11

AIR COMPRESSORS 11.1 Introduction Air compressors are used for supplying high-pressure air. There are many uses of high-pressure air in the industry. The main uses of high-pressure (compressed) air are : .. to drive compressed air engines (air motors) used in coal mines, .. to inject or spray fuel into the cylinder of a Diesel engine (air injection Diesel engine), to operate drills, hammers, air brakes for locomotives and railway carnages, water pumps and paint sprays, .. to start large (heavy) Diesel engines, .. to clean workshop machines, generators, automobile vehicles, etc., .. to operate blast furnaces, gas turbine plants, Bessemer convertors used in steel plants, etc., .. to cool large buildings and air crafts, and .. to supercharge I.C. engines. There are mainly two types of air compressors viz. reciprocating air compressors and rotary air compressors. Reciprocating air compressors are similar to reciprocating engines where a piston reciprocates inside a cylinder. In rotary air compressors, air is compressed due to rotation of impeller or blades inside a casing. Air compressors are driven by engines or electric motors. In this chapter the theory of reciprocating air compressors is discussed in details and principles of working of reciprocating compressed air motors and rotary compressors are explained in brief. 11.2 Reciprocating Air Compressors The principal parts of a reciprocating air compressor are the same as that for a engine. The reciprocating air compressor may be single-acting (air is admitted to one side of the piston only) or double-acting (air is admitted to each side of the piston alternatively), and may be single-stage or multi-stage. In a multi-stage compressor, the air is compressed in several stages instead of compressing the air fully in a single cylinder. This is equivalent to a number of compressors arranged in series. The pressure of air is increased in each stage. Single-stage compressors are used for delivery pressures upto 10 bar, three-stage compressors for pressure upto 200 bar and two-stage compressors for pressures in between 10 to 200 bar. The average piston speed of a reciprocating air compressor is limited to about 300 to 400 metres per minute to reduce friction wear. 11.3 Single-stage Air Compressor The sectional view of an air cooled, single-stage, single-acting reciprocating air compressor is shown in fig. 11-1. Both intake (suction) and discharge (delivery) valves

AIR COMPRESSORS

CyliMtr Piston

Fig. 11-1. Sectional view of a singleacting air cooled, reciprocating air compressor.

343 are disc type and are automatic in their action. They are opened and closed by difference in the air pressure acting on their two sides. When the pressures are equal on their two sides, they are kept closed by light springs. During the outward or suction stroke, the pressure in the cylinder falls below the atmospheric pressure as a result of which the intake valve opens and air is drawn from the atmosphere into the cylinder. During the inward or compression stroke, as a result of the piston action the pressure of the air in the cylinder gradually increases and reaches a value sufficiently above the receiver pressure. The high pressure of air, thus produced, overcomes the resistance of the spring on the discharge valve and causes the valve to open and discharge takes place from the cylinder to the receiver. The receiver is a simple vessel which acts as a storage tank. The compressor is driven by some form of prime mover (electric motor or engine). When the compressor is to be started against tank (receiver) pressure, the prime mover will have to supply very high starting torque. To avoid this, hand unloader (fig. 11-1) is used for releasing pressure from the compressor cylinder when the compressor is stopped.

11.3.1Indicator Diagram : The events described above can be conveniently represented by p -v diagram shown in fig. 11-2. The diagram is drawn for a compressor without clearance. During the suction stroke the charge of air is drawn into the cylinder along line 4-1 at constant pressure p i, which is slightly below that of the atmosphere. At point 1, the piston completes the suction (outward) stroke and starts on its return (compression) stroke. All valves being closed, the air is now compressed along the compression curve 1-2. At point 2, pressure P 2 is reached which is slightly higher than the pressure in the receiver. The discharge valve at this point opens and the delivery of the compressed air takes place along line 2-3 at pressure p2. The piston has now reached the left hand end of the cylinder and again starts on its suction stroke and the pressure in the cylinder will be lowered again to p i and the cycle of operations will be repeated. The net work required for compression and .delivery of the air per cycle is represented by the area Fig. 11-2. I heorotical indicator diagram of a single-stage air compressor. l-Z -o -4

iiOV (tig . 1 1 -2 ).

