ENERGY BALANCES AND ENERGY UTILIZATION
3.13. COMPRESSION AND EXPANSION OF GASES
The work term in an energy balance is unlikely to be significant unless a gas is expanded or compressed as part of the process. To compute the pressure work term
W¼ ðy2
y1
Pdy (3:5a)
a relationship between pressure and volume during the expansion is needed.
If the compression or expansion is isothermal (at constant temperature), then for unit mass of an ideal gas
Py¼constant (3:28)
and the work done, W¼P1y1lnP2
P1¼RT1
Mw
lnP2
P1 (3:29)
where
P1¼initial pressure;
P2¼final pressure;
y1¼initial volume;
Mw¼molecular mass (weight) of gas.
In industrial compressors or expanders the compression or expansion path will be
‘‘polytropic,’’ approximated by the expression:
Pyn¼constant (3:30)
The work produced (or required) is given by the general expression:
W¼P1y1 n n1
P2 P1
(n1)=n
1
" #
¼ZRT1 Mw
n n1
P2 P1
(n1)=n
1
" #
(3:31) where
Z¼compressibility factor (1 for an ideal gas);
R¼universal gas constant, 8:314 JK1mol1; T1¼inlet temperature, K;
W¼work done, J/kg.
The value ofnwill depend on the design and operation of the machine.
The energy required to compress a gas, or the energy obtained from expansion, can be estimated by calculating the ideal work and applying a suitable efficiency value.
For reciprocating compressors, the isentropic work is normally used (n¼g) (see Figure 3.7); and for centrifugal or axial machines, the polytropic work (see Figure 3.6 and Section 3.13.2).
3.13.1. Mollier Diagrams
If a Mollier diagram (enthalpy-pressure-temperature-entropy chart) is available for the working fluid, the isentropic work can be easily calculated.
W¼H1H2 (3:32)
where H1 is the specific enthalpy at the pressure and temperature corresponding to point 1, the initial gas conditions;H2is the specific enthalpy corresponding to point 2, the final gas condition.
Point 2 is found from point 1 by tracing a path (line) of constant entropy on the diagram.
The method is illustrated in Example 3.10.
Example 3.10
Methane is compressed from 1 bar and 290 K to 10 bar. If the isentropic efficiency is 0.85, calculate the energy required to compress 10,000 kg/h. Estimate the exit gas temperature.
Solution
From the Mollier diagram, shown diagrammatically in Figure 3.5:
H1¼4500 cal=mol;
H2¼6200 cal=mol (isentropic path);
Isentropic work¼62004500
¼1700 cal/mol
Entropy
Enthalpy Isentropic
path
Actual path
H1 = 4500 290 K H2 = 6200
460 K H⬘2= 6500
480 K
p = 10 p = 1
Figure 3.5. Mollier diagram, methane.
1.0 10 100 60
70 80 90
Axial - flow
Efficiency,Ep, % Centrifugal
Volumetric flow rate (suction conditions), m3/s
Figure 3.6. Approximate polytropic efficiencies of centrifugal and axial-flow compressors.
1 1.5 2.0 2.5 3.0 3.5 4.0
60 70 80 90 100
Range
Isentropic efficiency
Compression ratio
Figure 3.7. Typical efficiencies for reciprocating compressors.
For an isentropic efficiency of 0.85:
Actual work done on gas 1700
0:85 ¼2000 cal=mol So, actual final enthalpy
H20 ¼H1þ2000¼6500 cal=mol
From the Mollier diagram, if all the extra work is taken as irreversible work done on the gas, the exit gas temperature¼480 K
Molecular weight of methane¼16
Energy required¼(moles per hour)(specific enthalpy change)
¼10;000
16 2000103
¼1:25109 cal=h
¼1:251094:187 J=h
¼5:23109 J=h Power¼5:23109
3600 ¼1:45 MW
3.13.2. Polytropic Compression and Expansion
If no Mollier diagram is available, it is more difficult to estimate the ideal work in compression or expansion processes.
Equation 3.31 can be used if the compressibilityZand polytropic coefficientnare known. Compressibility can be plotted against reduced temperature and pressure, as shown in Figure 3.8.
At conditions away from the critical point n¼ 1
1m (3:33)
where
m¼(g1) gEp
for compression (3:34)
m¼(g1)Ep
g for expansion (3:35)
andEpis the polytropic efficiency, defined by
for compressionEp¼ polytropic work actual work required for expansionEp¼actual work obtained
polytropic work An estimate ofEpcan be obtained from Figure 3.6.
The outlet temperature can be estimated from T2¼T1 P2
P1
m
(3:36) Close to the critical conditions, these equations should not be used. The procedure for calculation of polytropic work of compression or expansion close to the critical point is more complex (Shultz, 1962), and it is easiest to make such calculations using process simulation programs.
