# Chapter 15 Axial Flow Air Compressors – Thermal Engineering

## Axial Flow Air Compressors

##### 15.1 ❐ INTRODUCTION

In an axial flow compressor, the air flows throughout the compressor parallel to its axis. Axial flow compressor has many merits over centrifugal compressor and is particularly suitable for super charging of internal combustion engines, turbojets, and gas turbines.

##### 15.2 ❐ CONSTRUCTIONAL FEATURES

The cross-section of a typical axial flow compressor is shown in Fig. 15.1. It consists of adjacent rows of moving and fixed blades. The moving blades are mounted on the rotating drum and the fixed blades are fixed to the casing or stator. One stage of the compressor comprises a row of moving blades, followed by a row of fixed blades.

The blades belong to the aerofoil section designed on the basis of aerodynamic theory. Such a design prevents losses due to shock and turbulence and is free from stalling problems. The blades are said to be stalled when the air stream fails to follow the blade profile.

Figure 15.1 Typical axial flow compressor

The blades of a reaction gas turbine have profile formed by a number of circular arcs. This occurs because the acceleration process carried out in the converging blade passages of a reaction turbine is much more efficient and stable as against the diffusing or decelerating process carried out in the diverging passage between the blades of an axial flow compressor.

To keep the flow velocity constant throughout the compressor length, the annular area is reduced from inlet to outlet of the compressor. In the diverging passage of the moving blades, there is a rise in temperature due to diffusion and the absolute velocity is also increased due to work input.

The fixed blades serve to convert a part of the kinetic energy of air into pressure energy by diffusion and also guide and redirect the air stream to enter the next stage without shock.

##### 15.3 ❐ WORKING PRINCIPLE

The axial compressor is generally driven by an internal combustion engine or a turbine. The work input to the rotor shaft is transferred by the rotor moving blades to the air, thus accelerating it. The space between the moving blades form diffuser passages so that the velocity of air relative to the blades is decreased as the air passes through them and consequently, the pressure rises. The air is further diffused in the fixed blades which also form diffuser passages. In the fixed blades, the air is turned through an angle so that its direction is such that it can pass to the next ring of moving blades without shock. Generally, 5 to 14 stages are used.

##### 15.4 ❐ SIMPLE THEORY OF AEROFOIL BLADING

When an object is placed within a fluid stream, a force is exerted on the object which may be inclined to the flow direction. The component of this force parallel to the direction of flow is called the drag and the normal component is called the lift, as shown in Fig. 15.2.

Consider an aerofoil section as shown in Fig. 15.3, which has an angle of attack α, the chord length C, and the length of the foil perpendicular to the plane of the paper, termed as the span, l. The resultant force F executed by the fluid on the aerofoil has two mutually perpendicular components: the lift FL and the drag FD. Consider a surface element of elementary area dA of aerofoil on which pressure p acts. Let the tangent to dA make an angle θ with the flow direction. The differential lift and drag on this area element are as follows:

dFL = pdA cos θ
dFD = pdA sin θ

Figure 15.2 Definition sketch of lift and drag

Figure 15.3 Pressure distribution on an aerofoil section

Lift, FL = ∫A pdAcosθ

Drag, FD = ∫A pdAsinθ

where ∫A represents the integration over the entire body surface area A.

We define the following lift and drag coefficients:

where ρ = density of air stream

U = free stream velocity of air

A = projected area of the body perpendicular to the oncoming flow, i.e., the frontal area of the body.

Let

= free air delivered by the compressor, m3/min

p1, T1 = ambient air conditions

Dm = blade ring mean diameter, m

N = rotor speed, rpm

CL, CD = lift and drag coefficients respectively

A = projected area of blades, m2

Then density of air,

= vaA
A = kbπDmh

Figure 15.4 Velocity triangle

From the velocity triangle as shown in Fig. 15.4, we get

Power input/stage,

where and ηc = efficiency of the compressor

##### 15.5 ❐ VELOCITY DIAGRAMS

Figure 15.5 shows the velocity triangles for one stage of an axial flow compressor. All angles are measured from the axial direction. The blade velocity u is taken to be same at blade entry and exit as the air enters and leaves the blades at almost equal radii. Due to the diffusion process in the moving blades, vr2 < vr1 but va2 > va1. The air leaves the fixed blades with velocity va3 at an angle α3 and is redirected to the next page. It is assumed that va1 = va3.

From the inlet velocity triangle, we have

Figure 15.5 Velocity diagrams for axial flow compressor

From the outlet velocity triangle, we have

Assuming 1 kg flow of air through the compressor stage, we have

Tangential force per kg of air = vw2vw1

Work required by the stage per kg of air,

Now vf1 = vf2 = vf

Also, isentropic efficiency of the stage,

##### 15.6 ❐ DEGREE OF REACTION

The degree of reaction, Rd, is defined as the ratio of pressure rise in the rotor blades to the pressure rise in the compressor stage.

Pressure rise in the rotor blades

Pressure rise in the compressor stage

For 50% degree of reaction,

From Eqs (15.8) and (15.9), we have

Therefore, with 50% degree of reaction, the blades are symmetrical. This reduces the friction losses considerably.

##### 15.7 ❐ PRESSURE RISE IN ISENTROPIC FLOW THROUGH A CASCADE

Consider the incompressible isentropic and steady flow through a cascade from condition 1 to condition 2. Using Bernoulli’s equation, we have

Now, v f1 = v f2 = v f

##### 15.8 ❐ POLYTROPIC EFFICIENCY

It is the isentropic efficiency of one stage of a multistage compressor. This small-stage efficiency is constant for all stages of a compressor with infinite number of stages.

Consider the compression process of a multistage compressor on a T-s diagram as shown in Fig. 15.6 from total pressure p01 to p02 in four stages of equal pressure ratio with intermediate pressure p0a, p0b, and p0c.

Overall isentropic stagnation efficiency of machine is,

Stagnation isentropic efficiency for the stage,

Total actual temperature rise,

Figure 15.6 Concept of polytropic efficiency

Now (ΔT0) m/c = (1 − a) + (ab) + (bc) + (c − 2)

Σ (dT0)st = (1 − a) + (a′b′′) + (b′c′′) + (c′ − 2′′)

On the T-s plot, the constant pressure lines diverge towards the right, therefore,

(a′b′′) > (ab)
(b′c′′) > (bc)

and so on.

Therefore, we can say that

Σ (dT0)st > (ΔT0) m/c

ηisen (st) > ηisen (m/c)

The small stage efficiency ηisen (st) which is constant for all stages is called polytropic efficiency and is denoted by ηp.

Let the law of compression for the irreversible adiabatic path 1 − 2′ be,

and for the isentropic path 1−2 is,

Eq. (15.20) can be written as:

Differentiating, we get

Differentiating, we get

Eq. (15.21) becomes

Similarly, for the ideal compression process we have

From Eq. (15.23), we have

Integrating between the end states 1 and 2′, we get

##### 15.9 ❐ FLOW COEFFICIENT, HEAD OR WORK COEFFICIENT, DEFLECTION COEFFICIENT, AND PRESSURE CO-EFFICIENT

Flow Coefficient: It is defined as the ratio of flow (or axial) velocity at inlet to the blade velocity. Mathematically, we can write

Flow coefficient,

Now u = vf1 (tan α1 + tan β1)

Also, vf1 = vf2 = vf

Head or Work Coefficient: It is defined as the ratio of actual work done to the kinetic energy corresponding to the mean peripheral velocity. One can write

Deflection Coefficient: ϕdef: It is defined as,

Pressure coefficient: It is defined as the ratio of isentropic work done to kinetic energy corresponding to the peripheral velocity.