The amount of work done on the air will depend upon the nature of the compression curve. If the compression occurs very rapidly in a non-conducting cylinder so that there is no heat transfer, the compression will be practically isentropic. If it is carried out slowly so that the heat of the compression is extracted from the air by the jacket cooling water, the compression will approach isothermal. However, in actual practice

344

ELEMENTS OF HEAT ENGINES Vol.l

neither of these conditions can be fulfilled and the actual compression will be between isentropic and isothermal. 11.3.2 Isothermal Compression Versus Isentropic Compression : The slope of the compression curve, represented by the law p t/1 = C, depends upon the value of the index n. A large value of n will give comparatively a steeper curve. The law for an isothermal or hyperbolic compression is pv = C, where the value of index n is unity.The law for an isentropic compression is pvy = C. Since the value of y for air is 1.4,the isentropic curve will be steeper than isothermal curve. Figure 11-3 shows curves representing an isentropic compression (1-2’) and an isothermal compression (1-2"). The middle curve (1-2) shows curve, which is obtained in actual practice. The curve is polytropic (pi/1 = C) having a value of n nearly equal to 1-3 for the water cooled cylinder. The isentropic work required to be done per cycle on the air is represented by the area 4-1-2’-3 (fig. 11-3). If the compression carried out had been isothermal, the slope of the compression curve would be less than that of isentropic and the isothermal work done would be represented by the area 4-1-2"-3 which is evidently less than the isentropic work done represented by the area 4-1-2-3. Therefore, it follows that an isothermal compression is economical and efficient, since less work is required to carry it out, while an isentropic compression requires more amount of work to be supplied. Compression curve with values of index n between 1 and 1.4 will fall within the isothermal and isentropic curves. Thus, it will be seen that the work required for compression and delivery of air per cycle decreases as the value of Fig. 11 3. Effect of nature of compression p d e c re a s e s , curve on work required for compression.

The theoretical indicator diagram for a single-stage compressor without clearance is shown in fig. 11-4. Let p i in N/m2 (newtons per square metre) or Pa (pascals) and vy in m*3 represent initial condition of the air before C o m p r e s s e d a ir compression, and pz in d isch arge to receiver N/m = Pa, the final PV -. co n stan t Piston driven delivery pressure after c fros omu er cx ete rn a l compression, Then, 0 ^cJuu^u

Q. V s / / / / / u r n

E==— 1

hY H

b

a - V ,-------

Volume Fig. 11-4. Single-stage air compressor without clearance.

(a) Work required to be done on the air W, per cycle assuming compression curve to be polytropic,i.e., pt/* = C, is given by area 1-2-3-4 of fig. 11-4.

Now area, 1-2-3-4 = area 0-a-2-3 plus area a-2-1-fc minus area £>-1-4-0 P2 V0 —Pi V] = p2V2 + ^ 2- ; ^ - p i V i

AIR COMPRESSORS

345 = 0P 2V2 - PzV2 + P2 V2 - P M - " P M +P1^1 n -1

o r

n ^ ~ n _'-\ (P2V2 - pivi) Joule per cycle

.. (1 i:ia )

mR(T2 - 7i) Joule per cycle

.. (11.1b)

w

=

From eqn. (11.1a) taking pjy* outside the bracket, n n - 1 P ivi pw \

Work required, W =

1

But for polytropic compression, p i v i n = p2V2n. Hence, — = w v\ S 's Substituting the value of £ in the above equation, v-1 Work required per cycle (or per revolution, if compressor is single-acting),

us n W = --—

1 ' k '1 n X -1 [p i j

p i vt W

-----------1

(p z )

1

P1V1

1 c

c

n —1

&

Joule per cycle

.. (11.2)

This equation gives the work required in Joules per cycle (or per revolution, if the compressor is single-acting) in compressing and delivering the air. It should be noted that units of pressure and volume in eqn. (11.2) are N/m2 or Pa and m3 respectively. . ^ W xN Indicated power of the compressor = — - - —

... J/sec. or W

...(11.3)

where W = work required in Joules per cycle, and N - No. of cycles performed per minute - r.p.m. for single-acting compressor, if p iv i in eqn. (11.2) is substituted by mRTi, then work required per cycle, n -1 c

(P z

Joule per cycle and

1

W = — mRTi n- 1

( p ij

1

Work required per kg of air, -

'P i

Joule

1

n RT, n- 1

c

W=

n- 1

1

[ p ij

...(11.4)

346

ELEMENTS OF HEAT ENGINES Vol.l Indicated power of the compressor = W x mass of air delivered per second J/sec. or W

... (11.5)

where W = work required in Joules per kg of air. (b) If the compression is isentropic (pvT = C), the index n will be replaced by y in eqn. (11.2) and eqn. (11.4), and then .1-1 1 pst 7 -1 Joule per cycle

Work required, W = — p\v\ y -1 r

... (11.6a)

Work required per kg of air,

n

W=

Y -1

Y-1

(11.6b)

Joule

VPi J (c) If the compression is isothermal (pv = C), then work required per cycle, W » area 1-2-3-4 (fig. 11-4) = area a-2-1-b plus area 0-a-2-3 minus area b-1-4-0 W =

P 1 V1

= P1V1

loge Y\ V2.

