acknowledgements / student learnings and outcome - … · 2017-10-16 · acknowledgements / student...
Post on 11-Mar-2020
35 Views
Preview:
TRANSCRIPT
1
ACKNOWLEDGEMENTS / STUDENT LEARNINGS AND OUTCOME
First and foremost this project would not have been possible without the chief expert experience and supervision of
PROFF.JANSSENS–P.ENG. Also this project would not have been so successful without the genuine passion and
love for teaching of thermodynamics by DR. ASSAD AL SALMAN – P.ENG from centennial college in
Scarborough Ont. And last but not least this project would not have been possible without the individual effort put
forth by STUART DAVIE student and author of this term project report (we love thermodynamics ty).
• Engage in simulations where critical thinking assists in mechanical engineering design problem
solving at an advanced level
• Understand the triple constraint of project management i.e. time, scope, and cost
• Understand specifications and requirements through an iterative process
• Analyze data to identify the performance outputs of a combined Brayton-Rankine Power Cycle
• Utilize software to express results in a clear comprehensible manner
• Reference properly external and internal sources
• Enhance technical writing in professional format
• Expand knowledge in thermal science by analysis on various control volume devices
• Prepare specific questions in advance for tutorial sessions
• Exercise professionalism
• Draw on prior mechanical engineering design technology knowledge to make conscience logical
assumptions to reinforce success
• Master precision with advanced calculations
• Demonstrate the ability be successful
• Strengthen individual confidence through self-accomplishment
• Learn advanced thermal science terminology, properties and application
• Master in class note taking
• Generate and apply methodologies
• Become familiar with the function of various thermodynamic cycles, processes, and properties
• Numerically evaluate thermal problems at an advanced level
• Discuss ideas, results, and observations in an effective insightful manner
• Make core relations to independent and dependent variables visually, graphically, and
scientifically
• Draft a mechanical schematic of a combined Brayton-Rankine Power Cycle
2
INTRODUCTION
Problem Statement: “There have been increasing demands on power generation. New
technologies have emerged to supply more power production and improve overall systems efficiency.
One of these technologies is the use of combined power cycles. In this term project, it is required to
model, analyze and simulate the steady state performance of a combined Brayton-Rankine cycle based on
the design conditions,” - (Lakehead University – Department of Mechanical Engineering Term Project
Handout 2016)
The three forms of energy can be classified as work, heat, and internal energy. For the course of
this project considering all of these forms for analysis is essential; however, the internal energy will be
embedded in the thermodynamic property called enthalpy h (kJ/kg) or better known as flow energy.
Another property of much interest is a property called entropy s (kJ/kg.K) or better known as the quality
of energy of a system. Entropy is a property that is always increasing on a global/universal scale. Let us
consider a match is as a system that system has a certain amount of quality of energy. Now consider the
match being lite, at first the match is bursting with highly intensified flames and the level of entropy is
small, but as the match burns and slowly dwindles away so does the quality of energy of that match.
When the reporters and the marketers promote conservation of energy they are essentially referring to the
quality of energy that needs to be conserved. Stating the laws of thermodynamics;
“The four laws of thermodynamics define fundamental physical quantities (temperature, energy,
and entropy) that characterize thermodynamic systems. The laws describe how these quantities
behave under various circumstances, and forbid certain phenomena (such as perpetual motion).
The four laws of thermodynamics are:[1][2][3][4][5]
• Zeroth law of thermodynamics: If two systems are in thermal equilibrium independently with a
third system, they must be in thermal equilibrium with each other. This law helps define the
notion of temperature.
• First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out
from a system, its internal energy changes in accord with the law of conservation of energy.
Equivalently, perpetual motion machines of the first kind are impossible.
• Second law of thermodynamics: In a natural thermodynamic process, the sum of the entropies
of the interacting thermodynamic systems increases. Equivalently, perpetual motion machines of
the second kind are impossible.
• Third law of thermodynamics: The entropy of a system approaches a constant value as the
temperature approaches absolute zero.[2] With the exception of non-crystalline solids (glasses)
3
the entropy of a system at absolute zero is typically close to zero, and is equal to
the logarithm of the multiplicity of the quantum ground states.
There have been suggestions of additional laws, but none of them achieves the generality of the four
accepted laws, and they are not mentioned in standard textbooks.[1][2][3][4][6][7]
The laws of thermodynamics are important fundamental laws in physics and they are applicable in
other natural sciences.”( https://en.wikipedia.org/wiki/Laws_of_thermodynamics)
The Brayton cycle is an open cycle that has a purpose as heat engine to generate an output power,
where the Rankine cycle is a closed system that is a steam heat engine that also has a purpose to generate
an output power. Together, the combined Brayton - Rankine Power Cycle is used to generate power and
this combined cycle is favored globally because the combined cycles is capable of increasing the
efficiency without increasing the initial cost, the Rankine cycle can be modified with regeneration, and
additional gas turbines can be added to reheat to the Brayton cycle to increase the temperature of the air
before entering the transitional process. The operating processes of the Brayton-Rankine Power Cycle are
as follows (see schematic in the appendices for nomenclature)
1-2 Isentropic compression (Pump)
2-3 Constant pressure heat addition (Heater)
3-4 Isentropic expansion (Gas Turbine)
4-5 Constant pressure heat rejection (Perfectly Insulated Heat Exchanger)
6-7 Constant pressure heat addition (Perfectly Insulated Heat Exchanger)
7-8 Isentropic Expansion (Steam Turbine)
8-9 Constant pressure heat rejection (Condenser)
10-11 Constant pressure heat rejection (Condenser)
9-6 Isentropic compression (Pump)
4
The formulas used in the sample calculations have been derived and modified through an
advanced formulation process, to show fine detail is beyond the scope of this project. See Schematic for
each individual control volume
METHODOLOGY
Arrangements of graphs, tables, results, key parameters and remarks are made in a clear
comprehensive manner for each simulation design condition where the assumptions listed have been
made exposing the data that has been revealed through an analysis of a Brayton-Rankine Combined
Power Cycle.
