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