Question 13.1: Air at 1 bar and 15°C enters a three stage axial compressor ......

The given data are $P_{01}=1 \text{ bar}, r_{\text{m}}=0.2 \text{ m},N=9000 \text{ rpm}, \eta_{\text{S}}=0.9, \beta_1-\beta_2$ = 25°, h = 0.05 m, Number of stages = 3, and $C_1 \equiv C_{\text{a}1}=120 \text{ m/s}$

1. Since the flow is axial, $\alpha_1=0, C_{\text{u}1}=0 ,\text{ and } \Delta C_{\text{u}} \equiv C_{\text{u}2}$ . A typical velocity triangle for this case is shown in Figure 13.15.

If the inlet total temperature is $T_{01}=15+273=288 \text{ K with }U_{\text{m}}=2\pi r_{\text{m}}N=188.496 \text{ m/s}$, then

$\tan \beta_1=\frac{U}{C_{\text{a}1}} \\ \beta_1=57.52^\circ=\text{constant for all stages} \\ \beta_2=\beta_1-25=32.52^\circ= \text{constant also for all stages}.$

Since

$\tan \beta_2=\frac{U-C_{\text{u}2}}{C_{\text{a}2}}$

then $C_{\text{u}2}=111.989 \text{ m/s} \equiv \Delta C_{\text{u}}$ .
The axial compressor is composed of three identical stages as for the three stages:

1. The flow enters axially.
2. The rotor turning angle is the same, $\beta_1-\beta_2$ = constant, thus $\Delta C_{\text{u}}$ = constant.
3. The polytropic efficiency is constant.
4. The compressor has a constant mean radius layout $U_{\text{m}}$ = constant.

Thus, the temperature rise per stage and the specific work are constant.
The specific work per stage is $\text{w}_{\text{S}}=U \Delta C_{\text{u}}$ = 21.11 kJ/kg ≡ Constant for all stages [or $\text{w}_{\text{S}}=U \Delta C_{\text{u}}=UC_{\text{a}}(\tan \beta_1 – \tan \beta_2)$ = 21.11 kJ/kg].
The specific work for the three-stage compressor is

$W_{\text{s}}=3\text{w}_{\text{s}}=3 \times 21.11=63.33 \text{ kJ/kg}$

2. The mass flow rate may be calculated from the relation $\dot{m}=\rho_1 C_{\text{a}1}A_1$ .
At the compressor inlet, the static temperature

$T_1=T_{01}-\frac{C_1^2}{2 Cp} =280.836 \text{ K}$

and the static pressure

$P_1=\frac{P_{01}}{(T_{01}/T_1)^{\gamma/(\gamma-1)}} =0.916 \text{ bar}$

The density

$\rho_1=\frac{P_1}{RT_1} =1.136 \text{ kg/m}^3$

The annulus area $A= 2\pi r_{\text{m}}h$ = 0.0628 m²

$\therefore \dot{m}= \rho_1 C_{\text{a}1}A=8.565 \text{ kg/s}$

3. The power necessary to drive the compressor is then

$\text{Power}=\dot{m}\text{w}_{\text{S}}=180.81 \text{ kW}$

4. The temperature rise per stage is $\Delta T_{0\text{S}}=\text{w}_{\text{S}}/Cp$ = 21 K ≡ constant for all stages.

Pressure ratio of stage # 1

$\pi_1=\left(1+ \eta_{\text{S}}\frac{\Delta T_{0\text{S}}}{T_{01}} \right) ^{\gamma/(\gamma-1)}=1.249$

Pressure ratio of stage # 2

$\pi_2=\left(1+ \eta_{\text{S}}\frac{\Delta T_{0\text{S}}}{T_{01}+\Delta T_{0\text{S}}} \right) ^{\gamma/(\gamma-1)}=1.231$

Pressure ratio of stage # 3

$\pi_3=\left(1+ \eta_{\text{S}}\frac{\Delta T_{0\text{S}}}{T_{01}+2\Delta T_{0\text{S}}} \right) ^{\gamma/(\gamma-1)}=1.215$

5. Overall pressure ratio $\pi_{\text{comp}}= \pi_1\times \pi_2 \times \pi_3=1.87$.
6. To calculate the blade height at the exit, the air density must be first calculated. The pressure at the outlet of the compressor is $(P_{03})_{\text{S}3}= \pi_{\text{C}} \times (P_{01})_{\text{S}1} =1.87 \text{ bar}$.
The temperature at the exit of the compressor is $(T_{03})_{\text{S}3}=(T_{01})_{\text{S}1}+3 \Delta T_{0\text{S}}=351 \text{ K}$.
With $C_3=C_1$

$(T_3)_{\text{S}3}=(T_{03})_{\text{S}3}-\frac{C_3^2}{2Cp} =343.836 \text{ K} \\ (P_3)_{\text{S}3}=\frac{(P_{03})_{\text{S}3}}{\left[\left(T_{03}/T_3\right)^{\gamma/(\gamma-1)} \right] _{\text{S}3}} =1.74 \text{ bar} \\ \rho_3=\left(\frac{P_3}{RT_3} \right) _{\text{S}3}=1.763 \text{ kg/m}^3$

Since $\dot{m}=(\rho_3 C_3 A_3)_{\text{S}3}=[\rho_3 C_3 (2\pi r_{\text{m}}h_3)]_{\text{S}3},$

$(h_3)_{\text{S}3}=3.2 \text{ cm}$

7. The degree of reaction is expressed as

$\Lambda =1-\frac{C_{\text{u}1}+C_{\text{u}2}}{2U} \\ =1-\frac{C_{\text{u}2}}{2U} =1-\frac{112}{2 \times 188.5} =0.7$