1.) Compressor pressure ratios rp =P2/P1=5,10,14,17,20
2.) Gas turbine, steam turbine, pump, and compressor isentropic efficiencies = 100%,90%,80%,
and 70%
The objective of each simulation is to identify and or study the required performance outputs:
• The mass flow rates of the air, steam, and cooling water (each in
kg/s)
• The net power delivered by the Brayton cycle (in MW)
• The back-work ratio (rbw) for the Brayton cycle
• The net power produced by the Rankin cycle (in MW)
• The rate of heat rejection by the cooling water at the condenser
(MW)
• The thermal efficiencies of the individual cycles (%)
• The thermal efficiency of the combined cycle in (%)
Logical Assumptions
• Steady state steady flow system
• No mechanical or volumetric losses due to friction, sounds, and internal leakage in pipes
• Fluid at the inlet of the pump is saturated liquid
• Assume mass flow rate of air and the mass flow of the steam to be constant
• The temperature at the exit of the pump is approximately higher than the inlet of the pump (for
illustration purposes this is exaggerated, most pumps and compressors are adiabatic)
• The heater is adiabatic and there are no heat losses from the device
• Fixed conditions remain fixed for throughout the iteration analysis
5
IDENTIFYING FIXED CONDITIONS:
• P1 = 100Kpa
• T1 = 25 ᵒC
• 𝑸�̇� = 50MW
• P2 = P3
• T3 = 1250 ᵒC
• P4 = P5 = 100kPa
• T5 = 200 ᵒC
• P7= P6 = 12.5 MPa
• T7 = 500 ᵒC
• P8=P9=10 Kpa
• T10 = 20 ᵒC
• T11 = 35 ᵒC
Identifying Required Tables:
• T, P, h values at all states ( and s wherever applicable) for isentropic efficiency for turbines,
pump, and compressor = 100,90,80,and 70 %
• Individual and combined plant thermal efficiencies values for pressure ratio rp =5, 10, 14, 17, 20
and isentropic efficiency for turbines , pump, and compressor = 80%
Identifying Required Graphs:
• Ẇcombined (MW) vs rp
• ἠcombined vs rp
• rbw vs rp
• �̇�𝒂𝒊𝒓 / �̇�𝒔𝒕𝒆𝒂𝒎 vs rp
• �̇�𝒄𝒘 (Kg/s) & 𝑸𝒄𝒐𝒏𝒅̇ (MW) vs rp
• T-S diagram for Rankine Cycle (isentropic efficiency 90%)
• T-S diagram for Brayton Cycle (isentropic efficiency 90%)
6
SAMPLE CALCULATIONS
Implementing the variation of specific heat method with data from ideal-gas properties of air
Table A17 & Table T4 -7 Properties of pure substance (Thermodynamics an Engineering Approach (Si Units Yunus
A.Cengel|Michael A.Boles 2011) .
Variable Specific Heat
Properties (1-2)
T1 = 298 K
h1 = 298.18 kJ/kg
Pr1 =1.3543
P2 = 14 * 100 = 1400 KPa
Pr2 = (P2/P1) * Pr1 = (14) * (1.3543) = 18.96
T2s = 624.05 K
T2a = 680 + (690 – 640) * [(693.32 - 691.82)/(702.52 – 691.82)]
T2a = 681.40 K
h2s = 632.32 kJ/kg
▪
Properties (3-4-5)
T3 = 1523 K
h3 = 1660.23 + (1684.51 – 1660.23) * [(1523 -1520) / (1540-1520)]
h3 = 1663.87 kJ/kg
P2 = P3 = 1400 KPa
Pr3 = 601.9 + (636.5 – 601.9) * [(1523 – 1520) / (1540 -1520)]
Pr3 = 607.09
Pr4 = (P4/P3) * Pr3 = (100/1400) * 607.09 = 43.36
T4s = 780 K
h4s = 800.03 kJ/kg
7
T4a = 860 + (880 – 860) * [(908.85 – 888.27) / (910.50 -888.27)]
T4a = 878.51
h5 = 472.24 + (482.49 -472.84) * [(473-470) / (480 – 470)]
h5 = 475.415 kJ/kg
T5 = 473 K
P5 = 100 Kpa
▪
Isentropic Analysis
Wsc = h2s - h1 = 632.35 - 298.18 = 334.18 kJ/kg
Wst = h3 - h4s = 1667.87 – 800.03 = 867.84 kJ/kg
rbw = wcompin / wturbout = 334.17 / 567.84 = 0.3850 * 100 = 38.50 %
qin = h3-h2 = 1663.87 - 632.35 = 1031.52 kJ/kg
wnet = wout – win = 867.84 – 334.17 = 533.17 kJ/kg
qout = h4s –h5 = 800.03 – 475.32 = 324 .71 kJ/kg
ἠth – wnet / qin = 533.17 / 1031.52 = 0.5173 * 100 = 57.13 %
▪
Actual Analysis
wcompin = wsc/ ἠc = 334.17 / 0.85 = 393.14 kJ/kg
wturbout = ἠt * wst = (o.87) * (867.84) = 755.02 kJ/kg
rbw = wcompin / wturbout = 393.14 / 755.02 = 0.52 * 100 = 52 %
