# Chapter 14 Centrifugal Air Compressors – Thermal Engineering

## Centrifugal Air Compressors

##### 14.1 ❐ INTRODUCTION

A centrifugal compressor is of rotodynamic type in which air flows continuously and steadily through various parts and the rise in pressure is primarily due to the centrifugal action. It is used to supply large quantities of air but at a lower pressure ratio.

##### 14.2 ❐ CONSTRUCTIONAL FEATURES

Essentially, a centrifugal compressor consists of four elements namely, inlet buckets, impeller, diffuser, and casing as shown in Fig. 14.1. The inlet buckets are attached to the shaft and rotate with it, which guide air on the impeller. The impeller consists of a disc on which radial blades are attached. The diffuser surrounds the impeller and provides diverging passages for air flow, thus increasing the air pressure. The air coming out from the diffuser is collected in the casing and taken out from the outlet of the compressor.

Figure 14.1 Centrifugal compressor

##### 14.3 ❐ WORKING PRINCIPLE

The air enters the eye of the compressor at atmospheric pressure and low velocity. The inlet buckets guide the air to the impeller where it moves radially outward and is guided by the impeller blades. The impeller increases the momentum of the air flowing through it, causing a rise in pressure and temperature of the air. The air leaving the impeller enters the diffuser where its velocity is reduced by providing more cross-sectional area for the flow of air. A part of the kinetic energy of air is converted into pressure energy and further increases the pressure of air.

##### 14.4 ❐ VARIATION OF VELOCITY AND PRESSURE

As the air flows through the impeller and diffuser, there is a variation of both velocity and pressure as shown in Fig. 14.2. Nearly half of the total pressure rise takes place in the impeller and the remaining half occurs in the diffuser. A pressure ratio of 4 can be achieved in single-stage compressors. For higher pressure ratios, multi-stage compressors are used. A pressure ratio of 12:1 is possible with multi-stage compressors.

Figure 14.2 Variation of velocity and pressure of air passing through impeller and diffuser

##### 14.5 ❐ TYPES OF IMPELLERS

Impellers are of two types—single-eye type and double-eyed type as shown in Fig. 14.3 (a) and (b), respectively. In a single-eye type, air enters into the compressor from one side only, whereas in a double-eyed type, air enters from both sides. A double-eyed type impeller sucks in more air and has the advantage of self-balancing over a single-eye impeller.

In multi-stage compressors, the output of the first stage is passed on to the second stage and so on, as shown in Fig. 14.3(c).

Figure 14.3 Types of impellers: (a) Single eye impellers, (b) Double eye impellers, (c) Multi-stage impellers

##### 14.6 ❐ COMPARISON OF CENTRIFUGAL AND RECIPROCATING COMPRESSORS

The comparison of centrifugal and reciprocating compressors is given in Table 14.1.

Table 14.1 Comparison of centrifugal and reciprocating compressors

##### 14.7 ❐ COMPARISON OF CENTRIFUGAL AND ROTARY COMPRESSORS

The comparison of centrifugal and rotary compressors is given in Table 14.2.

Table 14.2 Comparison of centrifugal and rotary compressors

##### 14.8 ❐ STATIC AND STAGNATION PROPERTIES

The velocities of air encountered in centrifugal compressors are very high as compared to that in reciprocating compressors. Therefore, total head quantities should be taken into account in the analysis of centrifugal compressors. The total head quantities account for the kinetic energy of the air passing through the compressor.

Consider a horizontal passage of varying area of cross-section, as shown in Fig. 14.4, through which air flows from left to right. Assuming no external heat transfer and work transfer to the system, the steady flow energy equation for one kg mass of air flow can be written as:

Temperature T is called the ‘static temperature’; it is the temperature of the air measured by a thermometer moving with the air velocity. If the moving air is brought to rest under reversible adiabatic conditions, the total kinetic energy of the air is converted into thermal energy, thereby increasing its temperature and pressure. This temperature and pressure of the air are known as ‘stagnation’ or ‘total head’ temperature and pressure. The stagnation quantities are denoted by a suffix notation o.

Figure 14.4 Passage of air through varying area of cross-section

where T0 = stagnation or total head temperature

where h0 = stagnation or total head enthalpy

The stagnation pressure can be obtained by using the following equation:

where p, T = static pressure and temperature

p0, T0 = stagnation pressure and temperature.

##### 14.9 ❐ ADIABATIC AND ISENTROPIC PROCESSES

For an adiabatic compression process, there is no heat exchange with the surroundings. If the adiabatic process is reversible (frictionless), then the process is called isentropic process in which the entropy of the system does not change. However, in an actual compressor, during adiabatic compression, there are losses due to friction in air and blade passages, eddies formation, and shocks at entry and exit. These factors cause internal heat generation and consequently, the maximum temperature reached would be higher than that for adiabatic compression. This results in a progressive increase in entropy. Such a process, although adiabatic, is not reversible adiabatic or isentropic.

The isentropic and adiabatic processes for the static and stagnation values are shown in Fig. 14.5 on the T-s diagram. Process 1-2 is the isentropic process for the static temperature and 01-02 for the total head temperature. Processes 1-2′ and 01-02′ are the adiabatic processes for the static and total head temperatures, respectively.

Figure 14.5 Isentropic and actual compression

#### 14.9.1 Isentropic Efficiency

Isentropic efficiency may be defined as the ratio of isentropic temperature rise to actual temperature rise.

where

During compression process, work has to be done on the impeller. The energy balance then gives,

cpT01 = cpT02 - W
##### 14.10 ❐ VELOCITY DIAGRAMS

The velocity diagrams at inlet and outlet of the impeller for the centrifugal compressor are shown in Fig. 14.6(a).

Figure 14.6 (a) Actual velocity diagrams

Let

u1 = mean blade velocity at entrance

u2 = mean blade velocity at exit

va1 = absolute velocity at inlet to rotor

va2 = absolute velocity at outlet to rotor

vr1 = relative velocity at inlet to rotor

vr2 = relative velocity at outlet to rotor

vw1 = velocity of whirl at inlet

vw2 = velocity of whirl at outlet

vf1 = velocity of flow at inlet

vf 2 = velocity of flow at inlet

D1, D2 = mean diameter of rotor at inlet and outlet, respectively

α1 = exit angle from stator or guide vanes at entrance

β1 = inlet angle to rotor in impeller blade angle at inlet

α2 = inlet angle to the diffuser or the stator

β2 = outlet angle from rotor or impeller blade angle at outlet.

If it is assumed that the entry of the air to the rotor is axial, then whirl component vw1 = 0 and va1 = vf1.

