Chapter 12 Reciprocating Air Compressors – Thermal Engineering

Chapter 12

Reciprocating Air Compressors

12.1 ❐ INTRODUCTION

Compressors are mechanical devices used for increasing the pressure of a gas. Compressors used for producing high pressure air are called air compressors. Air is drawn from the atmosphere by suction process, which is then compressed to the required pressure and delivered to the receiver. Air compressors may be classified as shown in Fig. 12.1.

Figure 12.1 Classification of compressors

If compression is done in a conventional cylinder with a closely fitted piston making reciprocating motion, then the compressor is called a reciprocating compressor. External work must be supplied from a prime mover to the compressor to achieve the required compression. The general arrangement of the compressor and prime mover used to run the compressor is shown in Fig. 12.2.

Figure 12.2 Energy flow in an air-compressor

12.2 ❐ USES OF COMPRESSED AIR IN INDUSTRY

The several uses of compressed air are as follows:

  1. To drive a compressed air engine
  2. For producing an air blast for a workshop
  3. For spraying the fuel (atomizing) into a boiler furnace
  4. For operating pneumatic drills and tools
  5. For operating pneumatic brakes for locomotives and rolling stock
  6. Pneumatic conveying and for pumping of water by compressed air
  7. For working compressed air engines especially in mines
  8. In gas turbine power plants and air-conditioning plants
12.3 ❐ WORKING PRINCIPLE OF SINGLE-STAGE RECIPROCATING COMPRESSOR

The schematic arrangement of a single-stage reciprocating compressor is shown in Fig. 12.3. It consists of a cylinder with cooling jacket, piston, connecting rod, crank, suction valve, and delivery valve. The piston will be making to-and-fro motions through the crankshaft arrangement and generally run by an electric motor, diesel engine, petrol engine, or steam engine. During the outward motion of the piston, the pressure inside the cylinder falls below the atmospheric pressure and the suction valve is opened due to the pressure difference. The atmospheric air is then drawn into the cylinder until the piston reaches the bottom dead centre position.

Figure 12.3 Single-stage reciprocating air-compressor

As the piston starts move inwards, the suction valve gets closed and the pressure starts increasing until the pressure inside the cylinder is more than the pressure of the delivery side. Then the delivery valve opens and high pressure air is delivered to the receiver till the piston reaches the top dead centre. At the end of the delivery stroke, the small volume of high pressure air left in the clearance volume expands as the piston starts moving outwards. Hence, the cycle is repeated.

12.4 ❐ TERMINOLOGY
  1. Single-acting compressor: If the air admission from the atmosphere is on only one side of the cylinder.
  2. Double-acting compressor: When, the air from the atmosphere is drawn on both sides of the piston.
  3. Single-stage compressor: If the total compression is done fully in one cylinder.
  4. Multi-stage compressor: If the compression is carried out in more than one cylinder, and every cylinder carries out a part of the compression.
  5. Free Air Delivered (FAD): It is the actual volume delivered at the stated pressure, reduced to intake pressure and temperature, and expressed in cubic metre/minute (m3/min). FAD is used to measure the capacity of a compressor.
  6. Displacement or Swept Volume of the Compressor: The volume displaced by piston movement between two dead centers is called displacement or swept volume. For a single acting compressor, swept volume,
    where d = diameter of piston and L = length of stroke
12.5 ❐ TYPES OF COMPRESSION

The theoretical p−V diagram for single acting compressor is shown in Fig.12.4. The line 4–1 represents the suction stroke. The air is then compressed adiabatically as shown by the curve 1–2 in Fig. 12.4. It is then forced out of the cylinder at constant pressure p2, as shown by the line 2–3. The work done is represented by the area 1–2–3–4–1.

Figure 12.4 Theoretical p-V diagram for single acting compressor

If the air had been compressed isothermally, as represented by the curve 1–2′, then the work done on the air would be the area 1–2′–3–4–1, which is considerably less than that due to adiabatic compression. However, it is not possible in practice to compress the air isothermally because in that case, the compressor would need to run extremely slow. In practice, the compressors are driven at fairly high speeds in order to compress as much air as possible in a given time. Hence, the compression of air will approximate to an adiabatic. The work saved by compressing isothermally is shown by the shaded area 1–2–2′–1.

12.5.1 Methods for Approximating Compression Process to Isothermal

The following practical methods are used to achieve nearly isothermal compression:

  1. Cold water spray: In this method, cold water is sprayed onto the cylinder during compression, thus reducing the temperature of the air. Without the cold water spray, the compression would have been adiabatic or pVγ = constant. This effect is shown in Fig. 12.5, where now the compression would be between adiabatic and isothermal or pVn = const., where 1 < n < 1.4. The work saved is represented by 1-2-2′′-1.

    Figure 12.5 Effect of type of compression on work done.

    Figure 12.6 Multi-stage compression

  2. Water jacketing: The water is circulated around the cylinder through the water jacket to cool the air during compression. This method is commonly used for all types of reciprocating compressors.
  3. Intercooling by using multi-stage compression: In a multi-stage compressor, the air is compressed in several stages. In principle, it is equivalent to number of compressors in series where the air passes from one cylinder to the next and the pressure increases in each cylinder. The pV diagram for a four-stage compressor is shown in Fig. 12.6. The dotted line 1–4′ is the isothermal. In the first stage, the air is compressed adiabatically to 2′ and cooled at constant pressure to 2′ in the intercooler, possibly to the initial temperature. For complete intercooling, the point 2′ is on the isothermal line 1–4′. The air is then drawn into the second cylinder for the second stage of compression and the process is repeated for the subsequent cylinders. Line 1–a represents the adiabatic compression in a single cylinder. The compressor work saved by intercooling is represented by the area 2–a–4–3′–3–2′–2.
  4. External fins: Small capacity compressors are provided with fins to increase the heat transfer from the surface of the cylinder.
12.6 ❐ SINGLE-STAGE COMPRESSION

12.6.1 Required Work

Without Clearance

Consider the theoretical p−V diagram for a single-stage air compressor as shown in Fig. 12.7. The work done on air per cycle is represented by the area 1–2–3–4–1.

Figure 12.7 p-V diagram without clearance

W = area (ab − 2 − 3 − a) + area (bc − 1 − 2 − b) − area (ac − 1 − 4 − a)

where m = mass of air delivered per cycle.