+ P2V2 -

loge

Joule per cycle

py

P 1 V1 rn P1V1

loge

(since P2V2 - pivi) ... (11.7a)

(11.7b) Work required per kg of air, W - RTi loge v Joule pi 11.3.3 Approximation of Isothermal Compression : Although isothermal compression is economical, it is not possible to achieve it in practice. To have an isothermal compression, the compressor will have to be run extremely slow, while in practice it is driven at high speed so that as much air as possible in compressed in a given time. Since, there is saving of work by compressing air isothermally, it is necessary to make an attempt to obtain approximately an isothermal compression. Three methods are adopted to achieve this object while still running the compressor at high speed. The three methods adopted are : — cooling the air during compression by spraying cold water into the cylinder, — cooling the air during compression by circulating cold water through the cylinder jacket, and — adopting multi-stage compression with inter-stage cooling. Cold water spray : In this method, cold water is sprayed into the cylinder during compression. The cooling, thus done, reduces the temperature of the air and the compression curve will be approximately of the form pv1 = C. This means that the compression is brought nearer to isothermal which results in the saving of work. Water jacket : In this method, the heat of compression is extracted by circulating cold water in the cylinder jacket thereby keeping the temperature rise as small as possible. This keeps the compression near to isothermal as shown in fig. 11-3.

347

AIR COMPRESSORS

Multi-stage compression : In this method, the compression of air is carried out in two ormore stages in separate cylinders. The pressure of the air is increased in each stage. It is a common practice to provide intercoolers between the cylinders of multi-stage compressor, for the purpose of cooling the compressed air to atmospheric (intake) temperature before entering the succeeding (next) stage. It is this cooling between the cylinders that keeps the compression very near to isothermal as shown in fig. 11-6. 11.4 Two-stage Air Compressor A two-stage air compressor with water jacketed cylinders and intercooler is shown in fig. 11-5(a). The suction in the L.P. cylinder (tig. 11-5b) ends at 1 and the air is drawn in the cylinder at pressure p i. The air is then compressed polytropically to 2’. The L.P. cylinder then discharges (delivers) the air along line 2-p2 to the intercooler where air is cooled at constant pressure pa to the original (intake) temperature corresponding to point 1 by the circulating cold water. When air is cooled in the intercooler to intake temperature corresponding to point 1, the cooling is perfect. The air in cooling at constant pressure suffers a reduction of volume from 2'-p 2 to p2- 2 . The cooled air is then drawn into the H.P. cylinder (fig. 11-5c) along line p2- 2 for the second stage compression, where it is compressed polytropically to the final pressure P3 along line 2-3, and then delivered to the receiver (not shown) at constant pressure P3 along line 3-p3. and

In fig. 11-6, the combined ideal indicator diagram is shown for the low pressure high pressure cylinders of a single-acting, two-stage air compressor with

p

iKBlilrit M

mh i Ii MRi«l rfttidr («) *

lb)

•

•«*

(a) Arrangement of cylinder*. (b) Indicator dia o n^ \ .M il.. 1 vOlvf

I*

LP.

t i.

it

p isto n o lff fW

if 3

V

A!

t fn

fp

f e■ucfton n

I

Dull*

•> - .

OfOtn cocks

Volvo

I J Volwmo Fig 11-9. Indicator diagram of three-stage compression with perfect intercooling.

Fig. 11-8. Arrangement of three-stage air compressor with intercoolers.