wcompin = h2a-h1 ;
h2a = h1 + wcompin = 298.18 + 393.14 = 693.32 kJ/kg
wturbout = h3 – h4a ;
h4a= h3 – wturbout = 1663.87 – 755.02
h4a = 908.85 kJ/kg
qin = h3 – h2a = 1663.87 – 697.32 = 970.55 kJ/kg
8
wnet = wout – win = 755.02 – 393.14 = 361.88 kJ/kg
ἠth = wnet / qin = (361.88 / 970.55) * 100 = 37.20 %6
▪
Properties (Rankine Cycle)
P9 = P8 = 10 KPa
v9 = vf = 0.001010 m3/kg
T9 = 45 ͦ C
u9 = uf = 191.79 kJ/kg
P7 = P6 = 12.5 MPa
s6s= s9 = sf = 0.6492 kJ/kg.K
h9 = h f = 191.81 kJ/kg @ 10 Kpa
h7 = 3343.6 kJ/kg
T7 = 500 ͦC
s 7s = s8s = 6.465.1 kJ/kg.k
u7 = 3023.2 kJ/kg
x = s7s – sf / sfg
x = (6.4651 - 0.6492) / 7.4996 = 0 .77
54
h8s = x * hfg + hf
h8s = (0.7754) * 2392.1 + 191.81 = 2046.87 kJ/kg
h11@35 ͦ C = 146.69 kJ/kg
h10@20 ͦ C = 83.915 kJ/kg
h5@100Kpa & 473 K = 475.24 kJ/kg.k
h4a = 908.85 kJ/kg
Assume T6 = 47.5 ͦC
▪
9
Energy Analysis
wps = v9 * (P6-P9) = 0.001010(12500 – 10 ) = 12.61 kJ/kg
h6 = h9 + wps = 191.81 + 12.61 = 204.42 kJ/kg
wts = (h7 - h8s) = 3343.6 – 2046.87 = 1296.73 kJ/kg
ἠpump = 100 % = (h6s - h9) / ( h6a – h9 ) wps / wa = ( 203.92 – 191.81 ) / ( h6a – 191.81) ;
h6a = 203.92 kJ/kg
ἠst = 90% = ( h7 – h8a) / (h7 – h8s) = wta/wts = ( 3343.6 – h8a) / (3343.6 – 2046.87 )
h8a = 2176.54 kJ/kg
qin = ( h7 – h6a) + (h4a – h5 ) = ( 3343.6 – 203.92) + ( 908.85 – 475.24) = 3573.29 kJ/kg
qout = (h11-h10) + (h8 – h9) = 146.69 – 83.96) + ( 2176.54 – 191.81) = 2047.47 kJ/kg
wturbout = (ἠst ) *(wts) = (0.90) * (1296.73) = 1167.05 kJ/kg
wpumpin = wps = 12.61 kJ/kg
wnet = wturbout – wpumpin = 1167.05 – 12.61 = 1154.44 kJ/kg
ἠth = wnet / qin = (1154.44 / 3563.29) * 100 = 32.3 %\
▪
Power Analysis Combined Gas-Vapor Power Cycle
�̇�𝒂𝒊𝒓 = �̇� / qin = 50,000 / 970.55 = 51.51 kg/s
�̇�𝒔𝒄 = �̇� * 334.17 = 17.213 MW
�̇�𝒔𝒕 = �̇� * 867.84 = 44.702 MW
�̇�𝒄𝒐𝒎𝒑𝒊𝒏 = �̇� * 393.14 = 20.250 MW
�̇�𝒕𝒖𝒓𝒃𝒐𝒖𝒕 = �̇� * 755.02 = 38.891 MW
�̇�𝑩𝒏𝒆𝒕 = �̇� * 361.88 = 18.640 MW
Considering transitional Process (4-5)
𝑸�̇� = �̇�𝑎𝑖𝑟 *(475.24) = 24.479 MW
�̇�𝒔𝒕𝒆𝒂𝒎 = 𝑄ℎ̇ / qin = 24,479 / 3139.68 = 7.80 kg/s
10
▪
�̇�𝒔𝒕𝒔 = �̇�𝑠 * 1296.73 = 10/115 MW
�̇�𝒕𝒐𝒖𝒕 = �̇�𝑠 1167.05 = 9.103 MW
�̇�𝒐𝒖𝒕 = �̇�𝑠 * (h8a – h9) = �̇�𝑠 (2176.54 – 191.81) = 15.480 MW
�̇�𝒄𝒘 = �̇�𝑜𝑢𝑡 / (h11 – h10) = 1548-.91 / (146.49 – 23.95) = 15,480.91 / 62.74 = 246.89 kg/s
�̇�𝑹𝒏𝒆𝒕 = �̇�𝑠 * 1154.44 = 9.006 MW
ἠcombined = (18,640.43 + 9004.63)/(50,000) = 55.96%
▪
Constant Specific Heat
Implementing the constant specific heat method using isentropic relationships and data from Table T4 -7
Properties of pure substance (Thermodynamics An Engineering Approach (Si Units Yunus A
Properties (Brayton Cycle)
P1 = 100 KPa
T1 = 298 K
rp = (P2/P1) = 14
P2 = (14)* (P1) = 1400 KPa
P2 = P3
T2s = (298) * (14)(1.4-1)/1.4 = 633.38 K
▪
Energy Analysis
wc = cp * (T2s –T1) = 1.005*(623.38-298) = 337.06 kJ/kg
wcompin = 337.06/(0.85) = 386.58 kJ/kg
qh = cp*(T3-T2)=1.004((1250+273) – (693)) = 833.32 kJ/kg
T4s = T3(P4/P3)(1.4 -1)/1.4 = 1523(100/1400) (1.4 -1)/1.4 = 716.53 K
wt = cp * ( T3 – T4) = 1.004*((1250+273) – 716.53)
wt = 809.70 kJ/kg
wturbout = (0.87)*(809.70) = 704.44 kJ/kg
11
wturbout = cp * (T3 – T4a); 704.44 = 1.004 ( 1523 – T4a) ; T4a = 822.06 K
qout = cp * ( T4a – T5) = 1.004 *( 822.06 – 473) = 350.80 kJ/kg
wbnet = wturbout- wcompin = 307.86 kJ/kg
▪
Power Analysis Combined Gas-Vapor Power Cycle
�̇�𝒂𝒊𝒓 = �̇� / qin = 50,000 / 833.32 = 60kg/s
�̇�𝒔𝒄 = �̇� * 337.06 = 20.223 MW
�̇�𝒔𝒕 = �̇� * 809.70 = 48.582 MW
�̇�𝒄𝒐𝒎𝒑𝒊𝒏 = �̇� * 396.58 = 23.794 MW
�̇�𝒕𝒖𝒓𝒃𝒐𝒖𝒕 = �̇� * 704.44 = 42.266 MW
�̇�𝑩𝒏𝒆𝒕 = �̇� * 376.86 = 22.612 MW
Considering transitional Process (4-5)
𝑸�̇� = �̇�𝑎𝑖𝑟 *(250.80) = 21.048 MW
�̇�𝒔𝒕𝒆𝒂𝒎 = 𝑄ℎ̇ / qin = 21,048 / 3490.48= 6.03 Kg/s
▪
�̇�𝒔𝒕𝒔 = �̇�𝑠 * 1296.73 = 7.819 MW