##### 15.10 ❐ PRESSURE RISE IN A STAGE AND NUMBER OF STAGES

The static pressure ratio from Eq. (15.15) is,

Pressure rise in stationary blades is,

Pressure increase in stage is,

The stagnation pressure ratio is,

If the work done per stage is assumed to be the same, then number of stage z is given by,

If pressure ratio per stage is the same, then

The overall pressure ratio is,

Note that the pressure rise per stage in axial-flow compressor is less than that of a centrifugal compressor. This is because of no centrifugal action in axial flow compressor.

##### 15.11 ❐ SURGING, CHOKING, AND STALLING
1. Surging: The delivery pressure vs mass flow rate in an axial flow compressor is shown in Fig. 15.7. The mass flow is zero when the discharge valve is closed, but the air entrapped in the blade area is compressed and pressure of air increases. The condition is represented by point ‘a’ in Fig. 15.7. As the valve is opened, the flow of air starts and the pressure rises upon point ‘b’.

Any increase in air mass flow rate after point ‘b’ is accompanied by a decrease in delivery pressure. This happens because the rate of increase in pressure loss due to friction is more than the rate of increase in pressure rise due to diffuser. Theoretically, the decrease in delivery pressure is continued up to point ‘e’. In practice, the maximum mass flow rate is limited by point ‘c’ because beyond point ‘c’, the mass flow exceeds design mass flow, the air angles are widely different from vane angles, and choking takes place.

If the compressor is working between points ‘a’ and ‘b’, as the mass flow is decreased, the delivery pressure also decreases. However, the pressure in the downstream side of the compressor does not fall quickly, resulting in reversal of flow from downstream side towards the resulting pressure gradient. When this occurs, the pressure to the downstream side of the compressor falls and the compressor will start delivering air and the cycle would be repeated continuously. Thus, the flow would be pulsating between point ‘a’ and ‘b’. This pulsating air flow phenomena is known as ‘surging’. The surging causes overheating and stress reversal in the blades and damages the compressor. Within the region bc, the flow is stable. A fall in mass flow rate will result in rise in delivery pressure to restore to fall.

Figure 15.7 Characteristic curve

2. Choking: The maximum mass flow rate possible in a compressor is known as choking. The point ‘e’ in Fig. 15.7 represents the choking of the compressor.
3. Stalling: It is a phenomena which occurs on the suction side of the compressor in which the flow breaks away from the aerofoil blading. It may be due to lesser flow rate than the designed value or due to non-uniformity in the profile of the blades. Thus, stalling precedes surging. Stalling is a local phenomenon, whereas surging is a complete system phenomenon.
##### 15.12 ❐ PERFORMANCE CHARACTERISTICS

The relationship between pressure rise and mass flow rate at various speeds is shown in Fig. 15.8. At a certain speed, efficiency increases as the mass flow rate increases and reaches a maximum value after which it decreases, as shown in Fig. 15.9. The power consumed increases as mass flow increases. Figure 15.10 shows the performance and constant efficiency curves. The performance curves are plotted with dimensionless parameters, , and At constant value of there is a considerable narrower range of mass flow rate than in centrifugal compressor. The surge point is reached in axial compressors much before the maximum pressure ratio for given value of . Since the design usually calls for the operating line to be near the maximum point on the curves, it follows that the operating line for axial flow compressors must be very near the surge line, thus narrowing the range of stable operation.

Figure 15.8 Pressure ratio v’s mass flow rate

Figure 15.9 Power and efficiency v’s mass flow rate

Figure 15.10 Pressure rise v’s volume flow

##### 15.13 ❐ COMPARISON OF AXIAL FLOW AND CENTRIFUGAL COMPRESSORS

The merits and demerits of axial flow compressors are given in Table 15.1.

Table 15.1 Merits and demerits of axial flow compressor over centrifugal compressor

##### 15.14 ❐ APPLICATIONS OF AXIAL FLOW COMPRESSORS

Axial flow compressors have wide applications in the following:

1. Jet engines
2. Gas turbines for power plants
3. Steel mills
4. Aircraft applications.

Note that the requirements of compressors for aircraft application:

1. High efficiency
2. Smaller frontal area
3. Suitability for multistaging
4. High starting torque
5. Lesser drag
##### 15.15 ❐ LOSSES IN AXIAL FLOW COMPRESSORS

The following losses occur in axial flow compressors:

1. Profile losses on the surface of blades
2. Skin friction loss on the annulus walls
3. Secondary flow losses

#### Example 15.1

An axial compressor is fitted with half-reaction blading, the blade inlet and outlet angles being 50° and 15° when measured from the axial direction. The mean diameter of a certain blade pair is 85 cm and the rotor runs at 5500 rpm. Calculate the necessary isentropic efficiency of the stage if the pressure ratio of compression is to be 1.4 and the inlet air temperature is 25°C. Take cp =1.005 kJ/kg. K, γ = 1.4.

[CSE, 1987]

Solution

β1 = 50°, β2 = 15°, vf1 = vf2 = vf, α1 = 15°, α2 = 50°

From the velocity triangles, (Fig. 15.11)

vw1 = 0.1826 u
vw2 = 0.8164 u
Δvw = vw2vw1 = 0.6328 u = 0.6328 × 244.78 = 155 m/s

Work done/kg, wa = u × ∆vw = 244.78 × 155 = 37940.9 N.m/kg

Isentropic work, wisen = cp (T2T1)

Figure 15.11 Velocity diagrams for axial flow compressor

wisen = 1.005 (328 − 298) = 30.15 kJ/kg

Isentropic efficiency,

#### Example 15.2

An axial flow compressor comprises a number of similar stages with equal work done per stage, and the velocity of flow is uniform throughout the compressor. The following data is given:

Overall stagnation pressure ratio = 4

Stagnation inlet temperature = 330 K

Relative air angle at rotor inlet = 130°

Relative air angle at rotor outlet = 100°

Degree of reaction = 0.5

Overall stagnation adiabatic efficiency = 0.86

Calculate (a) the stagnation outlet temperature, and (b) number of stages.

Solution

The velocity diagrams are shown in Fig. 15.12.

Figure 15.12 Velocity diagrams for axial flow compressor

T02′ = 516.5 K

For degree of reaction = 0.5, the velocity diagrams are similar.

uvw2 = vw1

vw1 = 0.1737 u
vw2 = uvw1 = 0.8263 u
Δvw = vw2vw1 = 0.6526 u

Work done per stage = u · ∆vw = 180 × 0.6526 × 180 = 21144 N.m/kg

Total work = cp (T02T01) = 1.005(490.38 − 330) = 161.182 kJ/kg

Number of stages = = 7.62 = 8

#### Example 15.3

Air at a temperature of 300 K enters a 10-stage axial flow compressor at the rate of 3.5 m/s. The pressure ratio is 6.0 and the isentropic efficiency is 90%. The process is adiabatic and the compressor has symmetrical stages. The axial velocity is uniform across the stage and equals to 120 m/s and the mean blade speed of each stage is 200 m/s. Assume that the temperature change is same in each stage.

