overview engineering cycles

Engineering Cycles

Dr. Md. Zahurul Haq

ProfessorDepartment of Mechanical Engineering

Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh

http://zahurul.buet.ac.bd/

ME 6101: Classical Thermodynamics

http://zahurul.buet.ac.bd/ME6101/

© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 1 / 63

Overview

1 Gas Power Cycles

Cycles for Engines

Cycles for Gas Turbines

2 Vapour Power Cycles

3 Refrigeration Cycles

Ideal Vapour Compression Refrigeration Cycle

Miscellaneous Refrigeration Systems


Gas Power Cycles Cycles for Engines

Air-Standard Cycle Assumptions

• The working fluid is air, which continuously circulates in a closed loop

and always behaves as an ideal gas.

• All the processes that make up the cycle are internally reversible.

• The combustion process is replaced by a heat-addition process from an

external source.

T223 T224

• The exhaust process is replaced by a heat-rejection process that

restores the working fluid to its initial state.

• Air has constant specific heats determined at room temperature.

⇒ A cycle for which the air-standard assumptions are applicable is

frequently referred to as an air-standard cycle.



Overview of Reciprocating (R/C) Engines��

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T625 T227

• TDC ≡ Top Dead Centre, BDC ≡ Bottom Dead Centre

• Compression Ratio ≡ r = VmaxVmin

=VBDCVTDC

• Displacement Volume ≡ Vd = π4 × S × B2 = Vmax − Vmin

• Mean Effective Pressure ≡ MEP = WnetVd



The Otto Cycle: Ideal Cycle for SI Engines

T228



T229

• 1 → 2 : Isentropic compression

• 2 → 3 : Reversible, Constant-volume heat addition

• 3 → 4 : Isentropic expansion

• 4 → 1 : Reversible, Constant-volume heat rejection

• Isentropic processes: 1 → 2 & 3 → 4. Also, V2 = V3 & V4 = V1

→ T1T2

=(

V2V1

)k−1=

(

V3V4

)k−1= T4

T3

T1T2

= 1rk−1 & T3

T2= T4

T1

• qin = u3 − u2 = cv (T3 − T2) : qout = u4 − u1 = cv (T4 − T1)

• ηth,Otto = wnet

qin= 1 − qout

qin= 1 − T2

T1

[T3/T2−1][T4/T1−1] = 1 − T2

T1= 1 − 1

rk−1

ηth,Otto = 1 − 1rk−1

r = compression ratio,

k = ratio of specific heats, cp/cv



T230

Thermal efficiency of the ideal Otto

cycle as a function of compression ratio

(k = 1.4).

T231

The thermal efficiency of the Otto cycle

increases with the specific heat ratio, k

of the working fluid.



Example: Otto cycle: ⊲

P1 = 0.1 MPa, T1 = 300 K , r = 10, qin = 1800 kJ/kg.

• P2/P1 = rk −→ P2 = 2.511 MPa

• T2/T1 = rk−1−→ T2 = 753.6 K

• v = RT/P ⇒ v1 = 0.861 m3/kg, v2 = 0.0861 m3/kg

• cp − cv = R cv = Rk−1

= 0.7175 kJ/kgK

• qin = cv (T3 − T2) → T3 = 3261.4 K

• v2 = v3, v4 = v1

• T4 = T3

(

v3v4

)k−1

= 1298.4 K

• wnet = qin − qout = 1083.4 kJ/kg

• ηth,Otto = 1 − 1

rk−1 = 0.602 ◭

• MEP = wnet

v1−v2= 1.398 MPa ◭



The Diesel Cycle: Ideal Cycle for CI Engines

T232T233 T275

• Isentropic processes: 1 → 2 & 3 → 4.

• Cut-off ratio, rc = V3V2

, and V4 = V1.

• qin = h3 − h2 = cp(T3 − T2) : qout = u4 − u1 = cv (T4 − T1)

• ηth,Diesel =wnet

qin= 1 − qout

qin= 1 − T2

kT1

[T3/T2−1][T4/T1−1] = 1 − 1

rk−1

[

rkc −1k(rc−1)

]

ηth,Diesel = 1 − 1rk−1

[

rkc −1k(rc−1)

]



T234

T235

• For same r : ηth,Otto > ηth,Diesel .

• Diesel engines operate at much higher

compression ratios, and thus are usually

more efficient than SI-engines.

Dual Cycle:

• Two heat transfer processes, one at

constant volume and one at constant

pressure.



Example: Diesel cycle: ⊲

P1 = 0.1 MPa, T1 = 300 K , r = 20, qin = 1800 kJ/kg.