�̇�𝒕𝒐𝒖𝒕 = �̇�𝑠 1167.05 = 7.037 MW
�̇�𝒐𝒖𝒕 = �̇�𝑠 * (h8a – h9) = �̇�𝑠 (2176.54 – 191.81) = 11.967 MW
�̇�𝒄𝒘 = �̇�𝑜𝑢𝑡 / (h11 – h10) = 11967.92 / 62.74 = 190.7 kg/s
�̇�𝑹𝒏𝒆𝒕 = �̇�𝑠 * 1154.44 = 6.961 MW
ἠcombined = (22,611.6+ 6961.27)/(50,000) = 59.15 %
▪
12
Comparing Results:
Quantity Variable Specific
Heat
Constant Specific Heat
�̇�𝑎𝑖𝑟 (kg/s) 51.51
60
�̇�𝑠𝑐 (MW)
17.213
20.224
�̇�𝑠𝑡 = (MW)
44.702
48.582
�̇�𝑐𝑜𝑚𝑝𝑖𝑛
(MW) 20.250
23.795
�̇�𝑡𝑢𝑟𝑏𝑜𝑢𝑡 (MW) 38.891
42.266
�̇�𝐵𝑛𝑒𝑡 (MW) 18.640
22.612
𝑄ℎ̇ (MW)
24,479
21.048
�̇�𝑠𝑡𝑒𝑎𝑚 (kg/s) 7.80
6.03
�̇�𝑠𝑡𝑠 (MW) 10.115
7.819
�̇�𝑡𝑜𝑢𝑡 (MW) 9.103
7.037
�̇�𝑜𝑢𝑡 (MW) 15.481
11.968
�̇�𝑐𝑤 (kg/s) 246.89
190.75
�̇�𝑅𝑛𝑒𝑡 (MW) 9.005
6.961
ἠcombined (%) 55.96
59.15
TABLE A
Tabulated in Table 1 are results of the variable specific heat and constant specific heat for each
one of the performance indicators. The variable specific heat method yields more accurate results.
¶
13
RESULTS & DISCUSSION
Identifying Key Parameters:
The key parameters listed below are mandatory for determining the 7 performance outputs for the
Brayton - Rankine Combined Power Cycle.
ἠgt, ἠst, ἠp, ἠc (%) T1 (K) rp = P2/P1 T2s (K) T2a (K)
100 298 5 471.98 768.0996016
90 298 5 471.98 768.0996016
80 298 5 471.9673918 768.0870437
70 298 5 471.9673918 768.0870437
100 298 10 575.328998 871.0368506
90 298 10 575.328998 871.0368506
80 298 10 575.328998 871.0368506
70 298 10 575.328998 871.0368506
100 298 14 633.3810182 928.8575879
90 298 14 633.3810182 928.8575879
80 298 14 633.3810182 928.8575879
70 298 14 633.3810182 928.8575879
100 298 17 669.5076148 964.8402538
90 298 17 669.5076148 964.8402538
80 298 17 669.5076148 964.8402538
70 298 17 669.5076148 964.8402538
100 298 20 701.3269596 996.5328282
90 298 20 701.3269596 996.5328282
80 298 20 701.3269596 996.5328282
70 298 20 701.3269596 996.5328282
TABLE 1a
Tabulated in Table 1a are the required state properties for the temperatures at state 1, and 2
corresponding to the efficiencies of the gas turbine, steam turbine, pump, and compressor at 100, 90, 80,
and 70 % with relation to the pressure ratio at 5,10,15,17, and 20. Notice how temperature T1 stays
constant at 298 K and both the isentropic temperature and actual temperature at state 2 remain constant
for each until the rp value varies where T2s has values from 471.98 -701.32 and T2a has values of
768.10 K to 996.532 K.
14
ἠgt, ἠst, ἠp, ἠc (%) P2=P3 (kPa) T3 (K) P4=P5 (kPa) T4s (K)
100 500 1523 100 959.452195
90 500 1523 100 959.452195
80 500 1523 100 959.452195
70 500 1523 100 959.452195
100 1000 1523 100 786.321052
90 1000 1523 100 786.321052
80 1000 1523 100 786.321052
70 1000 1523 100 786.321052
100 1400 1523 100 713.9150613
90 1400 1523 100 713.9150613
80 1400 1523 100 713.9150613
70 1400 1523 100 713.9150613
100 1700 1523 100 675.2086821
90 1700 1523 100 675.2086821
80 1700 1523 100 675.2086821
70 1700 1523 100 675.2086821
100 2000 1523 100 644.426878
90 2000 1523 100 644.426878
80 2000 1523 100 644.426878
70 2000 1523 100 644.426878
TABLE 1b
Tabulated in Table 1b are the required state properties for the temperatures at state 3, and 4 as
well as the pressures for state 2,3,4,and 5 corresponding to the efficiencies of the gas turbine, steam
turbine, pump, and compressor at 100,90,80, and 70 % with relation to the pressure ratio at 5,10,15,17,
and 20. Notice how temperature T3 stays constant at 1523 K, the pressure P4=P5 remains constant at
states 4, and 5at 100 kPA and both the isentropic temperature at state 4s and the pressures at state 2, and 3
remain constant for until the rp value varies where T4s has values from 959.45 -644.42 K and P2=P3
has values of 500 to 2000 kPa.
