Question 4.DE.16: The following design data apply to a double-sided centrifuga......

At eye root, C_a = 150 m/s
∴ C_1 = \frac{C_a} {\cos20°} = \frac{150} {\cos20°} = 159.63 m/s
and C_{w1} = 150 tan 20° = 54.6 m/s
Impeller speed at eye root
U_{er} = \frac{\pi D_{er}N} {60} = \frac{\pi × 0.18 × 15500} {60}
U_{er} = 146 m/s
From velocity triangle
tan β_{er} = \frac{C_a} {U_{er} – C_{w1}} = \frac{150} {146 – 54.6} = \frac{150} {91.4} = 1.641
i.e., β_{er} = 58.64°
At eye tip from Fig. 4.17(b)
U_{et} = \frac{\pi D_{et}N} {60} = \frac{\pi × 0.3175 × 15500} {60}
U_{er} = 258 m/s

tan α_{et} = \frac{150} {258 – 54.6} = \frac{150} {203.4} = 0.7375
i.e. α_{et} = 36.41°
Mach number will be maximum at the point where relative velocity is maximum.
Relative velocity at eye root is:
V_{er} = \frac{C_a} {\sin β_{er}} = \frac{150} {\sin 58.64°} = \frac{150} {0.8539}
V_{er} = 175.66 m/s
Relative velocity at eye tip is:
V_{et} = \frac{C_a} {\sin α_{er}} = \frac{150} {\sin 36.41°} = \frac{150} {0.5936}

V_{et} = 252.7 m/s
Relative velocity at the tip is maximum.
Static temperature at inlet:
T_1 = T_{01} = \frac{V^2 _{et}} {2C_p} = 288 – \frac{252.7^2} {2 × 1005} = 288 – 31.77
T_1 = 256.23 K
M_{max} = \frac{V_{et}} {(γRT_1)^{1/2}} = \frac{252.7} {(1.4 × 287 × 256.23)^{1/2}} = \frac{252.7} {320.86}
M_{max} = 0.788