Determine the direction of the air at entry to and exit from the rotor and the stator blades. Also, find the power given to the air. Take cp = 1.005 kJ/kg.K and γ = 1.4.

[IES, 2001]

Solution

The velocity diagrams are shown in Fig. 15.13.

z = 10, T1 = 300 K, = 3.5 kg/s, rp = 6, ηi = 0.9,
vf = 120 m/s, u = 200 m/s

Figure 15.13 Velocity diagrams for axial flow compressor

Increase in temperature per stage,

tan β1 = 1.2999, β1 = 52.43°, β2 = 20.14°

Power supplied = zṁcp (∆T)stage = 10 × 3.5 × 1.005 × 22.284 = 783.84 kW

#### Example 15.4

In an axial flow air compressor: u = 250 m/s, vf = 200 m/s, α1 = 50°, α2 = 15° and ρ = 1 kg/m3. Determine (a) the pressure rise, and (b) the work done per kg of air.

Solution

1. Pressure rise through a moving blade ring,
2. Work done per kg of air,
w = u (vw1vw2) = uvf (tan α1 − tan α2)
= 250 × 200 (tan 50° − tan 15°) × 10−3 = 46.19 kW

#### Example 15.5

An axial flow air compressor having eight stages and 50% degree of reaction compresses air in the pressure ratio of 4.5 : 1. Air enters the compressor at 27°C and flows through it with a constant speed of 100 m/s. The moving blades of compressor rotate with a mean speed of 200 m/s. The isentropic efficiency of the compressor is 85%. Calculate (a) the work done on the compressor and (b) blade angles. Take γ = 1.4 and cp = 1.005 kJ/kg.K for air.

Solution

z = 8, Rd = 0.5, rp = 4.5, T1 = 273 + 27 = 300 K, vf = 100 m/s, u = 200 m/s, ηisen = 0.85

1. Work required per kg of air,
w = cp (T2′T1) = 1.005 (489.4 − 300) = 190.36 kJ/kg
2. Total work done = zu (vw2vw1) = zu vf (tan α2 − tan α1)
190.36 = 8 × 200 × 100 (tan α2 − tan α1) × 10−3
tan α2 − tan α1 = 1.1897

For Rd = 0.5, α1 = β2 and α2 = β1

∴ 1.1897 = tan β1 − tan β2

or tan β1 − tan α1= 1.1897

Now, u = vf (tan α1 + tan β1)

Adding the above two equations, we get

2 tan β1 = 3.1897

or tan β1= 1.5949

or β1 = 57.9° = α2

tan α1 = tan β1 − 1.1897 = 0.4052

or α1 = 22° = β2

#### Example 15.6

The overall isentropic efficiency of an axial flow air compressor is 83%. It draws air at 25°C and 1 bar and compresses to 4 bar. Assuming 50% degree of reaction, blade velocity 180 m/s, exit angle from stator (α1) = 15°, inlet angle to rotor (β1) = 45° and work input factor as 0.80, calculate the flow velocity and the number of stages.

Solution

ηisen = 0.83, T1 = 273 + 25 = 298 K, p1 = 1 bar, p2 = 4 bar, Rd = 0.5, α1 = 15°, β1 = 45°, u = 180 m/s, ϕw = 0.80

Work required per kg of air

= cp (T2T1) = 1.005 (472.5 − 298) = 175.37 kJ/kg

From inlet velocity triangle (Fig.15.13), we have

For Rd = 0.5, α2 = β1 = 45° and α1 = β2 = 15°

vw2 = vf tan α2 = 141.96 tan 45° = 141.96 m/s
vw1 = vf tan α1 = 141.96 tan 15° = 38.04 m/s

Work done per kg per stage = ϕwu(vw2vw1)

= 0.80 × 180 (141.96 − 38.04) × 10−3 = 14.965 kJ/kg

Number of stages, z = = 11.72 = 12

#### Example 15.7

In a 10-stage axial flow air compressor, the overall stagnation pressure ratio is 4 : 1 with an overall isentropic efficiency of 90%. The inlet stagnation temperature and pressure are 293 K and 1 bar. The work is divided equally between the stages. The mean blade speed is 180 m/s and 50% reaction blading are used. The flow velocity throughout the compressor is 90 m/s. Calculate (a) the blade angles, and (b) the power required.

Solution

z = 10, rop = 4, ηisen = 0.9, T01 = 293 K, p01 = 1 bar, u = 180 m/s, Rd = 0.5, vf = 90 m/s

1. For Rd = 0.5, α1 = β2 and α2 = β1

With isentropic compression, the temperature of air leaving the compressor is,

Work required by the compressor = cp (T02′T01)

= 1.005 (451.2 − 293) = 159 kJ/kg
= zu (vw2vw1)
= zuvf (tan α2 − tan α1)

From inlet velocity triangle, we have,

Adding the above two equations, we get

2 tan β1 = 2.9815

or tan β1 = 1.4908

or β1 = 56.15° = α2

tan α1 = 2 − 1.4908 = 0.5092

or α1 = 26.98° = β2

2. Power required = cp(T02′T01) = 1 × 1.005 (451.2 − 293) = 159 kW

#### Example 15.8

The overall stagnation pressure ratio and isentropic efficiency for an axial flow air compressor are 5 and 82%, respectively. The inlet stagnation pressure and temperature are 1 bar and 303 K. The mean blade speed is 200 m/s and degree of reaction is 0.5. The relative air angles are 12° and 30° at the rotor inlet and outlet, respectively. The work input factor is 0.9. Calculate (a) the stagnation polytropic efficiency, (b) the number of stages, (c) the inlet temperature and pressure, and (d) the blade height in the first stage if the hub-tip ratio is 0.42 and mass flow rate is 20 kg/s.

Solution

rop = 5, ηisen = 0.82, p01 = 1 bar, T01 = 303 K, u = 200 m/s, Rd = 0.5, β1 = 30° = α2, β2 = 12° = α1, ϕw = 0.9, = 20 kg/s, hub-tip ratio = 0.42.

1. Polytropic efficiency,

2. From the inlet velocity triangle (Fig. 15.13), we have
vw2 = vf2 tan α2 = 253.2 tan 30° = 146.2 m/s
vw1 = vf tan α1 = 253.2 tan 12° = 53.8 m/s

Work consumed per stage = ϕw u(vw2vw1)

= 0.9 × 200 (146.2 − 53.8) × 10−3 = 16.632 kJ/kg

Total work consumed by the compressor = cp (T0zT01)

= 1.005 (518.7 − 303) = 216.78 kJ/kg

Number of stages,

3. = 258.86 m/s

Static temperature,

4. = 0.859 kg/m3
m = ρ1A1vf

or 20 = 0.859 × π [1 − (0.42)2] × 253.2

or r1 = 0.1885 m or 188.5 mm

#### Example 15.9

A multistage axial flow air compressor delivers 25 kg/s of air. The inlet stagnation condition is 1 bar and 20°C. The power consumed by the compressor is 4500 kW. Calculate (a) the delivery pressure, (b) the number of stages, and (c) the overall isentropic efficiency of the compressor.