ἠgt, ἠst, ἠp, ἠc (%) T4a (K) Tsat=T9 (K) P9=P8 (kPa) S9=S6=Sf (kJ/kg.K)
100 969.5432689 45.81 10 0.6492
90 1017.753984 45.81 10 0.6492
80 1072.165776 45.81 10 0.6492
70 1128.520054 45.81 10 0.6492
100 786.3210524 45.81 10 0.6492
90 859.9889472 45.81 10 0.6492
80 933.656842 45.81 10 0.6492
70 1007.324737 45.81 10 0.6492
15
100 713.9150613 45.81 10 0.6492
90 794.8235552 45.81 10 0.6492
80 875.7320491 45.81 10 0.6492
70 956.6405429 45.81 10 0.6492
100 675.2086821 45.81 10 0.6492
90 759.9878139 45.81 10 0.6492
80 844.7669457 45.81 10 0.6492
70 929.5460775 45.81 10 0.6492
100 644.4276878 45.81 10 0.6492
90 732.284919 45.81 10 0.6492
80 820.1421502 45.81 10 0.6492
70 907.9993815 45.81 10 0.6492
Table 1c
Tabulated in Table 1c are the required state properties for the temperatures at state 4, and 9 as
well as the pressures for state 9, and 8 as well as the entropy for state 9, and 6 corresponding to the
efficiencies of the gas turbine, steam turbine, pump, and compressor at 100, 90, 80, and 70 % with
relation to the pressure ratio at 5,10,15,17, and 20. Notice how temperature T9=T sat stays constant at
45.81ᵒC, and the pressure P9=P8 remains constant at 10 kPa and the Entropy S9=S6=Sf remain constant
at 0.6492 kJ/kg.K. The actual temperature at T4a has variation in values where the temperature increases
with the efficiencies for each of the pressure ratios.
ἠgt, ἠst, ἠp, ἠc (%)
h9 = hf
(kJ/kg) h6a (kJ/kg) P7=P6 (kPa) T7 (◦C)
100 191.81 204.42 12500 500
90 191.81 205.8211111 12500 500
80 191.81 207.5725 12500 500
70 191.81 209.8242857 12500 500
100 191.81 204.42 12500 500
90 191.81 205.8211111 12500 500
80 191.81 207.5725 12500 500
70 191.81 209.824857 12500 500
100 191.81 204.42 12500 500
90 191.81 205.8211111 12500 500
80 191.81 207.5725 12500 500
70 191.81 209.8242857 12500 500
100 191.81 204.42 12500 500
16
90 191.81 205.8211111 12500 500
80 191.81 207.5725 12500 500
70 191.81 209.8242857 12500 500
100 191.81 204.42 12500 500
90 191.81 205.8211111 12500 500
80 191.81 207.5725 12500 500
70 191.81 209.824857 12500 500
Table 1d
Tabulated in Table 1d are the required state properties for the temperatures at state 7 as well as
the pressures for state 7, and 6 as well as the enthalpy for state 9, and 6 corresponding to the efficiencies
of the gas turbine, steam turbine, pump, and compressor at 100, 90, 80, and 70 % with relation to the
pressure ratio at 5,10,15,17, and 20. Notice how enthalpy h9=hf stays constant at 191.81(kJ/kg), and the
pressure P7=P6 remains constant at 12500 kPa and the temperature T7 remains constant at 500ᵒC.The
actual enthalpy at h6a has variation in values 204.42,205.82,207.57, and 209.82 kJ/kg.
ἠgt, ἠst, ἠp, ἠc (%) h7 (kJ/kg) s7=s8s (kJ/kg.K) h8a (kJ/kg.K)
100 3343.6 6.4651 1957.47
90 3343.6 6.4651 2096.083
80 3343.6 6.4651 2234.696
70 3343.6 6.4651 2373.309
100 3343.6 6.4651 1957.47
90 3343.6 6.4651 2096.083
80 3343.6 6.4651 2234.696
70 3343.6 6.4651 2373.309
100 3343.6 6.4651 1957.47
90 3343.6 6.4651 2096.083
80 3343.6 6.4651 2234.696
70 3343.6 6.4651 2373.309
100 3343.6 6.4651 1957.47
90 3343.6 6.4651 2096.083
80 3343.6 6.4651 2234.696
70 3343.6 6.4651 2373.309
100 3343.6 6.4651 1957.47
90 3343.6 6.4651 2096.083
80 3343.6 6.4651 2234.696
70 3343.6 6.4651 2373.309
Table 1e
17
Tabulated in Table 1e are the required state properties for the enthalpy at state 7, and 8 as well as
the entropy at state 7, and 8 corresponding to the efficiencies of the gas turbine, steam turbine, pump, and
compressor at 100, 90, 80, and 70 % with relation to the pressure ratio at 5,10,15,17, and 20. Notice how
enthalpy h7 stays constant at 3343.6 (kJ/kg), and the entropy s7=s 8s remains constant at 6.4651 kPa. The
actual enthalpy at h6a has variation in values 1957.47, 2096.08, 2234.70, and2373.31 kJ/kg.
ἠgt, ἠst, ἠp, ἠc (%) T10 (◦C) h10 (kJ/kg) T11 (◦C) h11 (kJ/kg)
100 20 83.915 35 146.4
90 20 83.915 35 146.4
80 20 83.915 35 146.4
70 20 83.915 35 146.4
100 20 83.915 35 146.4
90 20 83.915 35 146.4
80 20 83.915 35 146.4
70 20 83.915 35 146.4
100 20 83.915 35 146.4
90 20 83.915 35 146.4
80 20 83.915 35 146.4
70 20 83.915 35 146.4
100 20 83.915 35 146.4
90 20 83.915 35 146.4
80 20 83.915 35 146.4
70 20 83.915 35 146.4
100 20 83.915 35 146.4
90 20 83.915 35 146.4
80 20 83.915 35 146.4
70 20 83.915 35 146.4
Table 1f
Tabulated in Table 1f are the required state properties for the enthalpy at state 10, and 11 as well
as the temperature at state 10, and 11 corresponding to the efficiencies of the gas turbine, steam turbine,
pump, and compressor at 100, 90, 80, and 70 % with relation to the pressure ratio at 5,10,15,17, and 20.
