Dinc Ibrahim. Refrigeration systems and applications 2th edition

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Heat Pumps 301

Inside coil

Vapor

Compressor

Outside coil

Compressed vapor

Liquid refrigerant

Figure 6.8 A vapor-compression heat pump running for heating (Courtesy of D&H Climate Control).

accomplished by reversing the ﬂow of refrigerant through a device found in heat pumps knows as

a “reversing valve”. This valve is automatically controlled through the thermostat when switched

to heat.

Figure 6.8 shows the heat pump in the heating mode of operation (e.g., in winter the heat

exchanger located outside the house functions as an evaporator, absorbing low-temperature heat

from the environment). Switching the heat pump from the cooling mode to the heating mode

is achieved simply by switching the direction of the refrigerant ﬂow. The difference between

Figures 6.7 and 6.8 is that the reversing valve (2) directs the compressed refrigerant to the inside

coil ﬁrst. This makes the inside coil the condenser and releases the heat energy (3-4). This heated

air is ducted to the home. The outside coil is used to collect the heat energy (3-7). This now

becomes the evaporator. Both heating and A/C modes do exactly the same thing. They pump

heat from one location to another. In these examples, the heat in the air is moved out of or into

the home.

There is usable heat in outdoor air at temperatures as low as −8.5

◦

C. As the temperature of the

outdoor air decreases, however, the heating capacity of the heat pump diminishes proportionately,

resulting in lower discharge air temperatures at the air registers and gradual cooling of the house.

To supplement the heating capacity of the heat pump, electric resistance heating elements are used,

which automatically engage via the thermostat when this condition occurs.

6.10.3 Single-Stage Vapor-Compression Heat Pump with Subcooler

A simple modiﬁcation can be made to the four-component cycle to make it, in some situa-

tions, considerably more efﬁcient. This is the addition of a refrigerant subcooler, as shown in

Figure 6.9a. This extra heat exchanger extracts heat from the hot liquid refrigerant before it

goes through the expansion valve. This leads to less ﬂash gas formation through the expansion

valve. Hence, the same compressor is doing more useful cooling and heating, with no extra power

consumption (note that the compressor is “unaware” of whether its suction vapors are formed

across the expansion valve or through boiling in the evaporator). Figure 6.9b shows the effect

of subcooling on a Mollier chart. In both cases illustrated, the primary heating is carried out

between 70 and 75

◦

C. In the case of a subcooler the hot refrigerant (80

◦

Cwhenleavingthe

condenser) is cooled to 30

◦

C by a stream of air. This extra free heat can successfully be used

by integrating an air space heating unit with the main hot water system. The ﬁnancial advantage

is considerable.

302 Refrigeration Systems and Applications

Evaporator

Condenser

Subcooler

Source Sink

log P

Subcooling

(a) (b)

Figure 6.9 (a) A single-stage vapor-compression heat pump with subcooler and (b) its log P −h diagram.

Table 6. 8 ARI standard rating conditions for variable capacity compressors and compressor units used in

heat pumps.

Rating

Test Point

Intended Use Suction Dew

Point

Temperature

(

◦

Discharge

Dew Point

Temperature

(

◦

Return Gas

Temperature

(

◦

Capacity Setting

A Air source (cooling) 7.2 56.4 18.3 Max.

B Air source (cooling) 7.2 46.1 18.3 Max.

C Air source (cooling and

heating)

7.2 37.8 18.3 Min.

D Air source (heating) –1.1 43.3 10.0 Max.

E Air source (heating) –15.0 35.0 –3.9 Max.

F Air source (cooling) 7.2 26.7 18.3 Min.

G Air source (heating) 1.7 32.2 12.8 Min.

H Water source (cooling and

heating)

7.2 48.9 18.3 Max. and Min.

Ratings based on 35

◦

C temperature surrounding compressor. If air ﬂow across the compressor is used to

determine ratings, it shall be speciﬁed by the compressor manufacturer.