#### 14.10.1 Theory of Operation

1. Ideal Velocity Diagrams

By Newton’s second law of motion, the rate of change of angular momentum of air is equal to the torque applied to the body causing that change. Considering 1 kg/s of mass flow rate of air,

where

For ideal case, let us assume that the impeller is radial vaned, and there is no frictional loss, no heat transfer and the air leaves the impeller with a tangential velocity vw2 = u2. Also for axial entry, vw1 = 0

Work done by impeller for 1 kg/s air flow rate,

Figure 14.6 (b) Ideal velocity diagrams

where ω = angular speed of rotor, rad/s

Now, r2ω = u2

Since air cannot leave the impeller at a velocity greater than the impeller tip velocity, Eq. (14.11) gives the maximum work capacity of the impeller.

Also, by steady flow energy equation, we have

In most practical problems,

va1 = va2
= h2h1

where = stagnation pressure ratio

where, = static pressure ratio

Comparing Eq. (14.11) and (14.14), we have

If va1 = va2, then

Total temperature increase across an impeller,

where = mass flow rate of air, kg/s, and u2 is in m/s.

Therefore, for a centrifugal compressor working under ideal condition, the power input depends on the following:

1. The mass flow rate of air
2. The total pressure ratio of the compressor which, in turn, depends on the square of the impeller velocity
3. The total inlet temperature
4. The total inlet temperature rise between the inlet and outlet
5. There is a maximum work capacity of an impeller depending on the tip velocity

The maximum pressure produced depends on the following:

1. Inlet temperature
2. Square of the impeller velocity
3. It is independent of the impeller diameter
2. Actual Velocity Diagrams

The actual velocity diagrams are shown in Fig. 14.7.

Theoretical torque, T = vw2r2 - vw1r1

Work done on 1 kg/s of air.

w = (vw2r2 - vw1r1)w

where, ω = angular speed of rotor, rad/s

Figure 14.7 Actual velocity diagrams for centrifugal compressor

= ΔKE + Δp due to diffusion action + Δp due to centrifugal action

Therefore, the fraction of kinetic energy imparted to air and converted into pressure energy in impeller is given by,

where ρ = density of air.

If va1 = diffuser outlet velocity, then

#### 14.10.2 Width of Blades of Impeller and Diffuser

Air mass flow rate per second,

where, vs = specific volume of air

b1, b2 = width or height of impeller blades at inlet and outlet, respectively

The width or height of the diffuser blades at the outlet is given by,

where suffix ‘d’ represents the quantities at the diffuser outlet. The width or height of the diffuser blades at the inlet is the same as that of the impeller blades at outlet.

##### 14.11 ❐ SLIP FACTOR AND PRESSURE COEFFICIENT

In the ideal velocity diagrams, we had assumed that vw2 = u2. However, in actual operation, this condition is not satisfied in actual practice due to secondary flow effects. Actually, vw2 < u2. The difference (u2 - vw2) is called the slip. The slip factor is defined as the ratio of actual whirl component to the ideal whirl component and is denoted by ϕs.

The total head pressure ratio is given by,

Substituting the value of (T02 - T01) from Eq. (10.5), we have

The actual work done per kg of air is given by:

cp (T02′ - T01) = u2vw2

The actual work done per kg of air by the compressor is always greater than u2vw2 due to fluid friction and windage losses. Therefore, the actual work is obtained by multiplying u2vw2 by the factor ϕw, called the work factor or power input factor.

cp (T02′ - T01) = ϕwu2vw2

Eq. (10.29) becomes

Now, vω2 = ϕsu2

The pressure coefficient is defined as the ratio of isentropic work to Euler work and is denoted by ϕp.

Using Eq. (14.5) and assuming radial vanes of impeller, i.e., vw2 = u2, we get

Using Eqs (14.28) and (14.30), we get

##### 14.12 ❐ LOSSES

There are two types of losses in a centrifugal compressor: internal and external. The internal losses sustained in the compressor are manifested by an increase of enthalpy of air. The internal losses which occur in compressor are as follows:

1. Friction between air and walls of flow passage
2. Disc friction
3. Leakage between impeller and casing
4. Turbulence
5. Shock

The compressor sustains an external loss in the form of bearing friction and windage loss. The friction losses are proportional to and hence proportional to . The incidence loss in terms of drag coefficient CD is proportional to

##### 14.13 ❐ EFFECT OF IMPELLER BLADE SHAPE ON PERFORMANCE

The following types of blade shapes are used for impellers of centrifugal compressors:

1. Backward-curved blades (β2 < 90°)
3. Forward-curved blades (β2 > 90°).

The characteristics of these blades are shown in Fig. 14.8. Centrifugal effects on the curved blades create a bending moment and produce increased stresses which limit the maximum speed at which the impeller can run.

Slightly backward-curved impeller blades (i.e., β2 < 90°) give optimum efficiency because the degree of reaction increases with decreasing blade tip angles. With increase in degree of reaction, the part of kinetic energy that is transferred into pressure energy within the impeller is also increased. By selecting very small tip blade angles β2, the length of impeller channel increases and consequently the friction loss increases considerably, thereby reducing the efficiency. The optimum blade angle is about 45°.

Radial blades are used for aircraft centrifugal compressors because they can be manufactured easily, have the lowest centrifugal stresses, are free from bending stresses, and have equal energy conversion in impeller and diffuser, giving higher pressure ratio and efficiency.

##### 14.14 ❐ DIFFUSER

The diffuser converts the kinetic energy imparted to air by the impeller into pressure rise. In the vaned diffuser, the vanes are used to remove the whirl of the fluid at a higher rate than is possible by a simple increase in radius, thereby reducing the length of flow path and diameter. A typical vaned type diffuser is shown in Fig. 14.9. A pressure gradient exists in the diffuser in the direction opposite to that of flow, and the flow streamlines will tend to break away from the diverging passage walls, reversing their direction and resulting in turbulence. About 20° is accepted as the maximum included angle of divergence for satisfactory diffusion.

The clearance between the impeller and the vanes leading edges constitutes a vaneless diffuser. This helps smoothen out velocity variation between the impeller tip and diffuser vanes. This also reduces the circumferential pressure gradient at the impeller tip.

The flow follows an approximately logarithmic spiral path to the vanes after which it is constrained by the diffuser channels.

Figure 14.9 Vaned diffuser

where suffices 3 and 4 denote upstream and downstream conditions of diffuser and ρ is the mass density of air.

##### 14.15 ❐ PRE-WHIRL

In order to avoid acceleration to sonic or supersonic velocities, the Mach number of flow entering the impeller eye should be kept below the value of 0.9. When the absolute velocity of approach is high enough or the static absolute temperature of the entering air is low enough to make the Mach number more than 0.9, the method of providing a pre-whirl and thereby reducing the relative velocity entering the impeller eye is quite helpful.