Work done per kg of air

If the compression had followed the law, pVn = c, then

Power required in driving the compressor

where N = number of complete cycles per minute

= rpm, if single acting

= number of strokes per minute, if double acting.

With Clearance Volume

The p-V diagram for a single-stage and single-acting air compressor is shown in Fig. 12.8 with clearance volume.

Let Vc = clearance volume

Vs = swept volume = V1Vc

Va = actual volume = V1V4

Let the compression and expansion processes follow the same low pV n = const.

Work done per cycle,

W = area (1–2–3–4–1)
= area (1–2–5–6–1) − area (3–4–6–5–3)

Now, p3 = p2 and p4 = p1

Figure 12.8 p-V diagram with clearance

where m1 = actual mass of air delivered per cycle.

12.6.2 Volumetric Efficiency

The volumetric efficiency of a reciprocating compressor is defined as the ratio of the actual free air delivered to the swept volume of the compressor. The free air delivered is (V1V4), whereas the swept volume is (V1Vc). Thus,

Let clearance ratio

When referred to ambient conditions,

Factors Affecting Volumetric Efficiency

The volumetric efficiency of a compressor can be lowered by any of the following conditions:

  1. Very high speed
  2. Leakage through piston seals
  3. Too large a clearance volume
  4. Obstruction at inlet valves
  5. Overheating of air by contact with cylinder walls
  6. Inertia effect of air

Figure 12.9 shows the variation in volumetric efficiency with clearance ratio (c), and the pressure ratio (p2/p1), and polytropic index of compression (n) (by changing one factor keeping the other two factors constant). The volumetric efficiency decreases with increase in both the clearance ratio (c), and the pressure ratio (p2/p1), whereas it increases with increase in the polytropic index of compression (n).

Figure 12.9 Variation of volumetric efficiency with clearance ratio, pressure ratio and polytropic index of compression

12.6.3 Isothermal Efficiency

The p-V and T- s diagrams for isothermal and polytropic compression of air in the compressor, respectively, are shown in Fig. 12.10.

Polytropic work done,

Figure 12.10 Isothermal and polytropic compression in a compressor

Isothermal work done

However, p1v1 = p2v2′

Work saved,   ∆W = Wp − Wi

The isothermal efficiency (ηi)is a measure of the degree to which isothermal compression has been achieved. It is defined as the ratio of isothermal work to that of actual indicated work and is given by

12.6.4 Adiabatic Efficiency

It is defined as the ratio of the work done on the compressor with reversible adiabatic compression to the work done with irreversible polytropic compression.

12.6.5 Calculation of Main Dimensions

The actual volume of air drawn in by the compressor per stroke,

Capacity of a single-acting compressor,

Also,

where d = piston diameter

L = length of stroke

Therefore, L and d can be determined.

12.7 ❐ MULTI-STAGE COMPRESSION

Single-stage compression suffers from many disadvantages such as (i) handling of very high pressure range in one cylinder resulting in leakage past the piston, (ii) ineffective cooling of the gas, and (iii) necessitating robust construction of the cylinder to withstand the high delivery pressure.

The volumetric efficiency of a single-stage compressor with fixed clearance decreases with an increase in pressure ratio and thus reduces the capacity. Thus the necessity of of multi-stage compression with intercooling between stages are listed below:

  1. The air can be cooled perfectly at pressures intermediate between the suction and delivery pressures resulting in less power required as compared to a single-stage compressor for the same pressure limits and quantity of free air delivered.
  2. The mechanical balance of the machine is better due to phasing of the operations.
  3. The pressure range and hence the temperature range in each stage can be kept within desirable limits. This results in:
    1. Less loss of air due to leakage past the piston.
    2. More perfect lubrication due to lower temperatures.
    3. Better volumetric efficiency.
    4. The lighter construction of the machine that reduces the cost.

12.7.1 Two-stage Compressor

The schematic arrangement of two-stage compressor with intercooler and pV diagram are shown in Fig. 12.11. The air is first taken into the low pressure (L.P.) cylinder at pressure p1. After compression to intermediate pressure p2, the air at condition 2 is passed through the intercooler and leaves it at point 3, where its temperature is reduced from T2 to T3. The air may be cooled to point 3′ such that T3′ = T1. Finally, the air is compressed in high pressure (H.P.) cylinder from condition 3 to 4 and is discharged to the receiver. Area 2–3–4–6–2 represents the work saved due to intercooling.

Figure 12.12 shows the pV diagram for two-stage compression with perfect intercooling. Process 1–2 represents the polytropic compression in the L.P. cylinder with pV n1 = constant. Process 2–3 represents intercooling from T2 to T3 = T1. Process 3–4 represents the polytropic compression in the H.P. cylinder with pV n2 = constant. Curve 1–3–4′ represents the isothermal compression.

Total work done on the air,

Let and

Figure 12.11 Two stage compression with intercooling: (a) Schematic arrangement, (b) p-V diagram

Figure 12.12 Two stage compression with perfect intercooling

For p1 and p3 to be constant, intermediate pressure p2 must be determined for minimum work.

Thus,

For n1 = n2 = n, and a = b.

For perfect intercooling, T3 = T1. Thus,

In general, with i number of stages, we have

The minimum indicated powers (I.P.), with imperfect intercooling is:

where m = mass of air delivered in kg per second

The number of stage is decided by the delivery pressure for a given inlet pressure (1 bar) as follows:

  1. For delivery pressure upto 5 bar: Single stage compressor
  2. For delivery pressure between 5 to 35 bar: Two stage compressor
  3. For delivery pressure between 35 to 85 bar: Three stage compressor
  4. For delivery pressure more than 85 bar: Four stage compressor

12.7.2 Heat Rejected to the Intercooler

Let m = mass of air in the cylinder

Then p1V1 = mRT1

or

From the compression 1–2, we have

From the constant pressure line 2–3, we have

12.7.3 Cylinder Dimensions

For steady-state flow, the mass of air passing through each cylinder per stroke must be same.

Let Va = actual volume of air per stroke taken during suction

ρ = density of air

Then, Va1 ρ1 = Va2 ρ2 = Va3 ρ3 = … = const.