Example 3.11
Estimate the power required to compress 5000 kmol/h of HCl at 5 bar, 158C, to 15 bar.
Solution
For HCl,Pc¼82 bar, Tc¼324:6 K
Cp¼30:300:72102Tþ12:5106T23:9109T3kJ=kmol K EstimateT2from equations 3.34 and 3.35.
For diatomic gasesg1:4.
Note:gcould be estimated from the relationship g¼Cp
Cv
At the inlet conditions, the flow rate in m3=s
¼5000
360022:4288 2731
5¼6:56 From Figure 3.6Ep¼0:73
From equations 3.34 and 3.35:
m¼ 1:41
1:40:73¼0:391 T2¼288 15
5 0:39
¼442 K Tr(mean)¼442þ288
2324:6 ¼1:12 Pr(mean)¼5þ15
282¼0:12 atT(mean), Cop¼29:14 kJ=kmol K Correction for pressure from Figure 3.2, 2 kJ/kmol K
Cp¼29:14þ231 kJ=kmol K
From Figure 3.8, at mean conditions:
Z¼0:98 From equation 3.33:
n¼ 1
10:391¼1:64 From equation 3.31:
W polytropic¼0:982888:314 1:64 1:641
15 5
(1:641)=1:64
1
!
¼3219 kJ=kmol
Actual work required¼polytropic work Ep
¼3219
0:73 ¼4409 kJ=kmol Power¼44095000
3600 ¼6124 kW, say 6:1 MW
3.13.3. Multistage Compressors
Single-stage compressors can be used only for low pressure ratios. At high pressure ratios, the temperature rise is too high for efficient operation.
To cope with the need for high pressure generation, the compression is split into a number of separate stages, with intercoolers between each stage. The interstage pressures are normally selected to give equal work in each stage.
For a two-stage compressor, the interstage pressure is given by Pi¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(P1P2)
p (3:37)
wherePiis the intermediate-stage pressure.
Example 3.12
Estimate the power required to compress 1000 m3=h air from ambient conditions to 700 kN=m2gauge, using a two-stage reciprocating compressor with an intercooler.
Solution
Take the inlet pressure,P1, as 1 atmosphere¼101:33 kN=m2, absolute.
Outlet pressure,P2,¼700þ101:33¼801:33 kN=m2, absolute.
For equal work in each stage, the intermediate pressure,Pi,
¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (1:01331058:0133105)
p ¼2:8495105N=m2
For air, take ratio of the specific heats,g, to be 1.4.
For equal work in each stage, the total work will be twice that in the first stage.
3
2
1.0 0.8 0.6
0.4 0.3
0.2
0.100.1 0.2 0.3 0.4 0.6 0.8 1.0 2 3 4 6 7 8 9 10 20 25 30
Compressibility factor, Z
0.70
0.75 0.80 0.85
0.90 0.95
1.0 1.01 1.03
1.05 1.1
1.15 1.2 1.3 1.4
1.6 1.8 2.0 1.5
Reduced temperature, Tr = 1.0
1.1 1.2 1.61.4 2.0 3 4 68 15 10
1.2 1.1 1.0
0.9 1.0
0.9
0.8
0.7
0 0.1 0.2 0.3 0.4
Low pressure range, Pr Reduced temperature,Tr
Reduced pressure,Pr
0.6 0.5
0.7
0.8
Figure 3.8. Compressibility factors of gases and vapors.
CHAPTER3FUNDAMENTALSOFENERGYBALANCESANDENERGYUTILIZATION
Take the inlet temperature to be 208C. At that temperature the specific volume is given by
y1¼ 29 22:4293
273¼1:39 m3=kg Isentropic work done, W¼21:01331051:39
1:4 1:41
2:8495 1:0133
(1:41)=1:4
1
!
¼338,844 J=kg¼339 kJ=kg
From Figure 3.7, for a compression ratio of 2.85, the efficiency is approximately 84%. So work required
¼339=0:84¼404 kJ=kg Mass flow rate¼ 1000
1:393600¼0:2 kg=s Power required¼4040:2¼80 kW
3.13.4. Electrical Drives
The electrical power required to drive a compressor (or pump) can be calculated from a knowledge of the motor efficiency:
Power¼Wmass flow-rate Ee
(3:38) where
W ¼work of compression per unit mass (equation 3.31);
Ee¼electric motor efficiency.
The efficiency of the drive motor will depend on the type, speed, and size. The values given in Table 3.1 can be used to make a rough estimate of the power required.
Table 3.1. Approximate Efficiencies of Electric Motors
Size (kW) Efficiency (%)
5 80
15 85
75 90
200 92
750 95
>4000 97