Figure 14.10 Effect of pre-whirl on inlet velocity triangle: (a) With pre-whirl (vw ≠ 0), (b) Without pre-whirl (vw = 0)

Pre-whirl is provided by intake guide vane as shown in Fig. 14.10, which shows the velocity triangles for blades without pre-whirl and with pre-whirl. It may be observed that the magnitude of the relative velocity with pre-whirl vane gets reduced at the entrance, but the flow enters with a large relative angle. Thus, the flow inlet Mach number is lowered by pre-whirl vane. However, since vw1 ≠ 0 with pre-whirl vane, the work done on the air must be greater for the same compressor inlet and outlet conditions.

##### 14.16 ❐ PERFORMANCE CHARACTERISTICS

Let us consider the following variables for the performance of the compressor.

p3 = pressure at the outlet of diffuser

p1 = inlet pressure

T1 = inlet temperature

D = diameter

G = mass rate of flow

N = r.p.m of rotor

Then, p3 = f (p1, T1, D, G, N)

The following dimensions are chosen:

M = mass (kg), L = length (m), t = time (s)

Number of variables, n = 6

Basic dimensions, m = 3

According to Buckingham’s Pi theorem, the number of dimensionless groupsm = nm = 6 − 3 = 3

The three dimensionless groups are formed by combining p1, T1, and D with each of the remaining variables, so that

Substituting the basic dimensions for the variables in π1, we have

π1 = (M1L−1t−2)a1 (L2t−2)b1 Lc1(M2L−1t−2)1

For M, 0 = a1 + 1

For L, 0 = − a1 + 2b1 + c1 −1

For t, 0 = − a1 − 2b1 −2

From which, a1 = −1, b1 = 0, c1 = 0

For a given compressor under consideration, D = const.

Figure 14.11

Similarly, the ratio can be formed to be function of the dimensionless groups of Eqs. (14.35) and (14.36). Similarly, if stagnation temperature and pressure are considered, the result will again be the same except that stagnation temperatures and pressures are substituted in place of static pressures and temperatures.

Figure 14.12 Pressure ratio v’s Volume flow rate

The performance of a given compressor or geometrically similar compressor can be plotted in terms of three dimensionless parameters provided dynamic similarity is maintained. Fig. 14.11 shows a plot of against and . The operating line is the locus of the points of maximum efficiency at various values of . The surge line represents the stable operation, the region to the right of the surge line being for stable conditions.

The performance of the compressor has been plotted in terms of volume flow rate and pressure ratio in Fig. 14.12.

##### 14.17 ❐ SURGING AND CHOKING

The delivery pressure v’s mass flow rate in a dynamic compressor is shown in Fig. 14.13. Suppose a discharge valve is put in the line for this compressor. The mass rate of flow will be zero when the discharge valve is closed and the static pressure developed is that delivered by the impeller to the air contained in the compressor. Such a situation is shown by point a in Fig. 14.13. If the valve is now opened, the flow of air takes place and the diffuser becomes effective in increasing the static pressure. This is shown by points b and c. The maximum delivery pressure is obtained at point c. As the valve is opened more, the mass flow rate increases beyond point c and the efficiency of compressor decreases with decrease in delivery pressure. When the designed mass rate is greatly exceeded, the incidence between the vane and air angle becomes so large that flow separation and shock occur, accompanied by rapid decrease in efficiency.

If the compressor operates at some point b to the left of point c, then a decrease in mass rate of flow is accompanied by decrease in pressure developed by the compressor. If the static pressure of the air at compressor outlet does not decrease as rapidly as the developed pressure, there is a natural tendency of the air to flow back into the compressor in the direction of pressure gradient. With the drop in pressure at the compressor outlet, the pressure gradient is reversed and also the direction of flow is reversed. Thus, this leads to unstable condition of cyclic reversal taking place at extremely high frequencies. This pulsating air flow phenomena is called ‘surging’. Such a situation is avoided by keeping the operating point to the right of point c, because in this region a decrease in mass flow rate is accompanied by increase in delivery pressure, leading to stability.

Figure 14.13 Surging and choking phenomena

After point c, any increase in mass flow rate is accompanied by decrease in delivery pressure. This happens because the rate of increase in pressure loss due to the friction is more than the rate of increase in pressure gain by diffuser. Theoretically, the decrease in delivery pressure is continued until the mass flow rate is represented by the point ‘f’. In practice, the maximum mass flow is limited by the point e because if the mass flow exceeds design mass flow, the air angles are widely different from vane angles and choking takes place. The point e represents the choking of the compressor, that is, the maximum mass flow rate condition.

Surging does not take place in the region ce as the reduction in mass flow rate is accompanied with the increase in delivery pressure and flow reversal is not possible and stability of operation is maintained.

#### Example 14.1

A centrifugal air compressor having a pressure ratio of 5, compresses air at a rate of 10 kg/s. If the initial pressure and temperature are 1 bar and 20°C respectively, find the final temperature of air and power required to drive the compressor. Take γ = 1.4 and cp = 1 kJ/kg.K.

Solution

Given: rp = 5, = 10 kg/s, p1 = 1 bar, T1 = 273 + 20 = 293 K, γ = 1.4, cp = 1 kJ/kg.K

Power required to drive the compressor,

P = cp (T2T1) = 10 × 1 × (464 − 293) = 1710 kW.

#### Example 14.2

A centrifugal compressor has a pressure ratio of 4:1 with an isentropic efficiency of 82% when running at 16,000 rpm. It takes in air at 17°C. Guide vanes at inlet give the air a pre whirl of 20° to the axial direction at all radii and the mean diameter of the eye is 200 mm, the absolute air velocity at inlet is 120 m/s. At exit, the blades are radially inclined and the impeller tip diameter is 550 mm. Calculate the slip factor of the compressor.

Figure 14.14 T – s Diagram

Solution

= 4, T1 = 273 + 17 = 290 K, N = 16,000 rpm, α1 = 90° − 20° = 70°

d1 = 200 mm, va1= 120 m/s, ηisen = 0.82, d2 = 550 mm, β2 = 90°

Temperature after isentropic compression (Fig. 14.14)

Isentropic temperature rise,

(∆T)isen = 430.94 - 390 = 140.94 K

Actual temperature rise,

Power input per unit mass flow rate

= cp × (∆T)a = 1.005 × 171.88 = 172.74 kJ/kg

An inlet to impeller (Fig. 14.15 (a)),

va1 = 120 m/s

Angle of pre-whirl = 20°

At exit of impeller (Fig. 14.15(b))

Figure 14.15 Velocity diagrams for centrifugal compressor: (a) Inlet, (b) Outlet

For radial discharge, vw2 = u2 = 460.77 m/s

Power input per unit mass flow rate

= u2vw2u1vw1
= 460.77 vw2 − 167.55 × 41.04 = 172.74 × 103
vw2 = 389.82 m/s

#### Example 14.3

A centrifugal compressor running at 15000 rpm takes in air at 15°C and compresses it through a pressure ratio of 4 with an isentropic efficiency of 80%. The blades are radially inclined and the slip factor is 0.85. Guide vanes at inlet give the air an angle of pre-whirl of 20° to the axial direction. The mean diameter of the impeller eye is 200 mm and the absolute air velocity at inlet is 120 m/s. Calculate the impeller tip diameter. Take cp = 1.005 kJ/kg.K and γ = 1.4.