For perfect intercooling, T1 = T2 =T3 =…

Va1 p1 = Va3 p2 = Va3 p3 =…

Vs1 ηv1 p1 = Vs2 ηv2 p2 = Vs3 ηv3 p3 =…

where ηv = volumetric efficiency of a cylinder

Vs = stroke volume of the cylinder

12.7.4 Intercooler and Aftercooler

Intercooler

An intercooler is a simple heat exchanger in which heat is removed from the air after it has been compressed and its temperature has risen as a result of compression. The intercooler commonly used is of the counter flow type as shown in Fig. 12.13 because it gives high effectiveness.

Effectiveness of intercooler,

A simple section showing the principles of construction of an intercooler is shown in Fig. 12.14. The coolant, which may be water or any other fluid, passes through the tubes secured between two end plates and the air circulates over the tubes through a system of baffles. Two passes are used for water flow and air is made to flow partly parallel and partly cross with the help of baffles. This type of intercooler gives better effectiveness than the ordinary counter-flow type. The purpose of the intercooler is to reduce the work done on the air.

Figure 12.13 Counter flow intercooler

Figure 12.14 Principles of construction of inter-cooler

Aftercooler

An aftercooler is used to cool the air coming out from the compressor before it enters the receiver. The air coming out from the compressor at pressure p3 is sufficiently hot at temperature T4 (Fig. 12.15). If this air is cooled in the aftercooler, then the pressure will remain p3, but the temperature will fall from T4 to T4′′. Therefore, the volume of air leaving the aftercooler will be given by,

However, p4′′ = p4.

As T4′′ < T4, v4′′ < v4

Therefore, the purpose of the aftercooler is to reduce the size of the receiver. The position of the intercooler and aftercooler is shown in Fig. 12.15.

Figure 12.15 Position of inter-cooler and after-cooler

12.8 ❐ INDICATED POWER OF A COMPRESSOR

where A = area of the cylinder, m2

L = length of stroke, m

N = rpm of compressor crank

pm = mean effective pressure of air, Pa (N/m2)

Theoretically, m.e.p. for a single-acting, single-stage compressor is,

where ηv = volumetric efficiency of compressor

Using indicator card, the m.e.p. is:

12.9 ❐ AIR MOTORS

The working of an air motor is similar to that of air compressor. High pressure air is admitted to the motor cylinder through a mechanically operated inlet valve and drives the piston in the forward direction. After a part of the stroke of the piston has been performed, the air supply is cut-off and the stroke is completed after decreasing pressure as the air expands in the cylinder. After the expansion stroke is completed, the air is allowed to escape into the atmosphere through a mechanically operated discharge valve. The return stroke is performed by compressed air acting on the other side of the piston in double-acting cylinder.

Figure 12.16 shows the pV diagram for the air motor with clearance. Process 4–1 is the intake with the high pressure p2 at cut-off point 1. From 1 to 2, the air expands from higher pressure p2 at 1 to atmospheric pressure p1 at 2, according to the polytropic law, pVn = c. The process 2–3 represents the exhaust.

Figure 12.16 p−V diagram for an air motor

Work obtained neglecting clearance,

Work done considering clearance,

where m1 = actual mass of air delivered per cycle.

12.10 ❐ INDICATOR DIAGRAM

The theoretical of pV diagram for the single-stage reciprocating compressor is represented by 1–2–3–4–1 in Fig. 12.17. The actual indicator diagram is 1–2′–3–4′–1. The difference between the actual and theoretical indicator diagrams is due to the intake and discharge losses. The intake losses include the friction losses in pipe, friction loss in inlet valve, and valve inertia loss. Theoretically, the inlet valve should open at 4, but actually, it opens little afterwards at 4′ due to the inertia of the valve, and the pressure inside the cylinder falls below the atmospheric pressure. The oscillating part of the curve indicates valve flutter due to vibration of spring loaded valve. On the delivery side, the discharge valve should open at 2, but actually, it opens little afterwards at 2′ due to the inertia of the spring loaded discharge valve. The effect of these losses is to increase the work required by the compressor.

Figure 12.17 Actual p−V diagram for single stage compressor

Figure 12.18 Actual p−V diagram for two-stage compressor

The actual indicator diagram for two-stage compressor with intercooler is shown in Fig. 12.18.

12.11 ❐ HEAT REJECTED

If the air is cooled to the initial temperature, then there is no change in internal energy per kg mass of air, and all the work done is rejected to the cooling medium party during the compression process and the remaining after compression at constant pressure.

Now, q1–2 = du + w1–2

However, du = 0

For a single-stage compressor, the heat rejected is given by work done - ∫ vdp.

= Heat rejected to the coolant in the intercooler + Heat rejected to the coolant during compression process
12.12 ❐ CONTROL OF COMPRESSOR

In order to balance the demand and supply of air, it is necessary to incorporate devices for the compressor control. The common methods of control are as follows:

  1. Throttle control
  2. Clearance control
  3. Blowing air to waste
  1. Throttle control: When the demand is less, there is build-up of pressure in the receiver, and the high pressure air from the receiver is led to piston and cylinder. The movement of piston is resisted by a spring. However, with excessive pressure, the piston depresses the spring, thus closing partly the suction valve. Therefore, during the suction stroke, the air intake is partly throttled. The reverse action takes place when the pressure in the receiver drops due to increase in demand.
  2. Clearance control: In this method, clearance pockets are provided for increasing or decreasing clearance. Therefore, the volumetric efficiency is reduced in proper proportion to control output.
  3. Blowing air to waste: In case of excessive pressure built-up in the receiver due to decrease in demand, a by-pass valve from the high pressure cylinder delivers air directly to atmosphere. When the pressure in the receiver drops, the relay piston closes the valve.

Example 12.1

A single-stage reciprocating air compressor is required to compress 72 m3 of air per minute from 15°C and 1.0 bar to 8 bar pressure. Find the temperature at the end of compression, work done, power, and heat rejected during each of the following processes: (a) isothermal, (b) adiabatic, and (c) polytropic compression following the law pV1.25 = constant.

Solution

  1. Isothermal compression:
  2. Adiabatic compression:

    Heat rejected = 0

  3. Polytropic compression:

Note: Example 9.1 illustrates that isothermal compression requires the minimum compression work, while isentropic compression requires the maximum for the same suction and delivery pressure.

Example 12.2

A double-acting reciprocating compressor with a piston displacement of 0.05 m3 per stroke operates at 500 rpm. The clearance is 5 per cent and it receives air at 100 kPa, discharges it at 600 kPa. The compression is polytropic, pV1.35 = c. Determine the power required and the air discharged in m3/s.

