Question

A centrifugal compressor is designed to operate with air. At the leading edge of the tip of the inducer, the blade angle is $45^\circ$, and the relative Mach number is 1.0. The stagnation temperature of the incoming air is 300 K. Consider $\gamma = 1.4$. Neglect pre-whirl and slip. The inducer tip speed is .............. m/s (rounded off to the nearest integer).

Hint:

For relative Mach number problems in compressors, use velocity triangle relations. The blade angle fixes the velocity ratio ($V_a/U$). Then apply $M_{rel} = V_{rel}/a$ with $a=\sqrt{\gamma RT}$ to find tip speed.

Answer 1

Step 1: Stagnation temperature relation.
Given $T_0 = 300 \, \text{K}$ and Mach number of relative velocity at the tip $M_{rel} = 1.0$. We use the relation: \[ T_0 = T \left(1 + \frac{\gamma - 1}{2} M^2 \right) \] For $M=1$: \[ T_0 = T \left(1 + \frac{0.4}{2}\right) = 1.2T \] \[ T = \frac{T_0}{1.2} = \frac{300}{1.2} = 250 \, \text{K} \] Step 2: Speed of sound at this $T$.
\[ a = \sqrt{\gamma R T} = \sqrt{1.4 \times 287 \times 250} \] \[ a \approx 317 \, \text{m/s} \] Since $M_{rel} = 1$, \[ V_{rel} = a = 317 \, \text{m/s} \] Step 3: Relation between velocities at blade inlet.
At inlet: no pre-whirl $\Rightarrow$ absolute velocity is purely axial ($V_a$). At blade tip: \[ \tan \beta = \frac{V_a}{U} \] where $\beta = 45^\circ$. So: \[ V_a = U \] Also, relative velocity: \[ V_{rel} = \sqrt{V_a^2 + U^2} = \sqrt{2U^2} = \sqrt{2} U \] Step 4: Solve for $U$.
\[ U = \frac{V_{rel}}{\sqrt{2}} = \frac{317}{1.414} \approx 224.2 \, \text{m/s} \] Wait! Let's recheck geometry: Actually, for relative velocity triangle at $\beta = 45^\circ$: \[ \tan \beta = \frac{V_a}{U} = 1 \Rightarrow V_a = U \] Then: \[ V_{rel} = \sqrt{V_a^2 + U^2} = \sqrt{U^2+U^2} = \sqrt{2}U \] So: \[ 317 = \sqrt{2}U \quad \Rightarrow \quad U = \frac{317}{1.414} \approx 224.2 \, \text{m/s} \] Step 5: Round to nearest integer.
\[ \boxed{224 \, \text{m/s}} \]