Assume that temperature rise in the first stage is 15° C, the polytropic efficiency of compression is 0.90 and the stage stagnation pressure ratio is constant.

Solution

= 25 kg/s, p01 = 1 bar, T01 = 273 + 20 = 293 K, P = 4500 kW, T02 = 293 + 15 = 308 K, np = 0.90

1. Polytropic efficiency

Power P = cp (T0zT01)

or 4500 = 25 × 1.005 (T0z − 293)

or T0z = 472.1 K

Again, polytropic efficiency can be expressed as

#### Example 15.10

An axial compressor receives 1000 m3/min of free air at 15°C and 0.9 bar. The blades are of aerofoil type having projected area and blade length as 19.25 cm2 and 6.75 cm, respectively. The blade ring mean diameter is 60 cm and the speed is 6000 rpm. On each blade ring, there are 50 blades and the blades occupy 10% of the axial area of flow. Value of CL and CD are 0.6 and 0.05 respectively at zero angle of incidence. Assuming isentropic compression, calculate the pressure rise per blade ring and the power input per stage. Assume axial inlet.

Solution

= 1000 m3/min, T1 = 273 + 15 = 288 K, p1 = 0.9 bar, A = 19.25 cm2, h = 6.75 cm, kb = 0.9, CL = 0.6, CD = 0.05, Dm = 60 cm, i = 50

Density of air,

Area across flow, Af = kb πDmh

= 0.9 × π × 0.6 × 6.75 × 10−2
= 0.1145 m2

For axial inlet, va1 = vf = 145.55 m/s

The velocity triangle at inlet is shown in Fig. 15.14.

Figure 15.14 Velocity diagram for axial flow compressor

Power input per stage,

p = (FL cos β1 + FD sin β1) ui
= (35.7 cos 52.325° + 2.975 sin 52.325°) × 188.5 × 50 × 10−3
= 227.839 kW

Mass flow rate,

p2 = 0.9 × 1.16 = 1.044 bar

Pressure rise = p2p1 = 1.044 − 0.9 = 0.144 bar

#### Example 15.11

An axial flow compressor gives a pressure rise of 4 : 1 and the total head isentropic efficiency is 86%. Stagnation inlet temperature is 17°C. The inlet and outlet air angles from the rotor blades are 45° and 10°, respectively. The rotor and stator blades are symmetrical. The mean blade and axial velocity remain constant. Assuming blade speed of 220 m/s and work input factor 0.86, find the polytropic efficiency, number of stages required and Mach number. Take R = 287 J/kg.K.

Solution

rp = 4, ηo isen = 0.86, T01 = 273 + 17 = 290 K, α1 = 45°, α2 = 10°, u = 220 m/s, ϕw = 0.86

Polytropic efficiency,

The velocity diagrams are shown in Fig. 15.15.

α1 = β2 = 10°, α2 = β1 = 45°
u = vf1 (tan α1 + tan β1)

Power required per kg of air = cp (T2T1) = ϕw uvf (tan α2 − tan α1)

Figure 15.15 Velocity diagrams at inlet and outlet of vane

Also,

T02 = 290 × 1.565 = 453.86 K

Total temperature rise = T02T01

= 453.86 − 290 = 163.86°C

Number of stages,

From the inlet velocity diagram,

#### Example 15.12

Air at 1.01325 bar and 288 K enters an axial flow compressor stage with an axial velocity of 150 m/s. There are no inlet guide vanes. The rotor stage has a tip diameter of 60 cm and a hub diameter of 50 cm and rotates at 100 rps. The air enters the rotor and leaves the stator in the axial direction with no change in velocity or radius. The air is turned through 30° as it passes through rotor.

Assuming constant specific heat and that air enters and leaves the blade at the blade angles, construct the velocity diagrams at mean diameter for this stage, and calculate (a) the mass flow rate, (b) the power required, and (c) the degree of reaction.

Solution

p1 = 1.01325 bar, T1 = 288 K, vf = 150 m/s, D2 = 60 cm, D1 = 50 cm, N = 100 rps, β2β1 = 30°

The velocity triangles are shown in Fig. 15.16.

Figure 15.16 Velocity diagrams at inlet and outlet of vane

or β1 = 49.04°

β2β1 = 30° or β1β2 = 30°

or β2 = 19.04° or 79.04°

vf2 tan β2 = u + vw2
150 tan 19.04° = 51.77 < 172.8 m/s

vw2 is −ve

Volume flow rate,

1. Mass flow rate,
= V̇ρ = 12.96 × 1.11 = 14.384 kg/s
2. Power required,
P = uvf = 14.384 × 172.8 × 150 × 10−3 = 372.85 kW
3. Degree of reaction,

#### Example 15.13

An axial flow compressor with pressure compression ratio 4 draws air at 20°C and delivers at 197°C. The mean blade speed and flow velocity are constant throughout the compressor. Assuming 50% reaction and blading velocity as 180 m/s, find the flow velocity and number of stages. Take work input factor = 0.82, α1 = 20°, β1 = 42°, cp = 1.005 kJ/kg K.

Solution

Given: rp = 4, T1 = 273 + 20 = 293 K, T2 = 273 + 197 = 470 K, u = u1 = u2 and vf1 = vf1 = vf, Rd = 0.5, u = 180 m/s, ϕw = 0.82, α1 = 20°, β1 = 42°, cp = 1.005 kJ/kg K

For 50% degree of reaction,

α2 = β1 = 42° and β2 = α1 = 20°

Work input factor,

Refer to Fig. 15.17

Δvw = vw2vw1 = vf (tan α2 − tan α1)
= vf (tan 42° − tan 20°) = 0.5364 vf

∴ 0.5364 vf = 73.8

Flow velocity, vf = 137.6 m/s

Work done per stage = u × ∆vw

= 180 × 73.8 = 13284 Nm/kg

Total work = cp (T2T1)

= 1.005 × 103 (470 − 293) = 177885 Nm/kg

Number of stages,

= 13.39 ≃ 14

Figure 15.17 Velocity diagrams for axial flow compressor

#### Example 15.14

An eight-stage axial flow compressor provides an overall pressure ratio of 6: 1 with an overall isentropic efficiency 90% when the temperature of air at inlet is 20°C. The work divided equally between the stages. A 50% reaction design is used with a mean blade speed 188 m/s and a constant axial velocity 100 m/s through the compressor. Estimate the power required and the blade angles. Take cp = 1.005 kJ/kg.K and γ = 1.4.

Solution

Given: z = 8, pz/p1 = 6, (ηisen)0 = 0.9, T1 = 273 + 20 = 293 K, Rd = 0.5, u = 188 m/s, vf = 100 m/s

Refer to Fig. 15.18.