• P2/P1 = rk −→ P2 = 6.629 MPa

• T2/T1 = rk−1−→ T2 = 994.3 K

• v = RT/P ⇒ v1 = 0.861 m3/kg, v2 = 0.0430 m3/kg

• qin = cp(T3 − T2) → T3 = 2785.6 K

• P2 = P3 → v3 = v2T3

T2

, v4 = v1

• T4 = T3

(

v3v4

)k−1

= 1270.0 K

• wnet = qin − qout = 1104.5 m3/kg

• ηth,Diesel =wnet

qin= 0.613 ◭

• rc = v3/v2 = 2.80

• ηth,Diesel = 1 − 1

rk−1

[

rkc −1

k(rc−1)

]

= 0.613 ◭

• MEP = wnet

v1−v2= 1.355 MPa ◭



Stirling & Ericsson Cycles

T276

• Stirling and Ericsson cycles involve an isothermal

heat-addition at TH and an isothermal heat-rejection at

TL. They differ from the Carnot cycle in that the two

isentropic processes are replaced by two

constant-volume regeneration processes in the Stirling

cycle and by two constant-pressure regeneration

processes in the Ericsson cycle.

• Both cycles utilize regeneration, a process during which

heat is transferred to a thermal energy storage device

(called a regenerator) during one part of the cycle and is

transferred back to the working fluid during another

part of the cycle.Stirling cycle is made up of four totally reversible processes:

1 → 2: Isothermal expansion (heat addition from the external source)

2 → 3: Isochoric regeneration (internal heat transfer from the working fluid to the

regenerator)

3 → 4: Isothermal compression (heat rejection to the external sink)

4 → 1: Isochoric regeneration (internal HT from regenerator back to fluid)



T236


Gas Power Cycles Cycles for Gas Turbines

Brayton Cycle: Ideal Cycle for Gas Turbines

T237

An open-cycle gas-turbine

T238

A closed-cycle gas turbine

T239



• qin = h3 − h2 = cp(T3 − T2)

• qout = h4 − h1 = cp(T4 − T1)

• rp ≡P2P1

⇒ T2T1

= r(k−1)/kp = T3

T4

• ηth,Brayton = 1 − qoutqin

= 1 −T1(T4/T1−1)T2(T3/T2−1)

⇒ ηth,Brayton = 1 − 1

r(k−1)/kp

• wnet = cp [(T3 − T4) + (T1 − T2)]

wnet

cp= T3

[

1 − 1

r(k−1)/kp

]

+ T1

[

1 − r(k−1)/kp

]

• For max. work: ∂wnet

∂rp= 0:

⇒ rp =(

T3T1

)k/2(k−1): for max. wnet

• If T3 = 1000 K , T1 = 300 K ⇒ rp = 8.2.

T240

T241



T626

T627

Moran 9.4, 9.6: ⊲ a) ηisen = 1.0, b) ηisen = 0.8

a) rp = 10, ηisen = 1.0:

• T2s = 543.4 K , T4s = 717.7 K

• wc,s = 244.6 kJ/kg, wt,s = 585.2 kJ/kg

• qin = 760.2 kJ/kg, ηth = 0.448 ◭

a) rp = 10, ηisen = 0.8:

• ηc =wc,s

wc,a≃

T2s−T1

T2−T1

: ηt =wt,a

wt,s≃

T3−T4

T3−T4s

• T2a = 604.3 K , T4a = 834.1 K

• wc,a = 305.7 kJ/kg, wt,a = 468.1 kJ/kg

• qin = 699.0 kJ/kg, ηth = 0.232 ◭



Brayton Cycle with Regeneration

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T628

• qregen,act = hx − h2 : qregen,max = h4 − h2

• Effectiveness, ǫ ≡qregen,actqregen,max

= hx−h2h4−h2

≃Tx−T2T4−T2

• ηth,regen = 1 −(

T1T3

)

r(k−1)/kp



Moran 9.7: ⊲

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T629

• ǫ ≡qregen,actqregen,max

≃Tx−T2

T4−T2

Tx = ǫ(T4 − T2) + T2 = 745.5 K

• wnet = wt + wc = (h3 − h4) + (h1 − h2) = Cp[(T3 − T4) + (T1 − T2)]

⇒ wnet= 427.4 kJ/kg

• qin = (h3 − hx) = Cp(T3 − Tx) = 753.2 kJ/kg

• η = wnet

qin= 56.7%



Turbines with Reheat, Compressors with Intercooling

T246

T250

T252



Moran 9.9: ⊲

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T631

• h1 = h(100 kPa, 300 K ), s1 = s(100 kPa, 300 K )

