT 4-5: CONSTANT PRESSURE AND REJECTION OF HEAT IN THE CONDENSER (P=C)
POINT 5-6: ADIABATIC PUMPING (S=C)
POINT 6,7 AND 8: CONSTANT PRESSURE (P=C)
POINT 8-9: ADIABATIC PUMPING (S=C)
POINT 9-1: CONSTANT PRESSURE, HEAT ADDITION TO THE BOILER (P=C)
THROTTLING PROCESS: (h=C) ISENTHALPIC, (S=C) ISENTROPIC/ADIABATIC, (T=C) ISOTHERMAL
P-h DIAGRAM
ting under the conditions shown in the following diagram:
h process involved in the cycle. The process analysis must include: schematic diagram of the process, thermodynamic analysis
type of coal as fuel using available references. Assume appropriate air-fuel ratio and other operating efficiency parameters.
y all values used and derive all equations used starting from the basic energy equation for a flow system and definition of each
mPlant/steam8.1.html
POINTS THAT LACKS THE GIVEN ACTUAL VALUES)
NDENSER (P=C)
ciency parameters.
and definition of each parameter.
ACTUAL PROCESS VALUES
POINT PRESSURE TEMPERATURE IN °C ENTHALPY IN KJ/Kg
1 10 MPa 500 3375.05844184641
2 1 MPa 200 2828.26753759806
3 1 MPa 500 3479.00367475891
4 10 KPa 50 2591.99361918512
5 10 KPa 40 167.543365678233
6 800 KPa 40.02 168.3
7 800 KPa 200 2828.26753759806
8 800 KPa 170 719.210731401668
9 10 MPa 171.2 729.5
Assuming that the process at points 1-2 is isentropic, the entropy at points 1 and 2 is the same,
process. Additionally, the mass entering the high-pressure turbine is equal to the mass exiting t
two parts, one for the work done to the turbine and the other for the circulating steam
BY MASS BALANCE
𝑚_𝑖𝑛=𝑚_𝑜𝑢𝑡
𝑚_1=𝑚_2+𝑚_7
1=(1−𝑦)+𝑦
y at points 1 and 2 is the same, indicating that the entropy is constant during this
ne is equal to the mass exiting the turbine. However, the energy will be divided into
or the circulating steam
BY ENERGY BALANCE
𝐸_𝑖𝑛 =𝐸_𝑜𝑢𝑡
〖𝑚 _1 ℎ 〗 _1=𝑚_2 ℎ_2+𝑚_7 ℎ_7+𝑊_1=3374.058442 𝐾𝐽
(1)(3374.058442)=(0.793)(2828.267)+(0.207)(2828.267)+ 𝑊_1
𝑾_𝑯𝑷𝑻=𝟓𝟒𝟓.𝟕𝟗𝟐𝟓 𝑲𝑱
PROPERTIES OF POINT 2 AND 3
The purpose of this process is to prevent any moisture from sticking in the turbine so that i
the steam and produce the best work. In this process, the steam will enter the reheater chamb
turbine.
POINT PRESSURE TEMPERATURE IN °C ENTHALPY IN KJ/Kg
2 1 Mpa 200 2828.26753759806
3 1 Mpa 500 3479.00367475891
BY MASS BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
𝑚_2=𝑚_3
1−𝑦=1−𝑦
BY ENERGY BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
(0.793)(2828.267)+𝑄_𝑎=(0.793)(3479.003675)
𝑸_𝑹𝑬𝑯𝑬𝑨𝑻𝑬𝑹=𝟓𝟏𝟔.𝟎𝟑𝟑𝟕 𝑲𝑱
PROPERTIES OF POINT 3 AND 4
Due to the adiabatic condition and isentropic process of the turbine in this process, neither hea
be sufficient to produce the work done to the turbine, and the exhaust steam will enter the con
BY MASS BALANCE
𝑚_𝑖𝑛=𝑚_𝑜𝑢𝑡
(1−𝑦)=(1−𝑦)
𝑦=0.207112 𝑓𝑟𝑜𝑚 𝑓𝑒𝑒𝑑𝑤𝑎𝑡𝑒𝑟 ℎ𝑒𝑎𝑡𝑒𝑟 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠
0.793112=0.793112
bine in this process, neither heat addition nor rejection will occur; yet, the energy will
xhaust steam will enter the condenser.
BY ENERGY BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
𝑚_3 ℎ_3=𝑚_4 ℎ_4+𝑊_2
(0.793)(3479.0037)=(0.793)(2591.994)+𝑊_2
𝑾_𝑳𝑷𝑻=𝟕𝟎𝟑.𝟑𝟗𝟖𝟕 𝑲𝑱
PROPERTIES OF POINT 4 AND 5
A significant change in energy will occur in the condenser during this process. heat rejection wi
absorb a huge amount of heat energy from the fluid exiting the turbine. this process will allow
POINT PRESSURE TEMPERATURE IN °C ENTHALPY IN KJ/Kg
4 10 Kpa 50 2591.99361918512
5 10 Kpa 40 167.543365678233
BY MASS BALANCE
g this process. heat rejection will cause cooling water from the cooling tower to
turbine. this process will allow the fluid to be condensed in the condenser.
ENTROPY IN KJ/Kg-K DENSITY IN Kg/m^3 STEAM CONDITION
8.17414407097184 0.0672613535554302 SATURATED LIQUID
0.572428587783723 992.184293629255 SUB-COOLED LIQUID
BY ENERGY BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
𝑚_4 ℎ_4=𝑚_5 ℎ_5+𝑄_𝑟
(0.793)(2591.994)=(0.793)(167.5434)+𝑄_𝑟
𝑸_𝒓=𝟏𝟗𝟐𝟐.𝟔 𝑲𝑱
PROPERTIES OF POINT 5 AND 6
However, the pumps may still exert some work on the fluid, and other properties may change,
mass will be assumed to be equal from the inlet to the outlet of the pump because in this proc
heat addition or heat rejection will occur.