The maximum and minimum capacity setting is the highest and lowest displacement capacity obtainable by

the compressor or compressor unit.

Source: ARI (2000).

6.10.4 Standard Rating Conditions for Compressors

The standard rating(s) of a compressor or compressor unit used in a heat pump is its compres-

sor rating(s) based on the tests performed at standard rating conditions at the test points from

Table 6.8.

Heat Pumps 303

6.10.5 ARI/ISO Standard 13256-1

This standard covers those heating and cooling systems usually referred to as water source heat

pumps. These electrically driven vapor-compression systems consist of one or more factory-made

matched assemblies which normally include an indoor conditioning coil with air moving means,

a compressor, and a refrigerant-to-liquid (water or brine) heat exchanger. A system may provide

cooling-only, heating-only, or both functions and is typically designed for use within one or more

of the following liquid heat source/sink applications: (i) water-loop heat pump using temperature-

controlled water circulating in a common piping loop, (ii) groundwater heat pump using water

pumped from a well, lake, or stream, and (iii) ground-loop heat pump using brine circulating

through a subsurface piping loop.

These three applications were previously separately covered by ARI Standards 320, 325, and

330. Rating and performance test conditions for ARI/ISO 13256-1 (Ellis, 2001), as compared to the

previous ARI standards, are summarized in Tables 6.9 and 6.10. As can be seen, the differences in

rating test temperatures are relatively minor, and consist mainly of rounding to the Celsius scale and

Table 6. 9 Comparison of ARI and ISO rating test conditions.

Rating Tests Water-Loop Ground-Loop

Heat Pumps Ground-Water Heat Pumps Heat Pumps

ARI/ISO ARI 320 ARI/ISO ARI 325 Hi ARI 325 Lo ARI/ISO ARI 330

Standard cooling:

Air dry bulb,

◦

Air wet bulb,

◦

Air ﬂow rate, l/s

Liquid full load,

◦

Liquid part load,

◦

Liquid ﬂow rate, l/s

per mfr

26.7

19.4

per mfr

29.4

23.9

5.6

◦

Crise

per mfr

26.7

19.4

per mfr

21.1

per mfr

26.7

19.4

per mfr

10.0

per mfr

26.7

19.4

per mfr

25.0

21.1

per mfr

Standard heating:

Air dry bulb,

◦

Air wet bulb,

◦

Air ﬂow rate, l/s

Liquid full load,

◦

Liquid part load,

◦

Liquid ﬂow rate, l/s

per mfr

21.1

15.6

std clg

21.1

23.9

std clg

per mfr

21.1

15.6

std clg

21.1

per mfr

21.1

15.6

std clg

10.0

per mfr

21.1

15.6

std clg

0.0

5.0

per mfr

External static:

Air, Pa

Liquid, kPa

25.0–75.0

150.0

25.0–75.0

150.0

25.0–75.0

50.0

per mfr: per manufacturer.

std clg: standard catalog.

Source: Ellis (2001).

304 Refrigeration Systems and Applications

Table 6. 10 Comparison of ARI and ISO performance test conditions.

Rating Tests Water-Loop Heat Pumps Ground-Water Heat Pumps Ground-Loop Heat Pumps

ARI/ISO ARI 320 ARI/ISO ARI 325 Hi ARI/ISO ARI 330

Maximum cooling:

Air dry bulb,

◦

Air wet bulb,

◦

Liquid,

◦

35.0

21.7

35.0

21.7

23.9

35.0

21.7

35.0

Maximum heating:

Air dry bulb,

◦

Liquid,

◦

26.7

32.3

26.7

23.9

26.7

23.9

Minimum cooling:

Air dry bulb,

◦

Air wet bulb,

◦

Liquid,

◦

19.4

13.9

18.3

26.7

19.4

0.0

Minimum cooling:

Air dry bulb,

◦

Liquid,

◦

15.6

7.2

−5

15.6

−3.9

Enclosure sweat:

Air dry bulb,

◦

Air wet bulb,

◦

Liquid,

◦

26.7

23.9

26.7

23.9

10.0

26.7

23.9

10.0

Source: Ellis (2001).

eliminating the dual rating points for ARI 325. Performance test temperatures vary more, but these

tests are concerned only with veriﬁcation of proper equipment operation under extreme conditions,

and results are not published as ratings.