For Rd = 0.5, α1 = β2, α2 = β1

From Fig. 15.19, we have vw1 = vf1 tan α1 and vw2 = vf2 tan α2

Work done per kg of air, w = zu (vw2vw1)

= 150.4 (tan α2 − tan α1)
= 150.4 (tan β1 − tan α1)
= cp (TzT1) = 0.24 × 4.187 (510.8 − 293) = 218.86

Figure 15.18 T-s diagram

Figure 15.19 Velocity triangles

∴ tan β1 − tan α1 = 1.455

After solving, we get

2 tan β1 = 3.335

or β1 = 59°

tan α1 = 1.88 − 1.6676 = 0.2124

or α1 = 12°

Power required = cp (TzT1) = 0.24 × 4.187 (510.8 − 293)

= 218.86 kW

#### Example 15.15

The first stage of an axial compressor is designed on free vortex principles with no inlet guide vanes. The rotational speed is 6000 rpm and the stagnation temperature rise is 20 K. The hub-tip ratio is 0.60, work done factor is 0.93 and isentropic efficiency of the stage is 0.89. Assuming an inlet velocity of 140 m/s and ambient conditions of 1.01 bar and 288 K, calculate (a) the tip radius and corresponding rotor air angles, if the Mach number relative to the tip is limited to 0.95, (b) the mass flow entering the stage, (c) the stage stagnation pressure ratio and power input, and (d) rotor air angles at the root section. Assume cp = 1.005 kJ/kgK and γ = 1.4.

Solution

Given: N = 6000 rpm, = 20 K, = 0.6, ϕw = 0.93, (ηisen)stage = 0.89, va1 = 140 m/s, p1 = 1.01 bar, T1 = 288 K, M = 0.95, cp = 1.005 kJ/kg.K, γ = 1.4

With no inlet guide vanes, α1 = 0° and vf1 = va1 = 140 m/s, vw1 = 0

For free vortex flow = vw1 × rh = vw2 × rt

The velocity diagrams are as shown in the Fig. 15.20(a).

Also, vf 1 = vf 2 = vf

∴ va2 = va1 and β1 = β2

Figure 15.20 (a) Velocity diagrams, (b) T-s diagram for axial flow compressor

T02 = T01 (T02′T01) ×(ηisen)stage

= 297.75 + 20 × 0.89 = 315.55 K

1. Mach number relative to tip, M = 0.95

or vr2 = 333.1 m/s

Tip diameter, Dt = 962 mm

Hub diameter, Dh = 0.6 Dt = 0.6 × 962 = 577.2 mm

2. Density of air at inlet,

Mass flow rate of air entering the stage,

3. α1 = α2 = 0°, β1 = β2 = 65.14°

#### Example 15.16

An axial flow compressor has an overall pressure ratio of 4 and mass flow rate of 3 kg/s. If the polytropic efficiency is 88% and the stagnation temperature rise per stage must not exceed 25 K, calculate the number of stages required and pressure ratio of first and last stages. Assume equal temperature rise in all stages. If the absolute velocity approaching the last rotor is 165 m/s at an angle of 20° from the axial direction, the work done factor is 0.83, the velocity diagram is symmetrical and the mean diameter of last stage rotor is 18 cm, calculate the rotational speed and length of the last stage rotor blade inlet to the stage. Ambient conditions are 1.01 bar and 288 K.

Solution

Given: = 4, a = 3 m/s, ηp = 0.88, (dT0η)stage ≤ 25K, (va1)z = 165 m/s, α1 = 20°, ϕw = 0.83, (dm)z = 0.18 m, p1 = 1.01 bar, T1 = 288 K

Figure 15.21 T – s diagram

Refer to Fig. 15.21.

Now, (va1)1 = (va1)z = 165 m/s

Pressure ratio of first and last stage,

or n = 1.48

T0z = 301.5(3.8)0.48/1.48 = 464.9 K

Number of stages,

Actual stagnation temperature rise per stage

Hence O.K.

For symmetrical velocity diagram, α1 = β2 (see Fig. 15.22)

Work done per stage, w = cp × (dT0)stage × ϕw

= 1.005 × 23.4 × 0.83 = 19.519 kJ/kg

Also, w = (vw2vw1)u

= (v − 2vw1)u = (u − 2 × va1 sin α1)u
= (u − 2 × 165 sin 200)u = (u − 112.87)u

or u2 − 112.87 u = 19519

or u2 − 112.87 u − 19519 = 0

Figure 15.22 Symmetrical velocity diagrams

Area across flow, Af = πdmh

Density of air, ρ1 = = 1.222 kg/m3

Velocity of flow, vf 1 = va1 cos α1 = 165 cos 20° = 155 m/s

a = ρ1Af vf 1
3 = 1.222 × π × 0.18 × h × 155
h = 0.028 m or 2.8 cm

Length of last stage rotor blade, h = 2.8 cm.

#### Example 15.17

An axial-flow compressor employed in gas turbine plant delivers air at the rate of 300 kg/s and develops a total pressure ratio of 20. The inlet stagnation conditions are 1 bar and 300 K and the blade speed is kept at 200 m/s to minimise noise generation. The stage degree of reaction at the mean blade height is 50%. The axial velocity of flow is 160 m/s. The work done factor is 0.88. The hub-tip diameter ratio is 0.8. Assume actual temperature rise in each stage. Show the process on T-s diagram and draw velocity diagram. Find (a) all the fluid angles of the first stage and (b) The hub and tip diameter including blade height. Take R = 0.287 kJ/kg-K and cp = 1.005 kJ/kg-K.

[IES, 2010]

Solution

Given: a = 300 kg/s, rp = 20, T01 = 300 K, p01 = 1 bar, ηisen = 0.87, z = 18, u = 200 m/s, Rd = 0.5, ϕw = 0.88, hub-tip ratio = 0.8, R = 0.287 kJ/kg-K, cp = 1.005 kJ/kg-K, vf = 160 m/s

The Ts and velocity diagrams are shown in Fig. 15.23.

1. Figure 15.23 (a) T-s diagram, (b) Velocity diagrams

Toz =766.7 K
vw2 = vf tan α2, vw1 = vf tan α1

For Rd = 0.5, α2 = β1, α1 = β2

Work consumed by compressor per kg of air

= u (vw2vw1) ϕw × z
= cp (TozT01)

200 (tan α2 − tan α1) 160 × 0.88 × 18 = 1.005 (766.7 − 300)

tan α2 − tan α1 = 9.25 × 10−4

or tan β1 − tan α1 = 9.25 × 10−4

Adding the above two equations, we get

2 tan β1 = 1.250925

or β1 = 32.02° = α2

or tan α1 = 1.25 − 0.62546 = 0.62453

α1 = 31.98° = β2
2. Density of approaching air to first stage.

Applying continuity equation,

ρ1Afvf = a

1.1614 × π [1 − (0.8)2] × 160 = 300

Tip radius, r1 = 1.194 m

Hub radius, rh = 0.8 r1 = 0.8 × 1.194 = 0.9558 m

h = r1rh = 1.194 − 0.9558 = 0.238 m.