Solution

Given: N = 15000 rpm, T1 = 273 + 15 = 288 K, rp = 4, ηisen = 80%, ϕs = 0.85, va1 = 120 m/s, d1 = 200 mm, α1 = 70°

Figure 14.16 (a) T-s diagram, (b) Velocity triangles.

From Fig. 14.16 (a), we have

Compressor work, wc = cp · ∆T = 1.005 × 175 = 175.875 kJ/kg

From Fig. 14.16 (b), vw1 = va1 cos 70°

= 120 cos 70° = 41.04 m/s

For radial discharge, β2 = 90° and

vw2 = u2

Power input = u2u1 vw1

175·875 × 103 = u2 × 0.85 u2 − 157 × 41.04

or u2 = 463.13 m/s

or d2 = 589.7 mm

#### Example 14.4

The free air delivered by a centrifugal compressor is 25 kg/min. The suction condition is 1 bar and 25°C. The velocity of air at inlet is 50 m/s. The isentropic efficiency of the compressor is 70%. If the total head pressure ratio of the compressor is 4, determine (a) the total head temperature of air at the exit of the compressor, and (b) brake power required to run the compressor assuming mechanical efficiency of 96%. Pressure and temperature of air at inlet are static. For air γ = 1.4 and R = 0.287 kJ/kg. K.

Solution

Figure 14.17 T–s diagram for centrifugal compressor

Refer to Fig. 14.17.

Brake power required to run the compressor =

#### Example 14.5

Air at a temperature of 20°C flows into the centrifugal compressor running at 21,000 rpm. The following data are given:

Slip factor = 0.85

Work input factor = 1.0

Isentropic efficiency = 72%

Outer diameter of blade tip = 60 cm

Assuming the absolute velocities of air entering and leaving the compressor are same, determine:

1. Temperature rise of air passing through compressor, and
2. Static pressure ratio

Take cp = 1.005 kJ/kg. K.

Solution

Work done per kg of air = u2vw2u1vw1

= (659.7)2 × 0.85 × 10−3 × 1 = 369.923 kJ/s

or T2 = 293 + 0.72 × 368.1 = 558 K

Static pressure ratio,

#### Example 14.6

A centrifugal compressor handles 180 kg/min of air. The suction pressure and temperature are 1 bar and 300 K. The suction velocity is 90 m/s. The delivery conditions are 2 bar, 400 K, and 240 m/s. Calculate (a) the isentropic efficiency, (b) the power required to drive the compressor, and (c) the overall efficiency of the unit. Take cp = 1.005 kJ/kg. K. Assume that entire kinetic energy gained in the impeller is converted into pressure in the diffuser.

Solution

= 3 kg/s, p1 = 1 bar, T1 = 300 K, va1 = 90 m/s, p2 = 2 bar, T2 = 350 K, va2 = 240 m/s

1. Isentropic efficiency,
T2 = 300 × 1.219 = 365.7 K

Isentropic work done, wisen = cp(T2T1) +

Actual work done in the impeller,

2. Power required to drive the compressor = × wa = 3 × 125.25 = 375.75 kW
3. Let T3 = temperature at the exit of diffuser

or 1.005 (T3 − 400) = 24.75

or T3 = 424.6 K

or p3 = 2 × 1.2326 = 2.465 bar

After isentropic compression, the delivery temperature from diffuser is,

#### Example 14.7

A centrifugal compressor handles 540 kg/min of air at 1 bar and 20°C ambient conditions. The air is compressed from 1 bar static pressure to 4.5 bar total pressure. The air enters the impeller eye with a velocity of 150 m/s with no pre-whirl. The ratio of whirl speed to tip speed is 0.9. Calculate (a) the rise in total temperature during compression if the change in kinetic energy is negligible, (b) the tip diameter of the impeller, (c) the power required, and (d) the eye diameter if the hub diameter is 130 mm.

Assume that the compressor runs to 20,000 rpm with isentropic efficiency of 82%.

Solution

= 9 kg/s, p01 = p1 = 1 bar, T1 = 273 + 20 = 293 K, N = 20,000 rpm, ηisen = 0.82,

Figure 14.18 T–s diagram

p02 = 4.5 bar, va1 = 150 m/s, = 0.9, dh = 130 mm = 0.13 m

The T - s diagram is shown in Fig. 14.18.

1. Isentropic rise in total temperature,

(∆T)isen = 450.3 − 304.2 = 146.13 K

Actual rise in total temperature,

2. Actual work consumed by the compressor,

wactual = cp (∆T)actual = 1.005 × 178.2 = 179.1 kJ/kg

Work consumed given by Euler’s equation without pre-whirl,

Tip diameter of impeller,

3. Power required = × wactual = 9 × 179.1 = 1611.9 kW
4. or D1 = 0.2848 m or 284.8 mm

#### Example 14.8

A centrifugal compressor delivers 720 m3/min of free air while running at 12,000 rpm. The ambient air conditions are 1 bar and 27°C. The air is compressed to 4.2 bar with an isentropic efficiency of 0.85. Blades are radial at the outlet of the impeller and the flow velocity is 60 m/s, which may be assumed constant throughout. The outer radius of impeller is twice the inner and the slip factor is 0.9. The blade area coefficient is 0.9 at inlet. Determine (a) the final temperature of air, (b) the theoretical power, (c) the inlet and outlet diameter of impeller, (d) the breadth of impeller at inlet, (e) the impeller blade angle at inlet, and (f) the diffuser blade angle at inlet.

Solution

p1 = 1 bar, T1 = 273 + 27 = 300 K, p2 = 4.2 bar, ηisen = 0.85, vf2 = 60 m/s, r2 = 2r1, ϕs = 0.9, kb = 0.9.