The ARI/ISO standard is not design prescriptive and provides a means for manufacturers

to specify unique air and liquid ﬂow rates for both heating and cooling, and for each step

of capacity, in each chosen application. Additionally, the ARI/ISO standard introduces the

concept of “effective power input” to the heat pump, which includes the power input of the

compressor and controls as well as the proportional power input of fans and pumps, whether

internal or external, and whether provided by the manufacturer or not. The power input of

fans and pumps is proportional in that it only includes that power required to transport air

and liquid through the heat pump, and again avoiding design prescription, does not include

arbitrary external static conditions for each application. Unlike the previous ARI standards,

the power input is calculated in a consistent manner, inclusive of fan and pump power, across

all applications.

Heat Pumps 305

(a) (b)

Condenser

Evaporator

Compressor

Expansion

valve

Figure 6.10 A vapor-compression heat pump system for analysis and its temperature–entropy diagram for

the ideal case.

6.11 Energy Analysis of Vapor-Compression Heat Pump Cycle

Energy analysis of a vapor-compression heat pump cycle is very similar to the energy analysis

of vapor-compression refrigeration cycle as given in Chapter 4. Applying conservation of energy

principle to each of the processes of the cycle as shown in Figure 6.10 for steady-ﬂow operation

with negligible kinetic and potential energy changes gives

Compressor:

W =˙m(h

− h

) (6.7)

Condenser:

=˙m(h

− h

) (6.8)

Expansion valve:

= h

(6.9)

Evaporator:

=˙m(h

− h

) (6.10)

An energy balance on the entire system gives

W +

(6.11)

A heat pump is used to supply heat to the high-temperature space. Therefore, the COP of the

heat pump cycle is deﬁned as

COP =

(6.12)

306 Refrigeration Systems and Applications

The temperature–entropy diagram of an ideal vapor-compression heat pump cycle is given in

Figure 6.10b. In this cycle, the refrigerant enters the compressor as a saturated vapor. It is com-

pressed isentropically in a compressor; it is cooled and condensed at constant pressure by rejecting

heat to high-temperature medium until it exists as a saturated vapor at the exit of the condenser.

The refrigerant is expanded in an expansion valve during which the enthalpy remains constant: it

is evaporated in the evaporator at constant pressure by absorbing heat from the refrigerated space,

and it leaves the evaporator as a saturated vapor.

6.12 Exergy Analysis of Vapor-Compression Heat Pump Cycle

Figure 6.10 is a schematic of a vapor-compression heat pump cycle operating between a low-

temperature medium (T

) and a high-temperature medium (T

). The maximum COP of a heat

pump cycle operating between temperature limits of T

and T

based on the Carnot heat pump

cycle was given in Chapter 1 as

COP

Carnot

− T

1 − T

(6.13)

This is the maximum COP that a heat pump operating between T

and T

can have. Equation 6.13

indicates that a smaller temperature difference between the heat sink and the heat source (T

− T

)

provides greater heat pump COP.