(b) When intercooling is perfect, ptV i = P2V2 - P3V3 Hfe 11-9). Substituting pivi lor p2v2 and P3V3 in eqn. (11.14a), work required per cycle, n -

n

W

n-1

p i v\

fpz

1

1

n -

n

+

[p i J

(f* )

n-

1

n

n

+

_

w

W

Joule

-3

... (11.15a)

If p/Vf in eqn. (11.15a) is substituted by mRTh then work required per kg air may be written as,

=

n~ . RT\ n- 1

' Pz' [p ij

n -1

n - 1

n

n

+

'pa' [pz]

(c) Work required is minimum when

n -1

+

f Pa '

Pz _ 03 _ Pa Pi “ Pz ~ P3

n -

3

Joule

.. (11.15b)

AIR COMPRESSORS Substituting

353

for ~

and ^

in eqn. (11.15a), n- 1

p i v'i ' b ? Ly",

Min. work required per cycle, W = Now, for minimum work, Substituting

J* P1

02

P3

P4

Pi

P2 “

P3

Joule

. (11.16a)

Joule

..(11.16b)

for —, ^ and — in eqn. (11.16a), p i P2 P3 7

Min. work required per cycle, W =

p iv i

n-1 3n

-1

A/

Minimum indicated power of the compressor = W x — J/sec. or. W 60

(11.16c)

where N = no. of cycles per min. = r.p.m. for single-acting compressor, and W = work required in Joule per cycle. If p i v i in eqns. (11.16a) and (11.16b) is substituted by mRTi, then minimum work required per kg of air compressed and delivered may be written as,

w = -^-n r, n —1 1

(- \ P2

n- 1

-1

(11.17a)

LV*, n-1

w = — - RTi n-1 1

E* . v Pi

3n

-1

Joule

(11.17b)

Minimum indicated power of the compressor = IV x mass of air delivered per second J/sec. or W

(11.17c)

where W = work required in Joule per kg of air (d) Equation (11.16b) can readily be extended to m stages. Then

n -1

while -

will be replaced by and similarly r J n- 1 7

3n

by ~ ~ mn

will always refer to delivery and suction pressures, i.e.

Pm4* 1 —.

Then minimum work required per cycle with complete intercooling,

354

ELEMENTS OF HEAT ENGINES Vol.l

All the expression derived above for work required refer to the work actually done or required to be done on the air, and the power derived from these expressions will be referred to as indicated power or air power. 11.5.1 Advantages of multi-stage compression : The advantages compression are as follows : -

of multi-stage

Reduction in power required to drive the compressor owing to compression being approximated to isothermal, Better mechanical balance of the whole unit and uniform torque, Increased volumetric efficiency as a result of the lower delivery pressure in the L.P. cylinder clearance space, Reduced leakage loss owing to reduced pressure difference on either sides of the piston and valves, Less difficulty in lubrication due to the lower working temperature, and Lighter cylinders.

11.6 A ir Compressor Terminology The following terminology should be well understood before attempting to estimate the performance of the air compressor. Free air delivered is the volume of air delivered under the conditions of temperature and pressure existing at the compressor intake, i.e., volume of air delivered at surrounding air temperature and pressure. In the absence of any given free air conditions, these are generally taken as 1 01325 bar and 15‘C. Capacity of a compressor is the quantity of the free air actually delivered by a compressor in cubic metres per minute. Piston displacement is the volume in cubic metre (m ) obtained asthe product of the piston area in m2 and the piston stroke in metre. Displacement per minute is the product of the piston displacement and working strokes per minute. For multi-stage compressors, the displacement is based on low-pressure cylinder only, since it determines the amount of air passing through the other cylinder. Indicated power or air power is the power determined from the actual indicator diagram taken during a test on the compressor. It is calculated in the same manner as is done in the case of a steam engine and internal combustion engine. Shaft or brake power is the power delivered to the shaft of the compressoror the power required to drive the compressor. The compressor may be driven by an engine or an electric motor. [Shaft or brake power] - [Air or indicated power] = [Friction power] , „ . . . „. . Air (indicated) power and Mechanicalefficiency, rim = o. - . —. - , -------1 1 Shaft (brake) power Isothermalpower of acompressor is calculated from the theoretical indicator diagram drawn on the basis of an assumption that the compression is isothermal. (a) Referring to eqn. (11.7a)

for a single-stage compressor without

Isothermal work required per cycle, W = p\ vi loge '( * Pi

Jou,e

clearance, .. (11.19a)

355

AIR COMPRESSORS Isothermal power = pw \ log©

x

.. (11.19b)

J/sec. or W

60 v *, where N = no. of cycles per minute.

