Chapter 6 Power and Refrigeration

Vapor Cycles
1. Introduction
2. The Rankine cycle
3. A possible steam Carnot cycle
4. Rankine cycle efficiency
5. The reheat cycle
6. The regenerative cycle
7. Effect of losses on power cycle efficiency
8. The vapor refrigeration cycle
9. The heat pump
 6.1 Introduction
• Steam power plants and vapor-compression
refrigerators adopt vapor cycles. This chapter discuss
the principles and efficiencies of some important ideal
vapor cycles.
• The ideal Carnot cycle is used as a model against
which all real and all other ideal cycles are compared.
• The efficiency of a Carnot cycle is the maximum
possible efficiency for any power cycle.
TL
 = 1−
TH
• The efficiency can be increased by increasing TH or
lowering TL.
 6-2 The Rankine Cycle
• The simplest vapor power cycle is the Rankine cycle.
• A major feature is that the pump requires very little work to
deliver high pressure water to the boiler.
• A possible disadvantage is that the expansion process in the
turbine usually enters the quality region, resulting in liquid
droplets that may damage the blades.
 The Rankine Cycle
• 1→2: isentropic compression in the 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.
• Pump is used to increase the pressure
• The boiler and condenser are heat exchangers.

Video: How Does a Thermal Power Plant Work

 The Rankine Cycle Efficiency
• If we neglect the KE and PE changes, the net work output is
the area under the T-s daigram, 1-2-3-4-1, and the heat
transfer to the working substance is the area a-2-3-b-a.
• Thus, the thermal efficiency is
area 1-2-3-4-1
=
area a-2-3-b-a
• η can be increase by ↑p2 or ↓p4.
 The Rankine Cycle Efficiency
• The Rankine cycle η is significantly less than that of
the Carnot cycle operating between T3 and T1,
because the heat transfer occurs across large
temperature difference, such process 2→3, which is
irreversible.
 The Rankine Cycle Efficiency
• Expression of the work and heat transfer through the devices.
This can be done by considering the 1st law for each device.
Since KE and PE are negligible in these cases, we have
Q − Ws = m(hout − hin )
So, 2 2 2
wp ,rev = h2 − h1 =  dh =  Tds +  vdp =v1 ( p2 − p1 ) (v1~v2)
1 1 1

qB = h3 − h2 , wT = h3 − h4 , qC = h4 − h1
 = ( wT − wp ) / qB  wT / qB (the pump work is small)

Fig. 6.1
SUPPLEMENT: Reversible Steady-flow Work (Ref: Cengel &
Boles, Thermodynamics)
Recall that the reversible (quasi-equilibrium) moving boundary
work associated with a closed system:
2
Wrev =  PdV
1
The reversible work for a steady-flow device
(usually no moving boundary) can be derived as:
 qrev −  wrev = dh + dke + dpe
But
 qrev = Tds = dh − vdP
so
− wrev = vdP + dke + dpe
If dke and dpe are negligible,
2
 wrev = −vdP and wrev = − vdP
1

The larger the v (e.g., gas), the greater the work

produced (or consumed) by a steady-flow device.
 EX 6.1: Determine the η of a Rankine-cycle power plant
A steam power plant is to
operate between the pressure
of 10 kPa and 2 MPa with a
maximum temperature of
400oC.

w p = v1 ( p2 − p1 ) = ( p2 − p1 ) /  = (2000 − 10) /1000 = 1.99 kJ/kg

h2 = h1 + w p = 191.8 + 1.99 = 193.8 kJ/kg
qB = h3 − h2 = 3248 − 193.8 = 3054 kJ/kg
s3 = s4 = 7.1279 = s f + x4 s fg → x4 = 0.8636, h4 = h f + x4h fg = 2259 kJ/kg
wT = h3 − h4 = 3248 − 2259 = 989 kJ/kg
wT − w p 989 − 1.99
= = = 0.3232
qB 3054
6.3 A possible steam Carnot cycle (practical?)
Possible steam Carnot cycles have some problems:
• (a) Excessive work to pump a mixture of liquid and vapor
• (b) Damage of turbine due to condensed droplet
• (c) damage of turbine relieved, but to maintain superheat vapor at
constant T, decrease of pressure in pipe required, which is
impractical or can be prohibitively expensive.
• (d) Rankine cycle: pump only liquid and p2 and T3(600oC) be quite
large