The aim in an exergy analysis is usually to determine the exergy destructions in each component

of the system and to determine exergy efﬁciencies. The components with greater exergy destructions

are also those with more potential for improvements. Exergy destruction in a component can be

determined from an exergy balance on the component. It can also be determined by ﬁrst calculating

the entropy generation and using

dest

= T

gen

(6.14)

where T

is the dead-state temperature or environment temperature. In a heat pump, T

is usu-

ally equal to the temperature of the low-temperature medium T

. Exergy destructions and exergy

efﬁciencies for major components of the cycle are as follows:

Compressor:

dest,1−2

W +

−

W − 

W −˙m

[

− h

− T

− s

)

]

W −

rev

(6.15)

dest,1−2

= T

gen,1−2

=˙mT

− s

) (6.16)

ex,Comp

rev

= 1 −

dest,1−2

(6.17)

Condenser:

dest,2−3

−

=˙m

[

− h

− T

− s

)

]

−



1 −



(6.18)

dest,2−3

= T

gen,2−3

=˙mT



− s



(6.19)

ex,Cond

−



1 −



˙m

[

− h

− T

− s

)

]

= 1 −

dest

−

(6.20)

Heat Pumps 307

Expansion valve:

dest,3−4

−

=˙m

[

− h

− T

− s

)

]

(6.21)

dest,3−4

= T

gen,3−4

=˙mT

− s

) (6.22)

ex,ExpValve

−

= 1 −

dest,3−4

−

= 1 −

−

(6.23)

Evaporator:

dest,4−1

= (

−

) −

(6.24)

=˙m

[

− h

− T

− s

)

]

−



−



1 −



dest,4−1

= T

gen,4−1

=˙mT



− s

−



(6.25)

ex,Evap

−



1 −



˙m

[

− h

− T

− s

)

]

= 1 −

dest,4−1

−

(6.26)

The total exergy destruction in the cycle can be determined by adding exergy destructions in

each component:

dest,total

dest,1−2

dest,2−3

dest,3−4

dest,4−1

(6.27)

It can be shown that the total exergy destruction in the cycle can also be expressed as the

difference between the exergy supplied (power input) and the exergy recovered (the exergy of the

heat transferred to the high-temperature medium):

dest,total

W −

(6.28)

where the exergy of the heat transferred to the high-temperature medium is given by



1 −



(6.29)

This is in fact the minimum power input to accomplish the required heating load

min

(6.30)

The second-law efﬁciency (or exergy efﬁciency) of the cycle is deﬁned as

min

= 1 −

dest,total

(6.31)

Substituting

W =

/COP and

(

1 − T

)

into the second-law efﬁciency equation,

(

1 − T

)

/COP



1 −



COP

/(T

− T

)

COP

Carnot

(6.32)

since T

= T

. Thus, the second-law efﬁciency is also equal to the ratio of actual and maximum

COPs for the cycle. This second-law efﬁciency deﬁnition accounts for irreversibilities within the

heat pump since heat transfers with the high- and low-temperature reservoirs are assumed reversible.

308 Refrigeration Systems and Applications

Example 6.1

A heat pump is used to keep a room at 25

◦

C by rejecting heat to an environment at 5

◦

C. The total

heat loss from the room to the environment is estimated to be 45,000 kJ/h and the power input to

the compressor is 4.5 kW. Determine (a) the rate of heat absorbed from the environment in kJ/h, (b)

the COP of the heat pump, (c) the maximum rate of heat supply to the room for the given power

input, and (d) the second-law efﬁciency of the cycle. (e) Also, determine the minimum power input

for the same heating load and the exergy destruction of the cycle.

Solution

(a) The rate of heat absorbed from the environment in kJ/h is

−

W = 45, 000 kJ/h − (4.5kW)



3600 kJ/h

1kW



= 28,800 kJ/h

(b) The COP of the heat pump is

COP =

(45, 000/3600) kW

4.5kW

= 2.78

rate of heat supply to the room for the given power input are

COP

Carnot

− T

298

298 − 278

= 14.9

H,max

W COP

Carnot

= (4.5kW)



3600 kJ/h

1kW



(14.9) = 241,380 kJ/h

(d) The second-law efﬁciency of the cycle is

COP

Carnot

2.78

14.9

= 0.186 = 18.6%

(e) The minimum power input for the same heating load and the exergy destruction of the

cycle are

min



1 −



= (45, 000 kJ/h)



1 −

278

298



= 3020 kJ/ h

dest

W −

min

= (4.5 × 3600) kJ/h − 3020 kJ/h = 13,180 kJ/h

The second-law efﬁciency may alternatively be determined from

min

3020 kJ/h

(4.5 × 3600) kJ/h

= 0.186 = 18.6%

The result is the same as expected.