1. Mass flow rate

Theoretical power,

= 13.94 × 1.005 (478.8 − 300) = 2504.4 kW

2. Work done =

3. Volume flow rate = π D1b1kbvf1
4. or β1= 15.03°

5. or α2 = 8.48°

#### Example 14.9

A centrifugal blower compresses 5.4 m3/s of air from 1 bar and 25°C to 1.5 bar. The index of compression is 1.5. The flow velocity of 70 m/s is constant throughout. The inlet and outlet diameters of impeller are 0.30 m and 0.60 m, respectively. The blower rotates at 9000 rpm. Determine (a) the blade angles at inlet and outlet of the impeller, (b) the absolute angle at the tip of the impeller, and (c) the breadth of blade at inlet and outlet.

Assume that no diffuser is employed and the whole pressure increase occurs in the impeller and the blades have negligible thickness.

Solution

1 = 5.4 m3/s, p1 = 1 bar, T1 = 273 + 25 = 298 K, n = 1.5, p2 = 1.5 bar, vf1 = vf2 = 70 m/s, D1 = 0.3 m, D2 = 0.6 m, N = 9000 rpm

u2 = 2u1 = 282.74 m/s
1. or β1 = 26.34°

or β2 = 28.38°

2. or α2 = 24.56°

3. 1 = πD1b1vf1

#### Example 14.10

A centrifugal compressor delivers 18 kg/s of air with a total head pressure ratio of 4 : 1. The speed of the compressor is 16000 rpm. Total head temperature at inlet is 22°C, slip factor 0.9, power input factor 1.05 and isentropic efficiency 0.82. Calculate (a) the overall diameter of the impeller and (b) the power input.

Solution

= 18 kg/s, rop = 4, N = 16000 rpm, T01 = 273 + 22 = 295 K, ϕs = 0.9, ϕw = 1.05, ηisen = 0.82

1. or u2 = 431.2 m/s

2. Power input, P = cp (T02T01)

= 18 × 1.005 (469.8 − 295) = 3162.1 kW

#### Example 14.11

A single-sided centrifugal compressor impeller fitted with a diffuser, without water cooling, requires 2000 kW when running at 16,500 rpm. The outlet diameter of impeller is 0.5 m and the width of the casing of the vortex chamber between the impeller and the diffuser is 50 mm. During a test on this compressor, static temperature and pressure were measured at a radius of 0.28 m and were found to be 120°C and 3 bar, respectively. Assuming whirl speed at the tip of the impeller to be 0.95 of the peripheral speed and neglecting friction in vortex chamber, calculate (a) the flow rate, (b) the resultant speed at the section given, and (c) the total temperature at the section given. Take cp = 1.005 kJ/kg, K, γ = 1.4 and intake pressure = 1 bar.

Solution

P = 2000 kW, N = 16500 rpm, D2 = 0.5 m, b = 50 mm, ϕs = 0.95, p1 = 1 bar

1. Work done = vw2u2
2. Using suffix 3 for the given section

Area of flow, Af 3 = 2πr3b3 = 2π × 0.28 × 50 × 10−3 = 0.088 m2

Volume rate of flow at the given section,

i.e., vw1r1 = vw2r2 = vw3r3

#### Example 14.12

A centrifugal compressor delivers 580 m3/min of free air when running at 8000 rpm. Use the following data: inlet pressure and temperature of air = 1.013 bar and 20°C, pressure ratio = 3.5, isentropic efficiency = 83 %, flow velocity throughout the compressor = 62 m/s, the blades are radial at the outlet of the impelier, tip diameter = 2 times eye diameter, blade area coefficient = 0.94. Find (a) the input power required to run the compressor, (b) the impeller diameters at inlet and outlet, (c) the breadth of impeller at inlet, and (d) the impeller blade angle at inlet.

Solution

Given that a = 580 m3/min, N = 800 rpm, p1 = 1.013 bar, T1 = 273 + 20 = 293 K, rp = 3.5, ηisen = 0.83, vf = 62 m/s, β2 = 90°, D2 = 2D1, kt = 0.94

Mass flow rate of air,

1. Input power required
= 1777.72 kW

3. a = πD1b1vfkt

or = π × 0.4665 × b1 × 62 × 0.94

or b = 0.113 m or 113 mm

4. β1 = 17.6°

#### Example 14.13

A centrifugal compressor runs at 16,000 rpm and 2300 kW of power is required to run the compressor. The outer diameter of the impeller is 50 cm. The uniform width of the casing of vortex chamber between impeller and diffuser is 4 cm. Static conditions at a radius of 27 cm were measured and it were equal to 2.4 bar and 390 K. Surrounding pressure is 1 bar. Assuming slip factor of 0.94, find (a) the mass flow rate of gas, (b) the resultant velocity and (c) the temperature at the given radius of 27 cm. Assume cp = 1.005 kJ/kgK and γ = 1.4.

Solution

Given that N = 16,000 rpm, P = 2300 kW, D2 = 50 cm, B = 4 cm, At R = 27 cm, p = 2.4 bar, T = 390 K, p1 = 1 bar, ϕs = 0.94, cp = 1.005 kJ/kg.K, γ = 1.4

1. Using suffix 3 for the given section,

Area of flow, Af3 = 2π R3B3 = 2π × 0.27 × 0.04 = 0.06786 m2

Volume rate of flow,

i.e., vw1r1 = vw2r2 = vw3r3

1. = 460.98 K = 461 K or 188°C

#### Example 14.14

A centrifugal compressor has to deliver 6kg/s of air with pressure ratio of 4:1 at 17500 rpm. Initial conditions are static air at 1 bar and 15°C. Assuming an adiabatic efficiency of 78%, ratio of whirl speed to tip speed 0.94 and neglecting all other losses, calculate the tip speed, diameter, and rise in total pressure. Determine the external diameter of eye for which the internal diameter is 12 cm and the axial velocity at inlet is 150 m/sec. Assume cp = 1.005 kJ/kgK and γ = 1.4.

Solution

Given that: a = 6 kg/s, = 4, N = 17500 rpm, p01= 1 bar, T01 = 273 + 15 = 288 K, ηisen = 0.78, = 0.94, dh = 12 cm, for radial inlet, α1 = 90° and vf1 = va1 = 150 m/s, cp = 1.005 kJ/kg.K, γ = 1.4

The T-s diagram is shown in Fig. 14.19(a) and the impeller is shown in Fig. 14.19(b)

Figure 14.19 Diagrams for centrifugal compressor: (a) T-s diagram, (b) Impeller

Let va1 = va2, then

Rise in total pressure, ∆p0 = p02p01 = 3.82 - 1 = 2.82 bar

Work required per kg of air, w = cp(T02′ − T01)

= 1.005(460.4 − 288) = 173.262 kJ/kg

Also, work required per kg of air,

or d2 = 0.468 m or 46.8 cm

#### Example 14.15

A single-sided centrifugal air compressor delivers 1800 kg of air per minute. The air enters the eye of the impeller axially at total pressure of 100 kPa and total temperature of 290 K. The overall diameter of the impeller is 700 mm and it rotates at 1600 rpm. The slip factor is 0.9 and the work input factor is 1.1. The isentropic efficiency is 85%. Calculate (a) power required by the compressor, (b) pressure coefficient, and (c) total pressure at delivery.