Fig. 6.2
 6.4 Rankine Cycle Efficiency
To increase η of Rankine cycle:
• Method 1: to increase boiler pressure p2 (Fig. 6.3)
– Similar work output, but decrease in added heat→ η↑
– But quality of steam exiting turbine may become too low
→blade damage

Fig. 6.3
 Rankine Cycle Efficiency
• Method 2: To increase T3 temperature (Fig. 6.4a)
– Higher percentage of added heat turning to work → η↑
– Quality of state 4 can be increased.
– The maximum temperature is limited by metallurgy (~600oC).
• Method 3: To decrease condenser pressure p4 (Fig. 6.4b)
– Net work increases→ η↑
– Quality of state 4 can be decreased.
– Minimum condenser pressure (or condenser temperature) is
limited by cooling water temperature available.

Fig. 6.4
 EX 6.2: η increase by increasing boiler pressure

A steam power plant is to

operate between the pressure
of 10 kPa and 4 MPa with a
maximum temperature of
400oC.

w p = v1 ( p2 − p1 ) = ( p2 − p1 ) /  = (4000 − 10) /1000 = 3.99 kJ/kg

h2 = h1 + w p = 191.8 + 3.99 = 195.8 kJ/kg
qB = h3 − h2 = 3214 − 195.8 = 3018.2 kJ/kg
s3 = s4 = 6.7698 = s f + x4 s fg → x4 = 0.8159, h4 = h f + x4h fg = 2144 kJ/kg
wT = h3 − h4 = 3214 − 2144 = 1070 kJ/kg
wT − w p 1070 − 3.99
= = = 0.3532 → 9.45% → 0.3232
qB 3018.2
 EX 6.3: η increase by increasing maximum temperature

A steam power plant is to operate between

the pressure of 10 kPa and 2 MPa with a
maximum temperature of 600oC.

w p = v1 ( p2 − p1 ) = ( p2 − p1 ) /  = (2000 − 10) /1000 = 1.99 kJ/kg

h2 = h1 + w p = 191.8 + 1.99 = 193.8 kJ/kg
qB = h3 − h2 = 3690 − 193.8 = 3496.2 kJ/kg
s3 = s4 = 7.7032 = s f + x4 s fg → x4 = 0.9403, h4 = h f + x4h fg = 2442 kJ/kg
wT = h3 − h4 = 3690 − 2442 = 1248 kJ/kg
wT − w p 1248 − 1.99
= = = 0.3564  10% → 0.3232
qB 3496.2
 EX 6.4: η increase by decreasing condenser pressure
A steam power plant is to operate
between the pressure of 4 kPa and 2
MPa with a maximum temperature of
400oC.

w p = v1 ( p2 − p1 ) = ( p2 − p1 ) /  = (2000 − 4) /1000 = 1.996 kJ/kg

h2 = h1 + w p = 121.4 + 1.99 = 123.4 kJ/kg
qB = h3 − h2 = 3248 − 123.4 = 3124.6 kJ/kg
s3 = s4 = 7.1279 = s f + x4 s fg → x4 = 0.8327, h4 = h f + x4h fg = 2147 kJ/kg
wT = h3 − h4 = 3248 − 2147 = 1101 kJ/kg
wT − w p 1101 − 1.99
= = = 0.3517 → 8% → 0.3232
qB 3124.6
 6.5 The Reheat Cycle
• For the Rankine cycle with high boiler pressure and low
condenser pressure, it is difficult to prevent liquid droplet from
forming in the low-pressure portion of the turbine.
• Furthermore, due to the metallurgical limitation, the temperature
can not exceed 600oC, reheat cycle is often used to prevent the
liquid droplet formation.
• The turbine is separated into a high-pressure turbine and a low-
pressure turbine.

Fig. 6.5
 The Reheat Cycle
• The reheat cycle does not significantly influence η of the cycle,
but it does result in a significant additional work output (Fig. 6.5)
• The reheat cycle demands substantial equipment investment, it
must be economically justified by the increased work output.
• If reheat is not used to avoid droplet formation, the condenser
pressure must be quite high, resulting low η. In this relative
sense, reheat significantly increases η.

