overview engineering cycles
TRANSCRIPT
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 2 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 3 / 63
Gas Power Cycles Cycles for 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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 4 / 63
Gas Power Cycles Cycles for Engines
The Otto Cycle: Ideal Cycle for SI Engines
T228
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 5 / 63
Gas Power Cycles Cycles for Engines
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 6 / 63
Gas Power Cycles Cycles for Engines
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 7 / 63
Gas Power Cycles Cycles for Engines
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 ◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 8 / 63
Gas Power Cycles Cycles for Engines
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)
]
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 9 / 63
Gas Power Cycles Cycles for Engines
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 10 / 63
Gas Power Cycles Cycles for Engines
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 ◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 11 / 63
Gas Power Cycles Cycles for Engines
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)
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 12 / 63
Gas Power Cycles Cycles for Engines
T236
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 13 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 14 / 63
Gas Power Cycles Cycles for Gas Turbines
• 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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 15 / 63
Gas Power Cycles Cycles for Gas Turbines
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 ◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 16 / 63
Gas Power Cycles Cycles for Gas Turbines
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 17 / 63
Gas Power Cycles Cycles for Gas Turbines
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%
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 18 / 63
Gas Power Cycles Cycles for Gas Turbines
Turbines with Reheat, Compressors with Intercooling
T246
T250
T252
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 19 / 63
Gas Power Cycles Cycles for Gas Turbines
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◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 20 / 63
Gas Power Cycles Cycles for Gas Turbines
Brayton Cycle with Intercooling, Reheating & Regeneration
T249
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 21 / 63
Gas Power Cycles Cycles for Gas Turbines
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 ◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 22 / 63
Vapour Power Cycles
Components of a Simple Vapour Power Plant
T193
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 23 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 24 / 63
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 ◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 25 / 63
Vapour Power Cycles
Deviation of Actual Vapour Power Cycle
T197
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 26 / 63
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%◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 27 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 28 / 63
Vapour Power Cycles
PB = 2.64 MPa, TH = 800 K
.
T217
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 29 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 30 / 63
Vapour Power Cycles
PB = 2.64 MPa, TL = 325 K
.
T216
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 31 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 32 / 63
Vapour Power Cycles
TH = 800 K, TL = 325 K
.
T215
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 33 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 34 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 35 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 36 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 37 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 38 / 63
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% ◭
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 39 / 63
Vapour Power Cycles
Closed Feed-water Heaters
T208
T221
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 40 / 63
Vapour Power Cycles
T209
A steam power plant with one open FWH and three closed FWH.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 41 / 63
Vapour Power Cycles
Combined Gas-Vapour Power Cycle
T214
Heat Exchanger → HRSG ≡ Heat Recovery Steam Generator
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 42 / 63
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.
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 43 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 44 / 63
Refrigeration Cycles
T266 T273
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 45 / 63
Refrigeration Cycles
Refrigerators, Air-conditioners & Heat Pumps
T270
• COPR = QL
WR: COPAC = QL
WAC: COPHP = QH
WHP
• COPHP = COPR + 1
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 46 / 63
Refrigeration Cycles
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 47 / 63
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
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 48 / 63
Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
Ideal Vapour Compression Refrigeration Cycle
T641 T640
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 49 / 63
Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
T255
T256
© Dr. Md. Zahurul Haq (BUET) Engineering Cycles ME6101 (2021) 50 / 63
Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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 ◭
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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.
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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% ◭
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Refrigeration Cycles Ideal Vapour Compression Refrigeration Cycle
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 ◭
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Refrigeration Cycles Miscellaneous Refrigeration Systems
Vapour Absorption Refrigeration System
T1388
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Refrigeration Cycles Miscellaneous Refrigeration Systems
T267
Ammonia absorption refrigeration system
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Refrigeration Cycles Miscellaneous Refrigeration Systems
Vapour-Compression Heat Pump (HP) System
T268
• Qout = Wc + Qin
• COP |HP = Qout
Wc= 1 + COP |Ref
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Refrigeration Cycles Miscellaneous Refrigeration Systems
Gas Refrigeration System: Brayton Cycle
T269
COP =Qin
Wc − Wt
=(h1 − h4)
(h2 − h1) − (h3 − h4)
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