Solution

The p–V diagram for the compressor is shown in Fig. 12.19.

Va = Vc + VsV4
Vc = 0.05 × 0.05 = 0.0025 m3

Figure 12.19 p–V diagram

Example 12.3

A double-acting reciprocating compressor with complete intercooling delivers air to the main at a pressure of 30 atm, the suction condition being 1 atm and 27°C. If both cylinders have the same stroke, then find the ratio of the diameters of the cylinders for the efficiency of compression to be a maximum. Assume the index of compression to be 1.3.

Solution

Volume of L.P. cylinder = V1

Volume of H.P. cylinder = V3

L1 = L3

For maximum efficiency,

From constant pressure process 2–3: (see Fig. 12.12)

Example 12.4

A double-acting reciprocating compressor with perfect intercooling takes in air at 1 bar and 27°C. The law of compression in both the stages is pV1.3 = constant. The compressed air is delivered at 9 bar from the H.P. cylinder to an air receiver. Calculate per kg of air (a) the minimum work done, (b) the heat rejected to the intercooler, and (c) the minimum work done in a three-stage compressor working under the same conditions. Take cp = 1.005 kJ/kg.K.

Solution

T1 = 273 + 27 = 300 k

For 1 kg of air, p1v1 = RT1,

or v1 = 0.861 m3/kg

  1. The minimum work required in two-stage compressor with perfect intercooling is,

    The intermediate pressure is found to be

  2. Heat rejected to intercooler,

    q = cp (T2T3) = cp (T2T1) [∴T2 = T3]
    = 1.005(386.6 − 300) = 87 kJ/kg of air
  3. The least work done in case of three-stage air compressor working between the same pressure limits is given by,

Example 12.5

A single-stage, single-acting reciprocating compressor delivers 15 m3/min of free air from 1 bar to 8 bar at 300 rpm. The clearance volume is 6% of the stroke volume, and compression and expansion follow the law pV1.3 = constant. Calculate the diameter and stroke of the compressor. Take L = 1.5 D. The temperature and pressure of air at suction are the same as atmospheric air.

Solution

Clearance ratio, c = 0.06

Volumetric efficiency,

D = 0.3817 m or 38.17 mm
L = 1.5 × 0.3817 = 0.5726 m or 57.26 mm

Example 12.6

A two-cylinder, single-acting reciprocating air compressor is to deliver 15 kg of air per minute at 6.5 bar from suction conditions 1 bar and 13°C. Clearance may be taken as 4 per cent of stroke volume and the index for both compression and re-expansion as 1.3. The compressor is directly coupled to a four-cylinder, four-stroke petrol engine that runs at 1800 rpm with b.m.e.p. of 6 bar. Assuming a stroke–bore ratio of 1.1 for both engine and compressor and a mechanical efficiency of 85% for the compressor, calculate the required cylinder dimensions. Take R = 287 J/kgK.

Solution

The p−V diagram for the compressor is shown in Fig. 12.20.

Amount of air delivered per cylinder

Now, p1(V1V4) = mRT1

From the expansion curve, we have

V3 = 0.04 Vs
V4 = 0.1688 Vs

Now, V1 = V3 + Vs = 0.04 Vs + Vs = 1.04 Vs

V1V4 = (1.04 − 0.1688) Vs = 0.8712 Vs = 3.42 × 10−3
Vs = 3.9256 × 10−3 m3
D = 0.1656 m or 165.6 mm
L = 1.1 D = 0.1822 m or 182.2 mm

Figure 12.20 p-V diagram

Example 12.7

A multistage reciprocating compressor has to be designed to supply air at 135 bar, while atmospheric condition is 1.03 bar and 15°C. The value of compression index may be assumed as 1.35. Due to practical reasons, the intercoolers are not able to cool the air below 45°C, while the maximum temperature allowable in the system is 120°C. Calculate the number of stages that are necessary in the compression and the rate of cooling water circulated per kg of air. Take cp = 1 kJ/kg. K.

Solution

The p−V diagram for the compressor is shown in Fig. 12.21.

T1 = 273 +15 = 288 K, p1 = 1.03 bar, n =1.35

Since the maximum allowable temperature is 120°C, the temperature after first stage of compression,

T2 = 273 + 120 = 393 K

The intercoolers cool the air from T2 = 393 K to T3 = 273 + 45 =318 K. For the second stage, the inlet temperature is T3 = 318 K and outlet temperature is T4 = 393 K (see Fig. 12.10b).

Figure 12.21 p-V diagram

Pressure ratio

The pressure ratio is same for all the subsequent stage till the pressure reaches 135 bar. Hence, the number of stage i can be calculated from,

i ln 2.26 = ln 3.952
i = 4.51

Therefore, the minimum number of stages required including the first stage

= 4.51 + 1 = 5.51 ~ 6

Heat received by stage coolers per kg of air compressed

= cp (T2T3) (i − 1) = 1(393 − 318) × 5 = 375 kJ

Cooling water circulated per kg of air

Example 12.8

A single-acting, two-stage reciprocating compressor with complete intercooling delivers 10 kg/min of air at 16 bar. The suction occurs at 1 bar and 27°C. The compression and expansion processes are reversible with polytropic index n = 1.3. The compressor runs at 450 rpm. Calculate the following:

  1. The power required to drive the compressor.
  2. The isothermal efficiency.
  3. The free air delivered.
  4. The heat transferred in intercooler.
  5. The swept and clearance volumes for each cylinder if the clearance ratios for L.P. and H.P. cylinders are 0.04 and 0.06, respectively.

Solution

The pV diagram is shown in Fig. 12.22.

p1 = bar, p3 = 16 bar, T1 = 273 + 27 = 300 K, N = 450 rpm, n = 1.3, c1 = 0.04, and c2 = 0.06

For perfect intercooling,

Figure 12.22 p − V diagram for two-stage compressor

  1. Work done in two stages with perfect intercooling,
  2. Isothermal power,
  3. Free air delivered
  4. Heat transferred in intercooler with perfect intercooling

  5. Clearance volume = c1 (V1V3) = c1Vs = 0.04 × 0.0207

    = 8.285 × 10−4 m3
  6. H.P. stage:

    Clearance volume = c2Vs = 0.06 × 5.4 × 10−3 = 3.24 × 10−4 m3

Example 12.9

A reciprocating compressor is to be designed to deliver 4.5 kg/min of air from 100 kPa and 27°C to compress through an overall pressure ratio of 9. The law of compression is pV1.3 = constant. Calculate the saving in power consumption and gain in isothermal efficiency, when a two-stage compressor with complete intercooling is used in place of a single-stage compressor. Assume equal pressure ratio in both the stages of the two-stage compressor. Take R = 0.287 kJ/kg K.