Notice how enthalpy h10 stays constant at 83.92 (kJ/kg), and the enthalpy h11 stays constant at 146.4
kJ/kg, and the temperature T10 stays constant at 20ᵒC, and the temperature T11 stays constant at 35ᵒC.
18
The above key parameters from Table 1(a-f) is also the required table (T, P, h values at all states (and
s wherever applicable) for isentropic efficiency for turbines, pump, and compressor = 100, 90, 80, and 70 %)
¶
ᾐdevices % ᾐR (%) ᾐB (%) ᾐcombined (%) rp
80 28.97 23.07 38.78 5
80 28.93 27.63 45.74 10
80 28.91 28.3 48.17 14
80 28.89 28.12 49.29 17
80 28.85 27.56 50.04 20
Table 2
Tabulated in Table 2 is the required table (Individual and combined plant thermal efficiencies values for
pressure ratio rp =5, 10, 14, 17, 20 and isentropic efficiency for turbines pump, and compressor = 80%) Notice how the
thermal efficiency of the Rankine cycle ᾐR slightly decreases from 28.97 – 28.85 % where the thermal
efficiency for the Brayton cycle ᾐB increases from 38.78 – 50.04 % with the variation of the pressure
ratio rp 5,10,14,17, and 20.
¶
19
Representing The Required Graphs:
Graph 1
Depicted in Graph 1 is a graphical representation of the Ẇcombined (MW) combined power
output for the Brayton – Rankine Combined Power Cycle at ᾐdevice (%) device isentropic thermal
efficiencies of 100, 90, 80, and 70 % for the rp values 5,10,14,17, and 20. Notice how the highest percent
efficiency generates the highest quantity of power. Also not how the lowest ᾐdevice efficiency of 70 %
decreases the power output with the increase of variation in rp values. Graph 1 is the required graph
(Ẇcombined (MW) vs rp).
REMARKS
ᾐdevices 100% 90% 80% 70% rp
Ẇcombined (MW)rp5 27.73 23.75 19.39 14.43 5
Ẇcombined (MW)rp7 34.74 29.12 22.87 15.58 10
Ẇcombined (MW)rp14 37.88 31.43 24.08 15.21 14
Ẇcombined (MW)rp17 39.71 32.33 24.64 14.64 17
Ẇcombined (MW)rp20 41.26 33.79 25.02 13.89 20
Table 3
Tabulated in Table 3 are the quantities that correspond to Graph 1 again, notice how the work
combined output decreases from 14.43 -13.89 MW for thermal efficiencies of 70 % compared to the work
05
1015202530354045
5 10 14 17 20
Ẇcombined (MW)
rp
Figure (R-1)
100%
90%
80%
70%
ᾐdevices
20
combined output for the isentropic thermal efficiencies of 100 % increases from 27.73 – 41.26 MW with
the increase of variation in rp values.
¶
Graph 2
Depicted in Graph 2 is a graphical representation of the ἠcombined (%) combined thermal
efficiency for the Brayton – Rankine Combined Power Cycle at ᾐdevice (%) device isentropic
thermal efficiencies of 100, 90, 80, and 70 % for the rp values 5,10,14,17, and 20. Notice how
the highest percent ᾐdevice (%) of 100% and combined efficiency ἠcombined (%) generates the
highest efficiencies in the system. Also not how the lowest ᾐdevice efficiency of 70 % decreases
the lowest efficiencies in the system with the increase of variation in rp values. Graph 2 is the
required graph (ἠcombined vs rp).
REMARKS
ᾐdevices 100% 90% 80% 70% rp
ᾐcombined(%) rp5 55.46 47.49 38.8 28.88 5
ᾐcombined(%) rp10 69.48 58.24 45.74 31.17 10
ᾐcombined(%) rp14 75.77 62.86 48.17 30.43 14
ᾐcombined(%) rp17 79.42 65.45 49.29 29.29 17
ᾐcombined(%)rp20 82.52 67.59 50.04 27.78 20
Table 4
Tabulated in Table 4 are the quantities that correspond to Graph 2 again, notice how the
combined efficiency decreases from 28.88 – 27.78 % for thermal efficiencies of 70 % compared to the
combined efficiency increase from 55.46-82.52% for isentropic thermal efficiencies of 100 % with the
increase of variation in rp values.
0
20
40
60
80
100
5 10 14 17 20
ᾐcombined(%)
rp
Figure (R-2)
100%
90%
80%
70%
21
¶
Graph 3
Depicted in Graph 3 is a graphical representation of the rbw (%) ratio back work for the Brayton
– Rankine Combined Power Cycle at ᾐdevice (%) device isentropic thermal efficiencies of 100, 90, 80,
and 70 % for the rp values 5,10,14,17, and 20. Notice how the highest ᾐdevice (%) at 100% generates
the lowest amount of ratio back work rbw (%) . Also notice how the lowest ᾐdevice (%) efficiency of 70
% generates the highest amount of ratio back work rbw (%) with the increase of variation in rp values.
Graph 3 is the required graph (rbw vs rp ).
REMARKS
ᾐdevices 100% 90% 80% 70% rp
rbw 31 38.25 48.51 63.72 5
rbw 37.65 46.48 58.82 76.83 10
rbw 41.45 51.18 64.77 84.6 14
rbw 43.82 54.11 68.47 89.43 17
rbw 45.91 56.78 71.73 93.69 20
Table 5
Tabulated in Table 5 are the quantities that correspond to Graph 3 again, notice how the back
work ratio increases from 63.72-93.69 % for isentropic thermal efficiencies of 70 % compared to the
back work ratio increase from 31 – 45.91% for thermal efficiencies of 100 % with the increase of
variation in rp values.
0
10
20
30
40
50
60
70
80
90
100
5 10 14 17 20
rbw(%)
rp
Figure (R-3)
100%
90%
80%
70%
22
¶
Graph 4
Depicted in Graph 4 is a graphical representation of the ₥air/₥steam (kg/s) ratio of mass flow
rate air to that of mass flow rate steam for the Brayton – Rankine Combined Power Cycle at ᾐdevice (%)
device isentropic thermal efficiencies of 100, 90, 80, and 70 % for the rp values 5,10,14,17, and 20.
