Question 4.DE.13: The following data refers to a single-sided centrifugal comp......

Let: r_h = hub radius
r_t = tip radius
The flow area of the impeller inlet annulus is:
A_1 = π (r^2_t – r^2_h) = \pi (0.125^2 – 0.0625^2) = 0.038 m²
Axial velocity can be determined from the continuity equation but since the inlet density (ρ_1) is unknown a trial and error method must be followed.
Assuming a density based on the inlet stagnation condition,
ρ_1 = \frac{P_{01}} {RT_{01}} = \frac{(1)(10^5)}{(287)(288)} = 1.21 kg/m³
Using the continuity equation,
C_a = \frac{\dot{m}} {ρ_1A_1} = \frac{5.5}{(1.21)(0.038)} = 119.6 m/s
Since the whirl component at the inlet is zero, the absolute velocity at the
inlet is C_1 = C_a.
The temperature equivalent of the velocity is:
\frac{C^2_1} {2C_p} = \frac{119.6^2} {(2)(1005)} = 7.12 K
Therefore:
T_1 = T_{01} – \frac{C^2_1 }{2C_p} = 288 – 7.12 = 280.9 K

Using isentropic P–T relationship,
\frac{P_1} {P_{01}} = \left(\frac{T_1} {T_{01}}\right)^{γ/(γ-1)}, or P_1 = 10^5\left(\frac{280.9}{288}\right)^{3.5} = 92 kPa
and:
ρ_1 = \frac{P_1} {RT_1} = \frac{(92)(10^3)}{(287)(280.9)} = 1.14 kg/m³, and
C_a = \frac{5.5}{(1.14)(0.038)} = 126.69 m/s
Therefore:
\frac{C^2_1} {2C_p} = \frac{(126.96)^2}{2(1005)} = 8.02 K
T_1 = 288 – 8.02 = 279.98° K
P_1 = 10^5 \left(\frac{279.98} {288}\right)^{3.5} = 90.58 kPa
ρ_1 =\frac{(90.58)(10^3)} {(287)(279.98)} = 1.13 kg/m³
Further iterations are not required and the value of ρ_1 = 1.13 kg/m³ may be taken as the inlet density and C_a = C_1 as the inlet velocity. At the eye tip:
U_{et} = \frac{2\pi r_{et}N} {60} = \frac{2\pi (0.125)(16,500)} {60} = 216 m/s
The blade angle at the eye tip:
β_{et} = \tan^{-1} \left(\frac{U_{et}} {C_a}\right) = \tan^{-1} \left(\frac{216} {126.96}\right) = 59.56°
At the hub,
U_{eh} = \frac{2\pi (0.0625)(16,500)} {60} = 108 m/s
The blade angle at the hub:
β_{eh} = \tan^{-1} \left(\frac{108} {126.96}\right) = 40.39°

The Mach number based on the relative velocity at the eye tip using the inlet velocity triangle is:
V_1 = \sqrt{C_a^2  +  U_1^2} = \sqrt{126.96^2  +  216^2}, or V_1 = 250.6 m/s

The relative Mach number
M = \frac{V_1}{\sqrt{γRT_1}} = \frac{250.6}{\sqrt{(1.4)(287)(279.98)}} = 0.747