Solution

Given: m = 4.5 kg/min, p1 = 100 kPa, T1 = 273 + 27 = 300 K, rp = 9, n = 1.3

Rate of work required in single-stage compression

Rate of work required in two-stage compression

Saving in power = 5.54 − 4.84 = 0.70 kW

Example 12.10

A double-acting, single-cylinder reciprocating air compressor has a piston displacement of 0.015 m3 per revolution, operates at 500 rpm, and has 5% clearance ratio. The air is received at 1 bar and delivered at 6 bar. The compression and expansion are polytropic with n = 1.3. If the inlet temperature of air is 20°C, determine (a) the volumetric efficiency, (b) the power required and (c) the heat transferred and its direction, during compression.

Solution

Given that Vs = 0.015 m3/rev, N = 500 rpm, c = Vc/Vs = 0.05

p1 = 1 bar, p2 = 6 bar, n = 1.3, T1 = 273 + 20 = 293 K

  1. Volumetric efficiency,
  2. Volumetric efficiency

    or Va = 0.015 × 0.8516 × 500 × 2 = 12.774 m3/min

    Power required,

    = 2834.54 kJ/min or 47.242 kW
  3. Mass of air sucked,

    Heat transferred during compression,

Example 12.11

Find the optimum intermediate pressure of a two-stage reciprocating compressor, if intercooling is done up to a temperature , which is greater than the inlet temperature.

Solution

Work done in the L.P. cylinder (Fig. 12.23)

Work done in the H.P. cylinder,

Total work done, W = W1 + W2

Figure 12.23 p − V diagram for two-stage compressor

For maximum work,

Example 12.12

A two-stage reciprocating air compressor takes in air at 1.013 bar and 15°C and delivers at 43.4 bar. The intercooler pressure is 7.56 bar. The intercooling is perfect and the index of compression is 1.3. Calculate the work done in compressing 1 kg of air. If both cylinders have the same stroke and the piston diameters are 9 cm and 3 cm and the volumetric efficiency of the compressor is 90%, will the intercooler pressure be steady or will rise or fall as the compressor continues working?

Solution

Given that p1 = 1.013 bar, T1 = 273 + 15 = 288 K

p3 = 43.4 bar, p2 = 7.56 bar, n = 1.3,

m = 1 kg, d1 = 9cm, d2 = 3 cm, ηv = 0.9

Input work per kg for perfect intercooling T3 = T1,

Compression in L.P. cylinder:

For constant pressure in the intercooler, the volume of air leaving the intercooler and entering the H.P. cylinder,

Ratio of effective cylinder volumes,

As more air is supplied to H.P. cylinder than it can hold and consequently the pressure in the intercooler will rise.

Example 12.13

Find the percentage saving in work done by compressing air in two stages from 1 bar to 7 bar instead of one stage. Assume compression index 1.35 in both the cases and optimum pressure and complete intercooling in two-stage compression.

Solution

Given that p1 = 1 bar, p3 = 7 bar, n1 = n2 = 1.35

For perfect intercooling, = 2.646 bar and T3 = T1

With compression in one stage (see Fig. 12.7) without clearance, work done per kg of air is given by,

Work done with perfect intercooling,

Example 12.14

A single-stage, double-acting reciprocating air compressor delivers 3 m3 of free air/min at 1.013 bar and 20°C to 8 bar with the following data:

Rotational speed = 300 rpm, mechanical efficiency ηmech = 0.9, pressure loss in passing through intake valve = 0.04 bar, temperature rise of air during suction stroke = 12°C, clearance volume = 5% of stroke volume. Index of compression and expression = 1.35, and length of stroke = 1.2 times of the cylinder diameter. Calculate: (a) volumetric efficiency, (b) cylinder dimension, (c) indicated power, and (d) isothermal efficiency of the compressor, take for air, R = 0.287 kJ/kg K.

Solution

Given that Va = 3 m3/min, pa = 1.013 bar, Ta = 273 + 20 = 293 K, p1 = 1.013 − 0.04 = 0.973 bar, T1 = Ta + 12 = 293 + 12 = 305 K, Vc = 0.05Vs, n = 1.35, L = 1.2D, R = 0.287 kJ/kg K, N = 300 rpm

The pV diagram is shown in Fig 12.24.

Mass of free air delivered,

Compression process 1–2:

Figure 12.24 p − V diagram for single stage compressor

Expansion process 3–4:

3 = c = 0.05s
14 = (c + 3) − 4 = (0.05s + s) − 0.238s = 0.812s

Corresponding value of F.A.D. per cycle,

  1. Volumetric efficiency,
  2. Volume of air inhaled per cycle,

    Stroke volume, Vs = = 0.94248D3 m3/cycle

    Also stroke volume = = 6.6756 ×10−3 m3/cycle

    ∴ 0.94248D3 = 6.6756 × 103

    D = 0.192 m or 192 mm

    L = 1.2 × 192 = 230.4 mm

  3. Indicated power, I.P. = ṁR (T2T1)
  4. Isothermal indicated power,

    Isothermal efficiency,

Example 12.15

The following data were obtained from a performance test of a 14 cm × 10 cm single stage reciprocating air compressor having 3 percent clearance: barometer 77 cm Hg, suction pressure 0 bar gauge, suction temperature 22°C, discharge pressure 4.05 bar gauge, discharge temperature 174°C, shaft rpm 1160, shaft power 350 kW, mass of air delivered per minute 1.75 kg. Determine (a) the actual volumetric efficiency; (b) the approximate indicated power; (c) the isothermal compression efficiency; (d) the mechanical efficiency and (e) the overall efficiency of the unit. Take patm = 1.031 bar.