• hc = h(300 kPa, s1), hd = h(300 kPa, 300 K )

• wc = wc1 + wc2 = (h1 − hc) + (hd − h2) = -234.9 kJ/kg◭

If single stage compression is done:

• h2 = h(1000 kPa, s1)

• wc = (h1 − h2) = -280.1 kJ/kg◭



Brayton Cycle with Intercooling, Reheating & Regeneration

T249



Cengel 9.8: ⊲ GT with reheating & intercooling: rp = 8, ηisen = 1.0, ǫ = 1.0

T251

• In case of staging, for min. compressor

work or for max. turbine work:

Pi =√

PminPmax

• wturb = (h6 − h7) + (h8 − h9)

• wcomp = (h2 − h1) + (h4 − h3)

• wnet = wturb − wcomp, bwr =wcomp

wturb

• Without regen: wturb = 685.28 kJ/kg ◭ wcomp = 208.29 kJ/kg ◭

• qin = (h6 − h4) + (h8 − h7) = 1334.30 kJ/kg ◭

⇒ ηth = 0.358 ◭ bwr = 0.304 ◭

• With regen: turbine and compressor works remains unchanged.

• qin = (h6 − h5) + (h8 − h7) = 685.28 kJ/kg ◭

⇒ ηth = 0.696 ◭ bwr = 0.304 ◭


Vapour Power Cycles

Components of a Simple Vapour Power Plant

T193


Vapour Power Cycles

Rankine Cycle

T195

1 → 2 : Isentropic compression in a pump

2 → 3 : Constant pressure heat addition in a boiler

3 → 4 : Isentropic expansion in a turbine

4 → 1 : Constant pressure heat rejection in a condenser


Vapour Power Cycles

Cengel 10.1: ⊲ Ideal Rankine Cycle: Steam enters the turbine at 3 MPa and

350oC, and condenser is at 75 kPa.

T196

⇒ Steady state ⇒ dEcv/dt = 0

⇒ Z2 = Z1 & V2 = V1, mi = me = m

⇒ 0 = q − w + (hi − he)

• Turbine: q = 0, w = wT = h3 − h4

• Pump: q = 0, w = wP = h1 − h2

• Boiler: w = 0, q = qB = h3 − h2

• Condenser: w = 0, q = qC

• h2 is in compressed state & difficult to estimate. Hence, wP = −∫

2

1vdP .

⇒ wnet = wT + wP = (713.0 − 3.03) kJ/kg = 710.0 kJ/kg. ◮ wT ≫ wP .

⇒ qin = qB = h3 − h2 = 2728.6 kJ/kg

⇒ Thermal efficiency, ηth = wnet

qin= 0.260 = 26.0% ◭

⇒ Back work ratio, bwr = |wP

wT| = 0.0043 ◭


Vapour Power Cycles

Deviation of Actual Vapour Power Cycle

T197


Vapour Power Cycles

Example: Actual Rankine Cycle: ⊲ Steam enters the turbine at 3 MPa and

350oC, and condenser is at 75 kPa and ηisen = 95% for pump and turbine.

T198

⇒ ηt =wt

wt |s= h3−h4a

h3−h4s

⇒ ηc =−wc |s−wc

= h2s−h1

h2a−h1

• wc = −∫

2s

1vdP = −v1(PB − PC )

• h2a = h1 +v1(PB−PC )

ηc

• q2a−3 = h3 − h2a

• qin = 2727.2 kJ/kg.

• wt = 676.97 kJ/kg.

• wp = 3.19 kJ/kg.

• ηth = 24.71%◭


Vapour Power Cycles

Effect of Lowering Condenser Pressure

T199 T218

• Pcond = Psat(Tcond ) : Tcond − Tatm ≃ 10 − 15oC.

• Pcond ↓ :⇒ wnet ↑, ηth ↑ & x4 ↓. Higher moisture decreases turbine

efficiency and erodes its blades. In general, x4 > 0.9 is maintained.

Lower Pcond promotes leakage.


Vapour Power Cycles

PB = 2.64 MPa, TH = 800 K

.

T217


Vapour Power Cycles

Effect of Super-heating Steam to Higher Temperature

T200 T219

• Tmax ↑ :⇒ wnet ↑, ηth ↑ & x4 ↑.• Higher average temperature of heat addition increases ηth. Tmax is

limited by metallurgical considerations. In general, Tmax = 620oC.


Vapour Power Cycles

PB = 2.64 MPa, TL = 325 K

.