[IES, 2003]

Solution

Given that a = 1800 kg/min, α1 = 90, p01 = 100 kPa, T01 = 290 K, D2 = 700 mm, N = 1600 rpm, ϕs = 0.9, ϕw = 1.1, ηisen = 85%

1. Pressure efficient, ϕp = ϕw ϕsηisen = 1.1 × 0.9 × 0.85 = 0.8415
2. or p02 = 100 × 1.00282 = 100.82 kPa

#### Example 14.16

A centrifugal compressor running at 18,000 rev/min takes in air at 25° C and compresses it through a pressure ratio of 4.0 with an isentropic efficiency of 80%. The guide vanes at inlet give the air an angle of pre-whirl of 20° to the axial direction. The mean diameter of impeller eye is 225 mm. The absolute air velocity at inlet is 130 m/s. At exit the blades are radially inclined. If the slip factor is 0.90, calculate the impeller tip diameter.

[IAS, 1999]

Solution

Given that N = 18000 rpm, T1 = 25 + 273 = 298 K, rp = 4, ηisen = 80%, pre-whirl at inlet = 20° or α1 = 90° − 20° = 70°, dm = 225 mm, va1 = 130 m/s, β2 = 90°, ϕs = 0.90

Temperature after isentropic compression,

Isentropic temperature rise,

(∆T)isen = T2T1 = 442.83 - 298 = 144.83 K

Actual temperature rise,

Power input per unit mass flow rate = cp × (∆T)act = 1.005 × 181 = 181.9 kJ/kg

The velocity diagrams are shown in Fig. 14.20

vw1 = va1 cos α1 = 130 cos 70° = 44.46 m/s

For radial discharge, vw2 = u2

Slip factor,

= 0.9 vw2 = 0.9 × 0.9425 d2 = 0.84825 × d2 m/s

Figure 14.20 Velocity diagrams: (a) Inlet, (b) Outlet

Power input per unit mass flow rate = u2 u1 vw1

or 181.9 × 103 = 0.9425 × d2 × 0.84825 × d2 − 212 × 44.46

= 0.7995 − 9425.52

or d2 = 488 mm

#### Example 14.17

A double-sided centrifugal compressor has root and tip diameters of 18 cm and 30 cm and is to deliver 16 kg of air per second at 16,000 rpm. The design ambient conditions are 15°C and 1 bar and the compressor has to be a part of a stationary power plant. Determine the following:

1. Suitable values for impeller vane angles at the root and tip of the eye if the air is given 20° of pre-whirl at all radii. The axial component of the velocity is constant throughout the impeller and is 150 m/s.
2. Power required if the power input factor is 1.05 and mechanical efficiency is 95%
3. Maximum Mach number at the eye

Take for air: cp = 1.005 kJ/kg K and γ = 1.4.

[IAS, 1998]

Solution

Given that r1 = 9 cm, r2 = 15cm, a = 16 kg/s, N = 16000 rpm, T1 = 273 + 15 = 288 K

p1 = 1 bar, α1 = α2 = 70°, vf1 = vf2 = 150 m/s, ϕs = 1.05, ηmech = 95%, cp = 1.005 kJ/kg K, γ = 1.4

1. Angular speed of rotor,

The velocity diagrams are shown in Fig. 14.21.

u1 = ωr1 = 1675.5 × 0.09 = 150.8 m/s

u2 = ωr2 = 1675.5 × 0.15 = 251.3 m/s

α1 = 90° - angle of pre-whirl = 70°

Figure 14.21 Velocity diagrams

Work done per kg of air, w = vw2 u2 - vw1 u1 = vw1 (u1u1)

= 54.6 (251.3 − 150.8) = 5487.3
2. Power required

#### Summary for Quick Revision

1. A centrifugal compressor is of rotodynamic type in which the pressure rise is primarily due to the centrifugal action.
2. Stagnation or total head values:
1. Stagnation temperature,
2. Stagnation enthalpy,
3. Stagnation pressure,
3. Isentropic efficiency,
4. Ideal velocity diagrams:
1. α1 = 90°, β2 = 90°, va1= vf1 and vr2 = vf2, i.e. vw1 = 0 and vw2 = u2
2. Torque, T = u2 r2
3. Work done for 1 kg/s of air flow,
4. From SFEE, w = cp (T02T01) = h02h01

= cp (T2 T1) = h2 - h1 for va1 = va2

5. Stagnation pressure ratio:
6. Total temperature increase across an impeller,
7. Indicated power,
5. Actual velocity diagrams:
1. Work done on 1 kg/s of air, w =
2. ∆KE

where kb = blades factor, b = width of height of blades, t = thickness of blades,

vs = specific volume

7. Slip factor,

Slip = u2vw2

8. Pressure coefficient,

9. Work factor,
10. Relationship between ϕp, ϕw, ϕs and ηisen:

ϕp = ϕs ϕw ηisen

1. There are three types of impeller blades:
12. Backward curved blades, β2 < 90°.
2. b. Forward curved blades, β2 > 90°.

Slightly backward curved impeller blades give optimum efficiency. Its consequences are:

3. a. Degree of reaction increases which results in increasing part of kinetic energy transferred into pressure energy within the impeller.
4. b. The length of impeller channel increases which increases the friction loss.
13. The diffuser converts the kinetic energy imparted to air by the impeller into pressure rise.
14. Pre-whirl is given to intake guide vanes in order to avoid acceleration to sonic or supersonic velocities to keep the Mach number below 0.9 of the air flow entering the impeller eye. This increases the work done on the air vw1 ≠ 0.
15. Performance characteristics:

For D = const

where G = mass rate of flow, η = rotor rpm, p1 = inlet pressure,

T1 = inlet temperature, p3 = pressure at diffuser outlet.

16. The pulsating air flow phenomena is called surging.

The maximum mass flow rate of air is called choking of the compressor.

