Page 167 - Fluid Mechanics and Thermodynamics of Turbomachinery

148 Fluid Mechanics, Thermodynamics of Turbomachinery
The pressure rise in a real stage (involving irreversible processes) can be deter-
mined if the stage efficiency is known. Defining the stage efficiency s as the ratio
of the isentropic enthalpy rise to the actual enthalpy rise corresponding to the same
finite pressure change, (cf. Figure 2.7), this can be written as
s D .h is //.h/ D .1/ /p/h.

Thus,
.1/ /p D s h D s Uc y . (5.27)

If c 1 D c 3 , then s is a very close approximation of the total-to-total efficiency
tt . Although the above expressions are derived for incompressible flow they are,
nevertheless, a valid approximation for compressible flow if the stage temperature
(and pressure) rise is small.

Pressure ratio of a multistage compressor

It is possible to apply the preceding analysis to the determination of multistage
compressor pressure ratios. The procedure requires the calculation of pressure and
temperature changes for a single stage, the stage exit conditions enabling the density
at entry to the following stage to be found. This calculation is repeated for each stage
in turn until the required final conditions are satisfied. However, for compressors
having identical stages it is more convenient to resort to a simple compressible flow
analysis. An illustrative example is given below.
EXAMPLE 5.1. A multistage axial compressor is required for compressing air
at 293 K, through a pressure ratio of 5 to 1. Each stage is to be 50% reaction
and the mean blade speed 275 m/s, flow coefficient 0.5, and stage loading factor
0.3, are taken, for simplicity, as constant for all stages. Determine the flow angles
and, the number of stages required if the stage efficiency is 88.8%. Take C p D
1.005 kJ/(kg ° C) and
D 1.4 for air.
Solution. From eqn. (5.14a) the stage load factor can be written as,

D .tan ˇ 1 tan ˇ 2 /.
From eqn. (5.11) the reaction is

R D .tan ˇ 1 C tan ˇ 2 /.
2
Solving for tan ˇ 1 and tan ˇ 2 gives

tan ˇ 1 D .R C /2// and tan ˇ 2 D .R /2// .
Calculating ˇ 1 and ˇ 2 and observing for R D 0.5 that the velocity diagram is
symmetrical,

ˇ 1 D ˛ 2 D 52.45 deg and ˇ 2 D ˛ 1 D 35 deg .