Solution

Atmospheric pressure corresponding to 77 cm Hg

p1 = 0 + 1.0447 = 1.0447 bar

and p2 = 4.05 + 1.0447465 = 5.0947 bar

Air entering the cylinder

Piston displacement

  1. Volumetric efficiency
  2. ∴ Compression index n = 1.353

  3. Isothermal power
  4. Compressor mechanical efficiency
  5. Overall efficiency = isothermal efficiency × mechanical efficiency
    = 0.8023 × 0.836 = 67.07%

Example 12.16

A two-stage reciprocating compressor is used to compress from 1.0 bar to 16 bar. The compression is as per the law pV1.25 = const. The temperature of air at inlet of compressor is 300 K. Neglecting the clearance and assuming perfect intercooling find out the minimum indicated power to deliver 5 m3/min air measured at inlet conditions and find the intermediate pressure also.

Solution

Given: p1 = 1 bar v1 = 5m3/min = 0.083 m3/s, p3 = 16 bar, n = 1.25, T1 = 300 K

For complete intercooling, the intercooler pressure,

Work done in compressing the air,

Power required to drive the compressor is 26.519 kW.

Summary for Quick Revision

  1. A reciprocating compressor is a device in which air is compressed in a conventional cylinder with a closely fitted piston making reciprocating motion.
  2. Free air delivered (FAD) is the actual volume delivered by the compressor at the stated pressure, reduced to intake pressure and temperature and expressed in m3/min.
  3. The work done on the compressor to compress a given volume of air at a given pressure shall be the least when the compression process is isothermal.
  4. Approximation to isothermal compression can be achieved by spraying cold water on the cylinder, water jacketing the cylinder, intercooling by using multi-stage compression and by using external fins of the cylinder surface.
  5. Single stage compression:
    1. Required work and power.
      1. Without clearance:

        Power required in driving the compressor,

        Where N = rpm, if single acting

        = number of strokes, if double acting.
      2. With clearance:
    2. When referred to ambient conditions,

      where = clearance ratio, Vc= clearance volume, Vs = swept volume.

    3. Isothermal efficiency,
    4. Adiabatic efficiency,
    5. Main dimensions,
  6. Two-stage compression:
    1. For perfect intercooling,
    2. Minimum compression work, (Wt)min
    3. In general, with i number of stages,
    4. Minimum shaft work,
    5. Minimum indicated power with imperfect intercooling,
    6. Heat rejected to intercooler = mcp (T2 T3)
    7. Cylinder dimensions:
    8. Effectiveness of intercooler,
  7. Indicated power,
  8. Mechanical efficiency,
  9. Heat rejected per kg of air,
  10. A compressor can be controlled by: Throttle control, clearance control or by blowing air to waste.

Multiple-choice Questions

  1. Which one of the following statements is correct? In reciprocating compressors, one should aim at compressing the air
    1. adiabatically
    2. isentropically
    3. isothermally
    4. polytropically
  2. Consider the following statements:
    1. Reciprocating compressors are best suited for high pressure and low volume capacity.
    2. The effect of clearance volume on power consumption is negligible of the same volume of discharge.
    3. iii. While the compressor is idling, the delivery value is kept open by the control circuit.
    4. Intercooling of air between the stages of compression helps to minimize losses. Of these statements
    1. i and ii are correct
    2. i and iii are correct
    3. ii and iv are correct
    4. iii alone is correct
  3. For two-stage compressor in which index of compression for low pressure stage is m and for high pressure stage in n. The load sharing with perfect intercooling in expressed as:
  4. For a two-stage reciprocating compressor, compression from pressure p1 to p3 is with perfect intercooling and no pressure losses. If compression in both cylinders follows the same polytropic process and the atmospheric pressure is pa then the intermediate pressure p2 is given by
    1. p2 = (p1 + p3)/2
  5. A large clearance volume in reciprocating compressor results in
    1. reduced volume flow rate
    2. increased volume flow rate
    3. lower suction pressure
    4. lower delivery pressure
  6. In a reciprocating air compressor, the compression work per kg of air
    1. increases as clearance volume increases
    2. decreases as clearance volume increases
    3. is independent of clearance volume
    4. increases with clearance volume only of multistage compressor
  7. Consider the following statements:

    When air is to be compressed to reasonably high pressure, it is usually carried out by multistage compressor with an intercooler between the stages because

    1. work supplied is saved.
    2. weight of compressor is reduced.
    3. more uniform torque is obtained leading to the reduction in the size of flywheel.
    4. volumetric efficiency is increased.

      Of the four statements listed above

    1. 1 alone is correct
    2. 2 and 4 are correct
    3. 1, 2 and 3 are correct
    4. 1, 2, 3 and 4 are correct
  8. Consider the following statements:

    The volumetric efficiency of a compressor depends upon

    1. clearance volume
    2. pressure ratio
    3. index of expansion

      Of these statements:

    1. 1 and 2 are correct
    2. 1 and 3 are correct
    3. 2 and 3 are correct
    4. 1, 2 and 3 are correct
  9. The heat rejection by a reciprocating air compressor during the reversible compression process AB, shown in the temperature-entropy diagram, is represented by the area:
    1. ABC
    2. ABDE
    3. ABFG
    4. ABFOE
  10. A 3-stage reciprocating compressor has suction pressure of 1 bar and delivery pressure of 27 bar. For minimum work of compression, the delivery pressure of 1st stage is
    1. 14 bar
    2. 9 bar
    3. 5.196 bar
    4. 3 bar
  11. Consider the following factors:
    1. Cylinder size
    2. Clearance ratio
    3. Delivery pressure
    4. Compressor shaft power

    The factors which affect the volumetric efficiency of a single-stage reciprocating air compressor would include

    1. 1 and 2
    2. 3 and 4
    3. 2 and 3
    4. 1 and 4
  12. A four-stage compressor with perfect intercooling between stages, compresses air from 1 bar to 16 bar. The optimum pressure in the last intercooler will be
    1. 6 bar
    2. 8 bar
    3. 10 bar
    4. 12 bar
  13. The clearance volume of a reciprocating compressor directly affects
    1. piston speed
    2. noise level
    3. volumetric efficiency
    4. temperature of air after compression
  14. The capacity of an air compressor is specified as 3m3/min. It means that the compressor is capable of
    1. supplying 3 m3 of compressed air per minute
    2. compressing 3 m3 of free air per minute
    3. supplying 3 m3 of compressed air at NTP
    4. compressing 3 m3 of standard air per minute
  15. A two-stage compressor takes in air at 1.1 bars and discharges at 20 bars. For minimum work input, the intermediate pressure is
    1. 10.55 bars
    2. 7.33 bars
    3. 5.5 bars
    4. 4.7 bars
  16. Consider the following statements:

    The volumetric efficiency of a reciprocating compressor can be enhanced by

    1. heating the intake air
    2. decreasing the clearance volume
    3. cooling the intake air

    Which of these statements is/are correct?