#### Example 15.18

Air enters an axial flow compressor at 25°C and undergoes a pressure increase of six times that at inlet. The mean velocity of rotor blades is 200 m/s. The inlet and exit angles of both the moving and fixed blades are 45° and 20° respectively. The degree of reaction at the mean diameter is 50 percent and there are 10 stages in the compressor. The isentropic efficiency of compressor is 85% and axial velocity is constant throughout. Calculate the work done factor of the compressor.

Solution

Given: T1 = 273 + 25 = 298 K, rp = 6, um = 200 m/s, β1 = 45°, β2 = 20°, Rd = 50%, z = 10, (ηc)isen = 85%, vf 1 = vf 2 = vf

For Rd = 50%, α2 = β1 and α1 = β2

Also vr1 = va2 and vr2 = va1

Figure 15.24 Velocity triangles

From Fig 15.24, we have

um = vr2 sin β2 + va2 sin α2
= va1 sin β2 + va1 cos α1 tan α2
200 = va1 [sin 20° + cos 20° tan 45°]
= 1.2817 va1

or va1 = 156 m/s

vw1 = va1 sin α1 = 156 sin 20° = 53.36 m/s
vw2 = va2 sin α2 = 207.3 sin 45° = 146.58 m/s
Δvw = vw2vw1 = 146.58 − 53.36 = 93.22 m/s

Work done on compressor, wc = cp × ∆T = 1.005 × 234.35 = 235.5 kJ/kg

Power input to compressor = ∆vw × um × ϕw × z

= 93.22 × 200 × ϕw × 10 = 186, 440 × ϕw

Work done factor,

#### Example 15.19

An axial flow compressor compresses the air up to overall stagnation pressure 10 bar with overall stagnation isentropic efficiency of 88%. The inlet stagnation pressure and temperature are 1 bar and 300 K. The mean blade speed is 200 m/s. The degree of reaction is 0.5 at the mean radius with air angles of 30° and 10° at rotor inlet and outlet with axial direction respectively. The work done factor is 0.88. The hub-tip ratio is 0.4. The mass flow rate is 50 kg/s. Show the compression process on T-s diagram and draw the inlet and outlet velocity triangles. Find (a) the stagnation polytropic efficiency, (b) the number of stages and (c) the blade height in first stage of the compressor.

[IAS, 2006]

Solution

Given: p1 = 1 bar, T01 = 300 K, poz = 10 bar, ηisen = 88%, um = 200 m/s, Rd = 0.5, β1 = 30°, β2 = 10°, ϕw = 0.88, , = 50 kg/s,

The T-s diagram is shown in Fig. 15.25(a)

The velocity triangles are shown in Fig.15.25(b).

1. Stagnation polytropic efficiency,
= 0.9117 or 91.17%
2. For Rd = 0.5

Figure 15.25 (a) T-s diagram, (b) Velocity triangles

vw2 = vf tan α2 = 265.4 tan 30° = 153.2 m/s
vw1 = vf tan α1 = 265.4 tan 10° = 46.8 m/s

Work consumed per stage = ϕw u (vw2vw1)

= 0.88 × 200 (153.2 − 46.8) × 10−3 = 18.726 kJ/kg

Total work consumed by compressor = cp (TozT01)

= 1.005 (617.3 − 300) = 318.89 kJ/kg.

Number of stages,

3. Static temperature,

a = ρ1A1vf
50 = 0.843 × π [1 − (0.4)2] × 265.4
r1 = 0.291 m or 291 mm
rh = 0.4 r1 = 0.4 × 291 = 116.4 mm

#### Example 15.20

An air compressor has eight stages of equal pressure ratio 1.35. The flow rate through the compressor and its overall efficiency are 50 kg/s and 82 per cent respectively. If the condition of air at entry are 1.0 bar and 40°C, determine:

1. the state of air at the compressor exit,
2. polytropic or small stage efficiency,
3. efficiency of each stage,
4. power required to drive the compressor assuming overall efficiency of the drive as 90%

Take cp = 1.005 kJ/kg K and γ = 1.4.

[IAS, 2005]

Solution

Given: z = 8, (rp)st = 1.35, = 50 kg/s, η0 = 82%, p1 = 1.0 bar, T1 = 313 K, ηmech = 90%, cp = 1.005 kJ/kg.K, γ = 1.4

1. Exit pressure, pa = 1.0 × 11.03 = 11.03 bar

2. Polytropic efficiency,
3. 3.0415 n − 3.0415 = n

or n = 1.49

Isentropic efficiency of each stage

4. Power required to drive the compressor,

#### Example 15.21

A small compressor has the following data:

Air flow rate = 1.5778 kg/s

Pressure ratio = 1.6

Rotational speed = 54,000 rpm

Efficiency = 85%

State of air at entry, p0p = 1.008 bar, T0p = 300 K

cp for air = 1.009 kJ/kg K.

1. Calculate the power required to drive this compressor.
2. A geometrically similar compressor of three times the size is constructed. Determine, for this compressor (i) mass flow rate, (ii) pressure ratio, (iii) speed, and (iv) the power required.

Assume same entry conditions and efficiency for the two compressors and also assume kinematic and dynamic similarities between the two machines.

[IAS, 2005]

Solution

Given: a = 1.5778 kg/s, rp = 1.6, N = 54,000 rpm, η = 85%, p01 = 1.008 bar, T01 = 300 K, cp = 1.009 kJ/kg.K, D2 = 3D1

1. Power required = acp(T02′T01) = 1.5778 × 1.009 (350.725 − 300) = 80.75 kW
2. = const., G = mass flow rate.
1. p2 = 21.24 × 1.009 (350.725 − 300) = 1087.6 kW

#### Example 15.22

Air at a temperature of 300 K enters a 10-stage axial flow compressor at the rate of 3.5 kg/s. The pressure ratio is 6.0 and the isentropic efficiency is 90%. The process is adiabatic and the compressor has symmetrical stages. The axial velocity of 120 m/s is uniform across the stages and the mean blade speed is 200 m/s. Assume that the temperature change is same in each stage.

Determine the direction of the air at entry to and exit from the rotor and stator blades. Also find the power given to the air.

For air, take cp = 1.005 kJ/kg.K and γ = 1.4.

