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8. Multi-pressure systems.

Example No. 1Calculate the power needed to compress 1.2 kg/s of ammonia from saturated vapor at 80 kPa to1000 kPa (a) by single-stage compression and (b) by two-stage compressiron with intercooling byliquid refrigerant at 300 kPa.

Given:Mass of ammonia = 1.2 kg/sFrom saturated vapor at 80 kPa to 1000 kPa.Intercooling at 300 kPaRequired:a. Power by single-stage compression.b. Power by two-stage compression with intercooling.Solution:

(a) Single stage compression (1-3-6)From properties of Ammonia.At state point 1. Saturated at 80 kPa

kgkJh 14101=

KkgkJs = 207261

.

At state point 3, 1000 kPa, s3= s1. (Superheated ammonia)kgkJh 1800

3=

Power required = ( ) ( ) kWhhw 468141018002113

== .

(b) By two-stage compression with intercooling.From properties of Ammonia.At state point 1. Saturated at 80 kPa

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kgkJh 14101=

KkgkJs = 207261

.

At state point 2, 300 kPa, s2= s1. (Superheated ammonia)

kgkJh 15882=

At state point 4, saturated at 300 kPa

kgkJh 14504 = KkgkJs = 74515

4.

At state point 5, 1000 kPa, s5= s4. (Superheated ammonia)

kgkJh 16285=

At state point 6, 1000 kPa, saturated liquid.

kgkJh 3166=

The flow rate of ammonia compressed in the high stage can be calculated by making a heat and amass balance about the intercooler, as shown in Figure 6.

Heat balance:

442266hwhwhw =+

( ) ( )( ) ( )kgkJwkgkJskgkgkJw 145015882131646

=+ .

Mass balance:

426www =+

4621 ww =+ .

Solving gives:

skgw 34614

.=

Power required = ( ) ( )454122hhwhhw +

Power required = ( ) ( ) kW24531450162834611410158821 ... =+=

Intercooling the ammonia with liquid refrigerant reduced the power requirement from 468 to 453.2kW.

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Example No. 2Calculate the power required by the two compressors in an ammonia system which serves a 250-kWevaporator at -25 C. The system uses two-stage compression with intercooling and removal of flashgas. The condensing temperature is 35 C.Given:Required:Solution:First sketch the schematic diagram of the system (Figure 10a) and the corresponding pressure-enthalpy diagram (Figure 10b). The functions of the intercooler and flash tank are combined in onevessel.

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Determine the intermediate pressure for optimum economy.ps= saturation pressure at -25 C = 152 kPapd= saturation pressure at 35 C = 1352 kPa

( )( ) kPappp dsi 4531352152 ===

The enthalpies at all points can now be determined from ammonia table of properties.At state point 1

h1 = hgat -25 C = 1430 kJ/kgs1 = 5.9813 kJ/kg-KAt state point 2, s2= s1h2= hat 453 kPa after isentropic compression = 1573 kJ/kgAt state point 3h3= hgat 453 kPas3= 5.6001 kJ/kgAt state point 4, s4= s3

h4 = h at 1352 kPa after isentropic compression = 1620 kJ/kg

At state point 5

h5= hfat 35 C = 366 kJ/kg

At state point 6

h6= h5 = 366 kJ/kgAt state point 7

h7= hfat 453 kPa = 202 kJ/kg

At state point 8

h8 = h7= 202 kJ/kg

Mass rates of flow:

Heat balance about the evaporator:

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skghh

Qm A 2040

2021430

250

81

1.=

=

=

skgmmmm 20408721

.====

Heat and mass balance about the intercooler:

33776622hmhmhmhm +=+

36 mm = and 27 mm =

( ) ( ) ( ) ( )146320220403661573204033

mm +=+ ..

skgm 25503

.=

Low-stage power: m1(h2 h1) = (0.204 kg/s)(1573 1430 kJ/kg) = 29.2 kW

High-stage power: m3(h4 h3) = (0.255 kg/s)(1620 1463 kJ/kg) = 40.0 kW

Total power: 29.2 + 40.0 = 69.2 kW

Example No. 3

In an ammonia system one evaporator is to provide 180 kW of refrigeration at -30 C and another

evaporator is to provide 200 kW at 5 C. The system uses two-stage compression with intercooling

and is arranged as in Figure 11a. The condensing temperature is 40 C. Calculate the power required

by the compressors.

Given:

180 kw of refrigeration at -30 C

200 kw of refrigeration at 5 C

Condensing temperature is 40 C

Required:

Power required by the compressor.

Solution:

Sketch the pressure-enthalpy diagram of the cycle as in Figure 11b.

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The discharge pressure of the low-stage compressor and the suction pressure of the high stage

compressor are the same as the pressure in the 5 C evaporator.

Enthalpies at the state points.

At state point 1

h1 = hg at -30 C = 1423 kJ/kg

s1= sg at -30 C = 6.0636 kJ/kg-K

At state point 2

h2 = h at 517 kPa after isentropic compression =1630 kJ/kg

At state point 3

h3 = hg at 5 C = 1467 kJ/kg

s3= sg at 5 C = 5,5545 kJ/kg-K

At state point 4

h4 = h at 1557 kPa after isentropic compression = 1652 kJ/kg

At state point 5

h5 = hfat 40 C = 390.6 kJ/kgAt state point 6

h6 = h5 = 390.6 kJ/kg

At state point 7

h7= hfat 5 C = 223 kJ/kg

At state point 8

h8 = h7= 223 kJ/kg

Mass rates of flow:

skgkgkJ

kW

hh

Qm A 1500

2231423

180

81

1

1.=

=

=

skgmmmm 15001287

.====

Probably the simplest way to calculate the mass rate of flow handled by the high-stage compressor

is to make a heat and mass balance about both the high temperature evaporator and the

intercooler, as shown in Figure 12.

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Heat balance:773322255hmhmhmQhm A +=++

Mass balance:

skgmm 150072

.==

Therefore:

35mm =

Combining gives:

( ) ( ) ( ) ( )2231500146716301500200639033

... +=++ mm

Solving leads to

skgm 38203

.=

The power required by the compressors can now be calculated:

Low-stage power: ( ) ( ) kWhhm 131142316301500343 .. ==

High-stage power: ( ) ( ) kWhhm 460146716253820121

.. ==

Total = 91.5 kW