If p i v i in eqn. (11.19a) is substituted by mRTi, then isothermal work required per kg of air may be written as, W = RTy log,

(11.20a)

Joule

Isothermal power » W x mass of air delivered per sec. J/sec. or W

(11.20b)

(b) For a two-stage compressor, Isothermal work required per cycle, W = pi vi loge

(11.21a)

Joule

V *, N x - - J/sec. or W

Isothermal power = pi v\ loge v

(11.21b)

60

vP1,

If p i v i in eqn. (11.21a) is substituted by mRTi, then isothermal work required per kg of air may be written as W = RTi \oge w

.. (11.22a)

Joule

V */

Isothermal power = W x mass of air delivered per sec. J/sec. or W

.. (11.22b)

(c) Similarly, for a three-stage compressor, Isothermal work required per cycle, W = pi vi log0

fp ^

Joule

S ', N Isothermal power = pi v\ ioge P4 x — J/sec. or W ,P1 v where N * no. of cycles per min. /_

,\

.. (11.23a) .. (11.23b)

Adiabatic power is calculated from a theoretical indicator diagram drawn on the basis of an assumption that the compression is an ideal adiabatic, i.e., isentropic. Adiabatic work required, W =

(pzvz-pivtf Joule per cyde

This equation for the adiabatic work required may be expressed in more convenient form by writing its equivalent, W=

mR(Tz - T O =

mRT,

f Tr

'

f r ' Iz l

Tz Since -=- = »1

v */

Adiabatic work required, W = —Y— mRT} y -1

i_ -1 fPz 7 -1 W

Joule per cycle

356

ELEMENTS OF HEAT ENGINES Vol.l Y~1

y -1

-1

PiV-1 l v

Joule per cyde

..(11.24a)

* \

Thus, for a single-stage air compressor, Y-1

Adiabatic (isentropic) power =

p i v\

P2 Y -1 [P lj

X

£

J/sec. or W

24b)

where A/ = no. of cydes per min. ■ . * __. „ . Isothermal efficiency =

Isothermal power in watts — , ---------- . Indicated or actural power in watts

(11.25a)

Isothermal power_ Shaft power or brake power required to] drive the compressor j

Overall isothermal efficiency or Compressor efficiency .

N „ *6 0 Shaft power or brake power required in watt to drive the compressor pi vi loge

(11.25b)

where N = no. of cydes per min. Isentropic efficiency or ideal adiabatic efficiency

Isentropic power in watt Shaft power required in watt Y-1

N „ 1 P1V1 P? X 60 Y-1 P1 Shaft power or Brake power required in watt to drive the compressor

(11.26)

where N = no. of cydes per min. Volumetric efficiency of an air compressor is the ratio of the adual volume of the free air at standard atmospheric conditions discharged in one delivery stroke, to the volume swept by the piston during the stroke. The standard atmospheric conditions (S.T.P.) is actually taken as pressure of 760 mm Hg (1 01325 bar) and temperature 15‘C in this connection. Thus, . Volume of free air delivered per stroke .. (11.27) Volumetric efficiency = — ;----------- zz— -------- r izr~ Volume swept by piston per stroke The value of volumetric efficiency varies between 70 to 85 per cent according to the type of compressor. The volumetric efficiency decreases as the dearance volume increases. Other factors that lower the volumetric efficiency are : - Valve leakage, specially at the inlet valve, Obstruction at inlet valves, Piston ring leakage, which allows air to pass from one side of the cylinder to the other, Heating of air by contact with hot cylinder walls, and

AIR COMPRESSORS -

357

Very high speed of rotation.

With the decrease of volumetric efficiency, the capacity (quantity of free air delivered) of the compressor decreases. Problem-1 : A single-cylinder, single-acting reciprocating air compressor has a cylinder of 24 cm diameter and linear piston speed of 100 metres per minute. It takes in air at 100 kPa (100 kN/m2) and delivers at 1 MPa ( 1 MN/rrP), Determine the indicated power of the compressor. Assume the law of compression to be pv 1.25 ■ constant. The temperature of air at inlet is 288 K. Neglect clearance effect. Given p i = 100 kPa « 100 x 10 Pa; p2 = 1 MPa = 1,000 kPa - 1,000 x 103 Pa. 1,000 x 10J 100 x 10J

P* Pi

1,000 = 10 100

Swept volume in m3/min. - -

x I x r.p.m.

(where, I - piston strokes in metre, and d= diameter of the cylinder in metre) f 24 ' 100 m3/min. .. Swept volume = ^ x x 100 / ( v piston speed = 2 x I x r.p.m. = 100 metres/min.) \2

It

_ ... 3. 2-261 3 . = 2-261 m /min. = m /sec. Referring to fig. 11-10 and using eqn. (11.2), Work required per sec., n- 1

n n-1

-1 J/sec. S' / [where, p i is pressure in Parcals (Pa) and volume of air compressed, vi is in m per sec.]