[IAS, 2004]

Solution

Given: T1 = 300 K, z = 10, a = 3.5 kg/s, rp = 6, ηisen = 90%, vf = 120 m/s, u = 200 m/s, cp = 1.005 kJ/kg.K, γ = 1.4

Symmetrical stages, same temperature change in each stage

The velocity diagrams are shown in Fig. 15.26

T)stage = (ΔT)overall/z

Increase in temperature per stage,

Figure 15.26 Velocity triangles

tan β1 = 1.2999, β1 = 52.43°, β2 = 20.14°

Power supplied = zacp(∆T)stage

= 10 × 3.5 × 1.005 × 22.284

= 783.84 kW

#### Example 15.23

An axial flow compressor of 50% reaction blading has isentropic efficiency of 82%. It draws air at 17°C and compresses in the pressure ratio of 4 : 1. The mean blade speed and flow velocity are constant throughout the compressor. The inlet and outlet angles of blades are 15° and 45° respectively (angles measured from axial direction). Blade speed = 180 m/s and work input factor = 0.84. Calculate (a) the flow velocity and (b) the number of stages. [IAS, 2002]

Solution

Given: Rd = 50%, ηisen = 82%, T1 = 17 + 273 = 290 K, rp = 4:1, um = const, vf = const., β1 = 15°, β2 = 45°, u = 180 m/s, ϕh = 0.84

For Rd = 50%, α1 = β2 = 45°, α2 = β1 = 15°

1. From Fig. 15.27, we get

∴ Flow velocity, vf = 142 m/s

wisen = cp (T2T1) = 1.005(430.9 − 290) = 141.6 kJ/kg

Figure 15.27 Velocity triangles

2. Work done per stage = u × ∆vw = 180 × 75.6 = 13.608 kJ/kg

Number of stages

#### Summary for Quick Revision

1. In axial flow compressors, the air flows parallel to its axis throughout.
2. Axial flow compressors are particularly suitable for supercharging of internal combustion engines, gas turbines, and turbojets.
3. Aerofoil theory:
1. Lift force,
2. Drag force,
4. Power input per stage,

5. Velocity relations:
6. where α1 = exit angle from stator, β1 = inlet angle to rotor, α2 = inlet angle to stator and β2 = outlet angle from rotor.

7. Work required per stage per kg of air, w = (vw2vw1)u = cp (T02′T01)

= vf (tan α2 − tan α1)u

= [uvf (tan α1 + tan β2]u

8. Stage isentropic efficiency,
9. Stagnation pressure ratio,
10. For Rd = 50%, the blades are symmetrical, i.e.,

α1 = β2 and α2 = β1

11. Isentropic pressure rise through a cascade,
12. Polytropic efficiency: It is the small stage efficiency (ηisen)stage, which is constant for all stages, and is denoted by ηp.
13. Flow coefficient,
15. Pressure coefficient,
16. Number of stages,
1. For same pressure ratio per stage,
17. The pulsating air flow phenomenon is known as surging. It causes overheating and stress reversal in the blades which damages the blades.
18. The maximum mass flow rate possible in a compressor is known as choking.
19. Stalling is a phenomenon in which the flow breaks away from the aerofoil blading. It occurs due to lesser flow rate than the designed value or due to non-uniformity of blade profile.

#### Multiple-choice Questions

1. A multistage compressor is to be designed for a given flow rate pressure ratio. If the compressor consists of axial flow stages followed by centrifugal instead of only axial flow stages, then the
1. overall diameter would be decreased
2. overall diameter would be increased
3. axial length of the compressor would be increased
4. axial length of the compressor would be decreased
2. What is the ratio of the isentropic work to Euler’s work known as?
1. Pressure coefficient
2. Slip factor
3. Work factor
4. Degree of reaction
3. In air-craft gas turbines, the axial flow compressor is preferred because
1. of high pressure rise
2. it is stall-free
3. of low frontal area
4. of higher thrust
4. In axial flow compressor, exit flow angle deviation from the blade angle is a function of
2. space-chord ratio
3. both blade camber and space-chord ratio
4. blade camber and incidence angle
5. High positive incidence in an axial compressor blade row leads to
1. suppression of separation of flow on the blade
2. choking of the flow
3. separation of flow on the pressure side of the blade
4. separation of flow on the suction side of the blade
6. Which one of the following is the correct expression for the degree of reaction for an axial-flow air compressor?
7. Which one of the following types of compressors is mostly used for supercharging of IC engines?
2. Axial flow compressor
3. Roots blower
4. Reciprocating compressor
8. Phenomenon of choking in compressor means
1. no flow of air
2. fixed mass flow rate regardless of pressure ratio
3. reducing mass flow rate with increase in pressure ratio
4. increased inclination of the chord with air steam
9. Degree of reaction in an axial compressor is defined as the ratio of static enthalpy rise in the
1. rotor to static enthalpy rise in the stator
2. stator to static enthalpy rise in the rotor
3. rotor to static enthalpy rise in the stage
4. stator to static enthalpy rise in the stage
10. The usual assumption in elementary compressor cascade theory is that
1. axial velocity through the cascade changes
2. for elementary compressor cascade theory, the pressure rise across the cascade is given by equation of state
3. axial velocity through the cascade does not change
4. with no change in axial velocity between inlet and outlet, the velocity diagram is formed
11. Match List 1 with List II (pertaining to blower performance) and select the correct answer using the codes given below the Lists:
List I List-II
A. Slip 1. Reduction of whirl velocity
B. Stall 2. Fixed mass flow rate regardless of pressure ratio
C. Choking 3. Flow separation
4. Flow area separation

Codes:

A   B   C

1.  4   3   2
2.  1   3   2
3.  4   1   3
4.  2   3   4
12. The critical value of Mach number for a subsonic airfoil is associated with a sharp increase in drag due to local shock formation and its interaction with the boundary layer. A typical value of this critical Mach number is of the order of
1. 0.4 to 0.5
2. 0.75 to 0.85
3. 1.1 to 1.3
4. 1.5 to 2.0
13. The inlet and exit velocity diagrams of a turbomachine rotor are shown in Fig. 15.28.

Figure 15.28

The turbomachine is

4. an axial compressor with forward curved blades
14. The turbomachine used to circulate refrigerant in large refrigeration plant is
1. A centrifugal compressor
3. An axial compressor
4. An axial turbine
15. The energy transfer process is
1. continuous in a reciprocating compressor and intermittent in an axial compressor
2. continuous in an axial compressor and intermittent in a reciprocating compressor
3. continuous in both reciprocating and axial compressors.
4. intermittent in both reciprocating and axial compressors
16. In an axial flow compressor stage, air enters and leaves the stage axially. If the whirl component of the air leaving the rotor is half the mean peripheral velocity of the rotor blades, then the degree of reaction will be
1. 1.00
2. 0.75
3. 0.50
4. 0.25
17. If an axial flow compressor is designed for a constant velocity through all stages, the area of the annulus of the succeeding stages will
1. remain the same
2. progressively decrease
3. progressively increase
4. depend upon the number of stages
18. If the static temperature rise in the rotor and stator, respectively, are ∆TA and ∆TB, the degree of reaction in an axial flow compressor is given by
19. Which one of the following pairs of features and compressors type is NOT correctly matched?
1. Intake and delivery ports compression is attained by back flow and internal compression cylindrical rotor set to eccentric casing: Vane compressor
2. Intermittent discharge requires receiver, produces high pressure, slow speed and lubrication problems : Reciprocating compressor
3. Continuous flow, radial flow, handles large volume, much higher speed and fitted into design of aero-engines : Centrifugal compressor
4. Successive pressure drops through contracting passages, blades are formed from a number of circular arcs, axial flow : Axial flow compressor
20. In an axial flow compressor design, velocity diagrams are constructed from the experimental data of aerofoil cascades. Which one of the following statements in this regard is correct?
1. Incidence angle of the approaching air is measured from the trailing edge of the blade
2. δ is the deviation angle between the angle of incidence and tangent to the camber line
3. The deflection ɛ of the gas stream while passing through the cascade is given by ɛ = α1α2
4. ɛ is the sum of the angle of incidence and camber less any deviation angle, i.e., ɛ = i + θδ
21. Consider the following statements regarding the axial flow in an air compressor:
1. Surging is a local phenomenon while stalling affects the entire compressor.
2. Stalling is a local phenomenon while surging affects the entire compressor.
3. The pressure ratio of an axial compressor stage is smaller than that of a centrifugal compressor stage

Of these statements

1. (i), (ii) and (iii) are correct
2. (i) and (ii) are correct
3. (ii) and (iii) correct
4. (i) and (iii) correct
22. Compared to axial compressors, centrifugal compressors are more suitable for
1. high head, low flow rate
2. low head, low flow rate
3. low head, high flow rate
4. high head, high flow rate
23. Stalling of blades in axial-flow compressor is the phenomenon of
1. air stream blocking the passage
2. motion of air at sonic velocity
3. unsteady, periodic and reversed flow
24. In an axial flow compressor:

α1 = exit angle from stator, β1 = inlet angle to rotor, α2 = inlet angle to stator, and β2 = outlet angle from rotor.