    1. 1 alone
    2. 1 and 2
    3. 2 and 3
    4. 3 alone
  17. Reciprocating compressors are provided with
    1. simple disc/plate valve
    2. poppet valve
    3. spring-loaded disc value
    4. solenoid valve
  18. If n is the polytropic index of compression and is the pressure ratio for a three-stage compressor with ideal intercooling, the expression for total work of three stage is
  19. Consider the following statements:

    Volumetric efficiency of a reciprocating air compressor increases with

    1. increase in clearance ratio
    2. decrease in delivery pressure
    3. multistaging

    Which of the statements given above is/are correct?

    1. Only 1 and 2
    2. Only 2 and 3
    3. Only 3
    4. 1, 2 and 3
  20. What is the preferred intercooler pressure for a two stage air compressor working between the suction pressure ps and the delivery pressure pd?
  21. Which of the following statements are correct for multi-staging in a reciprocating air compressor?
    1. It decreases the volumetric efficiency
    2. The work done can be reduced
    3. A small high-pressure cylinder is required.
    4. The size of flywheel is reduced.

    Select the correct answer using the code given below:

    1. 1, 2 and 3
    2. 2, 3 and 4
    3. 1, 3 and 4
    4. 1, 2 and 4
  22. For a two-stage reciprocating air compressor, the suction pressure is 1.5 bar and the delivery pressure is 54 bar. What is the value of the ideal intercooler pressure?
    1. 6 bar
    2. 9 bar
    3. 27.75 bar
  23. Consider the following statements:

    In a reciprocating compressor, clearance volume is provided

    1. so that piston does not hit and damage the valves
    2. to account for differential thermal expansion of piston and cylinder
    3. to account for machining tolerances
    4. to achieve isentropic compression

    Which of these statements are correct?

    1. 1, 2 and 3
    2. 1, 2 and 4
    3. 1, 3 and 4
    4. 2, 3 and 4
  24. Which of the following can be the cause/causes of an air-cooled compressor getting overheated during operation?
    1. Insufficient lubricating oil.
    2. Broken valve strip.
    3. Clogged intake filter.

    Select the correct answer using the code given below:

    1. Only 3
    2. Only 1 and 2
    3. Only 2 and 3
    4. 1, 2 and 3
  25. Performance of a reciprocating compressor is expressed by

Explanatory Notes

  1. 10. (d)
  2. 12. (b) Optimum pressure in the last intercooler
  3. 15. (d)
  4. 22. (b)

Review Questions

  1. List at least six uses of compressed air in industry.
  2. Explain the working of a single stage reciprocating air compressor.
  3. What do you mean by free air delivered (FAD)?
  4. Define volumetric efficiency.
  5. What are the factors on which volumetric efficiency depends?
  6. What are the methods generally adopted for approximating the compression process of a reciprocating air compressor as isothermal one?
  7. Define clearance ratio. How the volumetric efficiency is dependent on clearance ratio?
  8. Define isothermal efficiency and compressor efficiency.
  9. Differentiate between isothermal power and indicated power.
  10. Define adiabatic efficiency.
  11. What are the advantages of multi-stage compression?
  12. What is the condition for minimum compressor work with perfect intercooling for a multi-stage reciprocating air compressor?
  13. Distinguish between the function of intercooler and after-cooler.
  14. What is an air motor?
  15. How heat is rejected in air compressor?

Exercises

12.1 A single-stage single-acting reciprocating air compressor delivers 15 m3 of free air per minute from 1 bar to 8 bar at 300 rpm.The index of both compression and expansion is n = 1.3 for and clearance of swept volume, find the diameter and stroke of the compressor. Take L = 1.5 D.

[Ans. 383 mm, 587.5 mm]

12.2 A single-stage double-acting reciprocating air compressor delivers air at 7 bar. The pressure and temperature at the end of suction stroke are 1 bar and 27°C. It delivers 2 m3 of free air per minute when the compressor is running at 300 rpm. The clearance volume is 5% of the stroke volume. The ambient pressure and temperature are 1.03 bar and 20°C. Index of compression and expansion is 1.30 and 1.35 respectively. Calculate (a) the volumetric efficiency, (b) indicated power and brake power if mechanical efficiency is 80%, and (c) diameter and stroke of the cylinder if both are equal.

[Ans. 83.9%, 8.5 kW, 10.64 kW, 174.6 mm]

12.3 A two-stage single-acting reciprocating air compressor delivers air at 20 bar. The pressure and temperature of air before compression in L.P. cylinder are 1 bar and 27°C. The discharge pressure of L.P. cylinder is 4.7 bar. The pressure of air leaving the intercooler is 4.5 bar and the air is cooled to 27°C. The diameter and stroke of L.P. cylinder are 0.4 m and 0.5 m respectively. The clearance volume is 4% of stroke in both cylinders. The speed of compressor is 200 rpm. The index of compression and re-expansion in both cylinders is 1.3. Determine (a) the indicated power to run the compressor, and (b) the heat rejected to intercooler per minute.

[Ans. 68.9 kW, 1716 kJ/min]

12.4 A single-stage, double-acting reciprocating air compressor delivers 15 m3 of air per minute measured at 1.013 bar, 27°C and delivers at 7 bar. At the end of the suction stroke the pressure and temperature are 0.98 bar and 40°C. The clearance volume is 4% of the swept volume and the stroke is 1.3 times the bore. The compressor runs at 300 rpm. Calculate (a) the volumetric efficiency, (b) cylinder dimensions, (c) indicated power and (d) isothermal efficiency. n = 1.3 for both compression and expansion. Take R = 0.287 kJ/kg.K

[Ans. 79.6%; 313.3 mm, 407.3 mm; 65.83 kW; 78.92%]

12.5 A single-stage, double acting air reciprocating compressor takes air at 0.98 bar and 32°C and delivers at 6.32 bar. The clearance is 5% of the stroke volume. The compression and expansion follow the law pV1.32 = c. The air handled by the compressor is 17 m3/min when measured at 1 bar and 15°C. Determine the temperature of air delivered, the stroke volume and the indicated power of compressor if it turns at 500 rpm. Neglect the area of piston and take R = 0.287 kJ/kg.K.