Heat Pumps 309

Example 6.2

A heat pump operates on the ideal vapor-compression refrigeration cycle with refrigerant-134a as

the working ﬂuid. The refrigerant evaporates at −20

◦

C and condenses at 1200 kPa. The refrigerant

absorbs heat from ambient air at 4

◦

C and transfers it to a space at 24

◦

C. Determine (a) the work

input and the COP, (b) the exergy destruction in each component of the cycle and the total exergy

destruction in the cycle, (c) the minimum work input and the second-law efﬁciency of the cycle.

(d) Determine the COP, the minimum power input, the total exergy destruction, and the exergy

efﬁciency of the cycle if a ground-source heat pump is used with a ground temperature of 18

◦

The evaporating temperature in this case is −6

◦

C. Take everything else the same.

Solution

(a) Temperature–entropy diagram of the cycle is given in Figure 6.11.

−20° C

1.2 MPa

Figure 6.11 Temperature–entropy diagram of the cycle considered in Example 6.2.

From the refrigerant-134a tables (Tables B.3 through B.5)

=−20

◦

= 1



= 238.41 kJ/kg

= 0.9456 kJ/kg · K

= 1200 kPa

= s



= 284.43 kJ/kg

= 1200 kPa

= 0



= 117.77 kJ/kg

= 0.4244 kJ/kg · K

= h

= 117.77 kJ/kg

=−20

◦

= 117.77 kJ/kg



= 0.4691 kJ/kg · K

= h

− h

= 238.41 − 117.77 = 120.6kJ/kg

= h

− h

= 284.43 − 117.77 = 166.7kJ/kg

w = h

− h

= 284.43 − 238.41 = 46.0kJ/kg

310 Refrigeration Systems and Applications

The COP of the cycle is

COP =

166.7kJ/kg

46.0kJ/kg

= 3.62

(b) The exergy destruction in each component of the cycle is determined as follows:

Compressor:

gen,1−2

= s

− s

= 0

dest,1−2

= T

gen,1−2

= 0

Condenser:

gen,2−3

= s

− s

= (0.4244 − 0.9456) kJ/kg · K +

166.7kJ/kg

297 K

= 0.03991 kJ/kg · K

dest,2−3

= T

gen,2−3

= (277 K)(0.03991 kJ/kg · K) = 11.06 kJ/kg

Expansion valve:

gen,3−4

= s

− s

= 0.4691 − 0.4244 = 0.04473 kJ/kg · K

dest,3−4

= T

gen,3−4

= (277 K)(0.04473 kJ/kg · K) = 12.39 kJ/kg

Evaporator:

gen,4−1

= s

− s

−

= (0.9456 − 0.4691) kJ/kg · K −

120.6kJ/kg

277 K

= 0.04100 kJ/kg · K

dest,4−1

= T

gen,4−1

= (277 K)(0.04100 kJ/kg · K) = 11.36 kJ/kg

The total exergy destruction can be determined by adding exergy destructions in

each component:

dest,total

= ex

dest,1−2

+ ex

dest,2−3

+ ex

dest,3−4

+ ex

dest,4−1

= 0 + 11.06 + 12.39 + 11.36 = 34.8kJ/kg

= q



1 −



= (166.7 kJ/kg)



1 −

277

297



= 11.22 kJ/kg

Thus, the minimum work input is

min

= ex

= 11.22 kJ/kg

The second-law efﬁciency of the cycle is

11.22

46.0

= 0.244 = 24.4%

The total exergy destruction may also be determined from

dest,total

= w − ex

= 46.0 − 11.22 = 34.8kJ/kg

The result is identical as expected.