Notice how the highest ᾐdevice (%) at 100% has an increase in ₥air/₥steam (kg/s) ratio until an rp
value of 10 and then decreases to an rp value of 20. The90, 80, and 70 % ᾐdevice (%) has a decrease in
₥air/₥steam (kg/s) ratio with the increase of variation in rp values. Graph 4 is the required graph
( �̇�𝒂𝒊𝒓 / �̇�𝒔𝒕𝒆𝒂𝒎 vs rp).
REMARKS
ᾐdevices 100(%) 90(%) 80(%) 70(%) rp
₥air/₥steam (kg/s) 6.443537415 6.895714286 6.884561892 6.92069892 5
₥air/₥steam (kg/s) 6.869281046 6.858585859 6.847826087 6.82840909 10
₥air/₥steam (kg/s) 6.825609756 6.8168028 6.806167401 6.78556911 14
₥air/₥steam (kg/s) 6.784883721 6.774585635 6.763429752 6.74788335 17
₥air/₥steam (kg/s) 6.741935484 6.734243697 6.719844358 6.70891608 20
Table 6
Tabulated in Table 6 are the quantities that correspond to Graph 4 again, notice how the mass
flow rate of air to the mass flow rate of steam ratio increases from 6.44-6.86 as the pressure ratio
increases from 5-10. Again, notice how the mass flow rate of air to the mass flow rate of steam ratio
decreases from 6.87-6.74 with the increase in pressure ration from 10-20 for isentropic thermal
efficiencies of 100%. Notice how the behaviors of the mass flow rate of air to the mass flow rate of steam
6.2
6.4
6.6
6.8
7
5 10 14 17 20
₥air/₥steam (kg/s)
rp
Figure (R-4)
100(%)
90(%)
80(%)
70(%)
23
ratio for the isentropic thermal efficiencies 0f 90, 70, and 80 % decrease in similar fashion with the
increase of variation in rp values.
¶
Graph 5a
Graph 5b
Depicted in Graph 5a and depicted in Graph 5b is a graphical representation of the ₥cooling water
(kg/s) cooling water mass flow rate & Qcond (MW) for the Brayton – Rankine Combined Power Cycle
0
100
200
300
400
500
5 10 14 17 20
₥coolingwater (kg/s)
rp
Figure(R-5b)
100%
90%
80%
70%
0
5
10
15
20
25
30
5 10 14 17 20
Qcond (MW)
rp
Figure(R-5a)
100%
90%
80%
70%
24
at ᾐdevice (%) device isentropic thermal efficiencies of 100, 90, 80, and 70 % for the rp values
5,10,14,17, and 20. Notice how both graphs have similar behaviors and the highest thermal efficiency of
100 % has the lowest mass flow rate ₥coolingwater (kg/s) and Qcond (MW) heat rejection from the
condenser. Notice how both graphs have similar behaviors and the lowest thermal efficiency of 70 % has
the highest mass flow rate ₥coolingwater (kg/s) and Qcond (MW) heat rejection from the condenser.
Notice how and all efficiencies increase the ₥coolingwater (kg/s) and Qcond (MW) with the increase
of variation in rp values. Graph 5a and Graph 5b is the required graph( �̇�𝒄𝒘 (Kg/s) & 𝑸𝒄𝒐𝒏𝒅̇ (MW) vs
rp).
REMARKS
ᾐdevices 100% 90% 80% 70% rp
₥coolingwater
(kg/s) 217.37 219.28 241.15 266.22 5
₥coolingwater (kg/s)
222.85 248.29 278.01 314.69 10
₥coolingwater
(kg/s) 238.88 268.74 304.91 351.84 14
₥coolingwater (kg/s)
250.52 283.75 325.1 380.3 17
₥coolingwater
(kg/s) 261.81 298.47 345.016 409.41 20
Table 7a
ᾐdevices 100% 90% 80% 70% rp
Qcond (MW) 13.3 13.7 15.07 16.64 5
Qcond (MW) 13.93 15.51 17.37 19.66 10
Qcond (MW) 14.93 16.79 19.05 21.98 14
Qcond (MW) 15.65 17.73 20.31 23.76 17
Qcond (MW) 16.36 18.65 21.56 25.58 20
Table 7b
Tabulated in Table 7a and Table 7b are the quantities that correspond to Graph 5a and Graph 5b
again, notice how the mass flow rate of the cooling water and the heat rejection from the condenser have
lowest values of 217.37 -261.81 kg/s and 13.3-16.36 MW for the 100 % isentropic efficiencies as rp
increases in variation from 5, 10, 14, and 17 - 20. Notice how the mass flow rate of the cooling water and
the heat rejection from the condenser have highest values of 266.22-409.41 kg/s and 16.64 -25.58 MW
for the 70 % isentropic efficiencies as rp increases in variation from 5, 10, 14, and 17 – 20.
25
¶
Graph 6
Depicted in Graph 6 is a graphical representation of the behavior of temperature K with respect
to entropy kJ/kg.K. Notice how the entropy remains constant from state 2 to state 3(compressor) as well
from state 3 to state 4(gas turbine). This is the evidence of the isentropic process. Also notice how the
pressure is constant from state 3 to state 1 where the heat addition occurs through the heat exchanger.
Also notice how the pressure remains constant from state 4 to state 1 during the heat rejection process.
This is an open cycles, however for visualization effects state five is a dummy state not included. Also the
actual temperatures of state 4a and state 2a are disregarded respectfully. Graph 6 is required (T-S diagram
for Brayton Cycle isentropic efficiency90%).