T216


Vapour Power Cycles

Effect of Increasing Boiler Pressure

T201T220

• For fixed Tmax : PB ↑:⇒ ηth ↑ & x4 ↓. Higher ηth is achieved because

of higher average temperature of heat addition.


Vapour Power Cycles

TH = 800 K, TL = 325 K

.

T215


Vapour Power Cycles

Effects of Operating Parameters on Ideal Rankine Cycle Efficiency

Boiler Pressure [MPa] 3.0 3.0 3.0 15.0

Max. Temperature [oC] 350 350 600 600

Cond. Pressure [kPa] 75 10 10 10

Heat added [kJ/kg] 2729 2921 3488 3376

Turbine work [kJ/kg] 713 979 1302 1467

Pump work [kJ/kg] 3.03 3.02 3.02 15.1

Thermal efficiency [%] 26.0 33.4 37.3 43.0

x4 [-] 0.886 0.812 0.915 0.804


Vapour Power Cycles

T202

A supercritical Rankine cycle

• Some modern power plants operate at

supercritical pressure

(P ≈ 30 MPa > PC = 22.06 MPa) and

have ηth ∼ 40% for fossil-fuel plants and

ηth ∼ 34% for nuclear power plants.

• Lower ηth of nuclear power plants are

due to lower maximum temperatures

used due to safety reasons.


Vapour Power Cycles

Ideal Reheat Rankine Cycle

T203

• qin = q23 + q45 = (h3 − h2) + (h5 − h4)

• wturb = w34 + w56 = (h3 − h4) + (h5 − h6)

• Average temperature of heat addition is increased :⇒ ηth ↑. Moisture

quality at turbine exit is also improved.


Vapour Power Cycles

Preheat = 4.0 MPa

T205

Heat added = 3897.2 [kJ/kg]

Turbine work = 1768.5 [kJ/kg]

Pump work = -15.1 [kJ/kg]

Thermal efficiency = 45.0 [%]

Average temperature at which heat

is transferred during reheating

increases as the number of reheat

stages is increased.

T204


Vapour Power Cycles

Ideal Regenerative Rankine Cycle with Open FWH

T206

Tav(2 → 2 ′) is low ⇒ ηth ↓. A practical regeneration

process in steam power plants is accomplished by

extracting, or bleeding, steam from the turbine at various

points. The device where the feed-water is heated by

regeneration is called a regenerator, or a feed-water

heater (FWH).

T207


Vapour Power Cycles

Cengel 10.5: ⊲ Ideal Regenerative Rankine Cycle with Open FWH.

T210

• qin = h5 − h4 = 2769.2 kJ/kg

• Energy balance (at FWH): yh6 + (1 − y)h2 = 1 · h3 → y = 0.227.

• wt = (h5 − h6) + (1 − y)(h6 − h7) = 1299.1 kJ/kg

• wP = (1 − y)wP,I + wP,II = -16.58 kJ/kg

⇒ ηth = 46.3% ◭


Vapour Power Cycles

Closed Feed-water Heaters

T208

T221


Vapour Power Cycles

T209

A steam power plant with one open FWH and three closed FWH.


Vapour Power Cycles

Combined Gas-Vapour Power Cycle

T214

Heat Exchanger → HRSG ≡ Heat Recovery Steam Generator


Vapour Power Cycles

Cogeneration: CHP ≡ Combined Heating & Power

T211 T212

• Cogeneration is the production of more than one useful form of energy (such

as process heat and electric power) from the same energy source.

• Utilization factor, ǫu = Wnet+QP

Qin= 1 − Qout

Qin.


Vapour Power Cycles

Binary Vapour Cycle

• For Mercury: PC = 18 MPa, TC = 898oC.

• At 0.07 kPa, Tsat = 32oC & at 7.0 kPa, Tsat = 273oC.

• ηth > 50% is possible with binary-vapour cycle.

T213


Refrigeration Cycles

T266 T273



Refrigerators, Air-conditioners & Heat Pumps

T270

• COPR = QL

WR: COPAC = QL

WAC: COPHP = QH

WHP

• COPHP = COPR + 1



Refrigeration Capacity/Performance

• 1 ton refrigeration: heat absorbed by 1 ton (2000 lb) of ice melting

at 0oC in 24 hours.