Fig. 6.5
 EX 6.5: Determine cycle η with reheat

High pressure steam enters a turbine at 600 psia and 1000oF. It is

reheated at a pressure of 40 psia to 600oF and then expanded to 2
psia.
p3 = 600psia, T3 = 1000o F → h3 , s3
p4 = 40psia, s4 = s3 → h4
p5 = 40psia, T5 = 600o F → h5 , s5
p6 = 2psia, s6 = s5 → x6 = 0.9698, h6
p1 = 2psia → h1 ,s1 saturated liquid
p2 = 600psia, s2 = s1 → h2
qB = h5 − h4 + h3 − h2 , wT = h5 − h6 + h3 − h4
 = wT / qB = 0.365 → real process moves point 6 to the right
 6.6 The Regenerative Cycle
• For Rankine cycle, a considerable amount of the energy is used
to heat the high pressure from T2 to its saturation temperature.
• To reduce this energy, the water could be preheated before it
enters the boiler by intercepting some of the steam as it
expands in the turbine.
• A cycle that utilizes this type of heating is a regenerative cycle,
and the process is referred to as regeneration.

Fig. 6.6
 The Regenerative Cycle
• The steam is mixed with the water as it exits the first
of the pump, thereby preheating the water from T2 to
T6.
• This would avoids condensing all of the steam at the
condenser, thereby reducing the amount of energy lost
from the condenser.
• A feedwater heater (open or closed type) is needed.

Fig. 6.7
 Feedwater Heater
• Feedwater: the water entering the boiler.
• Feedwater heater: device used to mix the extracted steam and
the condenser water
• Open feedwater heater : requiring 2 pumps
m6 = m5 + m2 , m6h6 = m5h5 + m2h2
h6 − h2
m5 =  m6
h5 − h2
• A closed feedwater heater is more expensive and heat transfer
characteristic is not as desirable as in open heater.
 The Regenerative Cycle
• Temperature leaving turbine should lie between the
saturated stream in the boiler and the condenser
temperature.
• For several heaters, the temperature difference should
be divided as equally as possible.
• Feedwater heater improves efficiency.
• With a large number of heaters, it is possible to
approach the Carnot efficiency.
• However, feedwater heater is expensive.
• Small power plant may have two heaters, and large
power plants have as many as six.
 Combined Reheat-Regenerative Cycle
• The regenerative cycle is afflicted by the moisture
problem in the low pressure portion of the turbine,
thereby it is common to combine with the reheat
cycle to avoid the moisture problem and increase
the thermal efficiency.

Fig. 6.10
 Cogeneration
• The efficiency of the cycle can be improved by
recovering the rejected heat from the condenser to
heat or other industrial applications. cogeneration
• Often, one-half of the rejected heat can be
effectively used, almost doubling the efficiency of a
power plant.
• Since the steam can not be transported far, the
power plant must be located near an industrial area
or a densely populated area.
EX 6.6: Determine the η of a regenerative cycle
A steam power plant is to operate between the
pressure of 10 kPa and 2 MPa with a maximum
temperature of 600oC. A open feedwater heater is
used by extracting steam pressure of 200kPa.
Pump work is negligible.

h2  h1 = 192 kJ/kg, h6  h7 = 505 kJ/kg, h3 = 3690 kJ/kg, h4 = 2442 kJ/kg

p5 = 200 kPa → h5 = 2968 kJ/kg, qB = h3 − h7
For every kilogram of steam passing through the boiler
1 = m6 = m5 + m2 , m6h6 = m5h5 + m2h2
h6 − h2
m5 =  m6 = 0.1128, m2 = 0.8872, wT = h3 − h5 + (h5 − h4 )m2 = 1189 kJ
h5 − h2
wT 1189
= = = 0.3733  4.62% → 0.3568
qB 3185
 6.7 Effect of Losses on Power Efficiency
• The preceding sections dealt with ideal cycles,
assuming no pressure drops through pipes in boiler
and condenser, and no losses within turbine and pump.
• The losses in the combustion process and the heat
transfer in the fluid in the pipes of the boiler are not
included here, but included in the overall plant
efficiency.
• There is substantial piping loss in pressure, 10-20%,
in boiler, which can be compensated by increasing
pumping pressure, where the required pump work is
less than 1% of the turbine work output and is
negligible.
 Turbine efficiency
• Only one significant loss is accounted for: the loss in
the turbine due to friction and the steam separation
from the rear portion of the blade.
wa
• Turbine efficiency is defined as,  = w
s
• The efficiency is around 80 to 89 %.
• The turbine does reduce the
moisture content as shown in
point 4, as the losses tend to
act as a small reheater.