REMARKS
from table A17 Isentropic
State Entropy kJ/kg.K Temperature K
1 1.7 298
2 1.7 471.98
3 3.46 1523
0
200
400
600
800
1000
1200
1400
1600
0 1 2 3 4
Temperature K
S (kJ/kg.K)
Figure (R-7)
26
4 3.46 959.45
Table 8a
Tabulated in Table 8a are the values of the temperatures and entropies at the specified states for
rp =5. For values of rp 10, 14, 17, and 20 refer to table 1a-1f. Notice how the format of the shape of the
graph does not change and is aligned with the processes. See schematic for nomenclature.
¶
Graph 7
REMARKS
Depicted in Graph 7 is a graphical representation of the behavior of temperature ᵒC with respect
to entropy kJ/kg.K. Notice how the entropy remains constant from state 9 to state 6 (pump) as well from
state 7 to state 8(steam turbine). This is the evidence of the isentropic process. Also notice how the
pressure is constant from state 3 to state 1 where the heat addition occurs through the heat exchanger.
Also notice how the pressure remains constant from state 4 to state 1 during the heat rejection process.
This is a closed cycles, however for visualization effects states 10 and state 11 are not included. Also the
actual temperatures of state 6a and state 8a are disregarded respectfully. Graph 7 is required (T-S diagram
for Rankine Cycle isentropic efficiency90%).
0
100
200
300
400
500
600
0 2 4 6 8
Temperature o C
S (kJ/kg.K)
Figure (R-6)
P7=P6=12.5MPaP8=P9=10KPa
27
¶
CONCLUSION
Conducting mechanical engineering design analysis on the Brayton – Rankine Combined Power
Cycle that has a purpose to produce output power from the system under the stated fixed and simulation
design conditions
1.) Compressor pressure ratios rp =P2/P1=5, 10,14,17,20
2.) Gas turbine, steam turbine, pump, and compressor isentropic efficiencies = 100%, 90%, 80%,
and 70%
While utilizing property tables, making logical assumptions, and identifying key parameters has led to
accurate results in the form of the performance out puts:
• The mass flow rates of the air, steam, and cooling water (each in
kg/s)
• The net power delivered by the Brayton cycle (in MW)
• The back-work ratio (rbw) for the Brayton cycle
• The net power produced by the Rankine cycle (in MW)
• The rate of heat rejection by the cooling water at the condenser
(MW)
• The thermal efficiencies of the individual cycles (%)
• The thermal efficiency of the combined cycle in (%)
The performance outputs were needed to obtain the required tables and graphs for the variations
in quantities. Conscience remarks have been made and notices have been emphasized for each
requirement. Now places emphasis on each required table and graph.
Table 1a-1f
Identifies the key parameters needed such as state properties i.e. temperature, pressure, enthalpy,
entropy and pressure for states 1 – 11. Because of the isentropic processes and the required out puts states
2, 4, 6, and 9 have 2 states, one actual and one ideal. For calculations and determining the actual
28
performance out puts the actual enthalpy and temperature properties have been used in
calculations after implementing the isentropic efficiencies of the turbines, pumps, and compressor.
¶
Table 2
Represents the relationship in plant efficiencies individually and combined for the variation in
pressure ratio. The combined thermal plant efficiency is dramatically improved by increasing the pressure
ratio. This is evidence and reason why the Brayton – Rankine Combined Power Cycle generates more
power when operating with devices have higher isentropic efficiencies.
¶
Graph 1
Represents the relationships of the combined work power output with the variation in pressure
ratio. The combined work output is dramatically improved by having isentropic thermal efficiencies in the
devices maintained as high as possible. This is evidence and reason why the Brayton – Rankine
Combined Power Cycle generates more power when operating with devices have higher isentropic
efficiencies.
¶
Graph 3
Represents the relationships in the back work ratio with the variation in pressure ratio. The
Brayton cycle is an open system as stated and requires work from the turbine to operate the compressor.
Having low isentropic thermal efficiencies in the devices literally degrades the amount power output of
the Brayton-Rankine Combined Power Cycle. This is evidence and reason why the Brayton – Rankine
Combined Power Cycle generates more power when operating with devices have higher isentropic
efficiencies.
¶
29
Graph 4
Represents the relationships in the mass flow ratio of air and steam with the variation in pressure
ratio. Increasing the pressure ratio above 10 for all the isentropic efficiencies decreases the mass flow
ratio.
¶
Graph 5
Represents the relationships in the mass flow rate of cooling water and heat rejection in the
condenser with the variation in pressure ratio. Hat power rejection by the Rankine-Brayton Combined
Power Cycle is highest when isentropic efficiencies are lowest. Logically assuming as previously stated
that there are no energy losses in the pipes and the devices are adiabatic the amount of Heat rejected can
literally be transformed to output power by simply improving the isentropic thermal efficiencies. This is
evidence and reason why the Brayton – Rankine Combined Power Cycle generates more power when
operating with devices have higher isentropic efficiencies.
¶
Graph 6
Is the T-S diagram for the Rankine cycle at isentropic efficiency of 90%. This is an idyllic graph
as mentioned in introduction entropy can only increase in the universe and the Isentropic processed are
idealized processed used to compare actual processes. The properties are aligned and by integration it can
be proved and evidence can be presented why the Brayton – Rankine Combined Power Cycle generates
more power when operating with devices have higher isentropic efficiencies
¶
30
Graph 7
Is the T-S diagram for the Brayton Cycle at isentropic efficiency of 90%. This is an idyllic graph
as mentioned in introduction entropy can only increase in the universe and the Isentropic processed are
idealized processed used to compare actual processes. This graph differs from the Brayton Cycle because
a pure substance is used as means of heat transfer as appose to an ideal gas so a saturation dome is
present. The properties are aligned and by integration it can be proved and evidence can be presented why
the Brayton – Rankine Combined Power Cycle generates more power when operating with devices have
higher isentropic efficiencies
¶
31
DETAILED SCHEMATIC
32
Lakehead University – Department of Mechanical Engineering Term Project Handout 2016
https://en.wikipedia.org/wiki/Laws_of_thermodynamics
Thermodynamics An Engineering Approach Si Units Yunus A.Cengel|Michael A.Boles 2011
33
APPENDICES
34
35
36
37
38
39
top related