#### Multiple-choice Questions

1. The inlet and exit velocity diagrams of a turbomachine rotor are shown in Fig. 14.22.

This turbomachine is

1. An axial compressor with backward curved blades
4. An axial compressor with forward curved blades

Figure 14.22 Velocity triangles

2. It is recommended that the diffuser angle should be kept less than 18° because
1. Pressure decreases in flow direction and flow separation may occur
2. Pressure decreases in flow direction and flow may become turbulent
3. Pressure increases in flow direction and flow separation may occur
4. Pressure increases in flow direction and flow may become turbulent
3. When the outlet angle from the rotor of a centrifugal compressor is more than 90, then the blades are said to be
1. Forward curved
2. Backward curved
4. Either backward or forward curved
4. The degree of reaction of a turbomachine is defined as the ratio of the
1. Static pressure change in the rotor to that in the stator
2. Static pressure change in the rotor to that in the stage
3. Static pressure change in the stator to that in the rotor
4. Total pressure change in the rotor to that in the stage
5. If two geometrically similar impellers of a centrifugal compressor are operated at the same speed, their head, discharge, and power will vary with their diameter ratio d as
1. d, d2, and d3, respectively
2. d2, d3, and d5, respectively
3. d, d3, and d5, respectively
4. d2, d, and d3, respectively
6. Which one of the following velocity triangles (Fig. 14.23) represents the one at the exit of a radial impeller with forward curved blades?

(u2 = peripheral velocity, v2 = absolute velocity, w2 = relative velocity)

Figure 14.23 Velocity triangles

7. The stagnation pressure rise in a centrifugal compress or stage takes place
1. Only in the diffuser
2. In the diffuser and impeller
3. Only in the impeller
4. Only in the inlet guide vanes
8. For 15 m3/s air flow at 10 mm Hg head, which one of the following would be the best choice?
1. Centrifugal fan with forward curved blades
2. Axial fan with a large number of blades in rotor
3. Axial propeller fan with a few blades in rotor
4. Cross-flow fan
9. 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 fraction
10. Which of the following is the effect of blade shape on performance of centrifugal compressor?
1. Backward curved blade has poor efficiency
2. Forward curved blades have higher efficiency
4. Forward curved blades produce lower pressure ratio
11. Surging basically implies
1. Unsteady, periodic, and reversed flow
2. Forward motion of air at a speed above sonic velocity
3. The surging action due to the blast of air produced in a compressor
4. Forward movement of aircraft
12. Centrifugal compressors are suitable for large discharge and wider mass flow range, but at a relatively low discharge pressure of the order of 10 bars, because of
1. Low pressure ratio
2. Limitation of size of receiver
3. Large speed
4. High compression index
13. Given: vw2= velocity of whirl at outlet

u2 = peripheral velocity of the blade tips.

The degree of reaction in a centrifugal compressor is equal to

14. In a centrifugal compressor assuming the same overall dimensions, blade inlet angle, and rotational speeds, which of the following blades will give the maximum pressure rise?
4. All three types of bladings have the same pressure rise
15. Under which one of the following sets of conditions will a surpersonic compressor have the highest efficiency?
1. Rotor inlet velocity is supersonic and exit velocity subsonic; stator inlet velocity is subsonic and exit velocity is subsonic
2. Rotor inlet velocity is supersonic and exit velocity subsonic; stator inlet velocity is supersonic and exit velocity is subsonic
3. Rotor inlet velocity is supersonic and exit velocity supersonic; stator inlet is supersonic and exit velocity is subsonic
4. Rotor inlet velocity is supersonic and exit velocity supersonic; stator inlet velocity is subsonic and exit velocity is subsonic
16. The curve in Fig. 14.24 shows the variation of theoretical pressure ratio with mass of flow rate for a compressor running at a constant speed. The permissible operating range of the compressor is represented by the part of the curve from
1. A to B
2. B to C
3. B to D
4. D to E

Figure 14.24

17. In a centrifugal compressor, the highest Mach number leading to shockwave in the fluid flow occurs at
18. In the centrifugal air compressor design practice, the value of polytropic exponent of compression is generally taken as
1. 1.2
2. 1.3
3. 1.4
4. 1.5
19. For centrifugal compressors, which one of the following is the correct relationship between pressure coefficient (ϕp), slip factor (ϕs), work input factor (ϕn) and isentropic efficiency (ηa)?
20. At the eye tip of a centrifugal impeller, blade velocity is 200 m/s while the uniform axial velocity at the inlet is 150m/s. If the sonic velocity is 300 m/s, the inlet Mach number of the flow will be
1. 0.50
2. 0.66
3. 0.83
4. 0.87
21. What will be the shape of the velocity triangle at the exit of a radial bladed centrifugal impeller, taking into account slip?
1. Right-angled
2. Isosceles
3. All angles less than 90°
4. One angle greater than 90°
22. What does application of centrifugal air compressors lead to?
1. Large frontal area of aircraft
2. Higher flow rate through the engine
3. Higher aircraft speed
4. Lower frontal area of the aircraft
23. Consider the following statements:

In centrifugal compressors, there is a tendency of increasing surge when

1. the number of diffuser vanes is less than the number of impeller vanes
2. the number of diffuser vanes is greater than the number of impeller vanes
3. the number of diffuser vanes is equal to the number of impeller vanes
4. mass flow is greatly in excess of that corresponding to the design mass flow

Which of these statements is/are correct?

24. In centrifugal compressor terminology, vaneless space refers to the space between
1. The inlet and blade inlet edge
3. Diffuser exit and volute casing
4. Impeller tip and diffuser inlet edge
25. Match List-I with List-II (pertaining to blower performance) and select the correct answer using the codes given below the list:
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
26. The flow in the vaneless space between the impeller exit and diffuser inlet of a centrifugal compressor can be assumed as
1. Free vortex
2. Forced vortex
3. Solid body rotation
4. Logarithmic spiral
27. Which portion of the centrifugal compressor characteristics shown in Fig. 14.25 is difficult to obtain experimentally?

Figure 14.25

1. RS
2. ST
3. TU
4. UV
28. The pressure rise in the impeller of centrifugal compressor is achieved by
1. The decrease in volume and diffusion action
2. The centrifugal action and decrease in volume
3. The centrifugal and diffusion action
4. The centrifugal and push-pull action
29. In a radial blade centrifugal compressor, the velocity of blade tip is 400 m/s and slip factor is 0.9. Assuming the absolute velocity at inlet to be axial, what is the work done per kg of flow?
1. 36 kJ
2. 72 kJ
3. 144 kJ
4. 360 kJ
30. The power required to drive a turbo-compressor for a given pressure ratio decreases when
1. Air is heated at entry
2. Air is cooled at entry
3. Air is cooled at exit
4. Air is heated at exit
31. In a centrifugal compressor, how can the pressure ratio be increased?
1. Only by increasing the tip speed
2. Only by decreasing the inlet temperature
3. By both (a) and (b)
4. Only by increasing the inlet temperature

1. 20. (a)
2. 31. (c)

Work done

#### Review Questions

1. What are the constructional features of a centrifugal compressor?
2. Explain the principle of working of a centrifugal compressor.
3. How does the velocity and pressure of air vary through the impeller and the diffuser?
4. What are the various types of impellers?
6. Define isentropic efficiency.
7. Define stagnation pressure ratio.
8. Define slip factor.
9. What is work factor and pressure coefficient?
10. What is the relationship between ϕp, ϕw, ϕs, and ϕisen?
11. List the losses which generally occur in a centrifugal compressor.
12. What are the effects of impeller blade shapes on performance?
13. What is the role of a diffuser?
14. Explain pre-whirl and its role.
15. Explain the phenomena of surging and choking.