• 1 ton refrigeration (TR) = 3.516 kW = 12000 BTU/hr = 200

BTU/min

• Coefficient of Performance, COPR = Refrigeration EffectNet Work Required

• kW /ton ⇒ power required per ton of refrigeration

kW /ton = 3.516COP


Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle

Basic Refrigeration System using 2-Phase Refrigerant

R-134a

-50 -40 -30 -20 -10 0 10 20 30 40 50

0

200

400

600

800

1000

1200

1400

P sat

(kP

a)

T sat

( o C) T265 T264



Ideal Vapour Compression Refrigeration Cycle

T641 T640



T255

T256



Processes of Vapour Compression Refrigeration System

T257

QL = Q41 = m(h1 − h4)

QH = Q23 = m(h2 − h3)

Win = W12 = m(h2 − h1)

COP = QL/Win

1 → 2: Isentropic compression, Pevap → Pcond

2 → 3: Isobaric heat rejection, QH

3 → 4: Isenthalpic expansion, Pcond → Pevap

4 → 1: Isobaric heat extraction, QL



Example: ⊲ A theoretical single stage cycle using R134a as refrigerant operates

with a condensing temperature of 30oC and an evaporator temperature of -20oC.

The system produces 50 kW of refrigeration effect. Estimate:

1 Coefficient of performance, COP

2 Refrigerant mass flow rate, m

T257

QL = Q41 = m(h1 − h4) = 50 kW

m = 0.345 Kg/s ◭

Win = W12 = m(h2 − h1) = 12.5 kW

COP = QL/Win = 50.0/12.5 = 4.0 ◭



Effect of Evaporator Temperature

-50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0

0

1

2

3

4

5

6

7

8

9

10R134a: RE = 50 kW, T

cond = 30

oC

Ref. flow rate

COP

CO

P

Tevap

(oC)

0.30

0.31

0.32

0.33

0.34

0.35

0.36

0.37

0.38

0.39

0.40

Refr

igera

nt

flo

w r

ate

(kg

/s)

T259



Effect of Condenser Temperature

0 5 10 15 20 25 30 35 40 45 50

2

3

4

5

6

7

8

9

10

11

12R134a: RE = 50 kW, T

evap = -20

oC

Ref. flow rate

COP

CO

P

Tcond

(oC)

0.24

0.26

0.28

0.30

0.32

0.34

0.36

0.38

0.40

0.42

0.44

Refr

igera

nt

flo

w r

ate

(kg

/s)

T261



Effect of Evaporator & Condenser Temperatures

-50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0

0

1

2

3

4

5

6

7

8

9

10

11

12

Tcond

= 40oC

Tcond

= 30oC

Tcond

= 20oC

CO

P

Tevap

(oC)

T260



Deviations from Ideal Cycle

1 Refrigerant pressure drop in piping, evaporator, condenser, receiver

tank, and through the valves and passages.

2 Sub-cooling of liquid leaving the condenser.

3 Super-heating of vapour leaving the evaporator.

4 Compression process is not isentropic.

T262



Super-heating & Sub-cooling

T263

• Sub-cooling of liquid serves a desirable function of ensuring that 100%

liquid will enter the expansion device.

• Super-heating of vapour ensures no droplets of liquid being carried

over into the compressor.

• Even through refrigeration effect is increased, compression work is

greater & probably has negligible thermodynamic advantages.



Example: ⊲

T258

• h1 = h(P = 0.14 MPa,T = −10oC)

• h2 = h(P = 0.80 MPa,T = 50oC)

• h3 ≃ hf ,26oC

• h4 = h3

• QL = (h1 − h4) = 158.53 kJ/kg

• Win = (h2 − h1) = 40.33 kJ/kg

• COPR = QL

Win= 3.93 ◭

• ηc =h2s−h1

h2−h1

= 93.6% ◭



Example: ⊲ A dual-evaporator refrigeration system: TF = -18.0oC, TR = 4.0oC,

Tc = 30.0oC & ηc = 80%.

T274

• COPR = ? x5 = ?

• m = QR+QF

h1−h4h= 1.62 × 10−3 kg/s ◭

• Wc =m(h2s−h1)

ηc

• COPR = QR+QF

Wc= 3.37 ◭

• QR = m(h5 − h4h) → h5

• h5⇒ x5 = 0.56 ◭


Refrigeration Cycles Miscellaneous Refrigeration Systems

Vapour Absorption Refrigeration System

T1388



T267

Ammonia absorption refrigeration system



Vapour-Compression Heat Pump (HP) System

T268

• Qout = Wc + Qin

• COP |HP = Qout

Wc= 1 + COP |Ref



Gas Refrigeration System: Brayton Cycle

T269

COP =Qin

Wc − Wt

=(h1 − h4)

(h2 − h1) − (h3 − h4)


overview engineering cycles

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