Fig. 6.11
 6.8 The Vapor-Refrigeration Cycle

• Vapor refrigeration cycle--

similar to the reverse
Rankine cycle
• Drawback of Carnot cycle:
1: compressing mixture of
liquid and gas
2: isentropic expansion
difficult to achieve
• Ideal cycle:
1: saturated vapor compression
2: expansion by throttling
process is simple.
Fig. 6.12
 The vapor-refrigeration cycle
• The performance of the refrigeration cycle
• Refrigerator: COPref=Qin/Win～4
• Heat pump: COPhp=Qout/Win～5
• For refrigerator, the evaporation temperature must be low, -25oC,
common refrigerants are Amonia (NH3) and R134a.

Fig. 6.13
Video: A refrigeration system
 The actual vapor-refrigeration cycle
Deviation from the ideal cycle
– Pressure drop due to friction in connecting pipes
– Heat transfer occurs from or to the refrigerant through the pipes
connecting the components
– Pressure drops occur through the condenser and evaporator tubes
– Heat transfer occurs from the compressor
– Frictional effects and flow separation occur on the compressor blade.
– The vapor entering the compressor may be slightly superheated.
– The temperature of the liquid exiting the condenser may be below
the saturation temperature-increasing the refrigerant effect

Fig. 6.14
The arrangement of a window air-conditioner

From Turns, Thermal and

Fluid Science
 Size of an air-conditioning unit
• A “ton” of refrigeration is supposedly the heat rate
necessary to melt a ton of ice in 24 hours.
• By definition, 1 ton of refrigeration equals 3.52 kW
(12,000 Btu/hr)
• EER (energy efficiency rating): the output in BTU’s
divided by the energy purchased in Wh.
• EER=COP×3.412 (Different definitions in different
countries)
EX 6.9: Ideal vapor refrigeration cycle
R134a is used in an ideal vapor refrigeration cycle, operating
between saturation temperature of -26oC in the evaporator and
39.39oC in the condenser. Calculate the rate of refrigeration, the
COP and the rate in hp per ton if the refrigerant flows at 0.6 kg/s.
Calculate COP if used as a heat pump.

h1 = 231.6kJ/kg, s1 = 0.939 kJ/kgK = s2 → h2 , h4 = h3 = 105.3 kJ/kg

QE = m(h1 − h4 ) = 0.6(231.6 − 105.3) = 75.8 kJ/s
WC = m(h2 − h1 ) = 0.6(279 − 231.6) = 28.4 kJ/s
COPref = QE / WC = 75.8 / 28.4 = 2.67
hp/ton = (28.4 / 0.746) /(75.8 / 3.52) = 1.77
COPhp = QC / WC = (h2 − h3 ) /(h2 − h1 ) = 3.67
-26oC
EX 6.10: Actual vapor refrigeration cycle
• The refrigerant leaving the evaporator is 2’
superheated to -20oC.
• The refrigerant leaving the condenser is -26oC
subcooled to 40oC.
• The compressor is 80 percent efficient
T3 = 40o C → h4 = h3 = 106.2 kJ/kg
T1 = −20o C → p1 = 0.1MPa, h1 = 236.5kJ/kg, s1 = 0.96 kJ/kgK=s2'
s2' , p2 = 1 MPa → h2' = 285.8 kJ/kg (1→2’ isentropic)
ws
C = = (h2' − h1 ) /(h2 − h1 ) = 0.8 → h2 = 298.1 kJ/kg
wa
QE = m(h1 − h4 ) = 0.6(236.5 − 106.2) = 78.18 kJ/s
WC = m(h2 − h1 ) = 0.6(298.1 − 236.5) = 36.96 kJ/s
COPref = QE / WC = 78.18 / 36.96 = 2.12
 6.9 The heat pump
• The heat pump can be used to heat a house in cool weather or to
cool a house in warm weather, as in Fig. 6.15.
• Since the evaporator and the condenser are similar heat exchangers,
in actual situation valving is used to perform the desired switching
of the heat exchangers. Video: How heat pump works

Fig. 6.15
The arrangement of a home dehumidifier

From Turns, Thermal and Fluid Science

 Home evaporative cooler

From Turns, Thermal and Fluid Science

 6.10 Summary
• Rankine cycle
• Reheat cycle
• Regeneration cycle
• Cogeneration
• Refrigeration cycle