12.6 The free air delivered of a single cylinder single stage reciprocating air compressor is 2.5 m3/min. The ambient air is at 0°C and 1.013 bar and delivery pressure is 7 bar. The clearance ratio is 5% and law of compression and expansion is pv1.25 = c. If L =1.2 D and the compressor runs at 150 rpm, determine the size of the cylinder.

[Ans. 275.5 mm, 330 mm]

12.7 A single stage double acting reciprocating air compressor delivers air at 7 bar. The amount of free air delivered is 2 m3 at 300 rpm. The pressure and temperature at the end of suction stroke are 1 bar and 27°C. The ambient conditions are 1.03 bar and 20°C. The clearance is 5% stroke. The compression and re-expansion follow the law pV1.3 = const. Determine the brake power required to run the compressor if the mechanical efficiency is 80%, and the diameter and stroke of the cylinder if both are equal.

[Ans. 10.8kW, 179.5 mm]

12.8 A two-stage double acting reciprocating air compressor delivers air at the rate of 1.35 kg/s. The suction pressure is 1 bar and inter-stage pressure is 7 bar and delivery pressure 42 bar. Air enters the L.P. cylinder at 17°C and cooled in the intercooler to 32°C. The clearances in L.P. and H.P. cylinders are 6% and 8% of respective strokes. The law of compression and re-expansion is pV1.21 = c in both cylinders. The compressor runs at 500 rpm. Calculate the amount of cooling water required per minute in intercooler, if rise in temperature of water is limited to 20°C, power required, and size of cylinder if length of stroke and bore are equal.

[Ans. 2259 kg/min, 1068 kW, 483 mm]

12.9 Determine the size of a cylinder for a double acting reciprocating air compressor of 37 kW, in which air is drawn in at 1 bar, 15°C and compressed according to the law pV1.2 = const. to 6 bar. The compressor runs at 100 rpm with average piston speed of 152.5 m/min. Neglect clearance.

[Ans. 298.5 mm, 762.5 mm]

12.10 A single stage double acting reciprocating air compressor running at 500 rpm handles 17 m3/min of air, measured at 1 bar and 15°C. The pressure and temperature at the end of suction are 0.98 bar and 32°C. The air is delivered at 6.325 bar. Assuming a clearance factor of 5% and the compression and expansion processes to follow the law pV1.32 = const, determine the stroke volume of the compressor. Also calculate the indicated power of the compressor. Take R = 0.287 kJ/kg.K

[Ans. 0.2175 m3, 70.73 kW]

12.11 A single acting compressed air motor works on compressed air at 10.5 bar and 37°C supplied at the rate of 1 kg/min. Cut-off takes place at 25% of the stroke and the expansion follows the adiabatic frictionless law down to 1.0135 bar. Determine the mean effective pressure, the indicated power and the cylinder volume if the motor runs at 250 rpm. Neglect clearance.

[Ans. 3.37 bar, 1.94 kW, 0.001383 m3]

12.12 The cylinder of an air motor has a bore of 63.5 mm and a stroke of 114 mm. The supply pressure and temperature are 6.3 bar and 24°C and exhaust pressure is 1.013 bar. The clearance volume is 5% of the swept volume and the cut-off ratio is 0.5. The air is compressed by the returning piston after it has travelled through 0.95 of its stroke. The law of compression and expansion is pV1.3 = const. Calculate the temperature at the end of expansion and the indicated power of the motor which runs at 300 rpm. Also calculate the air supplied per minute. Take R = 0.287 kJ/kgK.

[Ans. 244.4 K, 0.746 kW, 0.418 kg/min]

12.13 A water cooled air compressor requires a work input of 200 kJ/kg of air delivered. The enthalpy of air leaving the compressor is 75 kJ/kg greater than that entering. Heat lost to the cooling water is 105 kJ/kg. From the first law analysis estimate the heat lost by compressor to atmosphere.

12.14 A two-stage single acting air reciprocating compressor equipped with an intercooler draws air at 1.013 bar and 30°C and delivers it at 20 bar. The mass of air delivered by the compressor is 10 kg/min. The ratio of clearance volume to swept volume in the low pressure and high pressure cylinders is 0.03 and 0.04 respectively. If the compressor has a mechanical efficiency of 82% and it operates under most efficient conditions, determine:

  1. power supplied at the shaft of the compressor;
  2. isothermal efficiency;
  3. ratio of cylinder diameters for identical stroke length; and
  4. heat rejected in the intercooler.

Take R = 0.287 kJ/kg.K, cp = 1.005 kJ/kgK and γ = 1.4.

12.15 A single acting two-stage reciprocating air compressor is to compress air from 1 bar and 30°C to 12 bar. The base of the low pressure cylinder is 30 cm. The stroke length of both the low and the high pressure cylinders is the same and is equal to 40 cm. The compressor runs at 180 rpm. The clearance volume in both the cylinders is 3% of the stroke volume. Index of compression and expansion is 1.3 in both the cylinders. Determine the shaft power required to drive the compressor when (a) the air is cooled to its original temperature before entering the H.P. cylinder, (ii) when the air is cooled to 45°C in the intercooler. Assume mechanical efficiency to be 85% in both cases. Take R = 0.287 kJ/kg.K.

12.16 A multi-stage reciprocating compressor has to be designed to supply air at 135 bar, while atmospheric condition is 1.03 bar and 15°C. The value of compression index may be assumed as 1.35. Due to practical reasons the intercoolers are not able to cool the air below 45°C, while the maximum temperature allowable in the system is 120°C. Calculate the number of stages that are necessary in the compression and the rate of cooling water circulated per kg of air. Take cp = 1 kJ/kgK.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS
  1. c
  2. b
  3. a
  4. b
  5. d
  6. a
  7. a
  8. d
  9. b
  10. d
  11. a
  12. b
  13. c
  14. b
  15. d
  16. c
  17. a
  18. d
  19. d
  20. c
  21. b
  22. b
  23. a
  24. c
  25. a