#### Exercises

14.1 A single eye, single-stage centrifugal compressor delivers 20 kg/s of air with a pressure ratio of 4 when running at 15,000 rpm. The pressure and temperature of air at the inlet are 1 bar and 17°C. Assume the following:

Slip factor = 0.9, work input factor = 1.04, isentropic efficiency = 75%.

Calculate (a) the IP required to drive the compressor and (b) the blade angles at the impeller eye if the root and tip diameters are 20 cm and 40 cm, respectively.

14.2 A centrifugal compressor delivers 600 m3/min of free air when running at 900 rpm. The following data is given:

Inlet pressure and temperature of air = 1 bar and 20°C

Compression ratio = 3.5

Isentropic efficiency = 83%

Flow velocity throughout the impeller = 60 m/s

Tip diameter = 2 × eye diameter

Calculate (a) the IP required to run the compressor, (b) impeller diameters at inlet and outlet, (c) breadth of impeller at inlet, and (d) impeller blade angle at inlet.

14.3 The first stage of a centrifugal compressor is single-inlet with an eye 60 cm, hub 20 cm and an overall diameter of 100 cm. When supplied with 20°C air at 1 bar and running at 6000 rpm, it compresses 30 kg/s of air through a compression ratio of 2.0.

Calculate (a) the isentropic efficiency of rotor based on impeller tip velocity with slip factor as 0.9, and (b) the rotor power and shaft power. Take ηmech = 0.97.

14.4 A single-sided centrifugal compressor is to deliver 14 kg/s of air when operating at a pressure ratio of 4:1 and a speed of 12000 rpm. The total head inlet conditions may be taken as 288 K and 1 bar. Assuming a slip factor of 0.9, a power input factor of 1.04 and an isentropic efficiency (based on the total head) of 80%, estimate the overall diameter of the impeller. If the Mach number is not to exceed unity at the impeller tip and 50% of the losses are assumed to occur in the impeller, find the minimum possible depth of the diffuser.

14.5 The following data refer to a single-sided centrifugal compressor:

Air mass flow rate = 9 kg/s, Eye tip diameter = 0.3 m, Slip factor = 0.9, Isentropic efficiency = 80%.

Eye root diameter = 0.15 m, overall diameter of impeller = 0.5 m, Power input factor = 1.04, Rotational speed = 18000 rpm, Inlet stagnation pressure and temperature = 1.1 bar, 295 K.

Calculate (a) pressure ratio of the compressor, (b) inlet angle of the impeller vane of the root and tip radii of the eye, and (c) the axial depth of the impeller channel at the periphery of the impeller.

14.6 An air-craft fitted with a single-sided centrifugal compressor flies with a speed of 850 km/h at an altitude where the pressure is 0.23 bar and the temperature is 217 K. The inlet duct of the impeller eye contains fixed vanes which give the air pre-whirl of 25° at all radii. These inner and outer diameters of the eye are 180 mm and 330 mm, respectively, the diameter of the impeller tip is 540 mm, and the rotational speed is 16,000 rpm. Estimate the stagnation pressure at the compressor outlet when the mass flow is 216 kg/min. Neglect losses in inlet duct and fixed vanes, and assume that the isentropic efficiency of the compressor is 0.80. Take the slip factor as 0.9 and the power input factor as 1.04.

14.7 A centrifugal compressor compresses 30 kg/s of air at a rotational speed of 1500 rpm. The air enters the compressor axially, and the conditions at the exit section are radius = 0.3 m, relative velocity of air at the tip = 100 m/s at an angle of 80°. Find the torque and power required to drive the compressor and work done.

14.8 Air at a temperature of 300 K flows in a centrifugal compressor running at 18,000 rpm. Slip factor = 0.8; isentropic total head efficiency = 0.75; outer diameter of blade tip = 0.5 m. Determine (a) the temperature rise of air and (b) static pressure ratio.

Assume that the velocities of air at inlet and exit of the compressor are the same.

14.9 A single inlet type centrifugal compressor delivers 8 kg/s of air. The ambient air conditions are 1 bar and 20°C. The compressor runs at 22,000 rpm with isentropic efficiency of 82%. The air is compressed in the compressor from 1 bar static pressure to 4.2 bar total head pressure. The air enters the impeller eye with a velocity of 150 m/s with no pre-whirl. Assuming that the ratio of whirl speed to tip speed is 0.9, calculate (a) the rise in total head temperature during compression if the change in kinetic energy is negligible, (b) the tip diameter of the impeller, (c) power required, and (d) the eye diameter if the hub diameter is 10 cm.

14.10 A centrifugal compressor is to have a pressure ratio of 3.5:1. The inlet eye of the impeller is 30 cm in diameter. The axial velocity at inlet is 130 m/s and the mass flow rate is 10 kg/s. The velocity in the delivery duct is 115 m/s. The tip speed of the impeller is 450 m/s and runs at 16,000 rpm with total head isentropic efficiency of 78% and pressure coefficient of 0.72. The ambient conditions are 1 bar and 15°C. Calculate: (a) the static pressure ratio, (b) the static pressure and temperature at inlet and outlet of compressor, (c) work of compressor per kg of air, and (d) the theoretical power required.

14.11 The following details refer to a centrifugal air compressor of a total head pressure ratio of 3 and isentropic efficiency of 70%. Find the power required to run the compressor.

Suction conditions:

Pressure = 1 bar,

Temperature = 20°C

Velocity of air at inlet = 60 m/s

Pressure and temperature are static values.

Free air delivered = 20 kg/min

Mechanical efficiency = 95%

Ratio of specific heats for air = 1.4

Specific gas constant for air = 0.287 kJ/kg.K

14.12 A centrifugal compressor has a pressure ratio of 4:1 with an isentropic efficiency of 82% when running at 16000 rpm. It takes in air at 17°C. Guide vanes at inlet give the air a pre-whirl of 20° to the axial direction at all radii and the mean diameter of the eye is 200 mm; the absolute air velocity at inlet is 120 m/s. At exit, the blades are radially inclined and the impeller tip diameter is 550 mm. Calculate the slip factor of the compressor.

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