The condition to have a 50% degree of reaction is

1. α1 = β2
2. α2 = β1
3. α1 = β2 and β1 = α2
4. α1 = α2 and B1 = β2
25. Consider the following statements with reference to supercharging of IC engines:
1. Reciprocating compressors are invariably used for high degree of super charging
2. Rotary compressors like roots blowers are quite suitable for low degree of supercharging
3. Axial flow compressors are most commonly employed for supercharging diesel engines used in heavy duty transport vehicles
4. Centrifugal compressors are used for turbo-charging

Which of the statements given above are correct?

1. (i) and (ii)
2. (ii) and (iii)
3. (i) and (iv)
4. (ii) and (iv)
26. While flowing through the rotor blades in an axial flow air compressor, the relative velocity of air
1. continuously decreases
2. continuously increases
3. first increases and then decreases
4. first decreases and then increases
27. In an axial flow compressor, when the degree of reaction is 50%, it implies that
1. work done in compression will be the least
2. 50% stages of the compressor will be ineffective
3. pressure after compression will be optimum
4. the compressor will have symmetrical blades
28. For a multistage compressor, the polytropic efficiency is
1. the efficiency of all stages combined together.
2. the efficiency of one stage.
3. constant throughout for all the stages
4. a direct consequence of the pressure ratio

#### Explanatory Notes

1. 16. (d) Degree of reaction,

#### Review Questions

1. How do axial flow compressors differ from centrifugal compressors?
2. Explain the working principle of axial flow compressors.
3. Define lift and drag coefficients.
4. Define degree of reaction and write down the mathematical expression for it.
5. For 50% degree of reaction, what is the condition on blade angles?
6. Write the expression for pressure rise in isentropic flow through a cascade.
7. Define polytropic efficiency and write down the mathematical expression for it.
8. Define flow coefficient and work coefficient.
9. Define deflection coefficient and pressure coefficient.
10. Differentiate between choking and stalling.
11. List the various losses in an axial flow compressors.
12. List the main applications of axial flow compressors.
13. List the merits and demerits of axial flow compressors.
14. Draw the efficiency and power consumption curves for the axial flow compressor.

#### Exercises

15.1 An axial flow air compressor of 50% degree of reaction has blades with inlet and outlet blade angles of 40° and 15°, respectively. The pressure ratio is 5 : 1 with an overall isentropic efficiency of 85% when the air inlet temperature is 30°C. The blade speed and axial velocity are constant throughout the compressor. Assuming a blade speed of 210 m/s, find the number of stages required when the work coefficient is 0.9 for all stages.

15.2 In an axial flow air compressor, the overall stagnation pressure ratio is 4.5 with overall stagnation isentropic efficiency 85%. The inlet stagnation pressure and temperature are 1 bar and 310 K. The mean blade speed is 180 m/s and degree of reaction is 0.5 at the mean radius. The relative air angles are 10° and 30° at rotor inlet and outlet, respectively. The work coefficient is 0.9. Calculate (a) the stagnation polytropic efficiency, (b) the number of stages, (c) the inlet temperature and pressure, and (d) the blade height in the first stage if the hub-tip ratio is 0.45 and mass flow rate is 20 kg/s.

15.3 A multi-stage axial flow compressor delivers 19 kg/s of air. The inlet stagnation condition is 1 bar and 25°C. The power consumed by the compressor is 4400 kW. Calculate (a) the delivery pressure, (b) the number of stages, and (c) the overall isentropic efficiency of the compressor.

Assume that temperature rise in the first stage is 21°C. The polytropic efficiency of compressor is 90% and stagnation pressure ratio is constant.

15.4 An axial flow compressor compresses the air up to overall stagnation pressure of 10 bar with overall stagnation isentropic efficiency of 88%. The inlet stagnation pressure and temperature are 1 bar and 300 K. The mean blade speed is 200 m/s. The degree of reaction is 0.5 at the mean radius with air angles of 30° and 10° at rotor inlet and outlet with axial direction respectively. The work done factor is 0.88. The hub-tip ratio is 0.4. The mass flow rate is 50 kg/s. Draw the inlet and outlet velocity triangles and show the compression process on T-s diagram. Find the following:

1. The stagnation polytropic efficiency
2. The number of stages
3. The blade height in the first stage of the compressor

15.5 An axial flow compressor provides a total head pressure ratio of 4 : 1 with an overall total head isentropic efficiency of 85%, when the inlet total head temperature is 290 K. The compressor is designed for 50% reaction with inlet and outlet air angles from the rotor blades of 45° and 10°, respectively. The mean blade speed and axial velocity are constant throughout the compressor. Assuming a value of 201.16 m/s for the blade speed and a work done factor of 0.86, find the number of stages required. What is the inlet Mach number relative to the rotor at the mean blade height of the first stage?

15.6 Air at a temperature of 300 K enters a 10-stage axial flow compressor at the rate of 3.5 kg/s. The pressure ratio is 6.0 and the isentropic efficiency is 90%. The compression process is adiabatic and the compressor has symmetrical stages. The axial velocity is uniform across the stages and equals to 120 m/s and the mean blade speed is 200 m/s. Assume that the temperature change is same in each stage. Determine the direction of the air at entry and exit from the rotor and stator blades. Also, find the power given to the air. For air, take cp = 1.005 kJ/kg.K and γ = 1.4.

15.7 An axial flow compressor consists of a number of similar stages with equal work done per stage. The velocity of flow is uniform throughout the compressor. The data given is as follows:

 Overall stagnation pressure ratio = 3.5 : 1 Stagnation inlet temperature = 60°C Relative air angle at rotor inlet = 130° Relative air angle at rotor outlet = 100° Blade velocity = 185 m/s Degree of reaction = 50% Overall stagnation adiabatic efficiency = 87%

Calculate (a) the stagnation outlet temperature, and (b) the number of stages.

1. b
2. a
3. c
4. c
5. a
6. a
7. b
8. b
9. c
10. c
11. b
12. c
13. a
14. c
15. c
16. d
17. a
18. b
19. c
20. a
21. d
22. a
23. c
24. c
25. d
26. a
27. d
28. a