W = pyV] x

L V

W = (too

x

to3)

x

2f l

X

'± \

1 2 5 -1 000) 1 25

100

-

1

/

= 1,88616 x (1 5848 - 1) - 11,030 J/sec. or 11,030 W .-. Indicated power of the compressor = 11,030 W i.e. 11-03 kW Problem-2- : A single-acting, single-stage air compressor developing indicated power of 11 kW, runs at 200 r.p.m. and has a linear piston speed of 100 metres per min. If the suction pressure and temperature are 100 kPa and 15‘C respectively and delivery pressure is 1,000 kPa, calculate the dimensions of the compressor cylinder. Assume the law of compression to be pv125 = constant. Neglect clearance effects. Referring to fig. 11-10, and considering polytropic compression 1-2, p iv in = p 2 V2n,

ELEMENTS OF HEAT ENGINES Vol.l

358 V 1 Vz

1 10' 1*26 = 6 31 1 V /

n

_

[p .j

P2VZ -

Using eqn. (11.1a), work done per cycle, W = pzvz + - — ^ p

P1V1

------- p iv i

Work done per cycle in kJ, W ^pa Displacement volume in m-3, vi pzva+

p z v z - p i vi

p in

V1

= 1,000 x £ “ -1 0 0 +

Vi - P i n- 1

P2 vz -P1 + V1

1,000 x

1

Piston stroke, I =

p zyz

1

631 0-25

-1 0 0 = 292 4 kPa 100 = 0-25 metre or 25 cm 2 x 200

piston speed per min. piston strokes per min. (2 x r.p.m.)

Indicated power of compressor = pm x I x a x n watt. where n = no. of cycles per sec. ^ i.e. 11 x 103 - 292 4 x 103 x 0 25

200N 60

x 0-7854 x

11 x 103 x 104 x 60

200 60

r 1-22 = 0-287 x 1 122-1 v = 290 kJ per kg of air

-1

-

1

f I 7.5 \ l -22 + 313 4078

o.ggo

Work done per sec. = W x m = 290 x ——- = 14-32 kJ/sec. 60 Air power = 14-32 kW

Fig. 11-13. Two-stage compression with imperfect intercooling

Fig. 11-14.

-

1

ELEMENTS OF HEAT ENGINES Vol.l

362 (c) Using eqn. (11.11),

For minimum work, intercooler pressure. p 2 = Vpip3 = V1 x 17 5 = 4-183 bar Since intercooling is perfect, Ti - T2. A p iv i • p2V2

i.e. 1 x vi * 4-183 x V2

vi 4-183 or — = —j— = 4-183

Ratio of cylinder diameters for minimum work, ~

°2

=

v2

= V4-183 = 2-045

( V /, = l2)

Problem-7 : It is desired to compress 16 m3 of air per minute from 1 bar (100 kPa) and 294 K to 105 bar (105 MPa). Calculate : (i) the minimum power required to drive the compressor with two-stage compression and compare it with the power required for single-stage compression, (ii) the maximum temperature in the two cases, (iii) the heat to be removed in the intercooler per minute, (iv) the amount of cooling water required per minute if the inlet and outlet temperatures of cooling water to and from the intercooler are 15'C and 40'C. Assume the value of index for compression process to be 135 for both cases. Also assume proper intercooler pressure for minimum work and perfect intercooling. Take R = 0287 kJ/kg K and kp - 10035 kJ/kg K for air. Given : p i P3

=

Pi

- 1 bar = 1x 10s Pa;pa » 10-5 10:5 * J 0 ! = 1 x 10s 1

bar * 10-5 x 105 Pa;

IQ-5

Using equation (11.11) for maximum efficiency or minimum work, intercooler pressure, Pz = pz " P1

Vpipa = V1 x

10-5= 3-241bar* 3-241 x 10s Pa

3 241 x 10s = 3 241 1 x 105 1

Referring to fig. 11-15 and using eqn. (11.12a), Minimum work required per sec. for two-stage compression.

Fig.

11-15. Two-stage compression with perfect intercooling.

Fig. 11-16. Single-stage compression.

AIR COMPRESSORS

363

[ p .J

1------------------------

* 1

n W = 2 x ------- x m vi n -1 r

1

1

C

'

J/sec.

(where, p* is pressure in pascals (Pa) and v-i is volume of air compressed in m per sec.) 135-1

3-241

o 135 /A