POINT PRESSURE TEMPERATURE IN °C ENTHALPY IN KJ/Kg
5 10 Kpa 40 167.543365678233
6 800 Kpa 40.02 168.3
BY MASS BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
𝑚_5 ℎ_5 〖 +𝑊 〗 _𝐶𝑃
(0.793)(167.5434)+𝑊
𝑾_𝑪𝑷=𝟎.𝟔 𝑲𝑱
other properties may change, but the difference will not be significant, and the fluid
he pump because in this process the pumps are considered adiabatic and so no
BY ENERGY BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
𝑚_5 ℎ_5 〖 +𝑊 〗 _𝐶𝑃=𝑚_6 ℎ_6
(0.793)(167.5434)+𝑊_𝐶𝑃=(0.793)(168.3)
𝑾_𝑪𝑷=𝟎.𝟔 𝑲𝑱
PROPERTIES OF POINT 6, 7, AND 8
As a result, the process in this section of the feed water heater is an isobaric process, and at po
the exception of the pressure due to the throttle process.
BY MASS BALANCE
𝑚_7+𝑚_6=𝑚_8
(1−𝑦)+𝑦=1
(1−0.207112)+0.207112=1
TRUE
(1−𝑦)(168.3)+(𝑦)(2828
𝒚=𝟎.𝟐
isobaric process, and at point seven all the properties of point two will be adjusted with
BY ENERGY BALANCE
𝐸_𝑖𝑛=𝐸_𝑜𝑢𝑡
𝐸_6+𝐸_7=𝐸_8
𝑚_6 ℎ_6+𝑚_7 ℎ_7=𝑚_8 ℎ_8
(1−𝑦)(168.3)+(𝑦)(2828.27)=(1)(719.211)
𝒚=𝟎.𝟐𝟎𝟕𝟏𝟏𝟐 kg
PROPERTIES OF POINT 8 AND 9
However, the pumps may still exert some work on the fluid, and other properties may change
will be assumed to be equal from the inlet to the outlet of the pump because in this process th
heat rejection will occur.
POINT PRESSURE TEMPERATURE IN °C ENTHALPY IN KJ/Kg
8 800 Kpa 170 719.210731401668
9 10 Mpa 171.2 729.5
𝑾_𝑭𝑷=𝟗.𝟕𝟖𝟗 𝑲𝑱
other properties may change, but the difference will not be significant, and the fluid mass
mp because in this process the pumps are considered adiabatic and so no heat addition or
heat rejection will occur.
ENTROPY IN KJ/Kg-K DENSITY IN Kg/m^3 STEAM CONDITION
2.04191316525499 897.459672825424 COMPRESSED LIQUID
2.04191316525499 901.7132552 COMPRESSED LIQUID
𝑜𝑢𝑡
〖 +𝑊 〗 _𝐹𝑃=𝑚_9 ℎ_9
.211)+𝑊_𝐹𝑃=(1)(729.5)
𝟗.𝟕𝟖𝟗 𝑲𝑱
PROPERTIES OF POINT 9 AND 1
This process is isobaric, therefore the pressure will remain constant while the fluid will receive
process for an additional cycle.
BY MASS BALANCE
𝑚_𝑖𝑛=𝑚_𝑜𝑢𝑡 𝑚_9 ℎ_9+ 𝑄_𝐵𝑂𝐼𝐿𝐸𝑅=𝑚_1 ℎ_1
(1)=(1)
(1)(729.5)+𝑄_𝐵𝑂𝐼𝐿𝐸𝑅=(1)(3375.05
BY ENERGY BALANCE
ℎ_9+ 𝑄_𝐵𝑂𝐼𝐿𝐸𝑅=𝑚_1 ℎ_1
729.5)+𝑄_𝐵𝑂𝐼𝐿𝐸𝑅=(1)(3375.058)
𝑾_𝑻𝑶𝑻𝑨𝑳=𝟏𝟐𝟒𝟗.𝟏𝟗𝟏𝟐 𝑲𝑱
𝑊_𝑁𝐸𝑇= 〖 (𝑊 〗 _𝐻𝑃𝑇+𝑊_𝐿𝑃𝑇)−(𝑊_𝐶𝑃+𝑊_𝐹𝑃)
𝑊_𝑇𝑂𝑇𝐴𝐿=(545.7925+703.3987)+(0.6+9.789)
𝑾_𝑻𝑶𝑻𝑨𝑳=𝟏𝟐𝟑𝟖. 𝟖𝟎𝟐𝟐 𝑲𝑱
ENTROPY IN KJ/Kg-K DENSITY IN Kg/m^3 STEAM CONDITION
6.59932253468914 30.4758533669238 SUPER HEATED VAPOR
6.69548824266664 4.85428292677589 SATURATED VAPOR
7.76396503855744 2.82397913044349 SUPER HEATED VAPOR
8.17414407097184 0.0672613535554302 SATURATED LIQUID
0.572428587783723 992.184293629255 SUB-COOLED LIQUID
0.572428587783723 993.04866 COMPRESSED LIQUID
6.69548824266664 4.087 COMPRESSED LIQUID
2.04191316525499 897.459672825424 COMPRESSED LIQUID
2.04191316525499 901.7132552 COMPRESSED LIQUID
𝑸_𝑻𝑶𝑻𝑨𝑳=𝟑𝟏𝟔𝟏.𝟓𝟗𝟏𝟕 𝑲𝑱
𝒆_𝑻𝑪=𝟑𝟗. 𝟏𝟖𝟑 %